A Beginner’s Guide to

the Theory of Viscosity Solutions

Shigeaki Koike∗

June 28, 2012

2nd edition

26 August 2010

∗Department of Mathematics, Saitama University, 255 Shimo-Okubo, Sakura, Saitama,338-8570 JAPAN

i

Preface for the 2nd edition

The first version of this text has been out of print since 2009. It seems thatthere is no hope to publish the second version from the publisher. Therefore,I have decided to put the files of the second version in my Home Page.

Although I corrected many errors in the first version, there must be somemistakes in this version. I would be glad if the reader would kindly informme errors and typos etc.

Acknowledgment for the 2nd edition

I would like to thank Professors H. Ishii, K. Ishii, T. Nagasawa, M. Ohta,T. Ohtsuka, and graduate students, T. Imai, K. Kohsaka, H. Mitake, S.Nakagawa, T. Nozokido for pointing out numerous errors in the First edition.

26 August 2010 Shigeaki Koike

ii

Preface

This book was originally written in Japanese for undergraduate studentsin the Department of Mathematics of Saitama University. In fact, the firsthand-written draft was prepared for a series of lectures on the viscosity so-lution theory for undergraduate students in Ehime University and HokkaidoUniversity.

The aim here is to present a brief introduction to the theory of viscositysolutions for students who have knowledge on Advanced Calculus (i.e. differ-entiation and integration on functions of several-variables) and hopefully, alittle on Lebesgue Integration and Functional Analysis. Since this is writtenfor undergraduate students who are not necessarily excellent, I try to give“easy” proofs throughout this book. Thus, if you do not feel any difficultyto read User’s guide [6], you should try to read that one.

I also try not only to show the viscosity solution theory but also to men-tion some related “classical” results.

Our plan of this book is as follows: We begin with our motivation insection 1. Section 2 introduces the definition of viscosity solutions and theirproperties. In section 3, we first show “classical” comparison principles andthen, extend them to viscosity solutions of first- and second-order PDEs,separately. We establish two kinds of existence results via Perron’s methodand representation formulas for Bellman and Isaacs equations in section 4.

We discuss boundary value problems for viscosity solutions in sections 5.Section 6 is a short introduction to the Lp-viscosity solution theory, on whichwe have an excellent book [4].

In Appendix, which is the hardest part, we give proofs of fundamentalpropositions.

In order to learn more on viscosity solutions, I give a list of “books”:A popular survey paper [6] by Crandall-Ishii-Lions on the theory of viscos-

ity solutions of second-order, degenerate elliptic PDEs is still a good choicefor undergraduate students to learn first. However, to my experience, itseems a bit hard for average undergraduate students to understand.

Bardi-Capuzzo Dolcetta’s book [1] contains lots of information on viscos-ity solutions for first-order PDEs (Hamilton-Jacobi equations) while Fleming-Soner’s [10] complements topics on second-order (degenerate) elliptic PDEswith applications in stochastic control problems.

Barles’ book [2] is also nice to learn his original techniques and Frenchlanguage simultaneously !

iii

It has been informed that Ishii would write a book [15] in Japanese onviscosity solutions in the near future, which must be more advanced thanthis.

For an important application via the viscosity solution theory, we referto Giga’s [12] on curvature flow equations. Also, I recommend the reader toconsult Lecture Notes [3] (Bardi-Crandall-Evans-Soner-Souganidis) not onlyfor various applications but also for a “friendly” introduction by Crandall,who first introduced the notion of viscosity solutions with P.-L. Lions in early80s.

If the reader is interested in section 6, I recommend him/her to attackCaffarelli-Cabre’s [4].

As a general PDE theory, although there are so many books on PDEs, Ionly refer to my favorite ones; Gilbarg-Trudinger’s [13] and Evans’ [8]. Alsoas a textbook for undergraduate students, Han-Lin’s short lecture notes [14]is a good choice.

Since this is a text-book, we do not refer the reader to original papersunless those are not mentioned in the books in our references.

Acknowledgment

First of all, I would like to express my thanks to Professors H. Morimotoand Y. Tonegawa for giving me the opportunity to have a series of lectures intheir universities. I would also like to thank Professors K. Ishii, T. Nagasawa,and a graduate student, K. Nakagawa, for their suggestions on the first draft.

I wish to express my gratitude to Professor H. Ishii for having given meenormous supply of ideas since 1980.

I also wish to thank the reviewer for several important suggestions.My final thanks go to Professor T. Ozawa for recommending me to publish

this manuscript. He kindly suggested me to change the original Japanese title(“A secret club on viscosity solutions”).

iv

Contents

1 Introduction 11.1 From classical solutions to weak solutions 21.2 Typical examples of weak solutions 41.2.1 Burgers’ equation 41.2.2 Hamilton-Jacobi equation 6

2 Definition 92.1 Vanishing viscosity method 92.2 Equivalent definitions 17

3 Comparison principle 233.1 Classical comparison principle 243.1.1 Degenerate elliptic PDEs 243.1.2 Uniformly elliptic PDEs 243.2 Comparison principle for first-order PDEs 263.3 Extension to second-order PDEs 313.3.1 Degenerate elliptic PDEs 333.3.2 Remarks on the structure condition 353.3.3 Uniformly elliptic PDEs 38

4 Existence results 404.1 Perron’s method 404.2 Representation formulas 454.2.1 Bellman equation 464.2.2 Isaacs equation 514.3 Stability 58

5 Generalized boundary value problems 625.1 Dirichlet problem 655.2 State constraint problem 685.3 Neumann problem 725.4 Growth condition at |x| → ∞ 75

v

6 Lp-viscosity solutions 786.1 A brief history 786.2 Definition and basic facts 816.3 Harnack inequality 846.3.1 Linear growth 856.3.2 Quadratic growth 876.4 Holder continuity estimates 896.5 Existence result 92

7 Appendix 957.1 Proof of Ishii’s lemma 957.2 Proof of the ABP maximum principle 1007.3 Proof of existence results for Pucci equations 1057.4 Proof of the weak Harnack inequlity 1087.5 Proof of the local maximum principle 113

References 118

Notation Index 120Index 121

vi

1 Introduction

Throughout this book, we will work in Ω (except in sections 4.2 and 5.4),where

Ω ⊂ Rn is open and bounded.

We denote by ⟨·, ·⟩ the standard inner product in Rn, and set |x| =√⟨x, x⟩ for x ∈ Rn. We use the standard notion of open balls: For r > 0

and x ∈ Rn,

Br(x) := y ∈ Rn | |x− y| < r, and Br := Br(0).

For a function u : Ω → R, we denote its gradient and Hessian matrix atx ∈ Ω, respectively, by

Du(x) :=

∂u(x)∂x1...

∂u(x)∂xn

,

D2u(x) :=

∂2u(x)∂x2

1· · · j-th · · · ∂2u(x)

∂x1∂xn

......

...

i-th · · · ∂2u(x)∂xi∂xj

· · · ......

......

∂2u(x)∂xn∂x1

· · · · · · · · · ∂2u(x)∂x2

n

.

Also, Sn denotes the set of all real-valued n × n symmetric matrices. Notethat if u ∈ C2(Ω), then D2u(x) ∈ Sn for x ∈ Ω.

We recall the standard ordering in Sn:

X ≤ Y ⇐⇒ ⟨Xξ, ξ⟩ ≤ ⟨Y ξ, ξ⟩ for ∀ξ ∈ Rn.

We will also use the following notion in sections 6 and 7: For ξ =t

(ξ1, . . . , ξn), η =t (η1, . . . , ηn) ∈ Rn, we denote by ξ ⊗ η the n × n matrixwhose (i, j)-entry is ξiηj for 1 ≤ i, j ≤ n;

ξ ⊗ η =

ξ1η1 · · · j-th · · · ξ1ηn...

......

i-th · · · ξiηj · · · ......

......

ξnη1 · · · · · · · · · ξnηn

.

1

We are concerned with general second-order partial differential equations(PDEs for short):

F (x, u(x), Du(x), D2u(x)) = 0 in Ω. (1.1)

We suppose (except in several sections) that

F : Ω×R×Rn × Sn → R is continuous

with respect to all variables.

1.1 From classical solutions to weak solutions

As the first example of PDEs, we present the Laplace equation:

−u = 0 in Ω. (1.2)

Here, we define u := trace(D2u). In the literature of the viscosity solutiontheory, we prefer to have the minus sign in front of .

Of course, since we do not require any boundary condition yet, all poly-nomials of degree one are solutions of (1.2). In many textbooks (particularlythose for engineers), under certain boundary condition, we learn how to solve(1.2) when Ω has some special shapes such as cubes, balls, the half-space orthe whole space Rn. Here, “solve” means that we find an explicit formula ofu using elementary functions such as polynomials, trigonometric ones, etc.

However, the study on (1.2) in such special domains is not applicablebecause, for instance, solutions of equation (1.2) represent the density of agas in a bottle, which is neither a ball nor a cube.

Unfortunately, in general domains, it seems impossible to find formulas forsolutions u with elementary functions. Moreover, in order to cover problemsarising in physics, engineering and finance, we will have to study more generaland complicated PDEs than (1.2). Thus, we have to deal with general PDEs(1.1) in general domains.

If we give up having formulas for solutions of (1.1), how do we investigatePDEs (1.1) ? In other words, what is the right question in the study of PDEs?

In the literature of the PDE theory, the most basic questions are as fol-lows:

2

(1) Existence: Does there exist a solution ?(2) Uniqueness: Is it the only solution ?(3) Stability: If the PDE changes a little,

does the solution change a little ?

The importance of the existence of solutions is trivial since, otherwise,the study on the PDE could be useless.

To explain the significance of the uniqueness of solutions, let us remem-ber the reason why we study the PDE. Usually, we discuss PDEs or theirsolutions to understand some specific phenomena in nature, engineerings oreconomics etc. Particularly, people working in applications want to knowhow the solution looks like, moves, behaves etc. For this purpose, it mightbe powerful to use numerical computations. However, numerical analysisonly shows us an “approximate” shapes, movements, etc. Thus, if thereare more than one solution, we do not know which is approximated by thenumerical solution.

Also, if the stability of solutions fails, we could not predict what will hap-pen from the numerical experiments even though the uniqueness of solutionsholds true.

Now, let us come back to the most essential question:

What is the “solution” of a PDE ?

For example, it is natural to call a function u : Ω → R a solution of(1.1) if there exist the first and second derivatives, Du(x) and D2u(x), forall x ∈ Ω, and (1.1) is satisfied at each x ∈ Ω when we plug them in the lefthand side of (1.1). Such a function u will be called a classical solution of(1.1).

However, unfortunately, it is difficult to seek for a classical solution be-cause we have to verify that it is sufficiently differentiable and that it satisfiesthe equality (1.1) simultaneously.

Instead of finding a classical solution directly, we have decided to choosethe following strategy:

(A) Find a candidate of the classical solution,(B) Check the differentiability of the candidate.

In the standard books, the candidate of a classical solution is called aweak solution; if the weak solution has the first and second derivatives, then

3

it becomes a classical solution. In the literature, showing the differentiabilityof solutions is called the study on the regularity of those.

Thus, with these terminologies, we may rewrite the above with mathe-matical terms:

(A) Existence of weak solutions,(B) Regularity of weak solutions.

However, when we cannot expect classical solutions of a PDE to exist,what is the right candidate of solutions ?

We will call a function the candidate of solutions of a PDE if it is a“unique” and “stable” weak solution under a suitable setting. In section 2,we will define such a candidate named “viscosity solutions” for a large classof PDEs, and in the proceeding sections, we will extend the definition tomore general (possibly discontinuous) functions and PDEs.

In the next subsection, we show a brief history on “weak solutions” toremind what was known before the birth of viscosity solutions.

1.2 Typical examples of weak solutions

In this subsection, we give two typical examples of PDEs to derive two kindsof weak solutions which are unique and stable.

1.2.1 Burgers’ equation

We consider Burgers’ equation, which is a model PDE in Fluid Mechanics:

∂u

∂t+

1

2

∂(u2)

∂x= 0 in R× (0,∞) (1.3)

under the initial condition:

u(x, 0) = g(x) for x ∈ R, (1.4)

where g is a given function.In general, we cannot find classical solutions of (1.3)-(1.4) even if g is

smooth enough. See [8] for instance.In order to look for the appropriate notion of weak solutions, we first

introduce a function space C10(R× [0,∞)) as a “test function space”:

C10(R× [0,∞)) :=

ϕ ∈ C1(R× [0,∞))

∣∣∣∣∣ there is K > 0 such thatsupp ϕ ⊂ [−K,K]× [0, K]

.

4

Here and later, we denote by supp ϕ the following set:

supp ϕ := (x, t) ∈ R× [0,∞) | ϕ(x, t) = 0.

Suppose that u satisfies (1.3). Multiplying (1.3) by ϕ ∈ C10(R × [0,∞))

and then, using integration by parts, we have∫R

∫ ∞

0

(u∂ϕ

∂t+u2

2

∂ϕ

∂x

)(x, t)dtdx+

∫Ru(x, 0)ϕ(x, 0)dx = 0.

Since there are no derivatives of u in the above, this equality makes sense ifu ∈ ∪K>0L

1((−K,K)×(0, K)). Hence, we may adapt the following propertyas the definition of weak solutions of (1.3)-(1.4).

∫R

∫ ∞

0

(u∂ϕ

∂t+u2

2

∂ϕ

∂x

)(x, t)dtdx+

∫Rg(x)ϕ(x, 0)dx = 0

for all ϕ ∈ C10(R× [0,∞)).

We often call this a weak solution in the distribution sense. As you no-ticed, we derive this notion by an essential use of integration by parts. Wesay that a PDE is in divergence form when we can adapt the notion ofweak solutions in the distribution sense. When the PDE is not in divergenceform, we say that it is in nondivergence form.

We note that the solution of (1.3) may have singularities even though theinitial value g belongs to C∞ by an observation via “characteristic method”.From the definition of weak solutions, we can derive the so-called Rankine-Hugoniot condition on the set of singularities.

On the other hand, unfortunately, we cannot show the uniqueness of weaksolutions of (1.3)-(1.4) in general while we know the famous Lax-Oleinikformula (see [8] for instance), which is the “expected” solution.

In order to obtain the uniqueness of weak solutions, for the definition, weadd the following property (called “entropy condition”) which holds for theexpected solution given by the Lax-Oleinik formula: There is C > 0 suchthat

u(x+ z, t)− u(x, t) ≤ Cz

t

for all (x, t, z) ∈ R× (0,∞)× (0,∞). We call u an entropy solution of (1.3)if it is a weak solution satisfying this inequality. It is also known that sucha weak solution has a certain stability property.

5

We note that this entropy solution satisfies the above mentioned impor-tant properties; “existence, uniqueness and stability”. Thus, it must be aright definition for weak solutions of (1.3)-(1.4).

1.2.2 Hamilton-Jacobi equations

Next, we shall consider general Hamilton-Jacobi equations, which arise inOptimal Control and Classical Mechanics:

∂u

∂t+H(Du) = 0 in (x, t) ∈ Rn × (0,∞) (1.5)

under the same initial condition (1.4).In this example, we suppose that H : Rn → R is convex, i.e.

H(θp+ (1− θ)q) ≤ θH(p) + (1− θ)H(q) (1.6)

for all p, q ∈ Rn, θ ∈ [0, 1].

Remark. Since a convex function is locally Lipschitz continuous in general,we do not need to assume the continuity of H.

Example. In Classical Mechanics, we often call this H a “Hamiltonian”.As a simple example of H, we have H(p) = |p|2.Notice that we cannot adapt the weak solution in the distribution sense

for (1.5) since we cannot use the integration by parts.

We next introduce the Lagrangian L : Rn → R defined by

L(q) = supp∈Rn

⟨p, q⟩ −H(p).

When H(p) = |p|2, it is easy to verify that the maximum is attained in theright hand side of the above.

It is surprising that we have a neat formula for the expected solution(called Hopf-Lax formula) presented by

u(x, t) = miny∈Rn

tL(x− y

t

)+ g(y)

. (1.7)

More precisely, it is shown that the right hand side of (1.7) is differentiableand satisfies (1.5) almost everywhere.

6

Thus, we could call u a weak solution of (1.5)-(1.4) when u satisfies (1.5)almost everywhere. However, if we decide to use this notion as a weak solu-tion, the uniqueness of those fails in general. We will see an example in thenext section.

As was shown for Burgers’ equation, in order to say that the “uniqueweak” solution is given by (1.7), we have to add one more property for thedefinition of weak solutions: There is C > 0 such that

u(x+ z, t)− 2u(x, t) + u(x− z, t) ≤ C|z|2 (1.8)

for all x, z ∈ R, t > 0. This is called the “semi-concavity” of u.We note that (1.8) is a hypothesis on the one-sided bound of second

derivatives of functions u.In 60s, Kruzkov showed that the limit function of approximate solutions

by the vanishing viscosity method (see the next section) has this property(1.8) when H is convex. He named u a “generalized” solution of (1.5) whenit satisfies (1.5) almost everywhere and (1.8).

To my knowledge, between Kruzkov’s works and the birth of viscositysolutions, there had been no big progress in the study of first-order PDEs innondivergence form.

Remark. The convexity (1.6) is a natural hypothesis when we consideronly optimal control problems where one person intends to minimize some“costs” (“energy” in terms of Physics). However, when we treat game prob-lems (one person wants to minimize costs while the other tries to maximizethem), we meet non-convex and non-concave (i.e. “fully nonlinear”)

Hamiltonians. See section 4.2.

In this book, since we are concerned with viscosity solutions of PDEs innondivergence form, for which the integration by parts argument cannot beused to define the notion of weak solutions in the distribution sense, we shallgive typical examples of such PDEs.

Example. (Bellman and Isaacs equations)We first give Bellman equations and Isaacs equations, which arise in

(stochastic) optimal control problems and differential games, respectively.As will be seen, those are extensions of linear PDEs.

Let A and B be sets of parameters. For instance, we suppose A and Bare (compact) subsets in Rm (for some m ≥ 1). For a ∈ A, b ∈ B, x ∈ Ω,

7

r ∈ R, p =t(p1, . . . , pn) ∈ Rn, and X = (Xij) ∈ Sn, we set

La(x, r, p,X) := −trace(A(x, a)X) + ⟨g(x, a), p⟩+ c(x, a)r,

La,b(x, r, p,X) := −trace(A(x, a, b)X) + ⟨g(x, a, b), p⟩+ c(x, a, b)r.

Here A(·, a), A(·, a, b), g(·, a), g(·, a, b), c(·, a) and c(·, a, b) are given functionsfor (a, b) ∈ A×B.

For inhomogeneous terms, we consider functions f(·, a) and f(·, a, b) in Ωfor a ∈ A and b ∈ B.

We call the following PDEs Bellman equations:

supa∈A

La(x, u(x), Du(x), D2u(x))− f(x, a) = 0 for x ∈ Ω. (1.9)

Notice that the supremum over A is taken at each point x ∈ Ω.Taking account of one more parameter set B, we call the following PDEs

Isaacs equations:

supa∈A

infb∈B

La,b(x, u(x), Du(x), D2u(x))− f(x, a, b) = 0 for x ∈ Ω (1.10)

and

infb∈B

supa∈A

La,b(x, u(x), Du(x), D2u(x))− f(x, a, b) = 0 for x ∈ Ω. (1.10′)

Example. (“Quasi-linear” equations)We say that a PDE is quasi-linear if the coefficients of D2u contains u

or Du. Although we will not study quasilinear PDEs in this book, we givesome of those which are in nondivergence form.

We first give the PDE of mean curvature type:

F (x, p,X) := −(|p|2trace(X)− ⟨Xp, p⟩

).

Notice that this F is independent of x-variables. We refer to [12] for appli-cations where this kind of operators appears.

Next, we show a relatively “new” one called L∞-Laplacian:

F (x, p,X) := −⟨Xp, p⟩.

Again, this F does not contain x-variables. We refer to Jensen’s work [16],where he first studied the PDE “−⟨D2uDu,Du⟩ = 0 in Ω” via the viscositysolution approach.

8

2 Definition

In this section, we derive the definition of viscosity solutions of (1.1) via thevanishing viscosity method.

We also give some basic properties of viscosity solutions and equivalentdefinitions using “semi-jets”.

2.1 Vanishing viscosity method

When the notion of viscosity solutions was born, in order to explain thereason why we need it, many speakers started in their talks by giving thefollowing typical example called the eikonal equation:

|Du|2 = 1 in Ω. (2.1)

We seek C1 functions satisfying (2.1) under the Dirichlet condition:

u(x) = 0 for x ∈ ∂Ω. (2.2)

However, since there is no classical solution of (2.1)-(2.2) (showing the non-existence of classical solutions is a good exercise), we intend to derive areasonable definition of weak solutions of (2.1).

In fact, we expect that the following function (the distance from ∂Ω)would be the unique solution of this problem (see Fig 2.1):

u(x) = dist(x, ∂Ω) := infy∈∂Ω

|x− y|.

-1 10

y

x

y=u(x)

Fig 2.1

1PSfrag replacements

−1

10y

y = u(x)

x

9

If we consider the case when n = 1 and Ω = (−1, 1), then the expectedsolution is given by

u(x) = 1− |x| for x ∈ [−1, 1]. (2.3)

Since this function is C∞ except at x = 0, we could decide to call u a weaksolution of (2.1) if it satisfies (2.1) in Ω except at finite points.

-1 0x

-1

1

y

Fig 2.2

PSfrag replacements

1−1

0xy

However, even in the above simple case of (2.1), we know that there areinfinitely many such weak solutions of (2.1) (see Fig 2.2); for example, −u isthe weak solution and

u(x) =

x+ 1 for x ∈ [−1,−1

2),

−x for x ∈ [−12, 12),

x− 1 for x ∈ [12, 1],

. . . etc.

Now, in order to look for an appropriate notion of weak solutions, weintroduce the so-called vanishing viscosity method; for ε > 0, we considerthe following PDE as an approximate equation of (2.1) when n = 1 andΩ = (−1, 1):

−εu′′ε + (u′ε)2 = 1 in (−1, 1),

uε(±1) = 0.(2.4)

The first term, −εu′′ε , in the left hand side of (2.4) is called the vanishingviscosity term (when n = 1) as ε→ 0.

By an elementary calculation, we can find a unique smooth function uεin the following manner: We first note that if a classical solution of (2.4)

10

exists, then it is unique. Thus, we may suppose that u′ε(0) = 0 by symmetry.Setting vε = u′ε, we first solve the ODE:

−εv′ε + v2ε = 1 in (−1, 1),vε(0) = 0.

(2.5)

It is easy to see that the solution of (2.5) is given by

vε(x) = − tanh(x

ε

).

Hence, we can find uε by

uε(x) = −ε log

cosh(xε

)cosh

(1ε

) = −ε log

(e

xε + e−

xε

e1ε + e−

1ε

).

It is a good exercise to show that uε converges to the function in (2.3)uniformly in [−1, 1].

Remark. Since uε(x) := −uε(x) is the solution ofεu′′ + (u′)2 = 1 in (−1, 1),

u(±1) = 0,

we have u(x) := limε→0 uε(x) = −u(x). Thus, if we replace −εu′′ by +εu′′,then the limit function would be different in general.

To define weak solutions, we adapt the properties which hold for the(uniform) limit of approximate solutions of PDEs with the “minus” vanishingviscosity term.

Let us come back to general second-order PDEs:

F (x, u,Du,D2u) = 0 in Ω. (2.6)

We shall use the following definition of classical solutions:

Definition. We call u : Ω → R a classical subsolution (resp.,supersolution, solution) of (2.6) if u ∈ C2(Ω) and

F (x, u(x), Du(x), D2u(x)) ≤ 0 (resp., ≥ 0, = 0) in Ω.

11

Remark. If F does not depend on X-variables (i.e. F (x, u,Du) = 0; first-order PDEs), we only suppose u ∈ C1(Ω) in the above in place of u ∈ C2(Ω).

Throughout this text, we also suppose the following monotonicity condi-tion with respect to X-variables:

Definition. We say that F is (degenerate) elliptic ifF (x, r, p,X) ≤ F (x, r, p, Y )

for all x ∈ Ω, r ∈ R, p ∈ Rn, X, Y ∈ Sn provided X ≥ Y.(2.7)

We notice that if F does not depend on X-variables (i.e. F = 0 is thefirst-order PDE), then F is automatically elliptic.

We also note that the left hand side F (x, r, p,X) = −trace(X) of theLaplace equation (1.2) is elliptic.

We will derive properties which hold true for the (uniform) limit (asε→ +0) of solutions of

−εu+ F (x, u,Du,D2u) = 0 in Ω (ε > 0). (2.8)

Note that since −εtrace(X) + F (x, r, p,X) is “uniformly” elliptic (see insection 3 for the definition) provided that F is elliptic, it is easier to solve(2.8) than (2.6) in practice. See [13] for instance.

Proposition 2.1. Assume that F is elliptic. Let uε ∈ C2(Ω) ∩ C(Ω)be a classical subsolution (resp., supersolution) of (2.8). If uε converges tou ∈ C(Ω) (as ε → 0) uniformly in any compact sets K ⊂ Ω, then, for anyϕ ∈ C2(Ω), we have

F (x, u(x), Dϕ(x), D2ϕ(x)) ≤ 0 (resp., ≥ 0)

provided that u− ϕ attains its maximum (resp., minimum) at x ∈ Ω.

Remark. When F does not depend on X-variables, we only need to sup-pose ϕ and uε to be in C1(Ω) as before.

Proof. We only give a proof of the assertion for subsolutions since theother one can be shown in a symmetric way.

12

Suppose that u−ϕ attains its maximum at x ∈ Ω for ϕ ∈ C2(Ω). Settingϕδ(y) := ϕ(y) + δ|y − x|4 for small δ > 0, we see that

(u− ϕδ)(x) > (u− ϕδ)(y) for y ∈ Ω \ x.

(This tiny technique to replace a maximum point by a “strict” one will appearin Proposition 2.2.)

Let xε ∈ Ω be a point such that (uε − ϕδ)(xε) = maxΩ(uε − ϕδ). Notethat xε also depends on δ > 0.

Since uε converges to u uniformly in Br(x) and x is the unique maximumpoint of u − ϕδ, we note that limε→0 xε = x. Thus, we see that xε ∈ Ω forsmall ε > 0. Notice that if we argue by ϕ instead of ϕδ, the limit of xε mightdiffer from x.

Thus, at xε ∈ Ω, we have

−εuε(xε) + F (xε, uε(xε), Duε(xε), D2uε(xε)) ≤ 0.

Since D(uε − ϕδ)(xε) = 0 and D2(uε − ϕδ)(xε) ≤ 0, in view of ellipticity, wehave

−εϕδ(xε) + F (xε, uε(xε), Dϕδ(xε), D2ϕδ(xε)) ≤ 0.

Sending ε→ 0 in the above, we have

F (x, u(x), Dϕδ(x), D2ϕδ(x)) ≤ 0.

Since Dϕδ(x) = Dϕ(x) and D2ϕδ(x) = D2ϕ(x), we conclude the proof. 2

Definition. We call u : Ω → R a viscosity subsolution (resp.,supersolution) of (2.6) if, for any ϕ ∈ C2(Ω),

F (x, u(x), Dϕ(x), D2ϕ(x)) ≤ 0 (resp., ≥ 0)

provided that u− ϕ attains its maximum (resp., minimum) at x ∈ Ω.We call u : Ω → R a viscosity solution of (2.6) if it is both a viscosity

sub- and supersolution of (2.6).

Remark. Here, we have given the definition to “general” functions butwe will often suppose that they are (semi-)continuous in Theorems etc.

In fact, in our propositions in sections 2.1, we will suppose that viscositysub- and supersolutions are continuous.

13

However, all the proposition in section 2.1 can be proved by replacing up-per and lower semi-continuity for viscosity subsolutions and supersolutions,respectively.

We will introduce general viscosity solutions in section 3.3.

Notation. In order to memorize the correct inequality, we will oftensay that u is a viscosity subsolution (resp., supersolution) of

F (x, u,Du,D2u) ≤ 0 (resp., ≥ 0) in Ω

if it is a viscosity subsolution (resp., supersolution) of (2.6).

Proposition 2.2. For u : Ω → R, the following (1) and (2) are equiva-lent:

(1) u is a viscosity subsolution (resp., supersolution) of (2.6),(2) if 0 = (u− ϕ)(x) > (u− ϕ)(x) (resp., < (u− ϕ)(x))

for ϕ ∈ C2(Ω), x ∈ Ω and x ∈ Ω \ x,then F (x, ϕ(x), Dϕ(x), D2ϕ(x)) ≤ 0 (resp., ≥ 0).

graph of u

graph of phi

hatxx

Fig 2.3

graphid

graphi1

PSfrag replacements

graph of ϕgraph of ϕδ

graph of ϕ(·) + (u− ϕ)(x)

xx

graph of u

Proof. The implication (1) ⇒ (2) is trivial.For the opposite implication in the subsolution case, suppose that u− ϕ

attains a maximum at x ∈ Ω. Set

ϕδ(x) = ϕ(x) + δ|x− x|4 + (u− ϕ)(x).

14

See Fig 2.3. Since 0 = (u− ϕδ)(x) > (u− ϕδ)(x) for x ∈ Ω \ x, (2) gives

F (x, ϕδ(x), Dϕδ(x), D2ϕδ(x)) ≤ 0,

which implies the assertion. 2

By the next proposition, we recognize that viscosity solutions are rightcandidates of weak solutions when F is elliptic.

Proposition 2.3. Assume that F is elliptic. A function u : Ω → Ris a classical subsolution (resp., supersolution) of (2.6) if and only if it is aviscosity subsolution (resp., supersolution) of (2.6) and u ∈ C2(Ω).

Proof. Suppose that u is a viscosity subsolution of (2.6) and u ∈ C2(Ω).Taking ϕ ≡ u, we see that u − ϕ attains its maximum at any points x ∈ Ω.Thus, the definition of viscosity subsolutions yields

F (x, u(x), Du(x), D2u(x)) ≤ 0 for x ∈ Ω.

On the contrary, suppose that u ∈ C2(Ω) is a classical subsolution of(2.6).

Fix any ϕ ∈ C2(Ω). Assuming that u − ϕ takes its maximum at x ∈ Ω,we have

D(u− ϕ)(x) = 0 and D2(u− ϕ)(x) ≤ 0.

Hence, in view of ellipticity, we have

0 ≥ F (x, u(x), Du(x), D2u(x)) ≥ F (x, u(x), Dϕ(x), D2ϕ(x)). 2

We introduce the sets of upper and lower semi-continuous functions: ForK ⊂ Rn,

USC(K) := u : K → R | u is upper semi-continuous in K,

and

LSC(K) := u : K → R | u is lower semi-continuous in K.

Remark. Throughout this book, we use the following maximum principlefor semi-continuous functions:

15

An upper semi-continuous function in a compact set attains its maximum.

We give the following lemma which will be used without mentioning it.Since the proof is a bit technical, the reader may skip it over first.

Proposition 2.4. Assume that u ∈ USC(Ω) (resp., u ∈ LSC(Ω)) is a viscos-ity subsolution (resp., supersolution) of (2.6) in Ω.

Then, for any open set Ω′ ⊂ Ω, u is a viscosity subsolution (resp., supersolution)of (2.6) in Ω′.

Proof.We only show the assertion for subsolutions since the other can be shownsimilarly.

For ϕ ∈ C2(Ω′), by Proposition 2.2, we suppose that for some x ∈ Ω′,

0 = (u− ϕ)(x) > (u− ϕ)(y) for all y ∈ Ω′ \ x.

For simplicity, we shall suppose x = 0.Choose r > 0 such that B2r ⊂ Ω′. We then choose ξk ∈ C∞(Rn) (k = 1, 2)

such that 0 ≤ ξk ≤ 1 in Rn, ξ1 + ξ2 = 1 in Rn,

ξ1 = 1 in Br, and ξ2 = 1 in Rn \B2r.

We define ψ = ξ1ϕ +Mξ2, where M = supΩ u + 1. Since it is easy to verify thatψ ∈ C2(Rn), and 0 = (u− ψ)(0) > (u− ψ)(x) for x ∈ Ω \ 0, we leave the proofto the reader. This concludes the proof. 2

2.2 Equivalent definitions

We present equivalent definitions of viscosity solutions. However, since wewill need those in the proof of uniqueness for second-order PDEs,

the reader may postpone this subsection until section 3.3.

First, we introduce “semi”-jets of functions u : Ω → R at x ∈ Ω by

J2,+u(x) :=

(p,X) ∈ Rn × Sn

∣∣∣∣∣∣∣∣∣u(y) ≤ u(x) + ⟨p, y − x⟩

+1

2⟨X(y − x), y − x⟩

+o(|y − x|2) as y ∈ Ω → x

16

and

J2,−u(x) :=

(p,X) ∈ Rn × Sn

∣∣∣∣∣∣∣∣∣u(y) ≥ u(x) + ⟨p, y − x⟩

+1

2⟨X(y − x), y − x⟩

+o(|y − x|2) as y ∈ Ω → x

.Note that J2,−u(x) = −J2,+(−u)(x).

Remark. We do not impose any continuity for u in these definitions.

We recall the notion of “small order o” in the above: For k ≥ 1,

f(x) ≤ o(|x|k) (resp., ≥ o(|x|k)) as x→ 0

⇐⇒

there is ω ∈ C([0,∞), [0,∞)) such that ω(0) = 0, and

supx∈Br\0

f(x)

|x|k≤ ω(r)

(resp., inf

x∈Br\0

f(x)

|x|k≥ −ω(|x|)

)

In the next proposition, we give some basic properties of semi-jets: (1)is a relation between semi-jets and classical derivatives, and (2) means thatsemi-jets are “defined” in dense sets of Ω.

Proposition 2.5. For u : Ω → R, we have the following:(1) If J2,+u(x) ∩ J2,−u(x) = ∅, then Du(x) and D2u(x) exist and,

J2,+u(x) ∩ J2,−u(x) = (Du(x), D2u(x)).

(2) If u ∈ USC(Ω) (resp., u ∈ LSC(Ω)), then

Ω =x ∈ Ω

∣∣∣∣ ∃xk ∈ Ω such that J2,+u(xk) = ∅, limk→∞

xk = x

(resp., Ω =

x ∈ Ω

∣∣∣∣ ∃xk ∈ Ω such that J2,−u(xk) = ∅, limk→∞

xk = x)

.

Proof. The proof of (1) is a direct consequence from the definition.We give a proof of the assertion (2) only for J2,+.Fix x ∈ Ω and choose r > 0 so that Br(x) ⊂ Ω. For ε > 0, we can choose

xε ∈ Br(x) such that u(xε)− ε−1|xε − x|2 = maxy∈Br(x)(u(y)− ε−1|y − x|2).

Since |xε − x|2 ≤ ε(maxBr(x)−u(x)), we see that xε converges to x ∈ Br(x)

17

as ε→ 0. Thus, we may suppose that xε ∈ Br(x) for small ε > 0. Hence, wehave

u(y) ≤ u(xε) +1

ε(|y − x|2 − |xε − x|2) for all y ∈ Br(x).

It is easy to check that (2(xε − x)/ε, 2ε−1I) ∈ J2,+u(xε). 2

We next introduce a sort of closure of semi-jets:

J2,±u(x) :=

(p,X) ∈ Rn × Sn

∣∣∣∣∣∣∣∃xk ∈ Ω and ∃(pk, Xk) ∈ J2,±u(xk)

such that (xk, u(xk), pk, Xk)→ (x, u(x), p,X) as k → ∞

.Proposition 2.6. For u : Ω → R, the following (1), (2), (3) are equiva-

lent.

(1) u is a viscosity subsolution (resp., supersolution) of (2.6).(2) For x ∈ Ω and (p,X) ∈ J2,+u(x) (resp., J2,−u(x)),

we have F (x, u(x), p,X) ≤ 0 (resp., ≥ 0).

(3) For x ∈ Ω and (p,X) ∈ J2,+u(x) (resp., J

2,−u(x)),

we have F (x, u(x), p,X) ≤ 0 (resp., ≥ 0).

Proof. Again, we give a proof of the assertion only for subsolutions.

Step 1: (2) =⇒ (3). For x ∈ Ω and (p,X) ∈ J2,+u(x), we can find (pk, Xk) ∈

J2,+u(xk) with xk ∈ Ω such that limk→∞(xk, u(xk), pk, Xk) = (x, u(x), p,X)and

F (xk, u(xk), pk, Xk) ≤ 0,

which implies (3) by sending k → ∞.Step 2: (3) =⇒ (1). For ϕ ∈ C2(Ω), suppose also (u−ϕ)(x) = max(u−ϕ).

Thus, the Taylor expansion of ϕ at x gives

u(y) ≤ u(x)+⟨Dϕ(x), y−x⟩+1

2⟨D2ϕ(x)(y−x), y−x⟩+o(|x−y|2) as y → x.

Thus, we have (Dϕ(x), D2ϕ(x)) ∈ J2,+u(x) ⊂ J2,+u(x).

Step 3: (1) =⇒ (2). For (p,X) ∈ J2,+u(x) (x ∈ Ω), we can find nonde-creasing, continuous ω : [0,∞) → [0,∞) such that ω(0) = 0 and

u(y) ≤ u(x) + ⟨p, y − x⟩+ 1

2⟨X(y − x), y − x⟩+ |y − x|2ω(|y − x|) (2.9)

18

as y → x. In fact, by the definition of o, we find ω0 ∈ C([0,∞), [0,∞)) suchthat ω0(0) = 0, and

ω0(r) ≥ supy∈Br(x)\x

1

|x− y|2u(y)− u(x)− ⟨p, y − x⟩ − 1

2⟨X(y − x), y − x⟩

,

we verify that ω(r) := sup0≤t≤r ω0(t) satisfies (2.9).Now, we define ϕ by

ϕ(y) := ⟨p, y − x⟩+ 1

2⟨X(y − x), y − x⟩+ ψ(|x− y|),

where

ψ(t) :=∫ √

3t

t

(∫ 2s

sω(r)dr

)ds ≥ t2ω(t).

It is easy to check that

(Dϕ(x), D2ϕ(x)) = (p,X) and (u− ϕ)(x) ≥ (u− ϕ)(y) for y ∈ Ω.

Therefore, we conclude the proof. 2

Remark. In view of the proof of Step 3, we verify that for x ∈ Ω,

J2,+u(x) =

(Dϕ(x), D2ϕ(x)) ∈ Rn × Sn

∣∣∣∣∣ ∃ϕ ∈ C2(Ω) such that u− ϕattains its maximum at x

,

J2,−u(x) =

(Dϕ(x), D2ϕ(x)) ∈ Rn × Sn

∣∣∣∣∣ ∃ϕ ∈ C2(Ω) such that u− ϕattains its minimum at x

.

Thus, we intuitively know J2,±u(x) from their graph.

Example. Consider the function u ∈ C([−1, 1]) in (2.3). From the graphbelow, we may conclude that J2,−u(0) = ∅, and J2,+u(0) = (1 × [0,∞)) ∪(−1 × [0,∞)) ∪ ((−1, 1)×R). See Fig 2.4.1 and 2.4.2.

We omit how to obtain J2,±u(0) of this and the next examples.We shall examine J2,± for discontinuous functions. For instance, consider

the Heaviside function:

u(x) :=

1 for x ≥ 0,0 for x < 0.

19

y=u(x)y=phi(x)

yy=x+1

Fig 2.4.1

x

PSfrag replacements

y = ϕ(x)

y = x+ 1

1y

y = u(x)

x1

−1

0

y=u(x)y=phi(x)

y

x

y=ax+1 (|a|<1)

Fig 2.4.2

PSfrag replacements

y = ϕ(x)

1y

y = u(x)

x1

−1

0y = ax+ 1(|a| < 1)

y=u(x)

0x

y=phi (x)

1

y=ax+1(a>0)

y

Fig 2.5

PSfrag replacements

y = ϕ(x)

1y

y = u(x)

x10

y = ax+ 1(a > 0)

20

We see that J2,−u(0) = ∅ and J2,+u(0) = (0 × [0,∞))∪ ((0,∞)×R). SeeFig 2.5.

In order to deal with “boundary value problems” in section 5, we preparesome notations: For a set K ⊂ Rn, which is not necessarily open, we definesemi-jets of u : K → R at x ∈ K by

J2,+K u(x) :=

(p,X) ∈ Rn × Sn

∣∣∣∣∣∣∣∣∣u(y) ≤ u(x) + ⟨p, y − x⟩

+1

2⟨X(y − x), y − x⟩

+o(|y − x|2) as y ∈ K → x

,

J2,−K u(x) :=

(p,X) ∈ Rn × Sn

∣∣∣∣∣∣∣∣∣u(y) ≥ u(x) + ⟨p, y − x⟩

+1

2⟨X(y − x), y − x⟩

+o(|y − x|2) as y ∈ K → x

,and

J2,±K u(x) :=

(p,X) ∈ Rn × Sn

∣∣∣∣∣∣∣∃xk ∈ K and ∃(pk, Xk) ∈ J2,±

K u(xk)such that (xk, u(xk), pk, Xk)→ (x, u(x), p,X) as k → ∞

.Remark. It is obvious to verify that

x ∈ Ω =⇒ J2,±Ω u(x) = J2,±

Ωu(x) and J

2,±Ω u(x) = J

2,±Ω u(x).

For x ∈ Ω, we shall simply write J2,±u(x) (resp., J2,±u(x)) for J2,±

Ω u(x) =

J2,±Ωu(x) (resp, J

2,±Ω u(x) = J

2,±Ω u(x)).

Example. Consider u(x) ≡ 0 in K := [0, 1]. It is easy to observe that

J2,+u(x) = J2,+K u(x) = 0 × [0,∞) provided x ∈ (0, 1). It is also easy to

verify thatJ2,+K u(0) = (0 × [0,∞)) ∪ ((0,∞)×R),

andJ2,−K u(0) = (0 × (−∞, 0]) ∪ ((−∞, 0)×R).

We finally give some properties of J2,±Ω

and J2,±Ω . Since the proof is easy,

we omit it.

21

Proposition 2.7. For u : Ω → R, ψ ∈ C2(Ω) and x ∈ Ω, we have

J2,±Ω

(u+ ψ)(x) = (Dψ(x), D2ψ(x)) + J2,±Ωu(x)

andJ2,±Ω (u+ ψ)(x) = (Dψ(x), D2ψ(x)) + J

2,±Ω u(x).

22

3 Comparison principle

In this section, we discuss the comparison principle, which implies the unique-ness of viscosity solutions when their values on ∂Ω coincide (i.e. under theDirichlet boundary condition). In the study of the viscosity solution theory,the comparison principle has been the main issue because the uniqueness ofviscosity solutions is harder to prove than existence and stability of them.

First, we recall some “classical” comparison principles and then, showhow to modify the proof to a modern “viscosity” version.

In this section, the comparison principle roughly means that

“Comparison principle”

viscosity subsolution uviscosity supersolution v

u ≤ v on ∂Ω

=⇒ u ≤ v in Ω

Modifying our proofs of comparison theorems below, we obtain a slightlystronger assertion than the above one:

viscosity subsolution uviscosity supersolution v

=⇒ max

Ω(u− v) = max

∂Ω(u− v)

We remark that the comparison principle implies the uniqueness of (con-tinuous) viscosity solutions under the Dirichlet boundary condition:

“Uniqueness for the Dirichlet problem”

viscosity solutions u and vu = v on ∂Ω

=⇒ u = v in Ω

Proof of “the comparison principle implies the uniqueness”.Since u (resp., v) and v (resp., u), respectively, are a viscosity subsolution

and supersolution, by u = v on ∂Ω, the comparison principle yields u ≤ v(resp., v ≤ u) in Ω. 2

In this section, we mainly deal with the following PDE instead of (2.6).

νu+ F (x,Du,D2u) = 0 in Ω, (3.1)

where we suppose thatν ≥ 0, (3.2)

andF : Ω×Rn × Sn → R is continuous. (3.3)

23

3.1 Classical comparison principle

In this subsection, we show that if one of viscosity sub- and supersolutionsis a classical one, then the comparison principle holds true. We call this the“classical” comparison principle.

3.1.1 Degenerate elliptic PDEs

We first consider the case when F is (degenerate) elliptic and ν > 0.

Proposition 3.1. Assume that ν > 0 and (3.3) hold. Assume alsothat F is elliptic. Let u ∈ USC(Ω) (resp., v ∈ LSC(Ω)) be a viscositysubsolution (resp., supersolution) of (3.1) and v ∈ LSC(Ω) ∩ C2(Ω) (resp.,u ∈ USC(Ω) ∩ C2(Ω)) a classical supersolution (resp., subsolution) of (3.1).

If u ≤ v on ∂Ω, then u ≤ v in Ω.

Proof. We only prove the assertion when u is a viscosity subsolution of(3.1) since the other one can be shown similarly.

Set maxΩ(u− v) =: θ and choose x ∈ Ω such that (u− v)(x) = θ.Suppose that θ > 0 and then, we will get a contradiction. We note that

x ∈ Ω because u ≤ v on ∂Ω.Thus, the definition of u and v respectively yields

νu(x) + F (x, Dv(x), D2v(x)) ≤ 0 ≤ νv(x) + F (x, Dv(x), D2v(x)).

Hence, by these inequalities, we have

νθ = ν(u− v)(x) ≤ 0,

which contradicts θ > 0. 2

3.1.2 Uniformly elliptic PDEs

Next, we present the comparison principle when ν = 0 but F is uniformlyelliptic in the following sense. Notice that if ν > 0 and F is uniformly ellip-tic, then Proposition 3.1 yields Proposition 3.3 below because our uniformellipticity implies (degenerate) ellipticity.

Throughout this book, we freeze the “uniform ellipticity” constants:

0 < λ ≤ Λ.

24

With these constants, we introduce the Pucci’s operators: For X ∈ Sn,

P+(X) := max−trace(AX) | λI ≤ A ≤ ΛI for A ∈ Sn,

P−(X) := min−trace(AX) | λI ≤ A ≤ ΛI for A ∈ Sn.

We give some properties of P±. We omit the proof since it is elementary.

Proposition 3.2. For X, Y ∈ Sn, we have the following:

(1) P+(X) = −P−(−X),(2) P±(θX) = θP±(X) for θ ≥ 0,(3) P+ is convex, P− is concave,

(4)

P−(X) + P−(Y ) ≤ P−(X + Y ) ≤ P−(X) + P+(Y )

≤ P+(X + Y ) ≤ P+(X) + P+(Y ).

Definition. We say that F : Ω×Rn ×Sn → R is uniformly elliptic(with the uniform ellipticity constants 0 < λ ≤ Λ) if

P−(X − Y ) ≤ F (x, p,X)− F (x, p, Y ) ≤ P+(X − Y )

for x ∈ Ω, p ∈ Rn, and X,Y ∈ Sn.

We also suppose the following continuity on F with respect to p ∈ Rn:There is µ > 0 such that

|F (x, p,X)− F (x, p′, X)| ≤ µ|p− p′| (3.4)

for x ∈ Ω, p, p′ ∈ Rn, and X ∈ Sn.

Proposition 3.3. Assume that (3.2), (3.3) and (3.4) hold. Assume alsothat F is uniformly elliptic. Let u ∈ USC(Ω) (resp., v ∈ LSC(Ω)) be aviscosity subsolution (resp., supersolution) of (3.1) and v ∈ LSC(Ω)∩C2(Ω)(resp., u ∈ USC(Ω)∩C2(Ω)) a classical supersolution (resp., subsolution) of(3.1).

If u ≤ v on ∂Ω, then u ≤ v in Ω.

Proof. We give a proof only when u is a viscosity subsolution and v aclassical supersolution of (3.1).

25

Suppose that maxΩ(u − v) =: θ > 0. Then, we will get a contradictionagain.

For ε > 0, we set ϕε(x) = εeδx1 , where δ := max(µ + 1)/λ, ν + 1 > 0.We next choose ε > 0 so small that

εmaxx∈Ω

eδx1 ≤ θ

2

Let x ∈ Ω be the point such that (u−v+ϕε)(x) = maxΩ(u−v+ϕε) ≥ θ.By the choice of ε > 0, since u ≤ v on ∂Ω, we see that x ∈ Ω.

From the definition of viscosity subsolutions, we have

νu(x) + F (x, D(v − ϕε)(x), D2(v − ϕε)(x)) ≤ 0.

By the uniform ellipticity and (3.4), we have

νu(x) + F (x, Dv(x), D2v(x)) + P−(−D2ϕε(x))− µ|Dϕε(x)| ≤ 0.

Noting that |Dϕε(x)| ≤ δεeδx1 and P−(−D2ϕε(x)) ≥ δ2ελeδx1 , we have

νu(x) + F (x, Dv(x), D2v(x)) + δε(λδ − µ)eδx1 ≤ 0. (3.5)

Since v is a classical supersolution of (3.1), by (3.5) and δ ≥ (µ + 1)/λ, wehave

ν(u− v)(x) + δεeδx1 ≤ 0.

Hence, we haveν(θ − ϕε(x)) ≤ −δεeδx1 ,

which gives a contradiction because δ ≥ ν + 1. 2

3.2 Comparison principle for first-order PDEs

In this subsection, without assuming that one of viscosity sub- and supersolu-tions is a classical one, we establish the comparison principle when F in (3.1)does not depend on D2u; first-order PDEs. We will study the comparisonprinciple for second-order ones in the next subsection.

In the viscosity solution theory, Theorem 3.4 below was the first surprisingresult.

Here, instead of (3.1), we shall consider the following PDE:

νu+H(x,Du) = 0 in Ω. (3.6)

26

We shall suppose thatν > 0, (3.7)

and that there is a continuous function ωH : [0,∞) → [0,∞) such thatωH(0) = 0 and

|H(x, p)−H(y, p)| ≤ ωH(|x− y|(1 + |p|)) for x, y ∈ Ω and p ∈ Rn. (3.8)

In what follows, we will call ωH in (3.8) a modulus of continuity. Fornotational simplicity, we use the following notation:

M := ω : [0,∞) → [0,∞) | ω(·) is continuous, ω(0) = 0.

Theorem 3.4. Assume that (3.7) and (3.8) hold. Let u ∈ USC(Ω) andv ∈ LSC(Ω) be a viscosity sub- and supersolution of (3.6), respectively.

If u ≤ v on ∂Ω, then u ≤ v in Ω.

Proof. Suppose maxΩ(u − v) =: θ > 0 as usual. Then, we will get acontradiction.

Notice that since both u and v may not be differentiable, we cannot usethe same argument as in Proposition 3.1.

Now, we present the most important idea in the theory of viscosity solu-tions to overcome this difficulty.

Setting Φε(x, y) := u(x) − v(y) − (2ε)−1|x − y|2 for ε > 0, we choose(xε, yε) ∈ Ω× Ω such that

Φε(xε, yε) = maxx,y∈Ω

Φε(x, y).

Noting that Φε(xε, yε) ≥ maxx∈ΩΦε(x, x) = θ, we have

|xε − yε|2

2ε≤ u(xε)− v(yε)− θ. (3.9)

Since Ω is compact, we can find x, y ∈ Ω, and εk > 0 such that limk→∞ εk = 0and limk→∞(xεk , yεk) = (x, y).

We shall simply write ε for εk (i.e. in what follows, “ε → 0” means thatεk → 0 when k → ∞).

Setting M := maxΩ u−minΩ v, by (3.9), we have

|xε − yε|2 ≤ 2εM → 0 (as ε→ 0).

27

Thus, we have x = y.Since (3.9) again implies

0 ≤ lim infε→0

|xε − yε|2

2ε≤ lim sup

ε→0

|xε − yε|2

2ε≤ lim sup

ε→0(u(xε)− v(yε))− θ

≤ (u− v)(x)− θ ≤ 0,

we have

limε→0

|xε − yε|2

ε= 0. (3.10)

Moreover, since (u − v)(x) = θ > 0, we have x ∈ Ω from the assumptionu ≤ v on ∂Ω. Thus, for small ε > 0, we may suppose that (xε, yε) ∈ Ω× Ω.

Furthermore, ignoring the left hand side in (3.9), we have

θ ≤ lim infε→0

(u(xε)− v(yε)). (3.11)

Taking ϕ(x) := v(yε) + (2ε)−1|x − yε|2, we see that u − ϕ attains itsmaximum at xε ∈ Ω. Hence, from the definition of viscosity subsolutions, wehave

νu(xε) +H(xε,

xε − yεε

)≤ 0.

On the other hand, taking ψ(y) := u(xε) − (2ε)−1|y − xε|2, we see thatv − ψ attains its minimum at yε ∈ Ω. Thus, from the definition of viscositysupersolutions, we have

νv(yε) +H(yε,

xε − yεε

)≥ 0.

The above two inequalities yield

ν(u(xε)− v(yε)) ≤ ωH

(|xε − yε|+

|xε − yε|2

ε

).

Sending ε→ 0 in the above together with (3.10) and (3.11), we have νθ ≤ 0,which is a contradiction. 2

Remark. In the above proof, we could show that limε→0 u(xε) = u(x) andlimε→0 v(yε) = v(x) although we do not need this fact. In fact, by (3.9), wehave

v(yε) ≤ u(xε)− θ,

28

which implies that

v(x) ≤ lim infε→0

v(yε) ≤ lim infε→0

u(xε)− θ ≤ lim supε→0

u(xε)− θ ≤ u(x)− θ,

and

v(x) ≤ lim infε→0

v(yε) ≤ lim supε→0

v(xε) ≤ lim supε→0

u(xε)− θ ≤ u(x)− θ.

Hence, since all the inequalities become the equalities, we have

u(x) = lim infε→0

u(xε) = lim supε→0

u(xε) and v(x) = lim infε→0

v(yε) = lim supε→0

v(yε).

We remark here that we cannot apply Theorem 3.4 to the eikonal equation(2.1) because we have to suppose ν > 0 in the above proof.

We shall modify the above proof so that the comparison principle forviscosity solutions of (2.1) holds.

To simplify our hypotheses, we shall consider the following PDE:

H(x,Du)− f(x) = 0 in Ω. (3.12)

Here, we suppose that H has homogeneous degree α > 0 with respect to thesecond variable; there is α > 0 such that

H(x, µp) = µαH(x, p) for x ∈ Ω, p ∈ Rn and µ > 0. (3.13)

To recover the lack of assumption ν > 0, we suppose the positivity of f ∈C(Ω); there is σ > 0 such that

minx∈Ω

f(x) =: σ > 0. (3.14)

Example. When H(x, p) = |p|2 (i.e. α = 2) and f(x) ≡ 1 ( i.e. σ = 1),equation (3.12) becomes (2.1).

The second comparison principle for first-order PDEs is as follows:

Theorem 3.5. Assume that (3.8), (3.13) and (3.14) hold. Let u ∈USC(Ω) and v ∈ LSC(Ω) be a viscosity sub- and supersolution of (3.12),respectively.

If u ≤ v on ∂Ω, then u ≤ v in Ω.

29

Proof. Suppose that maxΩ(u− v) =: θ > 0 as usual. Then, we will get acontradiction.

If we choose µ ∈ (0, 1) so that

(1− µ)maxΩ

u ≤ θ

2,

then we easily verify that

maxΩ

(µu− v) =: τ ≥ θ

2.

We note that for any z ∈ Ω such that (µu − v)(z) = τ , we may supposez ∈ Ω. In fact, otherwise (i.e. z ∈ ∂Ω), if we further suppose that µ < 1 isclose to 1 so that −(1 − µ)min∂Ω v ≤ θ/4, then the assumption (u ≤ v on∂Ω) implies

θ

2≤ τ = µu(z)− v(z) ≤ (µ− 1)v(z) ≤ θ

4,

which is a contradiction. For simplicity, we shall omit writing the dependenceon µ for τ and (xε, yε) below.

At this stage, we shall use the idea in the proof of Theorem 3.4: Considerthe mapping Φε : Ω× Ω → R defined by

Φε(x, y) := µu(x)− v(y)− |x− y|2

2ε.

Choose (xε, yε) ∈ Ω × Ω such that maxx,y∈ΩΦε(x, y) = Φε(xε, yε). Notethat Φε(xε, yε) ≥ τ ≥ θ/2.

As in the proof of Theorem 3.4, we may suppose that limε→0(xε, yε) =(x, y) for some (x, y) ∈ Ω× Ω (by taking a subsequence if necessary). Also,we easily see that

|xε − yε|2

2ε≤ µu(xε)− v(yε)− τ ≤Mµ := µmax

Ωu−min

Ωv. (3.15)

Thus, sending ε → 0, we have x = y. Hence, (3.15) implies that µu(x) −v(x) = τ , which yields x ∈ Ω because of the choice of µ. Thus, we see that(xε, yε) ∈ Ω× Ω for small ε > 0.

Moreover, (3.15) again implies

limε→0

|xε − yε|2

ε= 0. (3.16)

30

Now, taking ϕ(x) := (v(yε)+(2ε)−1|x−yε|2)/µ, we see that u−ϕ attainsits maximum at xε ∈ Ω. Thus, we have

H

(xε,

xε − yεµε

)≤ f(xε).

Hence, by (3.13), we have

H(xε,

xε − yεε

)≤ µαf(xε). (3.17)

On the other hand, taking ψ(y) = µu(xε) − (2ε)−1|y − xε|2, we see thatv − ψ attains its minimum at yε ∈ Ω. Thus, we have

H(yε,

xε − yεε

)≥ f(yε). (3.18)

Combining (3.18) with (3.17), we have

f(yε)− µαf(xε) ≤ H(yε,

xε − yεε

)−H

(xε,

xε − yεε

)≤ ωH

(|xε − yε|

(1 +

|xε − yε|ε

)).

Sending ε→ 0 in the above with (3.16), we have

(1− µα)f(x) ≤ 0,

which contradicts (3.14). 2

3.3 Extension to second-order PDEs

In this subsection, assuming a key lemma, we will present the comparisonprinciple for fully nonlinear, second-order, (degenerate) elliptic PDEs (3.1).

We first remark that the argument of the proof of the comparison principlefor first-order PDEs cannot be applied at least immediately.

Let us have a look at the difficulty. Consider the following simple PDE:

νu−u = 0, (3.19)

where ν > 0. As one can guess, if the argument does not work for this“easiest” PDE, then it must be hopeless for general PDEs.

31

However, we emphasize that the same argument as in the proof of The-orem 3.4 does not work. In fact, let u ∈ USC(Ω) and v ∈ LSC(Ω) be aviscosity sub- and supersolution of (3.19), respectively, such that u ≤ v on∂Ω. Setting Φε(x, y) := u(x) − v(y) − (2ε)−1|x − y|2 as usual, we choose(xε, yε) ∈ Ω× Ω so that maxx,y∈Ω Φε(x, y) = Φε(xε, yε) > 0 as before.

We may suppose that (xε, yε) ∈ Ω× Ω converges to (x, x) (as ε → 0) forsome x ∈ Ω such that (u − v)(x) > 0. From the definitions of u and v, wehave

νu(xε)−n

ε≤ 0 ≤ νv(yε) +

n

ε.

Hence, we only have

ν(u(xε)− v(yε)) ≤2n

ε,

which does not give any contradiction as ε→ 0.

How can we go beyond this difficulty ?

In 1983, P.-L. Lions first obtained the uniqueness of viscosity solutionsfor elliptic PDEs arising in stochastic optimal control problems (i.e. Bell-man equations; F is convex in (Du,D2u)). However, his argument heavilydepends on stochastic representation of viscosity solutions as “value func-tions”. Moreover, it seems hard to extend the result to Isaacs equations; Fis fully nonlinear.

The breakthrough was done by Jensen in 1988 in case when the coeffi-cients on the second derivatives of the PDE are constant. His argument reliespurely on “real-analysis” and can work even for fully nonlinear PDEs.

Then, Ishii in 1989 extended Jensen’s result to enable us to apply toelliptic PDEs with variable coefficients. We present here the so-called Ishii’slemma, which will be proved in Appendix.

Lemma 3.6. (Ishii’s lemma) Let u and w be in USC(Ω). For ϕ ∈C2(Ω× Ω), let (x, y) ∈ Ω× Ω be a point such that

maxx,y∈Ω

(u(x) + w(y)− ϕ(x, y)) = u(x) + w(y)− ϕ(x, y).

Then, for each µ > 1, there are X = X(µ), Y = Y (µ) ∈ Sn such that

(Dxϕ(x, y), X) ∈ J2,+

Ω u(x), (Dyϕ(x, y), Y ) ∈ J2,+

Ω w(y),

32

and

−(µ+ ∥A∥)(I 00 I

)≤(X 00 Y

)≤ A+

1

µA2,

where A = D2ϕ(x, y) ∈ S2n.

Remark. We note that if we suppose that u,w ∈ C2(Ω) and (x, y) ∈ Ω×Ωin the hypothesis, then we easily have

X = D2u(x), Y = D2w(y), and

(X 00 Y

)≤ A.

Thus, the last matrix inequality means that when u and w are only contin-uous, we get some error term µ−1A2, where µ > 1 will be large.

We also note that for ϕ(x, y) := |x− y|2/(2ε), we have

A := D2ϕ(x, y) =1

ε

(I −I−I I

)and ∥A∥ =

2

ε. (3.20)

For the last identity, since

∥A∥2 := sup

⟨A

(xy

), A

(xy

)⟩∣∣∣∣∣ |x|2 + |y|2 = 1

,

the triangle inequality yields ∥A∥2 = 2ε−2 sup|x − y|2 | |x|2 + |y|2 = 1 ≤4/ε2. On the other hand, taking x = −y (i.e. |x|2 = 1/2) in the supremumof the definition of ∥A∥2 in the above, we have ∥A∥2 ≥ 4/ε2.

Remark. The other way to show the above identity, we may use the factthat for B ∈ Sn, in general,

∥B∥ = max|λk| | λk is the eigen-value of B.

3.3.1 Degenerate elliptic PDEs

Now, we give our hypotheses on F , which is called the structure condition.

Structure conditionThere is an ωF ∈ M such that if X, Y ∈ Sn and µ > 1 satisfy

−3µ

(I 00 I

)≤(X 00 −Y

)≤ 3µ

(I −I−I I

),

then F (y, µ(x− y), Y )− F (x, µ(x− y), X)≤ ωF (|x− y|(1 + µ|x− y|)) for x, y ∈ Ω.

(3.21)

33

In section 3.3.2, we will see that if F satisfies (3.21), then it is elliptic.We first prove the comparison principle when (3.21) holds for F using

this lemma. Afterward, we will explain why assumption (3.21) is reasonable.

Theorem 3.7. Assume that ν > 0 and (3.21) hold. Let u ∈ USC(Ω)and v ∈ LSC(Ω) be a viscosity sub- and supersolution of (3.1), respectively.

If u ≤ v on ∂Ω, then u ≤ v in Ω.

Proof. Suppose that maxΩ(u− v) =: θ > 0 as usual. Then, we will get acontradiction.

Again, for ε > 0, consider the mapping Φε : Ω× Ω → R defined by

Φε(x, y) = u(x)− v(y)− 1

2ε|x− y|2.

Let (xε, yε) ∈ Ω×Ω be a point such that maxx,y∈ΩΦε(x, y) = Φε(xε, yε) ≥θ. As in the proof of Theorem 3.4, we may suppose that

limε→0

(xε, yε) = (x, x) for some x ∈ Ω (i.e. xε, yε ∈ Ω for small ε > 0).

Moreover, since we have (u− v)(x) = θ,

limε→0

|xε − yε|2

ε= 0, (3.22)

andθ ≤ lim inf

ε→0(u(xε)− v(yε)). (3.23)

In view of Lemma 3.6 (taking w := −v, µ := 1/ε, ϕ(x, y) = |x− y|2/(2ε))and its Remark, we find X, Y ∈ Sn such that(

xε − yεε

,X)∈ J2,+u(xε),

(xε − yεε

, Y)∈ J2,−v(yε),

and

−3

ε

(I 00 I

)≤(X 00 −Y

)≤ 3

ε

(I −I−I I

).

Thus, the equivalent definition in Proposition 2.6 implies that

νu(xε) + F(xε,

xε − yεε

,X)≤ 0 ≤ νv(yε) + F

(yε,

xε − yεε

, Y).

34

Hence, by virtue of our assumption (3.21), we have

ν(u(xε)− v(yε)) ≤ ωF

(|xε − yε|+

|xε − yε|2

ε

). (3.24)

Taking the limit infimum, as ε → 0, together with (3.22) and (3.23) in theabove, we have

νθ ≤ 0,

which is a contradiction. 2

3.3.2 Remarks on the structure condition

In order to ensure that assumption (3.21) is reasonable, we first present someexamples. For this purpose, we consider the Isaacs equation as in section1.2.2.

F (x, p,X) := supa∈A

infb∈B

La,b(x, p,X)− f(x, a, b),

where

La,b(x, p,X) := −trace(A(x, a, b)X) + ⟨g(x, a, b), p⟩ for (a, b) ∈ A×B.

If we suppose that A and B are compact sets in Rm (for some m ≥ 1),and that the coefficients in the above and f(·, a, b) satisfy the hypothesesbelow, then F satisfies (3.21).

(1) ∃M1 > 0 and ∃σij(·, a, b) : Ω → R such that Aij(x, a, b) =m∑k=1

σik(x, a, b)σjk(x, a, b), and |σjk(x, a, b)− σjk(y, a, b)| ≤M1|x− y|

for x, y ∈ Ω, i, j = 1, . . . , n, k = 1, . . . ,m, a ∈ A, b ∈ B,(2) ∃M2 > 0 such that |gi(x, a, b)− gi(y, a, b)| ≤M2|x− y| for x, y ∈ Ω,

i = 1, . . . , n, a ∈ A, b ∈ B,(3) ∃ωf ∈ M such that

|f(x, a, b)− f(y, a, b)| ≤ ωf (|x− y|) for x, y ∈ Ω, a ∈ A, b ∈ B.

We shall show (3.21) only when

F (x, p,X) := −n∑

i,j=1

m∑k=1

σik(x, a, b)σjk(x, a, b)Xij

35

for a fixed (a, b) ∈ A×B because we can modify the proof below to generalF .

Thus, we shall omit writing indices a and b.To verify assumption (3.21), we choose X, Y ∈ Sn such that(

X 00 −Y

)≤ 3µ

(I −I−I I

).

Setting ξk =t(σ1k(x), . . . , σnk(x)) and ηk =t(σ1k(y), . . . , σnk(y)) for anyfixed k ∈ 1, 2, . . . ,m, we have⟨(

X 00 −Y

)(ξkηk

),

(ξkηk

)⟩≤ 3µ

⟨(I −I−I I

)(ξkηk

),

(ξkηk

)⟩= 3µ|ξk − ηk|2≤ 3µnM2

1 |x− y|2.

Therefore, taking the summation over k ∈ 1, . . . ,m, we have

F (y, µ(x− y), Y )− F (x, µ(x− y), X) ≤n∑

i,j=1

(−Aij(y)Yij + Aij(x)Xij)

=m∑k=1

(−⟨Y ηk, ηk⟩+ ⟨Xξk, ξk⟩)

≤ 3µmnM21 |x− y|2. 2

We next give other reasons why (3.21) is a suitable assumption. Thereader can skip the proof of the following proposition if he/she feels that theabove reason is enough to adapt (3.21).

Proposition 3.8. (1) (3.21) implies ellipticity.(2) Assume that F is uniformly elliptic. If ω ∈ M satisfies that supr≥0 ω(r)/(r +1) <∞, and

|F (x, p,X)− F (y, p,X)| ≤ ω(|x− y|(∥X∥+ |p|+ 1)) (3.25)

for x, y ∈ Ω, p ∈ Rn, X ∈ Sn, then (3.21) holds for F .

Proof. For a proof of (1), we refer to Remark 3.4 in [6].For the reader’s convenience, we give a proof of (2) which is essentially used

in a paper by Ishii-Lions (1990). Let X,Y ∈ Sn satisfy the matrix inequality in(3.21). Note that X ≤ Y .

36

Multiplying

(−I −I−I I

)to the last matrix inequality from both sides, we

have (X − Y X + YX + Y X − Y

)≤ 12µ

(0 00 I

).

Thus, multiplying

(ξsη

)for s ∈ R and ξ, η ∈ Rn with |η| = |ξ| = 1, we see that

0 ≤ (12µ− ⟨(X − Y )η, η⟩)s2 − 2⟨(X + Y )ξ, η⟩s− ⟨(X − Y )ξ, ξ⟩.

Hence, we have

|⟨(X + Y )ξ, η⟩|2 ≤ |⟨(X − Y )ξ, ξ⟩|(12µ+ |⟨(X − Y )η, η⟩|),

which implies∥X + Y ∥ ≤ ∥X − Y ∥1/2(12µ+ ∥X − Y ∥)1/2.

Thus, we have

∥X∥ ≤ 1

2(∥X − Y ∥+ ∥X + Y ∥) ≤ ∥X − Y ∥1/2(6µ+ ∥X − Y ∥)1/2.

Since X ≤ Y (i.e. the eigen-values of X − Y are non-positive), we see that

F (y, p,X)− F (y, p, Y ) ≥ P−(X − Y ) ≥ λ∥X − Y ∥. (3.26)

For the last inequality, we recall Remark after Lemma 3.6.Since we may suppose ω is concave, for any fixed ε > 0, there is Mε > 0 such

that ω(r) ≤ ε+Mεr and ω(r) = infε>0(ε+Mεr) for r ≥ 0. By (3.25) and (3.26),since ∥X∥ ≤ 3µ and ∥Y ∥ ≤ 3µ, we have

F (y, p, Y )− F (x, p,X)

≤ ε+Mε|x− y|(|p|+ 1) + sup0≤t≤6µ

Mε|x− y|t1/2(6µ+ t)1/2 − λt

.

Noting that

Mε|x− y|t1/2(6µ+ t)1/2 − λt ≤ 3

λM2

ε µ|x− y|2,

we haveF (y, µ(x− y), Y )− F (x, µ(x− y), X)

≤ ε+Mε|x− y|(µ|x− y|+ 1) + 3λ−1M2ε µ|x− y|2,

which implies the assertion by taking the infimum over ε > 0. 2

37

3.3.3 Uniformly elliptic PDEs

We shall give a comparison result corresponding to Proposition 3.3; F isuniformly elliptic and ν ≥ 0.

Theorem 3.9. Assume that (3.2), (3.3), (3.4) and (3.21) hold. Assumealso that F is uniformly elliptic. Let u ∈ USC(Ω) and v ∈ LSC(Ω) be aviscosity sub- and supersolution of (3.1), respectively.

If u ≤ v on ∂Ω, then u ≤ v in Ω.

Remark. As in Proposition 3.3, we may suppose ν = 0.

Proof. Suppose that maxΩ(u− v) =: θ > 0.Setting σ := (µ+ 1)/λ, we choose δ > 0 so that

δmaxx∈Ω

eσx1 ≤ θ

2.

We then set τ := maxx∈Ω(u(x)− v(x) + δeσx1) ≥ θ > 0.Putting ϕ(x, y) := (2ε)−1|x − y|2 − δeσx1 , we let (xε, yε) ∈ Ω × Ω be the

maximum point of u(x)− v(y)− ϕ(x, y) over Ω× Ω.By the compactness of Ω, we may suppose that (xε, yε) → (x, y) ∈ Ω×Ω

as ε→ 0 (taking a subsequence if necessary). Since u(xε)−v(yε) ≥ ϕ(xε, yε),we have |xε−yε|2 ≤ 2ε(maxΩ u−minΩ v+2−1θ) and moreover, x = y. Hence,we have

u(x)− v(x) + δeσx1 ≥ τ,

which implies x ∈ Ω because of our choice of δ. Thus, we may suppose that(xε, yε) ∈ Ω× Ω for small ε > 0. Moreover, as before, we see that

limε→0

|xε − yε|2

ε= 0. (3.27)

Applying Lemma 3.6 to u(x) := u(x)+δeσx1 and−v(y), we findX, Y ∈ Sn

such that ((xε − yε)/ε,X) ∈ J2,+u(xε), ((xε − yε)/ε, Y ) ∈ J

2,−v(yε), and

−3

ε

(I OO I

)≤(X OO −Y

)≤ 3

ε

(I −I−I I

).

We shall simply write x and y for xε and yε, respectively.

38

Note that Proposition 2.7 implies(x− y

ε− δσeσx1e1, X − δσ2eσx1I1

)∈ J

2,+u(x),

where e1 ∈ Rn and I1 ∈ Sn are given by

e1 :=

10...0

and I1 :=

1 0 · · · 00 0 · · · 0...

......

0 0 · · · 0

.

Setting r := δσeσx1 , from the definition of u and v, we have

0 ≤ F(y,x− y

ε, Y)− F

(x,x− y

ε− re1, X − σrI1

).

In view of the uniform ellipticity and (3.4), we have

0 ≤ rµ+ σrP+(I1) + F(y,x− y

ε, Y)− F

(x,x− y

ε,X

).

Hence, by (3.21) and the definition of P+, we have

0 ≤ r(µ− σλ) + ωF

(|x− y|+ |x− y|2

ε

),

which together with (3.27) yields 0 ≤ δσeσx1(µ−σλ). This is a contradictionbecause of our choice of σ > 0. 2

39

4 Existence results

In this section, we present some existence results for viscosity solutions ofsecond-order (degenerate) elliptic PDEs.

We first present a convenient existence result via Perron’s method, whichwas established by Ishii in 1987.

Next, for Bellman and Isaacs equations, we give representation formulasfor viscosity solutions. From the dynamic programming principle below, wewill realize how natural the definition of viscosity solutions is.

4.1 Perron’s method

In order to introduce Perron’s method, we need the notion of viscosity solu-tions for semi-continuous functions.

Definition. For any function u : Ω → R, we denote the upper andlower semi-continuous envelope of u by u∗ and u∗, respectively, which aredefined by

u∗(x) = limε→0

supy∈Bε(x)∩Ω

u(y) and u∗(x) = limε→0

infy∈Bε(x)∩Ω

u(y).

We give some elementary properties for u∗ and u∗ without proofs.

Proposition 4.1. For u : Ω → R, we have

(1) u∗(x) ≤ u(x) ≤ u∗(x) for x ∈ Ω,(2) u∗(x) = −(−u)∗(x) for x ∈ Ω,(3) u∗(resp., u∗) is upper (resp., lower) semi-continuous in Ω, i.e.

lim supy→x

u∗(y) ≤ u∗(x), (resp., lim infy→x

u∗(y) ≥ u∗(x)) for x ∈ Ω,

(4) if u is upper (resp., lower) semi-continuous in Ω,then u(x) = u∗(x) (resp., u(x) = u∗(x)) for x ∈ Ω.

With these notations, we give our definition of viscosity solutions of

F (x, u,Du,D2u) = 0 in Ω. (4.1)

Definition. We call u : Ω → R a viscosity subsolution (resp., superso-lution) of (4.1) if u∗ (resp., u∗) is a viscosity subsolution (resp., supersolution)of (4.1).

40

We call u : Ω → R a viscosity solution of (4.1) if it is both a viscositysub- and supersolution of (4.1).

Remark. We note that we supposed that viscosity sub- and supersolu-tions are, respectively, upper and lower semi-continuous in our comparisonprinciple in section 3. Adapting the above new definition, we omit the semi-continuity for viscosity sub- and supersolutions in Propositions 3.1, 3.3 andTheorems 3.4, 3.5, 3.7, 3.9.

In what follows,

we use the above definition.

Remark. We remark that the comparison principle Theorem 3.7 impliesthe continuity of viscosity solutions.

“Continuity of viscosity solutions”

viscosity solution usatisfies u∗ = u∗ on ∂Ω

=⇒ u ∈ C(Ω)

Proof of the continuity of u. Since u∗ and u∗ are, respectively, a viscositysubsolution and a viscosity supersolution and u∗ ≤ u∗ on ∂Ω, Theorem 3.7yields u∗ ≤ u∗ in Ω. Because u∗ ≤ u ≤ u∗ in Ω, we have u = u∗ = u∗ in Ω;u ∈ C(Ω). 2

We first show that the “point-wise” supremum (resp., infimum) of viscos-ity subsolutions (resp., supersolution) becomes a viscosity subsolution (resp.,supersolution).

Theorem 4.2. Let S be a non-empty set of upper (resp., lower) semi-continuous viscosity subsolutions (resp., supersolutions) of (4.1).

Set u(x) := supv∈S v(x) (resp., u(x) := infv∈S v(x)). If supx∈K |u(x)| <∞ for any compact sets K ⊂ Ω, then u is a viscosity subsolution (resp.,supersolution) of (4.1).

Proof. We only give a proof for subsolutions since the other can be provedin a symmetric way.

For x ∈ Ω, we suppose that 0 = (u∗−ϕ)(x) > (u∗−ϕ)(x) for x ∈ Ω \ xand ϕ ∈ C2(Ω). We shall show that

F (x, ϕ(x), Dϕ(x), D2ϕ(x)) ≤ 0. (4.2)

41

Let r > 0 be such that B2r(x) ⊂ Ω. We can find s > 0 such that

max∂Br(x)

(u∗ − ϕ) ≤ −s. (4.3)

We choose xk ∈ Br(x) such that limk→∞ xk = x, u∗(x) − k−1 ≤ u(xk)and |ϕ(xk)−ϕ(x)| < 1/k. Moreover, we select upper semi-continuous uk ∈ Ssuch that uk(xk) + k−1 ≥ u(xk).

By (4.3), for 3/k < s, we have

max∂Br(x)

(uk − ϕ) < (uk − ϕ)(xk).

Thus, for large k > 3/s, there is yk ∈ Br(x) such that uk − ϕ attains itsmaximum over Br(x) at yk. Hence, we have

F (yk, uk(yk), Dϕ(yk), D2ϕ(yk)) ≤ 0. (4.4)

Taking a subsequence if necessary, we may suppose z := limk→∞ yk. Since

(u∗ − ϕ)(x) ≤ (uk − ϕ)(xk) +3

k≤ (uk − ϕ)(yk) +

3

k≤ (u∗ − ϕ)(yk) +

3

k

by the upper semi-continuity of u∗, we have

(u∗ − ϕ)(x) ≤ (u∗ − ϕ)(z),

which yields z = x, and moreover, limk→∞ uk(yk) = u∗(x) = ϕ(x). Therefore,sending k → ∞ in (4.4), by the continuity of F , we obtain (4.2). 2

Our first existence result is as follows.

Theorem 4.3. Assume that F is elliptic. Assume also that there area viscosity subsolution ξ ∈ USC(Ω) ∩ L∞

loc(Ω) and a viscosity supersolutionη ∈ LSC(Ω) ∩ L∞

loc(Ω) of (4.1) such that

ξ ≤ η in Ω.

Then, u(x) := supv∈S v(x) (resp., u(x) = infw∈S w(x)) is a viscosity solu-tion of (4.1), where

S :=

v ∈ USC(Ω)

∣∣∣∣∣ v is a viscosity subsolutionof (4.1) such that ξ ≤ v ≤ η in Ω

42

(resp., S :=

w ∈ LSC(Ω)

∣∣∣∣∣ w is a viscosity supersolutionof (4.1) such that ξ ≤ w ≤ η in Ω

).

Sketch of proof. We only give a proof for u since the other can be shownin a symmetric way.

First of all, we notice that S = ∅ since ξ ∈ S.Due to Theorem 4.2, we know that u is a viscosity subsolution of (4.1).

Thus, we only need to show that it is a viscosity supersolution of (4.1).Assume that u ∈ LSC(Ω). Assuming that 0 = (u − ϕ)(x) < (u − ϕ)(x)

for x ∈ Ω \ x and ϕ ∈ C2(Ω), we shall show that

F (x, ϕ(x), Dϕ(x), D2ϕ(x)) ≥ 0.

Suppose that this conclusion fails; there is θ > 0 such that

F (x, ϕ(x), Dϕ(x), D2ϕ(x)) ≤ −2θ.

Hence, there is r > 0 such that

F (x, ϕ(x) + t,Dϕ(x), D2ϕ(x)) ≤ −θ for x ∈ Br(x) ⊂ Ω and |t| ≤ r. (4.5)

First, we claim that ϕ(x) < η(x). Indeed, otherwise, since ϕ ≤ u ≤ η inΩ, η − ϕ attains its minimum at x ∈ Ω. See Fig 4.1.

y=phi(x)

y=eta_*(x)

y=u(x)

y=xi^*(x)

xhatx

O

Fig 4.1

PSfrag replacements

y = η(x)y = u(x)y = ξ(x)

xx

y = ϕ(x)

Ω

Hence, from the definition of supersolution η, we get a contradiction to(4.5) for x = x and t = 0.

43

We may suppose that ξ(x) < η(x) since, otherwise, ξ = ϕ = η at x.Setting 3τ := η(x)− u(x) > 0, from the lower and upper semi-continuity ofη and ξ, respectively, we may choose s ∈ (0, r] such that

ξ(x) + τ ≤ ϕ(x) + 2τ ≤ η(x) for x ∈ B2s(x).

Moreover, we can choose ε ∈ (0, s) and τ0 ∈ (0,minτ , r) such thatϕ(x) + 2τ0 ≤ u(x) for x ∈ Bs+ε(x) \Bs−ε(x).

If we can define a function w ∈ S such that w(x) > u(x), then we finishour proof because of the maximality of u at each point.

Now, we set

w(x) :=

maxu(x), ϕ(x) + τ0 in Bs(x),u(x) in Ω \Bs(x).

See Fig 4.2.

2s

x

y=eta_*(x)

y=u(x)

y=xi^*(x)

O

hatx

y=phi(x)+s

Fig 4.2y=w(x)

PSfrag replacements

y = η(x)y = u(x)y = ξ(x)y = w(x)

xx

y = ϕ(x) + τ0

Ω3τ

It suffices to show that w ∈ S. Because of our choice of τ0, s > 0, it iseasy to see ξ ≤ w ≤ η in Ω. Thus, we only need to show that w is a viscositysubsolution of (4.1).

To this end, we suppose that (w∗ − ψ)(x) ≤ (w∗ − ψ)(z) = 0 for x ∈ Ω,and then we will get

F (z, w∗(z), Dψ(z), D2ψ(z)) ≤ 0. (4.6)

If z ∈ Ω\Bs(x) =: Ω′, by Proposition 2.4, then u∗−ψ attains its maximumat z ∈ Ω′, we get (4.6).

44

If z ∈ ∂Bs(x), then (4.6) holds again since w = u in Bs+ε(x) \Bs−ε(x).It remains to show (4.6) when z ∈ Bs(x). Since ϕ + τ0 is a viscosity

subsolution of (4.1) in Bs(x), Theorem 4.2 with Ω := Bs(x) yields (4.6). 2

Correct proof, which the reader may skip first. Since we do not suppose thatu ∈ LSC(Ω) here, we have to work with u∗.

Suppose that 0 = (u∗−ϕ)(x) < (u∗−ϕ)(x) for x ∈ Ω\x for some ϕ ∈ C2(Ω),x ∈ Ω, θ > 0 and

F (x, ϕ(x), Dϕ(x), D2ϕ(x)) ≤ −2θ.

Hence, we get (4.5) even in this case.We also show that the w defined in the above is a viscosity subsolution of (4.1).

It only remains to check that supΩ(w − u) > 0.In fact, choosing xk ∈ B1/k(x) such that

u∗(x) +1

k≥ u(xk),

we easily verify that if 1/k ≤ minτ0/2, s and |ϕ(x)−ϕ(xk)| < τ0/2, then we have

w(xk) ≥ ϕ(xk) + τ0 > ϕ(x) +τ02

= u∗(x) +τ02

≥ u(xk). 2

4.2 Representation formula

In this subsection, for given Bellman and Isaacs equations, we present theexpected solutions, which are called “value functions”. In fact, via the dy-namic programming principle for the value functions, we verify that they areviscosity solutions of the corresponding PDEs.

Although this subsection is very important to learn how the notion ofviscosity solutions is the right one from a view point of applications in optimalcontrol and games,

if the reader is more interested in the PDE theory than these applications,he/she may skip this subsection.

We shall restrict ourselves to

investigate the formulas only for first-order PDEs

45

because in order to extend the results below to second-order ones, we needto introduce some terminologies from stochastic analysis. However, this istoo much for this thin book.

As will be seen, we study the minimization of functionals associated withordinary differential equations (ODEs for short), which is called a “deter-ministic” optimal control problem. When we adapt “stochastic” differentialequations instead of ODEs, those are called “stochastic” optimal controlproblems. We refer to [10] for the later.

Moreover, to avoid mentioning the boundary condition, we will work onthe whole domain Rn.

Throughout this subsection, we also suppose (3.7); ν > 0.

4.2.1 Bellman equation

We fix a control set A ⊂ Rm for some m ∈ N. We define A by

A := α : [0,∞) → A | α(·) is measurable.

For x ∈ Rn and α ∈ A, we denote by X(·; x, α) the solution ofX ′(t) = g(X(t), α(t)) for t > 0,

X(0) = x,(4.7)

where we will impose a sufficient condition on continuous functions g : Rn ×A→ Rn so that (4.7) is uniquely solvable.

For given f : Rn×A→ R, under suitable assumptions (see (4.8) below),we define the cost functional for X(·;x, α):

J(x, α) :=∫ ∞

0e−νtf(X(t;x, α), α(t))dt.

Here, ν > 0 is called a discount factor, which indicates that the right handside of the above is finite.

Now, we shall consider the optimal cost functional, which is called thevalue function in the optimal control problem;

u(x) := infα∈A

J(x, α) for x ∈ Rn.

Theorem 4.4. (Dynamic Programming Principle) Assume that(1) sup

a∈A

(∥f(·, a)∥L∞(Rn) + ∥g(·, a)∥W 1,∞(Rn)

)<∞,

(2) supa∈A

|f(x, a)− f(y, a)| ≤ ωf (|x− y|) for x, y ∈ Rn,(4.8)

46

where ωf ∈ M.For any T > 0, we have

u(x) = infα∈A

(∫ T

0e−νtf(X(t;x, α), α(t))dt+ e−νTu(X(T ;x, α))

).

Proof. For fixed T > 0, we denote by v(x) the right hand side of theabove.

Step 1: u(x) ≥ v(x). Fix any ε > 0, and choose αε ∈ A such that

u(x) + ε ≥∫ ∞

0e−νtf(X(t; x, αε), αε(t))dt.

Setting x = X(T ;x, αε) and αε ∈ A by αε(t) = αε(T + t) for t ≥ 0, we have∫ ∞

0e−νtf(X(t;x, αε), αε(t))dt =

∫ T

0e−νtf(X(t;x, αε), αε(t))dt

+e−νT∫ ∞

0e−νtf(X(t; x, αε), αε(t))dt.

Here and later, without mentioning, we use the fact that

X(t+ T ; x, α) = X(t; x, α) for T > 0, t ≥ 0 and α ∈ A,

whereα(t) := α(t+ T ) (t ≥ 0) and x := X(T ;x, α).

Indeed, the above relation holds true because of the uniqueness of solutionsof (4.7) under assumptions (4.8). See Fig 4.3.

Thus, taking the infimum in the second term of the right hand side of theabove among A, we have

u(x) + ε ≥∫ T

0e−νtf(X(t; x, α), α(t))dt+ e−νTu(x),

which implies one-sided inequality by taking the infimum over A since ε > 0is arbitrary.

Step 2: u(x) ≤ v(x). Fix ε > 0 again, and choose αε ∈ A such that

v(x) + ε ≥∫ T

0e−νtf(X(t; x, αε), αε(t))dt+ e−νTu(x),

47

xx

t=0

t=T

X(t;x,a)

X(t+T;x,a)=X(t;hatx,hata)

hatx=X(T;x,a)

Fig 4.3

PSfrag replacements

X(t; x, α)X(t+ T ;x, α) = X(t; x, α)

t = 0x

t = Tx = X(T ; x, α)

where x := X(T ;x, αε). We next choose α1 ∈ A such that

u(x) + ε ≥∫ ∞

0e−νtf(X(t; x, α1), α1(t))dt.

Now, setting

α0(t) :=

αε(t) for t ∈ [0, T ),α1(t− T ) for t ≥ T,

we see thatv(x) + 2ε ≥

∫ ∞

0e−νtf(X(t;x, α0), α0(t))dt,

which gives the opposite inequality by taking the infimum over α0 ∈ A sinceε > 0 is arbitrary again. 2

Now, we give an existence result for Bellman equations.

Theorem 4.5. Assume that (4.8) holds. Then, u is a viscosity solutionof

supa∈A

νu− ⟨g(x, a), Du⟩ − f(x, a) = 0 in Rn. (4.9)

Sketch of proof. In Steps 1 and 2, we give a proof when u ∈ USC(Rn)and u ∈ LSC(Rn), respectively.

Step 1: Subsolution property. Fix ϕ ∈ C1(Rn), and suppose that 0 =(u− ϕ)(x) ≥ (u− ϕ)(x) for some x ∈ Rn and any x ∈ Rn.

Fix any a0 ∈ A, and set α0(t) := a0 for t ≥ 0 so that α0 ∈ A.

48

For small s > 0, in view of Theorem 4.4, we have

ϕ(x)− e−νsϕ(X(s; x, α0)) ≤ u(x)− e−νsu(X(s; x, α0))

≤∫ s

0e−νtf(X(t; x, α0), a0)dt.

Setting X(t) := X(t; x, α0) for simplicity, by (4.7), we see that

e−νtνϕ(X(t))− ⟨g(X(t), α0), Dϕ(X(t))⟩ = − d

dt

(e−νtϕ(X(t))

). (4.10)

Hence, we have

0 ≥∫ s

0e−νtνϕ(X(t))− ⟨g(X(t), a0), Dϕ(X(t))⟩ − f(X(t), a0)dt.

Therefore, dividing the above by s > 0, and then sending s→ 0, we have

0 ≥ νϕ(x)− ⟨g(x, a0), Dϕ(x)⟩ − f(x, a0),

which implies the desired inequality of the definition by taking the supremumover A.

Step 2: Supersolution property. To show that u is a viscosity supersolu-tion, we argue by contradiction.

Suppose that there are x ∈ Rn, θ > 0 and ϕ ∈ C1(Rn) such that 0 =(u− ϕ)(x) ≤ (u− ϕ)(x) for x ∈ Rn, and that

supa∈A

νϕ(x)− ⟨g(x, a), Dϕ(x)⟩ − f(x, a) ≤ −2θ.

Thus, we can find ε > 0 such that

supa∈A

νϕ(x)− ⟨g(x, a), Dϕ(x)⟩ − f(x, a) ≤ −θ for x ∈ Bε(x). (4.11)

By assumption (4.8) for g, setting t0 := ε/(supa∈A ∥g(·, a)∥L∞(Rn)+1) > 0,we easily see that

|X(t; x, α)− x| ≤∫ t

0|X ′(s; x, α)|ds ≤ ε for t ∈ [0, t0] and α ∈ A.

Hence, by setting X(t) := X(t; x, α) for any fixed α ∈ A, (4.11) yields

νϕ(X(t))− ⟨g(X(t), α(t)), Dϕ(X(t))⟩ − f(X(t), α(t)) ≤ −θ (4.12)

49

for t ∈ [0, t0]. Since (4.10) holds for α in place of α0, multiplying e−νt in(4.12), and then integrating it over [0, t0], we obtain

ϕ(x)− e−νt0ϕ(X(t0))−∫ t0

0e−νtf(X(t), α(t))dt ≤ −θ

ν(1− e−νt0).

Thus, setting θ0 = θ(1 − e−νt0)/ν > 0, which is independent of α ∈ A, wehave

u(x) ≤∫ t0

0e−νtf(X(t), α(t))dt+ e−νt0u(X(t0))− θ0.

Therefore, taking the infimum over A, we get a contradiction to Theorem4.4. 2

Correct proof, which the reader may skip first.Step 1: Subsolution property. Assume that there are x ∈ Rn, θ > 0 and ϕ ∈

C1(Rn) such that 0 = (u∗ − ϕ)(x) ≥ (u∗ − ϕ)(x) for x ∈ Rn and that

supa∈A

νϕ(x)− ⟨g(x, a), Dϕ(x)⟩ − f(x, a) ≥ 2θ.

In view of (4.8), there are a0 ∈ A and r > 0 such that

νϕ(x)− ⟨g(x, a0), Dϕ(x)⟩ − f(x, a0) ≥ θ for x ∈ B2r(x). (4.13)

For large k ≥ 1, we can choose xk ∈ B1/k(x) such that u∗(x) ≤ u(xk) + k−1

and |ϕ(x)− ϕ(xk)| < 1/k. We will only use k such that 1/k ≤ r.Setting α0(t) := a0, we note that Xk(t) := X(t;xk, α0) ∈ B2r(x) for t ∈ [0, t0]

with some t0 > 0 and for large k.On the other hand, by Theorem 4.4, we have

u(xk) ≤∫ t0

0e−νtf(Xk(t), a0)dt+ e−νt0u(Xk(t0)).

Thus, we have

ϕ(xk)−2

k≤ ϕ(x)− 1

k≤ u(xk) ≤

∫ t0

0e−νtf(Xk(t), a0)dt+ e−νt0ϕ(Xk(t0)).

Hence, by (4.13) as in Step 1 of Sketch of proof, we see that

−2

k≤∫ t0

0e−νtf(Xk(t), a0) + ⟨g(Xk(t), a0), Dϕ(Xk(t))⟩ − νϕ(Xk(t))dt

≤ −θν(1− e−νt0),

50

which is a contradiction for large k.Step 2: Supersolution property. Assume that there are x ∈ Rn, θ > 0 and

ϕ ∈ C1(Rn) such that 0 = (u∗ − ϕ)(x) ≤ (u∗ − ϕ)(x) for x ∈ Rn and that

supa∈A

νϕ(x)− ⟨g(x, a), Dϕ(x)⟩ − f(x, a) ≤ −2θ.

In view of (4.8), there is r > 0 such that

νϕ(x)− ⟨g(x, a), Dϕ(x)⟩ − f(x, a) ≤ −θ for x ∈ B2r(x) and a ∈ A. (4.14)

For large k ≥ 1, we can choose xk ∈ B1/k(x) such that u∗(x) ≥ u(xk) − k−1

and |ϕ(x)− ϕ(xk)| < 1/k. In view of (4.8), there is t0 > 0 such that

Xk(t;xk, α) ∈ B2r(x) for all k ≥ 1

r, α ∈ A and t ∈ [0, t0].

Now, we select αk ∈ A such that

u(xk) +1

k≥∫ t0

0e−νtf(X(t;xk, αk), αk(t))dt+ e−νt0u(X(t0;xk, αk)).

Setting Xk(t) := X(t;xk, αk), we have

ϕ(xk) +3

k≥ ϕ(x) +

2

k≥ u(xk) +

1

k≥∫ t0

0e−νtf(Xk(t), αk(t))dt+ e−νt0ϕ(Xk(t)).

Hence, we have

3

k≥∫ t0

0e−νt⟨g(Xk(t), αk(t)), Dϕ(Xk(t))⟩+ f(Xk(t), αk(t))− νϕ(Xk(t))dt.

Putting (4.14) with αk in the above, we have

3

k≥ θ

∫ t0

0e−νtdt,

which is a contradiction for large k ≥ 1. 2

4.2.2 Isaacs equation

In this subsection, we study fully nonlinear PDEs (i.e. p ∈ Rn → F (x, p) isneither convex nor concave) arising in differential games.

51

We are given continuous functions f : Rn × A × B → R and g : Rn ×A×B → Rn such that

(1) sup(a,b)∈A×B

∥f(·, a, b)∥L∞(Rn) + ∥g(·, a, b)∥W 1,∞(Rn)

<∞,

(2) sup(a,b)∈A×B

|f(x, a, b)− f(y, a, b)| ≤ ωf (|x− y|) for x, y ∈ Rn,(4.15)

where ωf ∈ M.Under (4.15), we shall consider Isaacs equations:

supa∈A

infb∈B

νu− ⟨g(x, a, b), Du⟩ − f(x, a, b) = 0 in Rn, (4.16)

and

infb∈B

supa∈A

νu− ⟨g(x, a, b), Du⟩ − f(x, a, b) = 0 in Rn. (4.16′)

As in the previous subsection, we shall derive the expected solution.We first introduce some notations: While we will use the same notion A

as before, we set

B := β : [0,∞) → B | β(·) is measurable.

Next, we introduce the so-called sets of “non-anticipating strategies”:

Γ :=

γ : A → B

∣∣∣∣∣∣∣for any T > 0, if α1 and α2 ∈ A satisfythat α1(t) = α2(t) for a.a. t ∈ (0, T ),

then γ[α1](t) = γ[α2](t) for a.a. t ∈ (0, T )

and

∆ :=

δ : B → A

∣∣∣∣∣∣∣for any T > 0, if β1 and β2 ∈ B satisfythat β1(t) = β2(t) for a.a. t ∈ (0, T ),

then δ[β1](t) = δ[β2](t) for a.a. t ∈ (0, T )

.Using these notations, we will consider maximizing-minimizing problems

of the following cost functional: For α ∈ A, β ∈ B, and x ∈ Rn,

J(x, α, β) :=∫ ∞

0e−νtf(X(t; x, α, β), α(t), β(t))dt,

52

where X(·;x, α, β) is the (unique) solutions ofX ′(t) = g(X(t), α(t), β(t)) for t > 0,

X(0) = x.(4.17)

The expected solutions for (4.16) and (4.16′), respectively, are given by

u(x) = supγ∈Γ

infα∈A

∫ ∞

0e−νtf(X(t; x, α, γ[α]), α(t), γ[α](t))dt,

andv(x) = inf

δ∈∆supβ∈B

∫ ∞

0e−νtf(X(t; x, δ[β], β), δ[β](t), β(t))dt.

We call u and v upper and lower value functions of this differential game,respectively. In fact, under appropriate hypotheses, we expect that v ≤ u,which cannot be proved easily. To show v ≤ u, we first observe that u and vare, respectively, viscosity solutions of (4.16) and (4.16′). Noting that

supa∈A

infb∈B

νr−⟨g(x, a, b), p⟩−f(x, a, b) ≤ infb∈B

supa∈A

νr−⟨g(x, a, b), p⟩−f(x, a, b)

for (x, r, p) ∈ Rn×R×Rn, we see that u (resp., v) is a viscosity supersolution(resp., subsolution) of (4.16′) (resp., (4.16)). Thus, the standard comparisonprinciple implies v ≤ u in Rn (under suitable growth condition at |x| → ∞for u and v).

We shall only deal with u since the corresponding results for v can beobtained in a symmetric way.

To show that u is a viscosity solution of the Isaacs equation (4.16), we firstestablish the dynamic programming principle as in the previous subsection:

Theorem 4.6. (Dynamic Programming Principle) Assume that (4.15)hold. Then, for T > 0, we have

u(x) = supγ∈Γ

infα∈A

∫ T

0e−νtf(X(t;x, α, γ[α]), α(t), γ[α](t))dt

+e−νTu(X(T ;x, α, γ[α]))

.Proof. For a fixed T > 0, we denote by w(x) the right hand side of the

above.Step 1: u(x) ≤ w(x). For any ε > 0, we choose γε ∈ Γ such that

u(x)− ε ≤ infα∈A

∫ ∞

0e−νtf(X(t; x, α, γε[α]), α(t), γε[α](t))dt =: Iε.

53

For any fixed α0 ∈ A, we define the mapping T0 : A → A by

T0[α] :=

α0(t) for t ∈ [0, T ),α(t− T ) for t ∈ [T,∞)

for α ∈ A.

Thus, for any α ∈ A, we have

Iε ≤∫ T

0e−νtf(X(t;x, α0, γε[α0]), α0(t), γε[α0](t))dt

+∫ ∞

Te−νtf(X(t;x, T0[α], γε[T0[α]]), T0[α](t), γε[T0[α]](t))dt

=: I1ε + I2ε .

We next define γ ∈ Γ by

γ[α](t) := γε[T0[α]](t+ T ) for t ≥ 0 and α ∈ A.

Note that γ belongs to Γ.Setting x := X(T ;x, α0, γε[α0]), we have

I2ε = e−νT∫ ∞

0e−νtf(X(t; x, α, γ[α]), α(t), γ[α](t))dt.

Taking the infimum over α ∈ A, we have

u(x)− ε ≤ I1ε + e−νT infα∈A

∫ ∞

0e−νtf(X(t; x, α, γ[α]), α(t), γ[α](t))dt

=: I1ε + I2ε .

Since I2ε ≤ e−νTu(x), we have

u(x)− ε ≤ I1ε + e−νTu(x),

which implies u(x)− ε ≤ w(x) by taking the infimum over α0 ∈ A and then,the supremum over Γ. Therefore, we get the one-sided inequality since ε > 0is arbitrary.

Step 2: u(x) ≥ w(x). For ε > 0, we choose γ1ε ∈ Γ such that

w(x)− ε ≤ infα∈A

∫ T

0e−νtf(X(t; x, α, γ1ε [α]), α(t), γ

1ε [α](t))dt

+e−νTu(X(T ;x, α, γ1ε [α]))

.

54

For any fixed α0 ∈ A, setting x = X(T ; x, α0, γ1ε [α0]), we have

w(x)− ε ≤∫ T

0e−νtf(X(t; x, α0, γ

1ε [α0]), α0(t), γ

1ε [α0](t))dt+ e−νTu(x).

Next, we choose γ2ε ∈ Γ such that

u(x)− ε ≤ infα∈A

∫ ∞

0e−νtf(X(t; x, α, γ2ε [α]), α(t), γ

2ε [α](t))dt. =: I.

For α ∈ A, we define the mapping T1 : A → A by

T1[α](t) := α(t+ T ) for t ≥ 0.

Thus, we have

I ≤∫ ∞

0e−νtf(X(t; x, T1[α0], γ

2ε [T1[α0]]), T1[α0](t), γ

2ε [T1[α0]](t))dt =: I .

Now, for α ∈ A, setting

γ[α](t) :=

γ1ε [α](t) for t ∈ [0, T ),γ2ε [T1[α]](t− T ) for t ∈ [T,∞),

and X(t) := X(t; x, T1[α0], γ2ε [T1[α0]]), we have

I =∫ ∞

Te−ν(t−T )f(X(t− T ), T1[α0](t− T ), γ2ε [T1[α0]](t− T ))dt

= eνT∫ ∞

Te−νtf(X(t− T ), α0(t), γ[α0](t))dt.

Since

X(t;x, α0, γ[α0]) =

X(t; x, α0, γ

1ε [α0]) for t ∈ [0, T ),

X(t− T ) for t ∈ [T,∞),

we have

w(x)− 2ε ≤∫ ∞

0e−νtf(X(t; x, α0, γ[α0]), α0(t), γ[α0](t))dt.

Since α0 is arbitrary, we have

w(x)− 2ε ≤ infα∈A

∫ ∞

0e−νtf(X(t;x, α, γ[α]), α(t), γ[α](t))dt,

55

which yields the assertion by taking the supremum over Γ and then, bysending ε→ 0. 2

Now, we shall verify that the value function u is a viscosity solution of(4.16).

Since we only give a sketch of proofs, one can skip the following theorem.For a correct proof, we refer to [1], originally by Evans-Souganidis (1984).

Theorem 4.7. Assume that (4.15) holds.(1) Then, u is a viscosity subsolution of (4.16).(2) Assume also the following properties:

(i) A ⊂ Rm is compact for some integer m ≥ 1.(ii) there is an ωA ∈ M such that

|f(x, a, b)− f(x, a′, b)|+ |g(x, a, b)− g(x, a′, b)| ≤ ωA(|a− a′|)for x ∈ Rn, a, a′ ∈ A and b ∈ B.

(4.18)

Then, u is a viscosity supersolution of (4.16).

Remark. To show that v is a viscosity subsolution of (4.16′), instead of (4.18),we need to suppose the following hypotheses:

(i) B ⊂ Rm is compact for some integer m ≥ 1.(ii) there is an ωB ∈ M such that

|f(x, a, b)− f(x, a, b′)|+ |g(x, a, b)− g(x, a, b′)| ≤ ωB(|b− b′|)for x ∈ Rn, b, b′ ∈ B and a ∈ A,

(4.18′)

while to verify that v is a viscosity supersolution of (4.16′), we only need (4.15).

Sketch of proof.We shall only prove the assertion assuming that u ∈ USC(Rn)and u ∈ LSC(Rn) in Step 1 and 2, respectively.

To give a correct proof without the semi-continuity assumption, we need a bitcareful analysis similar to the proof for Bellman equations. We omit the correctproof here.

Step 1: Subsolution property. Suppose that the subsolution property fails; thereare x ∈ Rn, θ > 0 and ϕ ∈ C1(Rn) such that 0 = (u− ϕ)(x) ≥ (u− ϕ)(y) (for ally ∈ Rn) and

supa∈A

infb∈B

νu(x)− ⟨g(x, a, b), Dϕ(x)⟩ − f(x, a, b) ≥ 3θ.

We note that X(·;x, α, γ[α]) are uniformly continuous for any (α, γ) ∈ A × Γin view of (4.15).

56

Thus, we can choose that a0 ∈ A such that

infb∈B

νϕ(x)− ⟨g(x, a0, b), Dϕ(x)⟩ − f(x, a0, b) ≥ 2θ.

For any γ ∈ Γ, setting α0(t) = a0 for t ≥ 0, we simply write X(·) forX(·;x, α0, γ[α0]). Thus, we find small t0 > 0 such that

νϕ(X(t))− ⟨g(X(t), a0, γ[α0](t)), Dϕ(X(t))⟩ − f(X(t), a0, γ[α0](t)) ≥ θ

for t ∈ [0, t0]. Multiplying e−νt in the above and then, integrating it over [0, t0],we have

θ

ν(1− e−νt0) ≤ −

∫ t0

0

d

dt

(e−νtϕ(X(t))

)+ e−νtf(X(t), a0, γ[α0](t))

dt

= ϕ(x)− e−νt0ϕ(X(t0))−∫ t0

0e−νtf(X(t), a0, γ[α0](t))dt.

Hence, we have

u(x)− θ

ν(1− e−νt0) ≥

∫ t0

0e−νtf(X(t), a0, γ[α0](t))dt+ e−νt0u(X(t0)) =: I .

Taking the infimum over A, we have

I ≥ infα∈A

∫ t0

0e−νtf(X(t;x, α, γ[α]), α(t), γ[α](t))dt

+e−νt0u(X(t0;x, α, γ[α]))

.Therefore, since γ ∈ Γ is arbitrary, we have

u(x)− θ

ν(1− e−νt0) ≥ sup

γ∈Γinfα∈A

∫ t0

0e−νtf(X(t;x, α, γ[α]), α(t), γ[α](t))dt

+e−νt0u(X(t0;x, α, γ[α]))

,which contradicts Theorem 4.6.

Step 2: Supersolution property. Suppose that the supersolution property fails;there are x ∈ Rn, θ > 0 and ϕ ∈ C1(Rn) such that 0 = (u − ϕ)(x) ≤ (u − ϕ)(y)for y ∈ Rn, and

supa∈A

infb∈B

νu(x)− ⟨g(x, a, b), Dϕ(x)⟩ − f(x, a, b) ≤ −3θ.

For any a ∈ A, there is b(a) ∈ B such that

νu(x)− ⟨g(x, a, b(a)), Dϕ(x)⟩ − f(x, a, b(a)) ≤ −2θ.

57

In view of (4.18), there is ε(a) > 0 such that if |a− a′| < ε(a) and |x− y| < ε(a),then we have

νϕ(y)− ⟨g(y, a′, b(a)), Dϕ(y)⟩ − f(y, a′, b(a)) ≤ −θ.

From the compactness of A, we may select akMk=1 such that

A =M∪k=1

Ak,

whereAk := a ∈ A | |a− ak| < ε(ak).

Furthermore, we set A1 = A1, and inductively, Ak := Ak \ ∪k−1j=1Aj ; Ak ∩ Aj = ∅

for k = j. We may also suppose that Ak = ∅ for k = 1, . . . ,M .For α ∈ A, we define

γ0[α](t) := b(ak) provided α(t) ∈ Ak.

Now, setting X(t) := X(t;x, α, γ0[α]), we find t0 > 0 such that

νϕ(X(t))− ⟨g(X(t), α(t), γ0[α](t)), Dϕ(X(t))⟩ − f(X(t), α(t), γ0[α](t)) ≤ −θ

for t ∈ [0, t0]. Multiplying e−νt in the above and then, integrating it, we obtain

ϕ(x)− e−νt0ϕ(X(t0))−∫ t0

0e−νtf(X(t), α(t), γ0[α](t))dt ≤ − θ

ν(1− e−νt0).

Since α ∈ A is arbitrary, we have

u(x) +θ

ν(1− e−νt0) ≤ inf

α∈A

∫ t0

0e−νtf(X(t; x, α, γ0[α]), α(t), γ0[α](t))dt

+e−νt0u(X(t0; x, α, γ0[α]))

,which contradicts Theorem 4.6 by taking the supremum over Γ. 2

4.3 Stability

In this subsection, we present a stability result for viscosity solutions, whichis one of the most important properties for “solutions” as noted in section 1.Thus, this result justifies our notion of viscosity solutions.

However, since we will only use Proposition 4.8 below in section 7.3, thereader may skip the proof.

58

First of all, for possibly discontinuous F : Ω×R×Rn ×Sn → R, we areconcerned with

F (x, u,Du,D2u) = 0 in Ω. (4.19)

We introduce the following notation:

F∗(x, r, p,X) := limε→0

inf

F (y, s, q, Y )

∣∣∣∣∣ y ∈ Ω ∩Bε(x), |s− r| < ε,|q − p| < ε, ∥Y −X∥ < ε

,

F ∗(x, r, p,X) := limε→0

sup

F (y, s, q, Y )

∣∣∣∣∣ y ∈ Ω ∩Bε(x), |s− r| < ε,|q − p| < ε, ∥Y −X∥ < ε

.

Definition. We call u : Ω → R a viscosity subsolution (resp., super-solution) of (4.19) if u∗ (resp., u∗) is a viscosity subsolution (resp., super-soluion) of

F∗(x, u,Du,D2u) ≤ 0

(resp., F ∗(x, u,Du,D2u) ≥ 0

)in Ω.

We call u : Ω → R a viscosity solution of (4.19) if it is both a viscositysub- and supersolution of (4.19).

Now, for given continuous functions Fk : Ω×R×Rn × Sn → R, we set

F (x, r, p,X)

:= limk→∞

inf

Fj(y, s, q, Y )

∣∣∣∣∣∣∣|y − x| < 1/k, |s− r| < 1/k,|q − p| < 1/k, ∥Y −X∥ < 1/k

and j ≥ k

,F (x, r, p,X)

:= limk→∞

sup

Fj(y, s, q, Y )

∣∣∣∣∣∣∣|y − x| < 1/k, |s− r| < 1/k,|q − p| < 1/k, ∥Y −X∥ < 1/k

and j ≥ k

.Our stability result is as follows.

Proposition 4.8. Let Fk : Ω × R × Rn × Sn → R be continuousfunctions. Let uk : Ω → R be a viscosity subsolution (resp., supersolution)of

Fk(x, uk, Duk, D2uk) = 0 in Ω.

59

Setting u (resp., u) by

u(x) := limk→∞

sup(uj)∗(y) | y ∈ B1/k(x) ∩ Ω, j ≥ k(resp., u(x) := lim

k→∞inf(uj)∗(y) | y ∈ B1/k(x) ∩ Ω, j ≥ k

)for x ∈ Ω, then u (resp., u) is a viscosity subsolution (resp., supersolution)of

F (x, u,Du,D2u) ≤ 0 (resp., F (x, u,Du,D2u) ≥ 0) in Ω.

Remark. We note that u ∈ USC(Ω), u ∈ LSC(Ω), F ∈ LSC(Ω × R ×Rn × Sn) and F ∈ USC(Ω×R×Rn × Sn).

Proof. We only give a proof for subsolutions since the other can be shownsimilarly.

Given ϕ ∈ C2(Ω), we let x0 ∈ Ω be such that 0 = (u−ϕ)(x0) > (u−ϕ)(x)for x ∈ Ω \ x0. We shall show that F (x0, u(x0), Dϕ(x0), D

2ϕ(x0)) ≤ 0.We may choose xk ∈ Br(x0) (for a subsequence if necessary), where r ∈

(0,dist(x0, ∂Ω)), such that

limk→∞

xk = x0 and limk→∞

(uk)∗(xk) = u(x0). (4.20)

We select yk ∈ Br(x0) such that ((uk)∗ − ϕ)(yk) = supBr(x0)((uk)

∗ − ϕ).

We may also suppose that limk→∞ yk = z for some z ∈ Br(x0) (takinga subsequence if necessary). Since ((uk)

∗ − ϕ)(yk) ≥ ((uk)∗ − ϕ)(xk), (4.20)

implies

0 = lim infk→∞

((uk)∗ − ϕ)(xk) ≤ lim inf

k→∞((uk)

∗ − ϕ)(yk)

≤ lim infk→∞

(uk)∗(yk)− ϕ(z)

≤ lim supk→∞

(uk)∗(yk)− ϕ(z) ≤ (u− ϕ)(z).

Thus, this yields z = x0 and limk→∞(uk)∗(yk) = u(x0). Hence, we see that

yk ∈ Br(x0) for large k ≥ 1. Since (uk)∗ − ϕ attains a maximum over Br(x0)

at yk ∈ Br(x0), by the definition of uk (with Proposition 2.4 for Ω′ = Br(x0)),we have

Fk(yk, (uk)∗(yk), Dϕ(yk), D

2ϕ(yk)) ≤ 0,

which concludes the proof by taking the limit infimum with the definition ofF . 2

60

5 Generalized boundary value problems

In order to obtain the uniqueness of solutions of an ODE, we have to supposecertain initial or boundary condition. In the study of PDEs, we need toimpose appropriate conditions on ∂Ω for the uniqueness of solutions.

Following the standard PDE theory, we shall treat a few typical boundaryconditions in this section.

Since we are mainly interested in degenerate elliptic PDEs, we cannotexpect “solutions” to satisfy the given boundary condition on the wholeof ∂Ω. The simplest example is as follows: For Ω := (0, 1), consider the“degenerate” elliptic PDE

−dudx

+ u = 0 in (0, 1).

Note that it is impossible to find a solution u of the above such that u(0) =u(1) = 1.

Our plan is to propose a definition of “generalized” solutions for boundaryvalue problems. For this purpose, we extend the notion of viscosity solutionsto possibly discontinuous PDEs on Ω while we normally consider those in Ω.

For general G : Ω×R×Rn × Sn → R, we are concerned with

G(x, u,Du,D2u) = 0 in Ω. (5.1)

As in section 4.3, we define

G∗(x, r, p,X) := limε→0

inf

G(y, s, q, Y )

∣∣∣∣∣ y ∈ Ω ∩Bε(x), |s− r| < ε,|q − p| < ε, ∥Y −X∥ < ε

,

G∗(x, r, p,X) := limε→0

sup

G(y, s, q, Y )

∣∣∣∣∣ y ∈ Ω ∩Bε(x), |s− r| < ε,|q − p| < ε, ∥Y −X∥ < ε

.

Definition. We call u : Ω → R a viscosity subsolution (resp., superso-

lution) of (5.1) if, for any ϕ ∈ C2(Ω),

G∗(x, u∗(x), Dϕ(x), D2ϕ(x)) ≤ 0(

resp., G∗(x, u∗(x), Dϕ(x), D2ϕ(x)) ≥ 0

)provided that u∗ − ϕ (resp., u∗ − ϕ) attains its maximum (resp., minimum)at x ∈ Ω.

61

We call u : Ω → R a viscosity solution of (5.1) if it is both a viscositysub- and supersolution of (5.1).

Our comparison principle in this setting is as follows:

“Comparison principle in this setting”

viscosity subsolution u of (5.1)viscosity supersolution v of (5.1)

=⇒ u ≤ v in Ω

Note that

the boundary condition is contained in the definition.

Using the above new definition, we shall formulate the boundary valueproblems in the viscosity sense. Given F : Ω × R × Rn × Sn → R andB : ∂Ω×R×Rn×Sn → R, we investigate general boundary value problems

F (x, u,Du,D2u) = 0 in Ω,B(x, u,Du,D2u) = 0 on ∂Ω.

(5.2)

Setting G by

G(x, r, p,X) :=

F (x, r, p,X) for x ∈ Ω,B(x, r, p,X) for x ∈ ∂Ω,

we give the definition of boundary value problems (5.2) in the viscosity sense.

Definition. We call u : Ω → R a viscosity subsolution (resp., superso-lution) of (5.2) if it is a viscosity subsolution (resp., supersolution) of (5.1),where G is defined in the above.

We call u : Ω → R a viscosity solution of (5.2) if it is both a viscositysub- and supersolution of (5.2).

Remark. When F and B are continuous and G is given as above, G∗ andG∗ can be expressed in the following manner:

G∗(x, r, p,X) =

F (x, r, p,X) for x ∈ Ω,minF (x, r, p,X), B(x, r, p,X) for x ∈ ∂Ω,

G∗(x, r, p,X) =

F (x, r, p,X) for x ∈ Ω,maxF (x, r, p,X), B(x, r, p,X) for x ∈ ∂Ω.

62

It is not hard to extend the existence and stability results correspondingto Theorem 4.3 and Proposition 4.8, respectively, to viscosity solutions inthe above sense. However, it is not straightforward to show the comparisonprinciple in this new setting. Thus, we shall concentrate our attention tothe comparison principle, which implies the uniqueness (and continuity) ofviscosity solutions.

The main difficulty to prove the comparison principle is that we have to“avoid” the boundary conditions for both of viscosity sub- and supersolu-tions.

To explain this, let us consider the case when G is given by (5.2). Let uand v be, respectively, a viscosity sub- and supersolution of (5.1). We shallobserve that the standard argument in Theorem 3.7 does not work.

For ε > 0, suppose that (x, y) → u(x)− v(y)− (2ε)−1|x− y|2 attains itsmaximum at (xε, yε) ∈ Ω×Ω. Notice that there is NO reason to verify that(xε, yε) ∈ Ω× Ω.

The worst case is that (xε, yε) ∈ ∂Ω× ∂Ω. In fact, in view of Lemma 3.6,

we find X, Y ∈ Sn such that ((xε − yε)/ε,X) ∈ J2,+

Ω u(xε), ((xε − yε)/ε, Y ) ∈J2,−Ω v(yε), the matrix inequalities in Lemma 3.6 hold for X, Y . Hence, we

have

minF(xε, u(xε),

xε − yεε

,X), B

(xε, u(xε),

xε − yεε

,X)

≤ 0

and

maxF(yε, v(yε),

xε − yεε

, Y), B

(yε, v(yε),

xε − yεε

, Y)

≥ 0.

However, even if we suppose that (3.21) holds for F and B “in Ω”, we cannotget any contradiction when

F(xε, u(xε),

xε − yεε

,X)≤ 0 ≤ B

(yε, v(yε),

xε − yεε

, Y)

or

B(xε, u(xε),

xε − yεε

,X)≤ 0 ≤ F

(yε, v(yε),

xε − yεε

, Y).

It seems impossible to avoid this difficulty as long as we use |x− y|2/(2ε) as“test functions”.

63

Our plan to go beyond this difficulty is to find new test functions ϕε(x, y)(instead of |x− y|2/(2ε)) so that the function (x, y) → u(x)− v(y)−ϕε(x, y)attains its maximum over Ω × Ω at an interior point (xε, yε) ∈ Ω × Ω. Tothis end, since we will use several “perturbation” techniques, we suppose twohypotheses on F : First, we shall suppose the following continuity of F withrespect to (p,X)-variables.

There is an ω0 ∈ M such that|F (x, p,X)− F (x, q, Y )| ≤ ω0(|p− q|+ ∥X − Y ∥)

for x ∈ Ω, p, q ∈ Rn, X, Y ∈ Sn.(5.3)

The next assumption is a bit stronger than the structure condition (3.21):

There is ωF ∈ M such thatif X,Y ∈ Sn and µ > 1 satisfy

−3µ

(I 00 I

)≤(X 00 −Y

)≤ 3µ

(I −I−I I

), then

F (y, p, Y )− F (x, p,X) ≤ ωF (|x− y|(1 + |p|+ µ|x− y|))for x, y ∈ Ω, p ∈ Rn, X, Y ∈ Sn.

(5.4)

5.1 Dirichlet problem

First, we consider Dirichlet boundary value problems (Dirichlet problems forshort) in the above sense.

Assuming that viscosity sub- and supersolutions are continuous on ∂Ω,we will obtain the comparison principle for them.

We now recall the classical Dirichlet problemνu+ F (x,Du,D2u) = 0 in Ω,

u− g = 0 on ∂Ω.(5.5)

Note that the Dirichlet problem of (5.5) in the viscosity sense is as follows:

subsolution ⇐⇒

νu+ F (x,Du,D2u) ≤ 0 in Ω,minνu+ F (x,Du,D2u), u− g ≤ 0 on ∂Ω,

and

supersolution ⇐⇒

νu+ F (x,Du,D2u) ≥ 0 in Ω,maxνu+ F (x,Du,D2u), u− g ≥ 0 on ∂Ω.

64

We shall suppose the following property on the shape of Ω, which maybe called an “interior cone condition” (see Fig 5.1):

For each z ∈ ∂Ω, there are r, s ∈ (0, 1) such thatx− rn(z) + rξ ∈ Ω for x ∈ Ω ∩Br(z), r ∈ (0, r) and ξ ∈ Bs(0).

(5.6)

Here and later, we denote by n(z) the unit outward normal vector at z ∈ ∂Ω.

O

nz

x

-rnz

zhatr

rs

Fig 5.1

PSfrag replacements

x−rn(z)

n(z)

rsrzΩ

Theorem 5.1. Assume that ν > 0, (5.3), (5.4) and (5.6) hold. Forg ∈ C(∂Ω), we let u and v : Ω → R be, respectively, a viscosity sub- andsupersolution of (5.5) such that

lim infx∈Ω→z

u∗(x) ≥ u∗(z) and lim supx∈Ω→z

v∗(x) ≤ v∗(z) for z ∈ ∂Ω. (5.7)

Then, u∗ ≤ v∗ in Ω.

Remark. Notice that (5.7) implies the continuity of u∗ and v∗ on ∂Ω.

Proof. Suppose that maxΩ(u∗ − v∗) =: θ > 0. We simply write u and v

for u∗ and v∗, respectively.Case 1: max∂Ω(u− v) = θ. We choose z ∈ ∂Ω such that (u − v)(z) = θ.

We shall divide three cases:Case 1-1: u(z) > g(z). For ε, δ ∈ (0, 1), where δ > 0 will be fixed later,

setting ϕ(x, y) := (2ε2)−1|x− y− εδn(z)|2− δ|x− z|2, we let (xε, yε) ∈ Ω×Ωbe the maximum point of Φ(x, y) := u(x)− v(y)− ϕ(x, y) over Ω× Ω.

65

Since z−εδn(z) ∈ Ω for small ε > 0 by (5.6), Φ(xε, yε) ≥ Φ(z, z−εδn(z))implies that

|xε − yε − εδn(z)|2

2ε2≤ u(xε)−v(yε)−u(z)+v(z−εδn(z))−δ|xε−z|2. (5.8)

Since |xε− yε| ≤Mε, where M :=√2(maxΩ u−minΩ v−u(z)+ v(z)+1)1/2,

for small ε > 0, we may suppose that (xε, yε) → (x, x) and (xε − yε)/ε → zfor some x ∈ Ω and z ∈ Rn as ε → 0 along a subsequence (denoted by εagain). Thus, from the continuity (5.7) of v at z ∈ ∂Ω, (5.8) implies that

θ ≤ u(x)− v(x)− δ|x− z|2,

which yields x = z. Moreover, we have

limε→0

|xε − yε − εδn(z)|2

ε2= 0,

which implies that

limε→0

|xε − yε|ε

= δ. (5.9)

Furthermore, we note that yε = xε − εδn(z) + o(ε) ∈ Ω because of (5.6).Applying Lemma 3.6 with Proposition 2.7 to u(x)+ ε−1δ⟨n(z), x⟩− δ|x−

z|2 − 2−1δ2 and v(y) + ε−1δ⟨n(z), y⟩, we find X, Y ∈ Sn such that(xε − yεε2

− δ

εn(z) + 2δ(xε − z), X + 2δI

)∈ J

2,+

Ω u(xε), (5.10)

(xε − yεε2

− δ

εn(z), Y

)∈ J

2,−Ω v(yε), (5.11)

and

− 3

ε2

(I OO I

)≤(X OO −Y

)≤ 3

ε2

(I −I−I I

).

Putting pε := ε−2(xε − yε)− δε−1n(z), by (5.3), we have

F (xε, pε, X)− F (xε, pε + 2δ(xε − z), X + 2δI) ≤ ω0(2δ|xε − z|+ 2δ). (5.12)

Since yε ∈ Ω and u(xε) > g(xε) for small ε > 0 provided xε ∈ ∂Ω, in viewof (5.10) and (5.11), we have

ν(u(xε)− v(yε)) ≤ F (yε, pε, Y )− F (xε, pε + 2δ(xε − z), X + 2δI).

66

Combining this with (5.12) , by (5.4), we have

ν(u(xε)−v(yε)) ≤ ω0(2δ|xε−z|+2δ)+ωF

(|xε − yε|

(1 + |pε|+

|xε − yε|ε2

)).

Sending ε→ 0 together with (5.9) in the above, we have

νθ ≤ ω0(2δ) + ωF (2δ2),

which is a contradiction for small δ > 0, which only depends on θ and ν.Case 1-2: v(z) < g(z). To get a contradiction, we argue as above replacing

ϕ(x, y) by ψ(x, y) := (2ε2)−1|x − y + εδn(z)|2 − δ|x − z|2 so that xε = yε −εδn(z) + o(ε) ∈ Ω for small ε > 0. Note that we need here the continuity ofu on ∂Ω in (5.7) while the other one in (5.7) is needed in Case 1-1. (See alsothe proof of Theorem 5.3 below.)

Case 1-3: u(z) ≤ g(z) and v(z) ≥ g(z). This does not occur because 0 <θ = (u− v)(z) ≤ 0.

Case 2: sup∂Ω(u− v) < θ. In this case, using the standard test function

|x − y|2/(2ε) (without δ|x − z|2 term), we can follow the same argument asin the proof of Theorem 3.7. 2

Remark. Unfortunately, without assuming the continuity of viscosity so-lutions on ∂Ω, the comparison principle fails in general.

In fact, setting F (x, r, p,X) ≡ r and g(x) ≡ −1, consider the function

u(x) :=

0 for x ∈ Ω,−1 for x ∈ ∂Ω.

Note that u∗ ≡ 0 and u∗ ≡ u in Ω, which are respectively a viscosity sub- andsupersolution of G(x, u,Du,D2u) = 0 in Ω. Therefore, this example showsthat the comparison principle fails in general without assumption (5.7).

5.2 State constraint problem

The state constraint boundary condition arises in a typical optimal controlproblem. Thus, if the reader is more interested in the PDE theory, he/shemay skip Proposition 5.2 below, which explains why we will adapt the “stateconstraint boundary condition” in Theorem 5.3.

67

To explain our motivation, we shall consider Bellman equations of first-order.

supa∈A

νu− ⟨g(x, a), Du⟩ − f(x, a) = 0 in Ω.

Here, we use the notations in section 4.2.1.We introduce the following set of controls: For x ∈ Ω,

A(x) := α(·) ∈ A | X(t;x, α) ∈ Ω for t ≥ 0.

We shall suppose that

A(x) = ∅ for all x ∈ Ω. (5.13)

Also, we suppose that(1) sup

a∈A

(∥f(·, a)∥L∞(Ω) + ∥g(·, a)∥W 1,∞(Ω)

)<∞,

(2) supa∈A

|f(x, a)− f(y, a)| ≤ ωf (|x− y|) for x, y ∈ Ω,(5.14)

where ωf ∈ M.We are now interested in the following the optimal cost functional:

u(x) := infα∈A(x)

∫ ∞

0e−νtf(X(t;x, α), α(t))dt.

Proposition 5.2. Assume that ν > 0, (5.13) and (5.14) hold. Then, we have(1) u is a viscosity subsolution of

supa∈A

νu− ⟨g(x, a), Du⟩ − f(x, a) ≤ 0 in Ω,

(2) u is a viscosity supersolution of

supa∈A

νu− ⟨g(x, a), Du⟩ − f(x, a) ≥ 0 in Ω.

Remark. We often say that u satisfies the state constraint boundary conditionwhen it is a viscosity supersolution of

“F (x, u,Du,D2u) ≥ 0 in ∂Ω”.

Proof. In fact, at x ∈ Ω, it is easy to verify that the dynamic programmingprinciple (Theorem 4.4) holds for small T > 0. Thus, we may show Theorem 4.5replacing Rn by Ω.

68

Hence, it only remains to show (2) on ∂Ω. Thus, suppose that there are x ∈ ∂Ω,θ > 0 and ϕ ∈ C1(Ω) such that (u∗ − ϕ)(x) = 0 ≤ (u∗ − ϕ)(x) for x ∈ Ω, and

supa∈A

νϕ(x)− ⟨g(x, a), Dϕ(x)⟩ − f(x, a) ≤ −2θ.

Then, we will get a contradiction.Choose xk ∈ Ω ∩B1/k(x) such that u∗(x) + k−1 ≥ u(xk) and |ϕ(x)− ϕ(xk)| <

1/k. In view of (5.14), there is t0 > 0 such that for any α ∈ A(xk) and large k ≥ 1,we have

νϕ(Xk(t))− ⟨g(Xk(t), α(t)), Dϕ(Xk(t))⟩ − f(Xk(t), α(t)) ≤ −θ for t ∈ (0, t0),

where Xk(t) := X(t;xk, α). Thus, multiplying e−νt and then, integrating it over(0, t0), we have

ϕ(xk) ≤ e−νt0ϕ(Xk(t0)) +

∫ t0

0e−νtf(Xk(t), α(t))dt−

θ

ν(1− e−νt0).

Since we have

u(xk) ≤2

k+ e−νt0u(Xk(t0)) +

∫ t0

0e−νtf(Xk(t), α(t))dt−

θ

ν(1− e−νt0),

taking the infimum over A(xk), we apply Theorem 4.4 to get

0 ≤ 2

k− θ

ν(1− e−νt0),

which is a contradiction for large k. 2

Motivated by this proposition, we shall consider more general second-orderelliptic PDEs.

Theorem 5.3. Assume that ν > 0, (5.3), (5.4), (5.6) and (5.12) hold. Letu : Ω → R be, respectively, a viscosity sub- and supersolution of

νu+ F (x,Du,D2u) ≤ 0 in Ω,

andνv + F (x,Dv,D2v) ≥ 0 in Ω.

Assume also thatlim infx∈Ω→z

u∗(x) ≥ u∗(z) for z ∈ ∂Ω. (5.15)

69

Then, u∗ ≤ v∗ in Ω.

Remark. In 1986, Soner first treated the state constraint problems for deter-ministic optimal control (i.e. first-order PDEs) by the viscosity solution approach.

We note that we do not need continuity of v on ∂Ω while we need it forDirichlet problems. For further discussion on the state constraint problems, werefer to Ishii-Koike (1996).

We also note that the proof below is easier than that for Dirichlet problemsin section 5.1 because we only need to avoid the boundary condition for viscositysubsolutions.

Proof. Suppose that maxΩ(u∗ − v∗) =: θ > 0. We shall write u and v for u∗

and v∗, respectively, again.We may suppose that max∂Ω(u − v) = θ since otherwise, we can use the

standard procedure to get a contradiction.Now, we proceed the same argument in Case 1-2 in the proof of Theorem 5.1

(although it is not precisely written).For ε, δ > 0, setting ϕ(x, y) := (2ε2)−1|x− y + εδn(z)|2 + δ|x− z|2, where n is

the unit outward normal vector at z ∈ ∂Ω, we let (xε, yε) ∈ Ω × Ω the maximumpoint of u(x)− v(y)−ϕ(x, y) over Ω×Ω. As in the proof of Theorem 3.4, we have

limε→0

(xε, yε) = (z, z) and limε→0

|xε − yε|ε

= δ. (5.16)

Since xε = yε − εδn(z) + o(ε) ∈ Ω for small ε > 0, in view of Lemma 3.6 withProposition 2.7, we can find X,Y ∈ Sn such that(

xε − yεε2

+δ

εn(z) + 2δ(xε − z), X + 2δI

)∈ J

2,+

Ωu(xε),

(xε − yεε2

+δ

εn(z), Y

)∈ J

2,−Ωv(yε),

and

− 3

ε2

(I OO I

)≤(X OO −Y

)≤ 3

ε2

(I −I−I I

).

Setting pε := ε−2(xε − yε) + δε−1n(z), we have

ν(u(xε)− v(yε)) ≤ F (yε, pε, Y )− F (xε, pε + 2δ(xε − z), X + 2δI)

≤ ω0(2δ|xε − z|+ 2δ) + ωF

(|xε − yε|

(1 + |pε|+

|xε − yε|ε

)).

70

Hence, sending ε→ 0 with (5.16), we have

νθ ≤ ω0(2δ) + ωF (2δ2),

which is a contradiction for small δ > 0. 2

5.3 Neumann problem

In the classical theory and modern theory for weak solutions in the distribu-tion sense, the (inhomogeneous) Neumann condition is given by

⟨n(x), Du(x)⟩ − g(x) = 0 on ∂Ω,

where n(x) denotes the unit outward normal vector at x ∈ ∂Ω.In Dirichlet and state constraint problems, we have used a test function

which forces one of xε and yε to be in Ω. However, in the Neumann boundaryvalue problem (Neumann problem for short), we have to avoid the boundarycondition for viscosity sub- and supersolutions simultaneously. Thus, we needa new test function different from those in sections 5.1 and 5.2.

We first define the signed distance function from Ω by

ρ(x) :=

inf|x− y| | y ∈ ∂Ω for x ∈ Ωc,− inf|x− y| | y ∈ ∂Ω for x ∈ Ω.

In order to obtain the comparison principle for the Neumann problem,we shall impose a hypothesis on Ω (see Fig 5.2):

(1) There is r > 0 such thatΩ ⊂ (Br(z + rn(z)))c for z ∈ ∂Ω.

(2) There is a neighborhood N of ∂Ω such thatρ ∈ C2(N), and Dρ(x) = n(x) for x ∈ ∂Ω.

(5.17)

Remark. This assumption (1) is called the “uniform exterior sphere con-dition”. Since |x− z − rn(z)| ≥ r for z ∈ ∂Ω and x ∈ Ω, we have

⟨n(z), x− z⟩ ≤ |x− z|2

2rfor z ∈ ∂Ω and x ∈ Ω. (5.18)

It is known that when ∂Ω is “smooth” enough, (2) of (5.17) holds true.

71

O

z

pO

rnz

r

Fig 5.2

PSfrag replacements

rn(z)

∂ΩrzΩ

We shall consider the inhomogeneous Neumann problem:νu+ F (x,Du,D2u) = 0 in Ω,⟨n(x), Du⟩ − g(x) = 0 on ∂Ω.

(5.19)

Remember that we adapt the definition of viscosity solutions of (5.19) forthe corresponding G in (5.2).

Theorem 5.4. Assume that ν > 0, (5.3), (5.4) and (5.17) hold. Forg ∈ C(∂Ω), we let u and v : Ω → R be a viscosity sub- and supersolution of(5.19), respectively.

Then, u∗ ≤ v∗ in Ω.

Remark. We note that we do not need any continuity of u and v on ∂Ω.

Proof. As before, we write u and v for u∗ and v∗, respectively.As in the proof of Theorem 3.7, we suppose that maxΩ(u− v) =: θ > 0.

Also, we may suppose that max∂Ω(u− v) = θ.Let z ∈ ∂Ω be a point such that (u− v)(z) = θ. For small δ > 0, we see

that the mapping x ∈ Ω → u(x)− v(y)− δ|x− z|2 takes its strict maximumat z.

For small ε, δ > 0, where δ > 0 will be fixed later, setting ϕ(x, y) :=(2ε)−1|x − y|2 − g(z)⟨n(z), x − y⟩ + δ(ρ(x) + ρ(y) + 2) + δ|x − z|2, we let(xε, yε) ∈ Ω × Ω be the maximum point of Φ(x, y) := u(x) − v(y) − ϕ(x, y)over Ω ∩N × Ω ∩N , where N is in (5.17).

Since Φ(xε, yε) ≥ Φ(z, z), as before, we may extract a subsequence, whichis denoted by (xε, yε) again, such that (xε, yε) → (x, x). We may supposex ∈ ∂Ω. Since Φ(x, x) ≥ lim supε→0 Φ(xε, yε), we have

u(x)− v(x)− δ|x− z|2 ≥ θ,

72

which yields x = z. Moreover, we have

limε→0

|xε − yε|2

ε= 0. (5.20)

Applying Lemma 3.6 to u(x)− δ(ρ(x)+1)− g(z)⟨n(z), x⟩− δ|x− z|2 and−v(y)− δ(ρ(y) + 1) + g(z)⟨n(z), y⟩, we find X, Y ∈ Sn such that(

pε + δn(xε) + 2δ(xε − z), X + δD2ρ(xε) + 2δI)∈ J

2,+

Ω u(xε), (5.21)

(pε − δn(yε), Y − δD2ρ(yε)

)∈ J

2,−Ω v(yε), (5.22)

where pε := ε−1(xε − yε) + g(z)n(z), and

−3

ε

(I 00 I

)≤(X 00 −Y

)≤ 3

ε

(I −I−I I

).

When xε ∈ ∂Ω, by (5.18), we calculate in the following manner:

⟨n(xε), Dxϕ(xε, yε)⟩ = ⟨n(xε), pε + δn(xε) + 2δ(xε − z)⟩

≥ −|xε − yε|2

2rε+ g(z)⟨n(xε),n(z)⟩+ δ − 2δ|xε − z|.

Hence, given δ > 0, we see that

⟨n(xε), Dxϕ(xε, yε)⟩ − g(xε) ≥δ

2for small ε > 0.

Thus, by (5.21), this yields

νu(xε) + F (xε, pε + δn(xε) + 2δ(xε − z), X + δD2ρ(xε) + 2δI) ≤ 0. (5.23)

Of course, if xε ∈ Ω, then the above inequality holds from the definition.On the other hand, similarly, if yε ∈ ∂Ω, then

⟨n(yε),−Dyϕ(xε, yε)⟩ − g(yε) ≤ −δ2

for small ε > 0.

Hence, by (5.22), we have

νv(yε) + F (yε, pε − δn(yε), Y − δD2ρ(yε)) ≥ 0. (5.24)

73

Using (5.3) and (5.4), by (5.23) and (5.24), we have

ν(u(xε)− v(yε)) ≤ F (yε, pε, Y )− F (xε, pε, X) + 2ω0(δM)

≤ ωF

(|xε − yε|

(1 + |pε|+

|xε − yε|ε

))+ 2ω0(δM),

where M := 3 + supx∈N∩Ω(2|x − z| + |D2ρ(x)|). Sending ε → 0 with (5.20)in the above, we have

νθ ≤ 2ω0(δM),

which is a contradiction for small δ > 0. 2

5.4 Growth condition at |x| → ∞In the standard PDE theory, we often consider PDEs in unbounded domains,typically, in Rn. In this subsection, we present a technique to establish thecomparison principle for viscosity solutions of

νu+ F (x,Du,D2u) = 0 in Rn. (5.25)

We remind the readers that in the proofs of comparison results we alwayssuppose maxΩ(u − v) > 0, where u and v are, respectively, a viscosity sub-and supersolution. However, considering Ω := Rn, the maximum of u − vmight attain its maximum at “|x| → ∞”. Thus, we have to choose a testfunction ϕ(x, y), which forces u(x)− v(y)− ϕ(x, y) to takes its maximum ata point in a compact set.

For this purpose, we will suppose the linear growth condition (for sim-plicity) for viscosity solutions.

We rewrite the structure condition (3.21) for Rn:

There is an ωF ∈ M such that if X, Y ∈ Sn and µ > 1 satisfy

−3µ

(I 00 I

)≤(X 00 −Y

)≤ 3µ

(I −I−I I

),

then F (y, µ(x− y), Y )− F (x, µ(x− y), X)≤ ωF (|x− y|(1 + µ|x− y|)) for x, y ∈ Rn.

(5.26)

We will also need the Lipschitz continuity of (p,X) → F (x, p,X), whichis stronger than (5.3).

There is µ0 > 0 such that |F (x, p,X)− F (x, q, Y )|≤ µ0(|p− q|+ ∥X − Y ∥) for x ∈ Rn, p, q ∈ Rn, X, Y ∈ Sn.

(5.27)

74

Proposition 5.5. Assume that ν > 0, (5.26) and (5.27) hold. Let u andv : Rn → R be, respectively, a viscosity sub- and supersolution of (5.25).Assume also that there is C0 > 0 such that

u∗(x) ≤ C0(1 + |x|) and v∗(x) ≥ −C0(1 + |x|) for x ∈ Rn. (5.28)

Then, u∗ ≤ v∗ in Rn.

Proof. We shall simply write u and v for u∗ and v∗, respectively.For δ > 0, we set θδ := supx∈Rn(u(x)− v(x)− 2δ(1+ |x|2)). We note that

(5.28) implies that there is zδ ∈ Rn such that θδ = u(zδ)−v(zδ)−2δ(1+|zδ|2).Set θ := lim supδ→0 θδ ∈ R ∪ ∞.

When θ ≤ 0, since

(u− v)(x) ≤ 2δ(1 + |x|2) + θδ for δ > 0 and x ∈ Rn,

we have u ≤ v in Rn.Thus, we may suppose θ ∈ (0,∞]. Setting Φδ(x, y) := u(x) − v(y) −

(2ε)−1|x − y|2 − δ(1 + |x|2) − δ(1 + |y|2) for ε, δ > 0, where δ > 0 will befixed later, in view of (5.28), we can choose (xε, yε) ∈ Rn × Rn such thatΦδ(xε, yε) = max(x,y)∈Rn×Rn Φδ(x, y) ≥ θδ.

As before, extracting a subsequence if necessary, we may suppose that

limε→0

|xε − yε|2

ε= 0. (5.29)

By Lemma 3.6 with Proposition 2.7, putting pε := (xε − yε)/ε, we findX, Y ∈ Sn such that

(pε + 2δxε, X + 2δI) ∈ J2,+u(xε),

(pε − 2δyε, Y − 2δI) ∈ J2,−v(yε),

and

−3

ε

(I OO I

)≤(X OO −Y

)≤ 3

ε

(I −I−I I

).

Hence, we have

ν(u(xε)− v(yε))≤ F (yε, pε − 2δyε, Y − 2δI)− F (xε, pε + 2δxε, X + 2δI)≤ F (yε, pε, Y )− F (xε, pε, X) + 2δµ0(2 + |xε|+ |yε|)

≤ ωF

(|xε − yε|

(1 +

|xε − yε|ε

))+ νδ(2 + |xε|2 + |yε|2) + Cδ,

75

where C = C(µ0, ν) > 0 is independent of ε, δ > 0. For the last inequality,we used “2ab ≤ τa2 + τ−1b2 for τ > 0”.

Therefore, we have

νθ ≤ ωF

(|xε − yε|

(1 +

|xε − yε|ε

))+ Cδ.

Sending ε → 0 in the above together with (5.29), we get νθ ≤ Cδ, which isa contradiction for small δ > 0. 2

76

6 Lp-viscosity solutions

In this section, we discuss the Lp-viscosity solution theory for uniformlyelliptic PDEs:

F (x,Du,D2u) = f(x) in Ω, (6.1)

where F : Ω×Rn ×Sn → R and f : Ω → R are given. Since we will use thefact that u + C (for a constant C ∈ R) satisfies the same (6.1), we supposethat F does not depend on u itself. Furthermore, to compare with classicalresults, we prefer to have the inhomogeneous term (the right hand side of(6.1)).

The aim in this section is to obtain the a priori estimates for Lp-viscositysolutions without assuming any continuity of the mapping x → F (x, q,X),and then to establish an existence result of Lp-viscosity solutions for Dirichletproblems.

Remark. In general, without the continuity assumption of x→ F (x, p,X),even if X → F (x, p,X) is uniformly elliptic, we cannot expect the unique-ness of Lp-viscosity solutions. Because Nadirashvili (1997) gave a counter-example of the uniqueness.

6.1 A brief history

Let us simply consider the Poisson equation in a “smooth” domain Ω withzero-Dirichlet boundary condition:

−u = f in Ω,u = 0 on ∂Ω.

(6.2)

In the literature of the regularity theory for uniformly elliptic PDEs ofsecond-order, it is well-known that

“if f ∈ Cσ(Ω) for some σ ∈ (0, 1), then u ∈ C2,σ(Ω)”. (6.3)

Here, Cσ(U) (for a set U ⊂ Rn) denotes the set of functions f : U → R suchthat

supx∈U

|f(x)|+ supx,y∈U,x=y

|f(x)− f(y)||x− y|σ

<∞.

77

Also, Ck,σ(U), for an integer k ≥ 1, denotes the set of functions f : U → Rso that for any multi-index α = (α1, . . . , αn) ∈ 0, 1, 2, . . .n with |α| :=∑n

i=1 αi ≤ k, Dαf ∈ Cσ(U), where

Dαf :=∂|α|f

∂xα11 · · · ∂xαn

n

.

These function spaces are called Holder continuous spaces and the implicationin (6.3) is called the Schauder regularity (estimates). Since the PDE in(6.2) is linear, the regularity result (6.3) may be extended to

“if f ∈ Ck,σ(Ω) for some σ ∈ (0, 1), then u ∈ Ck+2,σ(Ω)”. (6.4)

Moreover, we obtain that (6.4) holds for the following PDE:

−trace(A(x)D2u(x)) = f(x) in Ω, (6.5)

where the coefficient A(·) ∈ C∞(Ω, Sn) satisfies that

λ|ξ|2 ≤ ⟨A(x)ξ, ξ⟩ ≤ Λ|ξ|2 for ξ ∈ Rn and x ∈ Ω.

Furthermore, we can obtain (6.4) even for linear second-order uniformlyelliptic PDEs if the coefficients are smooth enough.

Besides the Schauder estimates, we know a different kind of regularityresults: For a solution u of (6.5), and an integer k ∈ 0, 1, 2, . . .,

“if f ∈ W k,p(Ω) for some p > 1, then u ∈ W k+2,p(Ω)”. (6.6)

Here, for an open set O ⊂ Rn, we say f ∈ Lp(O) if |f |p is integrable in O,and f ∈ W k,p(O) if for any multi-index α with |α| ≤ k, Dαf ∈ Lp(O). Noticethat Lp(Ω) = W 0,p(Ω).

This (6.6) is called the Lp regularity (estimates). For a later con-venience, for p ≥ 1, we recall the standard norms of Lp(O) and W k,p(O),respectively:

∥u∥Lp(O) :=(∫

O|u(x)|pdx

)1/p

, and ∥u∥Wk,p(O) :=∑|α|≤k

∥Dαu∥Lp(O).

In Appendix, we will use the quantity ∥u∥Lp(Ω) even for p ∈ (0, 1) althoughthis is not the “norm” (i.e. the triangle inequality does not hold).

78

We refer to [13] for the details on the Schauder and Lp regularity theoryfor second-order uniformly elliptic PDEs.

As is known, a difficulty occurs when we drop the smoothness of Aij.An extreme case is that we only suppose that Aij are bounded (possibly

discontinuous, but still satisfy the uniform ellipticity). In this case, what canwe say about the regularity of “solutions” of (6.5) ?

The extreme case for PDEs in divergence form is the following:

−n∑

i,j=1

∂

∂xi

(Aij(x)

∂u

∂xj(x)

)= f(x) in Ω. (6.7)

De Giorgi (1957) first obtained Holder continuity estimates on weak so-lutions of (6.7) in the distribution sense; for any ϕ ∈ C∞

0 (Ω),∫Ω(⟨A(x)Du(x), Dϕ(x)⟩ − f(x)ϕ(x)) dx = 0.

Here, we set

C∞0 (Ω) :=

ϕ : Ω → R

∣∣∣∣∣ ϕ(·) is infinitely many times differentiable,and supp ϕ is compact in Ω

.

We refer to [14] for the details of De Giorgi’s proof and, a different proofby Moser (1960).

Concerning the corresponding PDE in nondivergence form, by a stochas-tic approach, Krylov-Safonov (1979) first showed the Holder continuity esti-mates on “strong” solutions of

−trace(A(x)D2u(x)) = f(x) in Ω. (6.8)

Afterward, Trudinger (1980) (see [13]) gave a purely analytic proof for it.Since these results appeared before the viscosity solution was born, they

could only deal with strong solutions, which satisfy PDEs in the a.e. sense.In 1989, Caffarelli proved the same Holder estimate for viscosity solutions

of fully nonlinear second-order uniformly elliptic PDEs.To show Holder continuity of solutions, it is essential to prove the follow-

ing “Harnack inequality” for nonnegative solutions. In fact, to prove theHarnack inequality, we split the proof into two parts:

79

weak Harnack inequalityfor “super”solutions

local maximum principlefor “sub”solutions

=⇒ Harnack inequality

for “solutions”

In section 6.4, we will show that Lp-viscosity solutions satisfy the (inte-rior) Holder continuous estimates.

6.2 Definition and basic facts

We first recall the definition of Lp-strong solutions of general PDEs:

F (x, u,Du,D2u) = f(x) in Ω. (6.9)

We will use the following function space:

W 2,ploc (Ω) := u : Ω → R | ζu ∈ W 2,p(Ω) for all ζ ∈ C∞

0 (Ω).

Throughout this section, we suppose at least

p >n

2

so that u ∈ W 2,ploc (Ω) has the second-order Taylor expansion at almost all

points in Ω, and that u ∈ C(Ω).

Definition. We call u ∈ C(Ω) an Lp-strong subsolution (resp., super-

solution, solution) of (6.9) if u ∈ W 2,ploc (Ω), and

F (x, u(x), Du(x), D2u(x)) ≤ f(x) (resp., ≥ f(x), = f(x)) a.e. in Ω.

Now, we present the definition of Lp-viscosity solutions of (6.9).

Definition. We call u ∈ C(Ω) an Lp-viscosity subsolution (resp., su-

persolution) of (6.9) if for ϕ ∈ W 2,ploc (Ω), we have

limε→0

ess. infBε(x)

(F (y, u(y), Dϕ(y), D2ϕ(y))− f(y)

)≤ 0

80

(resp. lim

ε→0ess. sup

Bε(x)

(F (y, u(y), Dϕ(y), D2ϕ(y))− f(y)

)≥ 0

)provided that u− ϕ takes its local maximum (resp., minimum) at x ∈ Ω.

We call u ∈ C(Ω) an Lp-viscosity solution of (6.9) if it is both an Lp-viscosity sub- and supersolution of (6.9).

Remark. Although we will not explicitly utilize the above definition, werecall the definition of ess. supA and ess. infA of h : A → R, where A ⊂ Rn

is a measurable set:

ess. supAh(y) := infM ∈ R | h ≤M a.e. in A,

andess. inf

Ah(y) := supM ∈ R | h ≥M a.e. in A.

Coming back to (6.1), we give a list of assumptions on F : Ω×Rn×Sn →R:

(1) F (x, 0, O) = 0 for x ∈ Ω,(2) x→ F (x, q,X) is measurable for (q,X) ∈ Rn × Sn,(3) F is uniformly elliptic.

(6.10)

We recall the uniform ellipticity condition of X → F (x, q,X) with the con-stants 0 < λ ≤ Λ from section 3.1.2.

For the right hand side f : Ω → R, we suppose that

f ∈ Lp(Ω) for p ≥ n. (6.11)

We will often suppose the Lipschitz continuity of F with respect to q ∈Rn;

there is µ ≥ 0 such that |F (x, q,X)− F (x, q′, X)| ≤ µ|q − q′|for (x, q, q′, X) ∈ Ω×Rn ×Rn × Sn.

(6.12)

Remark. We note that (1) in (6.10) and (6.12) imply that F has the lineargrowth in Du;

|F (x, q, O)| ≤ µ|q| for x ∈ Ω and q ∈ Rn.

Remark. We note that when x → F (x, q,X) and x → f(x) are continu-ous, the definition of Lp-viscosity subsolution (resp., supersolution) of (6.1)

81

coincides with the standard one under assumption (6.10) and (6.12). For aproof, we refer to a paper by Caffarelli-Crandall-Kocan-Swiech [5].

In this book, we only study the case of (6.11) but most of results can beextended to the case when p > p⋆ = p⋆(Λ, λ, n) ∈ (n/2, n), where p⋆ is theso-called Escauriaza’s constant (see the references in [4]).

The following proposition is obvious but it will be very convenient tostudy Lp-viscosity solutions of (6.1) under assumptions (6.10), (6.11) and(6.12).

Proposition 6.1. Assume that (6.10), (6.11) and (6.12) hold. If u ∈C(Ω) is an Lp-viscosity subsolution (resp., supersolution) of (6.1), then it isan Lp-viscosity subsolution (resp., supersolution) of

P−(D2u)− µ|Du| ≤ f in Ω(resp., P+(D2u) + µ|Du| ≥ f in Ω

).

We recall the Aleksandrov-Bakelman-Pucci (ABP for short) maximumprinciple, which will play an essential role in this section (and also Appendix).

To this end, we introduce the notion of “upper contact sets”: For u :O → R, we set

Γ[u,O] :=

x ∈ O

∣∣∣∣∣ there is p ∈ Rn such thatu(y) ≤ u(x) + ⟨p, y − x⟩ for all y ∈ O

.

Proposition 6.2. (ABP maximum principle) For µ ≥ 0, there is C0 :=C0(Λ, λ, n, µ, diam(Ω)) > 0 such that if for f ∈ Ln(Ω), u ∈ C(Ω) is anLn-viscosity subsolution (resp., supersolution) of

P−(D2u)− µ|Du| ≤ f in Ω+[u]

(resp., P+(D2u) + µ|Du| ≥ f in Ω+[−u]),

thenmaxΩ

u ≤ max∂Ω

u+ + diam(Ω)C0∥f+∥Ln(Γ[u,Ω]∩Ω+[u])(resp., max

Ω(−u) ≤ max

∂Ω(−u)+ + diam(Ω)C0∥f−∥Ln(Γ[−u,Ω]∩Ω+[−u])

),

82

y=u(x)

O

G[u,O]

x

Fig 6.1

PSfrag replacements

Γ[u,Ω]y = u(x)

Ωx

whereΩ+[u] := x ∈ Ω | u(x) > 0.

The next proposition is a key tool to study Lp-viscosity solutions, partic-ularly, when f is not supposed to be continuous. The proof will be given inAppendix.

Proposition 6.3. Assume that (6.11) holds for p ≥ n. For any µ ≥ 0,there are an Lp-strong subsolution u and an Lp-strong supersolution v ∈C(B1) ∩W 2,p

loc (B1), respectively, ofP+(D2u) + µ|Du| ≤ f in B1,

u = 0 on ∂B1,and

P−(D2v)− µ|Dv| ≥ f in B1,

v = 0 on ∂B1.

Moreover, we have the following estimates: for w = u or w = v, and smallδ ∈ (0, 1), there is C = C(Λ, λ, n, µ, δ) > 0 such that

∥w∥W 2,p(Bδ) ≤ Cδ∥f∥Lp(B1).

Remark. In view of the proof (Step 2) of Proposition 6.2, we see that−C∥f−∥Ln(B1) ≤ w ≤ C∥f+∥Ln(B1) in B1, where w = u, v.

6.3 Harnack inequality

In this subsection, we often use the cubeQr(x) for r > 0 and x =t(x1, . . . , xn) ∈Rn;

Qr(x) := y =t(y1, . . . , yn) | |xi − yi| < r/2 for i = 1, . . . , n,

83

and Qr := Qr(0). Notice that

Br/2(x) ⊂ Qr(x) ⊂ Br√n/2(x) for r > 0.

We will prove the next two propositions in Appendix.

Proposition 6.4. (Weak Harnack inequality) For µ ≥ 0, there are p0 =p0(Λ, λ, n, µ) > 0 and C1 := C1(Λ, λ, n, µ) > 0 such that if u ∈ C(B2

√n) is a

nonnegative Lp-viscosity supersolution of

P+(D2u) + µ|Du| ≥ 0 in B2√n,

then we have∥u∥Lp0 (Q1) ≤ C1 inf

Q1/2

u.

Remark. Notice that p0 might be smaller than 1.

Proposition 6.5. (Local maximum principle) For µ ≥ 0 and q > 0,there is C2 = C2(Λ, λ, n, µ, q) > 0 such that if u ∈ C(B2

√n) is an L

p-viscositysubsolution of

P−(D2u)− µ|Du| ≤ 0 in B2√n,

then we havesupQ1

u ≤ C2∥u+∥Lq(Q2).

Remark. Notice that we do not suppose that u ≥ 0 in Proposition 6.5.

6.3.1 Linear growth

The next corollary is a direct consequence of Propositions 6.4 and 6.5.

Corollary 6.6. For µ ≥ 0, there is C3 = C3(Λ, λ, n, µ) > 0 such that ifu ∈ C(B2

√n) is a nonnegative Lp-viscosity sub- and supersolution of

P−(D2u)− µ|Du| ≤ 0 and P+(D2u) + µ|Du| ≥ 0 in B2√n,

respectively, then we have

supQ1

u ≤ C3 infQ1

u.

84

In order to treat inhomogeneous PDEs, we will need the following corol-lary:

Corollary 6.7. For µ ≥ 0 and f ∈ Lp(B3√n) with p ≥ n, there is

C4 = C4(Λ, λ, n, µ) > 0 such that if u ∈ C(B3√n) is a nonnegative Lp-

viscosity sub- and supersolution of

P−(D2u)− µ|Du| ≤ f and P+(D2u) + µ|Du| ≥ f in B2√n,

respectively, then we have

supQ1

u ≤ C4

(infQ1

u+ ∥f∥Lp(B3√

n)

).

Proof.According to Proposition 6.3, we find v, w ∈ C(B3√n)∩W 2,p

loc (B3√n)

such that P+(D2v) + µ|Dv| ≤ −f+ a.e. in B3

√n,

v = 0 on ∂B3√n,

and P−(D2w)− µ|Dw| ≥ f− a.e. in B3

√n,

w = 0 on ∂B3√n.

In view of Proposition 6.3 and its Remark, we can choose C = C(Λ, λ, n, µ) >0 such that

0 ≤ −v ≤ C∥f+∥Lp(B3√

n)in B3

√n, ∥v∥W 2,p(B2

√n)

≤ C∥f+∥Lp(B3√

n),

and

0 ≤ w ≤ C∥f−∥Lp(B3√

n)in B3

√n, ∥w∥W 2,p(B2

√n)

≤ C∥f−∥Lp(B3√n).

Since v, w ∈ W 2,p(B2√n), it is easy to verify that u1 := u + v and

u2 := u+ w are, respectively, an Lp-viscosity sub- and supersolution of

P−(D2u1)− µ|Du1| ≤ 0 and P+(D2u2) + µ|Du2| ≥ 0 in B2√n.

Since v ≤ 0 in B3√n, applying Proposition 6.5 to u1, for any q > 0, we find

C2(q) > 0 such that

supQ1

u ≤ supQ1

u1 + C∥f+∥Lp(B3√

n)

≤ C2(q)∥(u1)+∥Lq(Q2) + C∥f+∥Lp(B3√

n)

≤ C2(q)∥u∥Lq(Q2) + C∥f+∥Lp(B3√

n).

(6.13)

85

On the other hand, applying Proposition 6.4 to u2, there is p0 > 0 suchthat

∥u∥Lp0(Q2) ≤ ∥u2∥Lp0 (Q2) ≤ C1 infQ1

u2 ≤ C1

(infQ1

u+ C∥f−∥Lp(B3√

n)

). (6.14)

Therefore, combining (6.14) with (6.13) for q = p0, we can find C4 > 0 suchthat the assertion holds. 2

Corollary 6.8. (Harnack inequality, final version) Assume that (6.10),(6.11) and (6.12) hold. If u ∈ C(Ω) is an Lp-viscosity solution of (6.1), andif B3

√nr(x) ⊂ Ω for r ∈ (0, 1], then

supQr(x)

u ≤ C4

(inf

Qr(x)u+ r2−

np ∥f∥Lp(Ω)

),

where C4 > 0 is the constant in Corollary 6.7.

Proof. By translation, we may suppose that x = 0.Setting v(x) := u(rx) for x ∈ B3

√n, we easily see that v is an Lp-viscosity

subsolution and supersolution of

P−(D2v)− µ|Dv| ≤ r2f and P+(D2v) + µ|Dv| ≥ −r2f , in B3√n,

respectively, where f(x) := f(rx). Note that ∥f∥Lp(B3√

n)= r−

np ∥f∥Lp(B3

√nr)

.Applying Corollary 6.7 to v and then, rescaling v to u, we conclude the

assertion. 2

6.3.2 Quadratic growth

Here, we consider the case when q → F (x, q,X) has quadratic growth. Werefer to [10] for applications where such quadratic nonlinearity appears.

We present a version of the Harnack inequality when F has a quadraticgrowth in Du in place of (6.12);

there is µ ≥ 0 such that |F (x, q,X)− F (x, q′, X)|≤ µ(|q|+ |q′|)|q − q′| for (x, q, q′, X) ∈ Ω×Rn ×Rn × Sn,

(6.15)

which together with (1) of (6.10) implies that

|F (x, q, O)| ≤ µ|q|2 for (x, q) ∈ Ω×Rn.

86

The associated Harnack inequality is as follows:

Theorem 6.9. For µ ≥ 0 and f ∈ Lp(B3√n) with p ≥ n, there is

C5 = C5(Λ, λ, n, µ) > 0 such that if u ∈ C(B3√n) is a nonnegative Lp-

viscosity sub- and supersolution of

P−(D2u)− µ|Du|2 ≤ f and P+(D2u) + µ|Du|2 ≥ f in B3√n,

respectively, then we have

supQ1

u ≤ C5e2µλM(infQ1

u+ ∥f∥Lp(B3√

n)

),

where M := supB3√nu.

Proof. Set α := µ/λ. Fix any δ ∈ (0, 1).We claim that v := eαu − 1 and w := 1 − e−αu are, respectively, a non-

negative Lp-viscosity sub- and supersolution of

P−(D2v) ≤ α(eαM + δ)f+ and P+(D2w) ≥ −α(1 + δ)f− in B3√n.

We shall only prove this claim for v since the other for w can be obtainedsimilarly.

Choose ϕ ∈ W 2,ploc (B3

√n) and suppose that u−ϕ attains its local maximum

at x ∈ B3√n. Thus, we may suppose that v(x) = ϕ(x) and v ≤ ϕ in Br(x),

where B2r(x) ⊂ B3√n. Note that 0 ≤ v ≤ eαM − 1 in B3

√n.

For any δ ∈ (0, 1), in view of W 2,p(Br(x)) ⊂ Cσ0(Br(x)) with someσ0 ∈ (0, 1), we can choose ε0 ∈ (0, r) such that

−δ ≤ ϕ ≤ v + δ in Bε0(x).

Setting ψ(y) := α−1 log(ϕ(y) + 1) for y ∈ Bε0(x) (extending ψ ∈ W 2,p inB3

√n \Bε0(x) if necessary), we have

limε→0

ess infBε(x)

(P−(D2ψ)− µ|Dψ|2 − f+

)≤ 0.

Since

Dψ =Dϕ

α(ϕ+ 1)and D2ψ =

D2ϕ

α(ϕ+ 1)− Dϕ⊗Dϕ

α(ϕ+ 1)2,

87

the above inequality yields

limε→0

ess infBε(x)

(P−(D2ϕ)

α(ϕ+ 1)− f+

)≤ 0.

Since 0 < 1− δ ≤ ϕ+ 1 ≤ eαM + δ in Bε0(x), we have

limε→0

ess infBε(x)

(P−(D2ϕ)− α(eαM + δ)f+

)≤ 0.

Since αu ≤ v ≤ αueαM and αue−αM ≤ w ≤ αu, using the same argumentto get (6.13) and (6.14), we have

supQ1

u ≤ 1

αsupQ1

v ≤ C6

∥v∥Lp0 (Q2) + (eαM + δ)∥f+∥Lp(B3

√n)

≤ C7

e2αM∥w∥Lp0 (Q2) + (eαM + δ)∥f+∥Lp(B3

√n)

≤ C8

e2αM inf

Q1

w + (e2αM + δ)∥f∥Lp(B3√

n)

≤ C9e

2αM

infQ1

u+ (1 + δ)∥f∥Lp(B3√

n)

.

Since Ck (k = 6, . . . , 9) are independent of δ > 0, sending δ → 0, we concludethe proof. 2

Remark. We note that the same argument by using two different trans-formations for sub- and supersolutions as above can be found in [14] foruniformly elliptic PDEs in divergence form with the quadratic nonlinearity.

6.4 Holder continuity estimates

In this subsection, we show how the Harnack inequality implies the Holdercontinuity.

Theorem 6.10. Assume that (6.10), (6.11) and (6.12) hold. For eachcompact set K ⊂ Ω, there is σ = σ(Λ, λ, n, µ, p, dist(K, ∂Ω), ∥f∥Lp(Ω)) ∈(0, 1) such that if u ∈ C(Ω) is an Lp-viscosity solution of (6.1), then there isC = C(Λ, λ, n, µ, p, dist(K, ∂Ω),maxΩ |u|, ∥f∥Lp(Ω)) > 0

|u(x)− u(y)| ≤ C|x− y|σ for x, y ∈ K.

Remark. We notice that σ is independent of supΩ |u|.

88

In our proof below, we may relax the dependence maxΩ |u| in C by

sup|u(x)| | dist(x,K) < ε for small ε > 0.

Proof. Setting r0 := min1,dist(K, ∂Ω)/(3√n) > 0, we may suppose

that there is C4 > 1 such that if w ∈ C(Ω) is a nonnegative Lp-viscositysub- and supersolution of

P−(D2w)− µ|Dw| ≤ f and P+(D2w) + µ|Dw| ≥ f in Ω,

respectively, then we see that for any r ∈ (0, r0] and x ∈ K (i.e. B3√nr(x) ⊂

Ω),

supQr(x)

w ≤ C4

(inf

Qr(x)w + r2−

np ∥f∥Lp(Ω)

).

For simplicity, we may suppose x = 0 ∈ K.Now, we set

M(r) := supQr

u, m(r) := infQr

u and osc(r) :=M(r)−m(r).

It is sufficient to find C > 0 and σ ∈ (0, 1) such that

M(r)−m(r) ≤ Crσ for small r > 0.

We denote by S(r) the set of all nonnegative w ∈ C(B3√nr), which is,

respectively, an Lp-viscosity sub- and supersolution of

P−(D2w)− µ|Dw| ≤ |f | and P+(D2w) + µ|Dw| ≥ −|f | in B3√nr.

Setting v1 := u−m(r) and w1 :=M(r)−u, we see that v1 and w1 belongto S(r). Hence, setting C10 := maxC4∥f∥Lp(Ω), C4, 4 > 3, we have

supQr/2

v1 ≤ C10

(infQr/2

v1 + r2−np

)and sup

Qr/2

w1 ≤ C10

(infQr/2

w1 + r2−np

).

Thus, setting β := 2− np> 0, we have

M(r/2)−m(r) ≤ C10

(m(r/2)−m(r) + (r/2)β

),

M(r)−m(r/2) ≤ C10

(M(r)−M(r/2) + (r/2)β

).

89

Hence, adding these inequalities, we have

(C10 + 1)(M(r/2)−m(r/2)) ≤ (C10 − 1)(M(r)−m(r)) + 2C10(r/2)β.

Therefore, setting θ := (C10−1)/(C10+1) ∈ (1/2, 1) and C11 := 2C10/(C10+1), we see that

osc(r/2) ≤ θosc(r) + C11(r/2)β.

Moreover, changing r/2k−1 for integers k ≥ 2, we have

osc(r/2k) ≤ θkosc(r) + C11rβ

k∑j=1

2−βj

≤ θkosc(r0) +C11

2β − 1rβ ≤ C12(θ

k + rβ),

where C12 := maxosc(r0), C11/(2β − 1).

For r ∈ (0, r0), by setting s = rα, where α = log θ/(log θ−β log 2) ∈ (0, 1),there is a unique integer k ≥ 1 such that

s

2k≤ r <

s

2k−1,

which yieldslog(s/r)

log 2≤ k <

log(s/r)

log 2+ 1.

Hence, recalling θ ∈ (1/2, 1), we have

osc(r) ≤ osc(s/2k−1) ≤ C12(θk + (2s)β) ≤ 2βC12

(θ(α−1) log r/ log 2 + rβα

).

Setting σ := (α− 1) log θ/ log 2 ∈ (0, 1) (because θ ∈ (1/2, 1)), we have

θ(α−1) log r/ log 2 = rσ and rβα = rσ.

Thus, setting C13 := 2βC12, we have

osc(r) ≤ C13rσ. 2 (6.16)

Remark. We note that we may derive (6.16) when p > n/2 by takingβ = 2− n

p> 0.

We shall give the corresponding Holder continuity for PDEs with quadraticnonlinearity (6.15). Since we can use the same argument as in the proof of

90

Theorem 6.1 using Theorem 6.9 instead of Corollaries 6.7 and 6.8, we omitthe proof of the following:

Corollary 6.11. Assume that (6.10), (6.11) and (6.15) hold. For eachcompact set K ⊂ Ω, there are C = C(Λ, λ, n, µ, p, dist(K, ∂Ω), supΩ |u|) > 0and σ = σ(Λ, λ, n, µ, p, dist(K, ∂Ω), supΩ |u|) ∈ (0, 1) such that if an Lp-viscosity solution u ∈ C(Ω) of (6.1), then we have

|u(x)− u(y)| ≤ C|x− y|σ for x, y ∈ K.

Remark. Note that both of σ and C depend on supΩ |u| in this quadraticcase.

6.5 Existence result

For the existence of Lp-viscosity solutions of (6.1) under the Dirichlet condition,we only give an outline of proof, which was first shown in a paper by Crandall-Kocan-Lions-Swiech in [7] (1999).

Theorem 6.12. Assume that (6.10), (6.11) and (6.12) hold. Assume also that(1) of (5.17) holds.

For given g ∈ C(∂Ω), there is an Lp-viscosity solution u ∈ C(Ω) of (6.1) suchthat

u(x) = g(x) for x ∈ ∂Ω. (6.17)

Remark. We may relax assumption (1) of (5.17) so that the assertion holds forΩ which may have some “concave” corners. Such a condition is called “uniformexterior cone condition”.

Sketch of proof.Step1: We first solve approximate PDEs, which have to satisfy a sufficient

condition in Step 3; instead of (6.1), under (6.17), we consider

Fk(x,Du,D2u) = fk in Ω, (6.18)

where “smooth” Fk and fk approximate F and f , respectively. In fact, Fk and fkare given by F ∗ρ1/k and f ∗ρ1/k, where ρ1/k is the standard mollifier with respectto x-variables. We remark that F ∗ ρ1/k means the convolution of F (·, p,X) andρ1/k.

91

We find a viscosity solution uk ∈ C(Ω) of (6.18) under (6.17) via Perron’smethod for instance. At this stage, we need to suppose the smoothness of ∂Ω toconstruct viscosity sub- and supersolutions of (6.18) with (6.17). Remember thatif F and f are continuous, then the notion of Lp-viscosity solutions equals to thatof the standard ones (see Proposition 2.9 in [5]).

In view of (1) of (5.17) (i.e. the uniform exterior sphere condition), we canconstruct viscosity sub- and supersolutions of (6.18) denoted by ξ ∈ USC(Ω) andη ∈ LSC(Ω) such that ξ = η = g on ∂Ω. To show this fact, we only note that wecan modify the argument in Step 1 in section 7.3.

Step 2: We next obtain the a priori estimates for uk so that they converge to

a continuous function u ∈ C(Ω), which is the candidate of the original PDE.For this purpose, after having established the L∞ estimates via Proposition

6.2, we apply Theorem 6.10 (interior Holder continuity) to uk in Step 1 because(6.10)-(6.12) hold for approximate PDEs with the same constants λ,Λ, µ.

We need a careful analysis to get the equi-continuity up to the boundary ∂Ω.See Step 1 in section 7.3 again.

Step 3: Finally, we verify that the limit function u is the Lp-viscosity solutionvia the following stability result, which is an Lp-viscosity version of Proposition4.8.

To state the result, we introduce some notations: For B2r(x) ⊂ Ω with r > 0and x ∈ Ω, and ϕ ∈W 2,p(Br(x)), we set

Gk[ϕ](y) := Fk(y,Dϕ(y), D2ϕ(y))− fk(y),

andG[ϕ](y) := F (y,Dϕ(y), D2ϕ(y))− f(y)

for y ∈ Br(x).

Proposition 6.13. Assume that Fk and F satisfy (6.10) and (6.12) withλ,Λ > 0 and µ ≥ 0. For f, fk ∈ Lp(Ω) with p ≥ n, let uk ∈ C(Ω) be an Lp-viscositysubsolution (resp., supersolution) of (6.18). Assume also that uk converges to uuniformly on any compact subsets of Ω as k → ∞, and that for any B2r(x) ⊂ Ωwith r > 0 and x ∈ Ω, and ϕ ∈W 2,p(Br(x)),

limk→∞

∥(G[ϕ]−Gk[ϕ])+∥Lp(Br(x)) = 0

(resp., lim

k→∞∥(G[ϕ]−Gk[ϕ])

−∥Lp(Br(x)) = 0

).

Then, u ∈ C(Ω) is an Lp-viscosity subsolution (resp., supersolution) of (6.1).

92

Proof of Proposition 6.13. We only give a proof of the assertion for subsolu-tions.

Suppose the contrary: There are r > 0, ε > 0, x ∈ Ω and ϕ ∈ W 2,p(B2r(x))such that B3r(x) ⊂ Ω, 0 = (u− ϕ)(x) ≥ (u− ϕ)(y) for y ∈ B2r(x), and

u− ϕ ≤ −ε in B2r(x) \Br(x), (6.19)

andG[ϕ](y) ≥ ε a.e. in B2r(x). (6.20)

For simplicity, we shall suppose that r = 1 and x = 0.It is sufficient to find ϕk ∈W 2,p(B2) such that limk→∞ supB2

|ϕk| = 0, and

Gk[ϕ+ ϕk](y) ≥ ε a.e. in B2.

Indeed, since uk − (ϕ + ϕk) attains its maximum over B2 at an interior pointz ∈ B2 by (6.19), the above inequality contradicts the fact that uk is an Lp-viscosity subsolution of (6.18).

Setting h(x) := G[ϕ](x) and hk(x) := Gk[ϕ](x), in view of Proposition 6.3, wecan find ϕk ∈ C(B2) ∩W 2,p

loc (B2) such thatP−(D2ϕk)− µ|Dϕk| ≥ (h− hk)

+ a.e. in B2,ϕk = 0 on ∂B2,

0 ≤ ϕk ≤ C∥(h− hk)+∥Lp(B2) in B2,

∥ϕk∥W 2,p(B1) ≤ C∥(h− hk)+∥Lp(B2).

We note that our assumption together with the third inequality in the above yieldslimk→∞ supB2

|ϕk| = 0.Using (6.10), (6.12) and (6.20), we have

Gk[ϕ+ ϕk] ≥ P−(D2ϕk)− µ|Dϕk|+ hk≥ (h− hk)

+ + ε− (h− hk)≥ ε a.e. in B2. 2

93

7 Appendix

In this appendix, we present proofs of the propositions, which appeared in theprevious sections. However, to prove them, we often need more fundamentalresults, for which we only give references. One of such results is the following“Area formula”, which will be employed in sections 7.1 and 7.2. We refer to[9] for a proof of a more general Area formula.

Area formulaξ ∈ C1(Rn,Rn),g ∈ L1(Rn),

A ⊂ Rn measurable

=⇒∫ξ(A)

|g(y)|dy ≤∫A|g(ξ(x))||det(Dξ(x))|dx

We note that the Area formula is a change of variable formula when|det(Dξ)| may vanish. In fact, the equality holds if |det(Dξ)| > 0 and ξ isinjective.

7.1 Proof of Ishii’s lemma

First of all, we recall an important result by Aleksandrov. We refer to theAppendix of [6] and [10] for a “functional analytic” proof, and to [9] for a“measure theoretic” proof.

Lemma 7.1. (Theorem A.2 in [6]) If f : Rn → R is convex, then fora.a. x ∈ Rn, there is (p,X) ∈ Rn × Sn such that

f(x+ h) = f(x) + ⟨p, h⟩+ 1

2⟨Xh, h⟩+ o(|h|2) as |h| → 0.

(i.e., f is twice differentiable at a.a. x ∈ Rn.)

We next recall Jensen’s lemma, which is a version of the ABP maximumprinciple in 7.2 below.

Lemma 7.2. (Lemma A.3 in [6]) Let f : Rn → R be semi-convex(i.e. x → f(x) + C0|x|2 is convex for some C0 ∈ R). Let x ∈ Rn be a strictmaximum point of f . Set fp(x) := f(x)− ⟨p, x⟩ for x ∈ Rn and p ∈ Rn.

Then, for r > 0, there are C1, δ0 > 0 such that

|Γr,δ| ≥ C1δn for δ ∈ (0, δ0],

94

where

Γr,δ :=x ∈ Br(x)

∣∣∣∃p ∈ Bδ such that fp(y) ≤ fp(x) for y ∈ Br(x).

Proof. By translation, we may suppose x = 0.For integers m, we set fm(x) = f ∗ ρ1/m(x), where ρ1/m is the mollifier.

Note that x→ fm(x) + C0|x|2 is convex.Setting

Γmr,δ =

x ∈ Br

∣∣∣∃p ∈ Bδ such that fmp (y) ≤ fm

p (x) for y ∈ Br

,

where fmp (x) = fm(x)−⟨p, x⟩, we claim that there are C1, δ0 > 0, independent

of large integers m, such that

|Γmr,δ| ≥ C1δ

n for δ ∈ (0, δ0].

We remark that this concludes the assertion. In fact, setting Am := ∪∞k=mΓ

kr,δ,

we have ∩∞m=1Am ⊂ Γr,δ. Because, for x ∈ ∩∞

m=1Am, we can select pk ∈ Bδ

and mk such that limk→∞mk = ∞, and

maxBr

fmkpk

= fmkpk

(x).

Hence, sending k → ∞ (along a subsequence if necessary), we find p ∈ Bδ

such that maxBrfp = fp(x), which yields x ∈ Γr,δ.

Therefore, we have

C1δn ≤ lim

m→∞|Am| = | ∩∞

m=1 A| ≤ |Γr,δ|.

Now we shall prove our claim. First of all, we notice that x → fm(x) +C0|x|2 is convex.

Since 0 is the strict maximum of f , we find ε0 > 0 such that

ε0 = f(0)− maxB4r/3\Br/3

f.

Fix p ∈ Bδ0 , where δ0 = ε0/(3r). For m ≥ 3/r, we note that

fm(x)− ⟨p, x⟩ ≤ f(0)− ε0 + δ0r ≤ f(0)− 2ε03

in Br \B2r/3.

95

On the other hand, for large m, we verify that

fm(0) ≥ f(0)− ωf (m−1) > f(0)− ε0

3,

where ωf denotes the modulus of continuity of f . Hence, in view of theseobservations, for any p ∈ Bδ0 , if maxBr

fmp = fm

p (x) for x ∈ Br, then x ∈ Br.In other words, we see that

Bδ = Dfm(Γmr,δ) for δ ∈ (0, δ0].

Thanks to the Area formula, we have

|Bδ| =∫Dfm(Γm

r,δ)dy ≤

∫Γmr,δ

|detD2fm|dx ≤ (2C0)n|Γm

r,δ|.

Here, we have employed that −2C0I ≤ D2fm ≤ O in Γmr,δ. 2

Although we can find a proof of the next proposition in [6], we recall theproof with a minor change for the reader’s convenience.

Proposition 7.3. (Lemma A.4 in [6]) If f ∈ C(Rm), B ∈ Sm, ξ →f(ξ) + (λ/2)|ξ|2 is convex and maxξ∈Rmf(ξ) − 2−1⟨Bξ, ξ⟩ = f(0), thenthere is an X ∈ Sm such that

(0, X) ∈ J2,+f(0) ∩ J2,−

f(0) and − λI ≤ X ≤ B.

Proof. For any δ > 0, setting fδ(ξ) := f(ξ)− 2−1⟨Bξ, ξ⟩− δ|ξ|2, we noticethat the semi-convex fδ attains its strict maximum at ξ = 0.

In view of Lemmas 7.1 and 7.2, there are ξδ, qδ ∈ Bδ such that ξ →fδ(ξ) + ⟨qδ, ξ⟩ has a maximum at ξδ, at which f is twice differentiable.

It is easy to see that Df(ξδ) → 0 (as δ → 0) and, moreover, from theconvexity of ξ → f(ξ) + (λ/2)|ξ|2,

−λI ≤ D2f(ξδ) ≤ B + 2δI.

Noting (Df(ξδ), D2f(ξδ)) ∈ J2,+f(ξδ) ∩ J2,−f(ξδ), we conclude the assertion

by taking the limit as δ → 0. 2

We next give a “magic” property of sup-convolutions. For the reader’sconvenience, we put the proof of [6].

96

Lemma 7.4. (Lemma A.5 in [6]) For v ∈ USC(Rn) with supRn v < ∞and λ > 0, we set

v(ξ) := supx∈Rn

(v(x)− λ

2|x− ξ|2

).

For η, q ∈ Rn, Y ∈ Sn, and (q, Y ) ∈ J2,+v(η), we have

(q, Y ) ∈ J2,+v(η + λ−1q) and v(η) +|q|2

2λ= v(η + λ−1q).

In particular, if (0, Y ) ∈ J2,+v(0), then (0, Y ) ∈ J

2,+v(0).

Proof. For (q, Y ) ∈ J2,+v(η), we choose y ∈ Rn such that

v(η) = v(y)− λ

2|y − η|2.

Thus, from the definition, we see that for any x, ξ ∈ Rn,

v(x)− λ

2|ξ − x|2 ≤ v(ξ) ≤ v(η) + ⟨q, ξ − η⟩

+1

2⟨Y (ξ − η), ξ − η⟩+ o(|ξ − η|2)

= v(y)− λ

2|y − η|2 + ⟨q, ξ − η⟩

+1

2⟨Y (ξ − η), ξ − η⟩+ o(|ξ − η|2).

Taking ξ = x− y + η in the above, we have (q, Y ) ∈ J2,+v(y).To verify that y = η + λ−1q, putting x = y and ξ = η − ε(λ(η − y) + q)

for ε > 0 in the above again, we have

ε|λ(η − y) + q|2 ≤ o(ε),

which concludes the proof. 2

Proof of Lemma 3.6. First of all, extending upper semi-continuous func-tions u,w in Ω into Rn by −∞ in Rn \Ω, we shall work in Rn ×Rn insteadof Ω× Ω.

By translation, we may suppose that x = y = 0, at which u(x) + w(y)−ϕ(x, y) attains its maximum.

97

Furthermore, replacing u(x), w(y) and ϕ(x, y), respectively, by

u(x)− u(0)− ⟨Dxϕ(0, 0), x⟩, w(y)− w(0)− ⟨Dyϕ(0, 0), y⟩

andϕ(x, y)− ϕ(0, 0)− ⟨Dxϕ(0, 0), x⟩ − ⟨Dyϕ(0, 0), y⟩,

we may also suppose that ϕ(0, 0) = u(0) = w(0) = 0 and Dϕ(0, 0) = (0, 0) ∈Rn ×Rn.

Since ϕ(x, y) =

⟨A

2

(xy

),

(xy

)⟩+o(|x|2+|y|2), whereA := D2ϕ(0, 0) ∈

S2n, for each η > 0, we see that the mapping (x, y) → u(x) + w(y) −1

2

⟨(A+ ηI)

(xy

),

(xy

)⟩attains its (strict) maximum at 0 ∈ R2n.

We will show the assertion for A+ηI in place of A. Then, sending η → 0,we can conclude the proof. Therefore, we need to prove the following:

Simplified version of Ishii’s lemma.For upper semi-continuous functions u and w in Rn, we suppose that

u(x) + w(y)−⟨A

2

(xy

),

(xy

)⟩≤ u(0) + w(0) = 0 in Rn ×Rn.

Then, for each µ > 1, there are X, Y ∈ Sn such that (0, X) ∈ J2,+u(0),

(0, Y ) ∈ J2,+w(0) and −(µ+ ∥A∥)

(I OO I

)≤(X OO Y

)≤ A+

1

µA2.

Proof of the simplified version of Lemma 3.6. Since Holder’s inequality im-plies ⟨

A

(xy

),

(xy

)⟩≤

⟨(A+

1

µA2

)(ξη

),

(ξη

)⟩+(µ+ ∥A∥)(|x− ξ|2 + |y − η|2)

for x, y, ξ, η ∈ Rn and µ > 0, setting λ = µ+ ∥A∥, we have

u(x)− λ

2|x− ξ|2 + w(y)− λ

2|y − η|2 ≤ 1

2

⟨(A+

1

µA2

)(ξη

),

(ξη

)⟩.

Using the notation in Lemma 7.4, we denote by u and w the sup-convolutionof u and w, respectively, with the above λ > 0. Thus, we have

u(ξ) + w(η) ≤ 1

2

⟨(A+

1

µA2

)(ξη

),

(ξη

)⟩for all ξ, η ∈ Rn.

98

Since u(0) ≥ u(0) = 0 and w(0) ≥ w(0) = 0, the above inequality impliesu(0) = w(0) = 0.

In view of Proposition 7.3 with m = 2n, f(ξ, η) = u(ξ) + w(η) and

B = A+ µ−1A2, there is Z ∈ S2n such that (0, Z) ∈ J2,+f(0, 0)∩ J2,−

f(0, 0)and −λI ≤ Z ≤ B.

Hence, from the definition of J2,±

, it is easy to verify that there areX, Y ∈Sn such that (0, X) ∈ J

2,+u(0)∩J2,−

u(0), (0, Y ) ∈ J2,+w(0)∩J2,−

w(0), and

Z =

(X OO Y

).

Applying the last property in Lemma 7.4 to u and w, we see that

(0, X) ∈ J2,+u(0) and (0, Y ) ∈ J

2,+w(0). 2

7.2 Proof of the ABP maximum principle

First of all, we remind the readers of our strategy in this and the next sub-sections.

We first show that the ABP maximum principle holds under f ∈ Ln(Ω)∩C(Ω) in Steps 1 and 2 of this subsection. Next, using this fact, we estab-lish the existence of Lp-strong solutions of “Pucci” equations in the nextsubsection when f ∈ Lp(Ω).

Employing this existence result, in Step 3, we finally prove Proposition6.2; the ABP maximum principle when f ∈ Ln(Ω).

ABP maximum principle for f ∈ Ln(Ω) ∩ C(Ω) (Section 7.2)⇓

Existence of Lp-strong solutions of Pucci equations (Section 7.3)⇓

ABP maximum principle for f ∈ Ln(Ω) (Section 7.2)

Proof of Proposition 6.2. We give a proof in [5] for the subsolution asser-tion of Proposition 6.2.

By scaling, we may suppose that diam(Ω) ≤ 1.Setting

r0 := maxΩ

u−max∂Ω

u+,

we may also suppose that r0 > 0 since otherwise, the conclusion is obvious.

99

We first introduce the following notation: For u : Ω → R and r ≥ 0,

Γr :=x ∈ Ω

∣∣∣∃p ∈ Br such that u(y) ≤ u(x) + ⟨p, y − x⟩ for y ∈ Ω.

Recalling the upper contact set in section 6.2, we note that

Γ[u,Ω] =∪r>0

Γr.

Step 1: u ∈ C2(Ω) ∩ C(Ω). We first claim that for r ∈ (0, r0),(i) Br = Du(Γr),(ii) D2u ≤ O in Γr.

(7.1)

To show (i), for p ∈ Br, we take x ∈ Ω such that u(x) − ⟨p, x⟩ =maxx∈Ω(u(x) − ⟨p, x⟩). Since u(x) − u(x) ≤ r < r0 for x ∈ Ω, taking themaximum over Ω, we have x ∈ Ω. Hence, we see p = Du(x), which concludes(i).

For x ∈ Γr, Taylor’s formula yields

u(y) = u(x) + ⟨Du(x), y − x⟩+ 1

2⟨D2u(x)(y − x), y − x⟩+ o(|y − x|2).

Hence, we have 0 ≥ ⟨D2u(x)(y − x), y − x⟩+ o(|y − x|2), which shows (ii).

Now, we introduce functions gκ(p) :=(|p|n/(n−1) + κn/(n−1)

)1−nfor κ > 0.

We shall simply write g for gκ.Thus, for r ∈ (0, r0), we see that∫

Du(Γr)g(p)dp ≤

∫Γr

g(Du(x))|det(D2u(x))|dx

=∫Γr

(|Du|n/(n−1) + κn/(n−1)

)1−n|detD2u(x)|dx.

Recalling (7.1), we utilize |detD2u| ≤ (−trace(D2u)/n)n in Γr to findC > 0 such that∫

Br

g(p)dp ≤ C∫Γr

(|Du|n/(n−1) + κn/(n−1)

)1−n(−trace(D2u))ndx. (7.2)

Thus, since (µ|Du|+f+)n ≤ g(Du)−1(µn+κ−n(f+)n) by Holder’s inequality,we have ∫

Br

g(p)dp ≤ C∫Γr

(µn +

(f+

κ

)n)dx. (7.3)

100

On the other hand, since (|p|n + κn)−1 ≤ g(p), we have

log((

r

κ

)n

+ 1)≤ C

∫Br

1

|p|n + κndp ≤ C

∫Br

g(p)dp.

Hence, noting Γr ⊂ Ω+[u] for r ∈ (0, r0), by (7.3), we have

r ≤ κ

[exp

C∫Γ[u,Ω]∩Ω+[u]

(µn +

(f+

κ

)n)dx

− 1

]1/n. (7.4)

When ∥f+∥Ln(Γ[,Ω]∩Ω+[u]) = 0, then sending κ→ 0, we get a contradiction.Thus, we may suppose that ∥f+∥Ln(Γ[,Ω]∩Ω+[u]) > 0.

Setting κ := ∥f+∥Ln(Γ[u,Ω]∩Ω+[u]) and r := r0/2, we can find C > 0,independent of u, such that r0 ≤ C∥f+∥Ln(Γ[u,Ω]∩Ω+[u]).

Remark. We note that we do not need to suppose f to be continuous inStep 1 while we need it in the next step.

Step 2: u ∈ C(Ω) and f ∈ Ln(Ω) ∩ C(Ω). First of all, because of f ∈C(Ω), we remark that u is a “standard” viscosity subsolution of

P−(D2u)− µ|Du| ≤ f in Ω+[u].

(See Proposition 2.9 in [5].)Let uε be the sup-convolution of u for ε > 0;

uε(x) := supy∈Ω

u(y)− |x− y|2

2ε

.

Note that uε is semi-convex and thus, twice differentiable a.e. in Rn.We claim that for small ε > 0, uε is a viscosity subsolution of

P−(D2uε)− µ|Duε| ≤ f ε in Ωε, (7.5)

where f ε(x) := supf+(y) | |x − y| ≤ 2(∥u∥L∞(Ω)ε)1/2 and Ωε := x ∈

Ω+[u] | dist(x, ∂Ω+[u]) > 2(∥u∥L∞(Ω)ε)1/2. Indeed, for x ∈ Ωε and (q,X) ∈

J2,+uε(x), choosing x ∈ Ω such that uε(x) = u(x)− (2ε)−1|x− x|2, we easilyverify that |q| = ε−1|x − x| ≤ 2

√∥u∥L∞(Ω)/ε. Thus, by Lemma 7.4, we see

that (q,X) ∈ J2,+u(x+ εq). Hence, we have

P−(X)− µ|q| ≤ f+(x+ εq) ≤ f ε(x).

101

We note that for small ε > 0, we may suppose that

rε := maxΩε

uε −max∂Ωε

(uε)+ > 0. (7.6)

Here, we list some properties on upper contact sets: For small δ > 0, weset

Ωδ := x ∈ Ω | dist(x, ∂Ω) > δ.

Lemma 7.5. Let vδ ∈ C(Ωδ) and v ∈ C(Ω) satisfy that vδ → v uniformly

on any compact sets in Ω as δ → 0. Assume that r := maxΩ v−max∂Ω v+ > 0.

Then, for r ∈ (0, r), we have the following properties:(1) Γr[v,Ω] is a compact set in Ω+[v],(2) lim sup

δ→0Γr[vδ,Ω

δ] ⊂ Γr[v,Ω],

(3) for small α > 0, there is δα such that ∪0≤δ<δα Γr[vδ,Ωδ] ⊂ Γα

r ,

where Γαr := x ∈ Ω | dist(x,Γr[v,Ω]) < α.

Proof of Lemma 7.5. To show (1), we first need to observe that for r ∈(0, r), dist(Γr[v,Ω], ∂Ω) > 0. Suppose the contrary; if there is xk ∈ Γr[v,Ω]such that xk ∈ Ω → x ∈ ∂Ω, then there is pk ∈ Br such that v(y) ≤v(xk) + ⟨pk, y − xk⟩ for y ∈ Ω. Hence, sending k → ∞, we have

maxΩ

v −max∂Ω

v+ ≤ r < r,

which is a contradiction. Thus, we can find a compact set K ⊂ Ω such thatΓr[v,Ω] ⊂ K.

Moreover, if v(x) ≤ 0 for x ∈ Γr[v,Ω], then we get a contradiction:

r ≤ maxΩ

v ≤ r < r.

Next, choose x ∈ lim supδ→0 Γr[vδ,Ωδ]. Then, for any k ≥ 1, there are

δk ∈ (0, 1/k) and pk ∈ Br such that

vδk(y) ≤ vδk(x) + ⟨pk, y − x⟩ for y ∈ Ωδk .

We may suppose pk → p for some p ∈ Br taking a subsequence if necessary.Sending k → ∞ in the above, we see that x ∈ Γr[v,Ω].

102

If (3) does not hold, then there are α0 > 0, δk ∈ (0, 1/k) and xk ∈Γr[vδk ,Ω

δk ]\ Γα0r . We may suppose again that limk→∞ xk = x for some x ∈ Ω.

When x ∈ ∂Ω, since there is pk ∈ Br such that vδk(y) ≤ vδk(xk)+ ⟨pk, y−xk⟩for y ∈ Ω, we have r < r, which is a contradiction. Thus, we may supposethat x ∈ Ω and, then x ∈ Γr[v,Ω]. Thus, there is k0 ≥ 1 such that xk ∈ Γα0

r

for k ≥ k0, which is a contradiction. 2

For δ > 0, we set uεδ := uε ∗ ρδ, where ρδ is the standard mollifier. We setΓε,δr := Γr[u

εδ,Ωε] for r ∈ (0, rεδ), where r

εδ := maxΩε

uεδ−max∂Ωε(uεδ)

+. Noticethat for small δ > 0, rεδ > 0.

In view of the argument to derive (7.2) in Step 1, we have∫Br

g(p)dp ≤ C∫Γε,δr

(|Duεδ|n/(n−1) + κn/(n−1)

)1−n(−trace(D2uεδ))

ndx

for small r > 0.Also, by the same argument for (ii) in (7.1), we can show that D2uεδ(x) ≤

O in Γε,δr . Furthermore, from the definition of uε, we verify that −ε−1I ≤

D2uεδ(x) in Ωε.Hence, sending δ → 0 with Lemma 7.5 (3), we have∫Br

g(p)dp ≤ C∫Γr[uε,Ωε]

(|Duε|n/(n−1) + κn/(n−1)

)1−n(−trace(D2uε))ndx

≤ C∫Γr[uε,Ωε]

(µn +

(f ε

κ

)n)dx.

Therefore, sending ε→ 0 (again with Lemma 7.5 (3)), we obtain (7.4), whichimplies the conclusion.

Remark. Using the ABP maximum principle in Step 2 (i.e. f ∈ C(Ω)), wecan give a proof of Proposition 6.3, which will be seen in section 7.3. Thus,in Step 3 below, we will use Proposition 6.3.

Step 3: u ∈ C(Ω) and f ∈ Ln(Ω). Let fk ∈ C(Ω) be nonnegative func-tions such that ∥fk − f+∥Ln(Ω) → 0 as k → ∞.

In view of Proposition 6.3, we choose ϕk ∈ C(Ω) ∩W 2,nloc (Ω) such that

P+(D2ϕk) + µ|Dϕk| = fk − f+ a.e. in Ω,ϕk = 0 on ∂Ω,

∥ϕk∥L∞(Ω) ≤ C∥fk − f+∥Ln(Ω).

103

Setting wk := u+ ϕk −∥ϕk∥L∞(Ω), we easily verify that wk is an Ln-viscositysubsolution of

P−(D2wk)− µ|Dwk| ≤ fk in Ω.

Note that Ω+[wk] ⊂ Ω+[u].Thus, by Step 2, we have

maxΩ

wk ≤ max∂Ω

wk + C∥(fk)+∥Ln(Γr[wk,Ω]∩Ω+[u]).

Therefore, sending k → ∞ with Lemma 7.5 (2), we finish the proof. 2

7.3 Proof of existence results for Pucci equations

We shall solve Pucci equations under the Dirichlet condition in Ω. For sim-plicity of statemants, we shall treat the case when Ω is a ball though we willneed the existence result in smooth domains later. To extend the result forgeneral Ω with smooth boundary, we only need to modify the function vz inthe argument below.

For µ ≥ 0 and f ∈ Lp(Ω) with p ≥ n,P−(D2u)− µ|Du| ≥ f in B1,

u = 0 on ∂B1,(7.7)

and P+(D2u) + µ|Du| ≤ f in B1,

u = 0 on ∂B1.(7.8)

Note that the first estimate of (7.10) is valid by Proposition 6.2 when theinhomogeneous term is continuous.

Sketch of proof. We only show the assertion for (7.8).

Step 1: f ∈ C∞(B1). We shall consider the case when f ∈ C∞(B1).Set Sλ,Λ := A := (Aij) ∈ Sn | λI ≤ A ≤ ΛI. We can choose a countable

set S0 := Ak := (Akij) ∈ Sλ,Λ∞k=1 such that S0 = Sλ,Λ.

Noting that µ|q| = max⟨b, q⟩ | b ∈ ∂Bµ for q ∈ Rn, we choose B0 :=bk ∈ ∂Bµ∞k=1 such that B0 = ∂Bµ.

According to Evans’ result in 1983, we can find classical solutions uN ∈C(Ω) ∩ C2(Ω) of max

k=1,...,N

−trace(AkD2u) + ⟨bk, Du⟩

= f in B1,

u = 0 on ∂B1.(7.9)

104

Moreover, we find σ = σ(ε) ∈ (0, 1), Cε > 0 (for each ε ∈ (0, 1)) and C1 > 0,which are independent of N ≥ 1, such that

∥uN∥L∞(B1) ≤ C1∥f∥Ln(B1) and ∥uN∥C2,σ(B1−ε) ≤ Cε. (7.10)

Note that the first estimate of (7.10) is valid by Proposition 6.2 when theinhomogeneous term is continuous.

More precisely, by the classical comparison principle, Proposition 3.3, wehave

uN ≤ u1 in B1. (7.11)

Furthermore, we can construct a subsoluion of (7.9) for any N ≥ 1 inthe following manner: Fix z ∈ ∂B1. Set vz(x) := α(e−β|x−2z|2 − e−β), whereα, β > 0 (independent of z ∈ ∂B1) will be chosen later. We first note thatvz(z) = 0 and vz(x) ≤ 0 for x ∈ B1.

Setting Lkw(x) := −trace(AkD2w(x)) + ⟨bk, Dw(x)⟩, we verify that

Lkvz(x) ≤ 2αβe−β|x−2z|2(Λn− 2βλ|x− 2z|2 + µ|x− 2z|)≤ 2αβe−9β(Λn− 2βλ+ 3µ).

Thus, fixing β := (Λn+3µ+1)/(2λ), we have Lkvz(x) ≤ −2αβe−9β. Hence,taking α > 0 large enough so that 2αβe−9β ≥ ∥f∥L∞(B1), we have

maxk=1,2,...,N

Lkvz(x) ≤ f(x) in B1.

Now, putting V (x) := supz∈∂B1vz(x), in view of Theorem 4.2, we see that

V is a viscosity subsolution of

maxk=1,2,...,N

Lku(x)− f(x) ≤ 0 in B1.

Moreover, it is easy to check that V ∗(x) = 0 for x ∈ ∂B1. Thus, by Proposi-tion 3.3 again, we obtain that

V ≤ uN in B1. (7.12)

Therefore, in view of (7.10)-(7.12), we can choose a sequence Nk andu ∈ C2(B1) such that limk→∞Nk = ∞,

(uNk , DuNk , D2uNk) → (u,Du,D2u) uniformly in B1−ε

105

for each ε ∈ (0, 1), andV ≤ u ≤ u1 in B1. (7.13)

We note that (7.13) implies that u∗ = u∗ on ∂B1.By virtue of the stability result (Proposition 4.8), we see that u is a

viscosity solution of

P+(D2u) + µ|Du| − f = 0 in B1

since supk≥1−trace(AkX) + ⟨bk, p⟩ = P+(X) + µ|p|. Hence, Theorem 3.9yields u ∈ C(B1).

Therefore, by Proposition 2.3, we see that u ∈ C(B1) ∩ C2(B1) is aclassical solution of (7.8).

Step 2: f ∈ Lp(B1). (Lemma 3.1 in [5]) Choose fk ∈ C∞(B1) such that∥fk − f∥Lp(Ω) → 0 as k → ∞.

Let uk ∈ C(B1) ∩ C2(B1) be a classical solution of

P+(D2u) + µ|Du| − fk = 0 in B1

such that uk = 0 on ∂B1.We first claim that uk∞k=1 is a Cauchy sequence in L∞(B1). Indeed,

since (1) and (4) of Proposition 3.2 imply that

P−(D2(uj − uk))− µ|D(uj − uk)|≤ P+(D2uj) + P−(−D2uk) + µ|Duj| − µ|Duk|= fj − fk≤ P+(D2uj)− P+(D2uk) + µ|D(uj − uk)|≤ P+(D2(uj − uk)) + µ|D(uj − uk)|,

using Proposition 6.2 when the inhomogeneous term is continuous, wehave

maxB1

|uj − uk| ≤ C∥fj − fk∥Ln(B1).

Recalling p ≥ n, we thus have

∥uj − uk∥L∞(B1) ≤ C∥fj − fk∥Lp(B1).

Hence, we find u ∈ C(B1) such that uk converges to u uniformly in B1 ask → ∞.

106

Therefore, by the standard covering and limiting arguments with weaklyconvergence in W 2,p locally, it suffices to find C > 0, independent of k ≥ 1,such that

∥uk∥W 2,p(B1/2) ≤ C.

Moreover, we see that −C∥f−∥Lp(B1) ≤ u ≤ C∥f+∥Lp(B1) in B1.For ε ∈ (0, 1/2), we select η := ηε ∈ C2(B1) such that

(i) 0 ≤ η ≤ 1 in B1,(ii) η = 0 in B1 \B1−ε,(iii) η = 1 in B1−2ε,(iv) |Dη| ≤ C0ε

−1, |D2η| ≤ C0ε−2 in B1,

where C0 > 0 is independent of ε ∈ (0, 1/2).Now, we recall Caffarelli’s result (1989) (see also [4]): There is a universal

constant C > 0 such that

∥D2(ηuk)∥Lp(B1−ε) ≤ C∥P+(D2(ηuk))∥Lp(B1−ε).

Hence, we find C1 > 0 such that for 0 < ε < 1/4,

∥D2uk∥Lp(B1−2ε) ≤ ∥D2(ηuk)∥Lp(B1−ε) ≤ C∥P+(D2(ηuk))∥Lp(B3/4)

≤ C1

(∥fk∥Lp(B1−ε) + ε−1∥Duk∥Lp(B1−ε) + ε−2∥uk∥Lp(B1−ε)

)Multiplying ε2 > 0 in the above, we get

ε2∥D2uk∥Lp(B1−2ε) ≤ C1(∥fk∥Lp(B1) + ϕ1(uk) + ϕ0(uk)),

where ϕj(uk) := sup0<ε<1/2 εj∥Djuk∥Lp(B1−ε) for j = 0, 1, 2.

Therefore, in view of the “interpolation” inequality (see [13] for example),i.e. for any δ > 0, there is Cδ > 0 such that

ϕ1(uk) ≤ δϕ2(uk) + Cδϕ0(uk),

we find C3 > 0 such that

ϕ2(uk) ≤ C3

(∥fk∥Lp(B1) + ϕ0(uk)

).

On the other hand, since we have L∞-estimates for uk, we conclude theproof. 2

Remark. It is possible to show that the uniform limit u in Step 2 isan Lp-viscosity solution of (7.8) by Proposition 6.13. Moreover, since it isknown that if Lp-viscosity supersolution of (7.8) belongs to W 2,p

loc (B1), thenit is an Lp-strong supersolution (see [5]), u satisfies P+(D2u)+µ|Du| = f(x)a.e. in B1.

107

7.4 Proof of the weak Harnack inequality

We need a modification of Lemma 4.1 in [4] since our PDE (7.14) below hasthe first derivative term.

Lemma 7.6. (cf. Lemma 4.1 in [4]) There are ϕ ∈ C2(B2√n) and

ξ ∈ C(B2√n) such that

(1) P−(D2ϕ)− µ|Dϕ| ≥ −ξ in B2

√n,

(2) ϕ(x) ≤ −2 for x ∈ Q3,(3) ϕ(x) = 0 for x ∈ ∂B2

√n,

(4) ξ(x) = 0 for x ∈ B2√n \B1/2.

Proof. Set ϕ0(r) := A1− (2√n/r)α for A,α > 0 so that ϕ0(2

√n) = 0.

r

1/2 2sqrtnQ3

-2y=phi

y=phi0

0

-1/2-2sqrtn

Fig 7.1

PSfrag replacements

y = ϕ(x)y = ϕ0(x)

2√n

−2√n

−2

0Q3

1/2

r−1/2

Since Dϕ0(|x|) = A(2

√n)αα|x|−α−2x,

D2ϕ0(|x|) = A(2√n)αα|x|−α−4|x|2I − (α + 2)x⊗ x,

we caluculate in the following way: At x = 0, we have

P−(D2ϕ0(|x|))− µ|Dϕ0(|x|)| ≥ A(2√n)αα|x|−α−2(α + 2)λ− nΛ− µ|x|.

108

Setting α := λ−1(nΛ + 2µ√n) − 2 so that α > 0 for n ≥ 2, we see that

the right hand side of the above is nonnegative for x ∈ B2√n \ 0. Thus,

taking ϕ ∈ C2(B2√n) such that ϕ(x) = ϕ0(|x|) for x ∈ B2

√n \ B1/2 and

ϕ(x) ≤ ϕ0(3√n/2) for x ∈ B3

√n/2, we can choose a continuous function ξ

satisfying (1) and (4). See Fig 7.1.Moreover, taking A := 2/(4/3)α − 1 so that ϕ0(3

√n/2) = −2, we see

that (2) holds. 2

We now present an important “cube decomposition lemma”.We shall explain a terminology for the lemma: For a cube Q := Qr(x) with

r > 0 and x ∈ Rn, we call Q a dyadic cube of Q if it is one of cubes Qk2n

k=1

so that Qk := Qr/2(xk) for some xk ∈ Q, and ∪2n

k=1Qk ⊂ Q ⊂ ∪2n

k=1Qk.

Lemma 7.7. (Lemma 4.2 in [4]) Let A ⊂ B ⊂ Q1 be measurable setsand 0 < δ < 1 such that

(a) |A| ≤ δ,

(b) Assume that if a dyadic cube Q of Q ⊂ Q1 satisfies |A ∩Q| > δ|Q|,then Q ⊂ B.

Then, |A| ≤ δ|B|.

Proof of Proposition 6.4. Assuming that u ∈ C(B2√n) is a nonnegative

viscosity supersolution of

P+(D2u) + µ|Du| ≥ 0 in B2√n, (7.14)

we shall show that for some constants p0 > 0 and C1 > 0,

∥u∥Lp0 (Q1) ≤ C1 infQ1/2

u.

To this end, it is sufficient to show that if u ∈ C(B2√n) satisfies that

infQ1/2u ≤ 1, then we have ∥u∥Lp0 (Q1) ≤ C1 for some constants p0, C1 > 0.

Indeed, by taking v(x) := u(x)(infQ1/2

u+ δ)−1

for any δ > 0 in place of u,

we have ∥v∥Lp0 (Q1) ≤ C1, which implies the assertion by sending δ → 0.

Lemma 7.8. There are θ > 0 and M > 1 such that if u ∈ C(B2√n) is a

nonnegative Lp-viscosity supersolution of (7.14) such that

infQ3

u ≤ 1, (7.15)

109

then we have|x ∈ Q1 | u(x) ≤M| ≥ θ.

Remark. In our setting of proof of Proposition 7.4, assumption (7.15) isautomatically satisfied.

Proof of Lemma 7.8. Choose ϕ ∈ C2(B2√n) and ξ ∈ C(B2

√n) from Lemma

7.6. Using (4) of Proposition 3.2, we easily see that w := u + ϕ is an Ln-viscosity supersolution of

P+(D2w) + µ|Dw| ≥ −ξ in B2√n.

Since infQ3 w ≤ −1 and w ≥ 0 on ∂B2√n by (2) and (3) in Lemma 7.6,

respectively, by Proposition 6.2, we find C > 0 such that

1 ≤ supQ3

(−w) ≤ supB2

√n

(−w) ≤ C∥ξ∥Ln(Γ[−w,B2√

n]∩B+2√n[−w]). (7.16)

In view of (4) of Lemma 7.6, (7.16) implies that

1 ≤ CmaxB1/2

|ξ||x ∈ Q1 | (u+ ϕ)(x) < 0|.

Sinceu(x) ≤ −ϕ(x) ≤ max

B2√

n

(−ϕ) =:M for x ∈ B2√n.

Therefore, setting θ = (C supQ1|ξ|)−1 > 0 and M = supB2

√n(−ϕ) ≥ 2, we

haveθ ≤ |x ∈ Q1 | u(x) ≤M|. 2

We next show the following:

Lemma 7.9. Under the same assumptions as in Lemma 7.8, we have

|x ∈ Q1 | u(x) > Mk| ≤ (1− θ)k for all k = 1, 2, . . .

Proof. Lemma 7.8 yields the assertion for k = 1.Suppose that it holds for k − 1. Setting A := x ∈ Q1 | u(x) > Mk and

B := x ∈ Q1 | u(x) > Mk−1, we shall show |A| ≤ (1− θ)|B|.

110

Since A ⊂ B ⊂ Q1 and |A| ≤ |x ∈ Q1 | u(x) > M| ≤ δ := 1 − θ, inview of Lemma 7.8, it is enough to check that property (b) in Lemma 7.7holds.

To this end, let Q := Q1/2j(z) be a dyadic cube of Q := Q1/2j−1(z) (forsome z, z ∈ Q1 and j ≥ 1) such that

|A ∩Q| > δ|Q| = 1− θ

2jn. (7.17)

It remains to show Q ⊂ B.Assuming that there is x ∈ Q such that x /∈ B; i.e. u(x) ≤Mk−1.Set v(x) := u(z + 2−jx)/Mk−1 for x ∈ B2

√n. Since |xi − zi| ≤ 3/2j+1, we

see that infQ3 v ≤ u(x)/Mk−1 ≤ 1. Furthermore, since z ∈ Q1, z + 2−jx ∈B2

√n for x ∈ B2

√n.

hatz tildeQ

Q

tildex

2j

2j-1 z+2jQ3

Fig 7.2

zPSfrag replacements

xQQ

z12j1

2j−1

z + 12jQ3

Thus, since v is an Lp-viscosity supersolution of

P+(D2v) + µ|Dv| ≥ 0,

Lemma 7.8 yields |x ∈ Q1 | v(x) ≤M| ≥ θ. Therefore, we have

|x ∈ Q | u(x) ≤Mk| ≥ θ

2jn= θ|Q|.

Thus, we have |Q \ A| ≥ θ|Q|. Hence, in view of (7.17), we have

|Q| = |A ∩Q|+ |Q \ A| > δ|Q|+ θ|Q| = |Q|,

111

which is a contradiction. 2

Back to the proof of Proposition 6.4. A direct consequence of Lemma 7.9

is that there are C, ε > 0 such that

|x ∈ Q1 | u(x) ≥ t| ≤ Ct−ε for t > 0. (7.18)

Indeed, for t > M , we choose an integer k ≥ 1 so that Mk+1 ≥ t > Mk.Thus, we have

|x ∈ Q1 | u(x) ≥ t| ≤ |x ∈ Q1 | u(x) > Mk| ≤ (1− θ)k ≤ C0t−ε,

where C0 := (1− θ)−1 and ε := − log(1− θ)/ logM > 0.Since 1 ≤ M εt−ε for 0 < t ≤ M , taking C := maxC0,M

ε, we obtain(7.18).

Now, recalling Fubini’s theorem,∫Q1

up0(x)dx ≤∫x∈Q1 | u(x)≥1

up0(x)dx+ 1

= p0

∫ ∞

1tp0−1|x ∈ Q1 | u(x) ≥ t|dt+ 1,

(see Lemma 9.7 in [13] for instance), in view of (7.18), for any p0 ∈ (0, ε), wecan find C(p0) > 0 such that ∥u∥Lp0 (Q1) ≤ C(p0). 2

7.5 Proof of the local maximum principle

Although our proof is a bit technical, we give a modification of Trudinger’sproof in [13] (Theorem 9.20), in which he observed a precise estimate for“strong” subsolutions on the upper contact set. Recently, Fok in [11] (1996)gave a similar proof to ours.

We note that we can find a different proof of the local maximum principlein [4] (Theorem 4.8 (2)).

Proof of Proposition 6.5. We give a proof only when q ∈ (0, 1] because itis immediate to show the assertion for q > 1 by Holder’s inequality.

Let x0 ∈ Q1 be such that maxQ1u = u(x0). It is sufficient to show that

maxB1/4(x0)

u ≤ C2∥u+∥Lq(B1/2(x0))

112

since B1/2(x0) ⊂ Q2. Thus, by considering u((x − x0)/2) instead of u(x), itis enough to find C2 > 0 such that

maxB1/2

u ≤ C2∥u+∥Lq(B1).

We may suppose thatmaxB1

u > 0 (7.19)

since otherwise, the conclusion is trivial.Furthermore, by the continuity of u, we can choose τ ∈ (0, 1/4) such that

1− 2τ ≥ 1/2 andmaxB1−2τ

u > 0.

We shall consider the sup-convolution of u again: For ε ∈ (0, τ),

uε(x) := supy∈B1

u(y)− |x− y|2

2ε

.

By the uniform convergence of uε to u, (7.19) yields

maxB1−τ

uε > 0 for small ε > 0. (7.20)

For small ε > 0, we can choose δ := δ(ε) ∈ (0, τ) such that limε→0 δ = 0,and

P−(D2uε)− µ|Duε| ≤ 0 a.e. in B1−δ.

Putting ηε(x) := (1− δ)2 − |x|2β for β := 2n/q ≥ 2, we define vε(x) :=ηε(x)uε(x). We note that

rε := maxB1−δ

vε > 0.

Fix r ∈ (0, rε) and set Γεr := Γr[v

ε, B1−δ]. By (1) in Lemma 7.5, we seeΓεr ⊂ B+

1−δ[vε].

For later convenience, we observe that

Dvε(x) = −2βxη(x)(β−1)/βuε(x) + η(x)Duε(x), (7.21)

D2vε(x) = −2βη(x)(β−1)/βuε(x)I + x⊗Duε(x) +Duε(x)⊗ x+4β(β − 1)η(x)(β−2)/βuε(x)x⊗ x+ η(x)D2uε(x).

(7.22)

113

Since uε is twice differentiable almost everywhere, we can choose a mea-surable set Nε ⊂ B1−δ such that |Nε| = 0 and uε is twice differentiable atx ∈ B1−δ \Nε. Of course, vε is also twice differentiable at x ∈ B1−δ \Nε.

By (7.22), we have

P−(D2vε) ≤ ηP−(D2uε) + 2βη(β−1)/βΛnuε − P−(x⊗Duε +Duε ⊗ x)

in B+1−δ[v

ε]. By using (7.21), the last term in the above can be estimatedfrom above by

Cη−2/β(vε)+ + η−1/β|Dvε|.

Moreover, using (7.21) again, we have

P−(D2uε) ≤ µ|Duε| ≤ µη−1|Dvε|+ Cη−1/β(uε)+.

Hence, we find C > 0 such that

P−(D2vε) ≤ Cη−1/β|Dvε|+ Cη−2/β(vε)+ =: gε in B1−δ \Nε. (7.23)

We next claim that there is C > 0 such that

|Dvε(x)| ≤ Cη−1/β(x)vε(x) for x ∈ Γεr \Nε. (7.24)

First, we note that at x ∈ Γεr \ Nε, v

ε(y) ≤ vε(x) + ⟨Dvε(x), y − x⟩ fory ∈ B1−δ.

To show this claim, since we may suppose |Dvε(x)| > 0 to get the esti-mate, setting y := x − tDvε(x)|Dvε(x)|−1 ∈ ∂B1−δ for t ∈ [1 − δ − |x|, 1 −δ + |x|], we see that

0 = vε(y) ≤ vε(x)− t|Dvε(x)|,

which implies|Dvε(x)| ≤ Cvε(x)η−1/β(x) in Γε

r \Nε. (7.25)

Here, we use Lemma 2.8 in [5], which will be proved in the end of thissubsection for the reader’s convenience:

Lemma 7.10. Let w ∈ C(Ω) be twice differentiable a.e. in Ω, and satisfy

P−(D2w) ≤ g a.e. in Ω,

114

where g ∈ Lp(Ω) with p ≥ n. If −C1I ≤ D2w(x) ≤ O a.e. in Ω for someC1 > 0, then w is an Lp-viscosity subsolution of

P−(D2w) ≤ g in Ω. (7.26)

Since uε is Lipschitz continuous in B1−δ, by (7.22), we see that vε is anLn-viscosity subsolution of

P−(D2vε) ≤ gε in B1−δ.

Noting (7.25), in view of Proposition 6.2, we have

maxB1−δ

vε ≤ C∥η−2/β(vε)+∥Ln(Γεr)

≤ C

(maxB1−δ

(vε)+)β−2

β

∥((uε)+)2/β∥Ln(B1−δ),

which together with our choice of β yields

maxB1−δ

vε ≤ C∥(uε)+∥Lq(B1−δ).

Therefore, by (7.20), we have

maxB1/2

uε ≤ CmaxB1−δ

vε ≤ C∥(uε)+∥Lq(B1−δ),

Therefore, sending ε→ 0 in the above, we finish the proof. 2

Proof of Lemma 7.10. In order to show that w ∈ C(Ω) is an Lp-viscositysubsolution of (7.26), we suppose the contrary; there are ε, r > 0, x ∈ Ω andϕ ∈ W 2,p

loc (Ω) such that 0 = (w − ϕ)(x) = maxΩ(w − ϕ), B2r(x) ⊂ Ω, and

P−(D2ϕ)− g ≥ 2ε a.e. in Br(x).

We may suppose that x = 0 ∈ Ω. Setting ψ(x) := ϕ(x) + τ |x|4 for smallτ > 0, we observe that

h := P−(D2ψ)− g ≥ ε a.e. in Br.

Notice that 0 = (w − ψ)(0) > (w − ψ)(x) for x ∈ Br \ 0.

115

Moreover, we observe

P−(D2(w − ψ)) ≤ −ε a.e. in Br. (7.27)

Consider wδ := w ∗ ρδ, where ρδ is the standard mollifier for δ > 0. Fromour assumption, we see that, as δ → 0,

(1) wδ → w uniformly in Br,(2) D2wδ → D2w a.e. in Br.

By Lusin’s Theorem, for any α > 0, we find Eα ⊂ Br such that |Br\Eα| < α,∫Br\Eα

(1 + |P−(−D2ψ)|)pdx < α,

andD2wδ → D2w uniformly in Eα (as δ → 0).

Setting hδ := P−(D2(wδ − ψ)), we find C > 0 such that

hδ ≤ C + P−(−D2ψ)

because of our hypothesis. Hence, we have

∥(hδ)+∥pLp(Br)≤ C

∫Br\Eα

(1 + |P−(−D2ψ)|)pdx+∫Eα

|(hδ)+|pdx.

Sending δ → 0 in the above, by (7.27), we have

lim supδ→0

∥(hδ)+∥Lp(Br) ≤ C∥(1 + |P−(−D2ψ)|)∥Lp(Br\Eα) ≤ Cα. (7.28)

On the other hand, in view of Proposition 6.2, we see that

maxBr

(wδ − ψ) ≤ max∂Br

(wδ − ψ) + C∥(hδ)+∥Lp(Br).

Hence, by sending δ → 0, this inequality together with (7.28) implies that

0 = maxBr

(w − ψ) ≤ max∂Br

(w − ψ) + Cα for any α > 0.

This is a contradiction since max∂Br(w − ψ) < 0. 2

116

References

[1] M. Bardi & I. Capuzzo Dolcetta, Optimal Control and ViscositySolutions of Hamilton-Jacobi-Bellman Equations, Systems & Control,Birkauser, 1997.

[2] G. Barles, Solutions de Viscosite des Equations de Hamilton-Jacobi,Mathematiques et Applications 17, Springer, 1994.

[3] M. Bardi, M. G. Crandall, L. C. Evans, H. M. Soner & P. E.Souganidis, Lecture Notes in Mathematics, 1660, Springer, 1997.

[4] L. A. Caffarelli & X. Cabre, Fully Nonlinear Elliptic Equations,Colloquium Publications 43, AMS, 1995.

[5] L. A. Caffarelli, M. G. Crandall, M. Kocan & A. Swiech,On viscosity solutions of fully nonlinear equations with measurable in-gredients, Comm. Pure Appl. Math. 49 (1996), 365-397.

[6] M. G. Crandall, H. Ishii & P.-L. Lions, User’s guide to viscos-ity solutions of second order partial differential equations, Bull. Amer.Math. Soc., 27 (1992), 1-67.

[7] M. G. Crandall, M. Kocan, P.-L. Lions & A. Swiech, Existenceresults for boundary problems for uniformly elliptic and parabolic fullynonlinear equations, Electron. J. Differential Equations, 1999 (1999),1-22.

[8] L. C. Evans, Partial Differential Equations, GSM 19, Amer. Math.Soc., 1998.

[9] L. C. Evans & R. F. Gariepy, Measure Theory and Fine Propertiesof Functions, Studies in Advanced Mathematics, CRC Press, 1992.

[10] W. H. Fleming & H. M. Soner, Controlled Markov Processes andViscosity Solutions, Applications of Mathematics 25, Springer, 1992.

[11] P.-K. Fok, Some maximum principles and continuity estimates for fullynonlinear elliptic equations of second order, Dissertation, University ofCalifornia at Santa Barbara, 1996.

117

[12] Y. Giga, Surface Evolution Equations – a level set method – Preprint,2002.

[13] D. Gilbarg & N. S. Trudinger, Elliptic Partial Differential Equa-tions of Second Order, 2nd edition, GMW 224, Springer, 1983.

[14] Q. Han & F. H. Lin, Elliptic Partial Differential Equations, CourantInstitute Lecture Note 1, Amer. Math. Soc., 1997.

[15] H. Ishii, Viscosity Solutions, (in Japanese).

[16] R. Jensen, Uniqueness of Lipschitz extensions: Minimizing the supnorm of the gradient, Arch. Rational Mech. Anal., 123 (1993), 51-74.

118

Notation Index

A, 46B, 52Ck,σ, 79Cσ, 78C∞

0 , 80∆, 52ess. inf, 80ess. sup, 80G∗, G

∗, 62Γ, 52Γ[u,O], 83J2,±, 17, 22

J2,±, 18, 22

, 2Lp, 79LSC, 15M, 27n, 66o, 17P±, 25supp, 5USC, 15u∗, u

∗, 40W 2,p, 79

W 2,ploc , 81

119