ma3g8 functional analysis ii - warwick

MA3G8 Functional Analysis II

June 10, 2013

Contents

0 Introduction 1

1 Basics 11.1 Normed Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Linear Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Duality and the separation theorems 42.1 1-Codimensional Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Separation Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Weak and Weak-? Topologies . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 The Category Theorems 143.1 Open Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2 Closed Graph Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.3 Application to Fourier Analysis . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Unbounded Operators 184.1 Closed Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.2 The Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5 The Laplacian and Quantum Harmonic Oscillator. 255.1 Quantum Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . 26

These notes are based on the 2012 MA3F6 Functional Analysis II course, taughtby Keith Ball, typeset by Matthew Egginton.No guarantee is given that they are accurate or applicable, but hopefully they will assistyour study.Please report any errors, factual or typographical, to [email protected]

i

MA3G8 Functional Analysis II Lecture Notes Spring 2013

0 Introduction

At the beginning of the 20th century.

1 Basics

1.1 Normed Spaces

Definition 1.1 A norm || · || on a vector space V is a map x 7→ ||x|| ∈ R satisfying

1. ||x|| ≥ 0 for all x and ||x|| = 0 ⇐⇒ x = 0.

2. ||λx|| = |λ|||x|| for all x ∈ V and λ ∈ R

3. ||x+ y|| ≤ ||x||+ ||y|| for x, y ∈ V .

A norm gives us a way to measure distance: d(x, y) = ||x− y||. The unit ball B plays animportant role (geometry of space is invariant under translation) and it is a convex set.In other words, if x, y ∈ B and λ ∈ [0, 1] then (1− λ)x+ λy ∈ B

Lemma 1.2 If N : V → R satisfies

1. N(x) ≥ 0 for all x and N(x) = 0 ⇐⇒ x = 0

2. N(λx) = |λ|N(x) for λ ∈ R and x ∈ V

3. x ∈ V : N(x) ≤ 1 is convex

then N is a norm.

The proof essentially uses a simple trick.Proof We need to check the triangle inequality. Suppose x, y ∈ V . If N(x) = 0 thenx = 0 and then x + y = y and so N(x + y) = N(y) = N(x) + N(y). Thus supposea = N(x), b = N(y) and are strictly positive. Then N(a−1x) = a−1N(x) = 1 and soa−1x ∈ x : N(x) ≤ 1 and the same for b−1y ∈ x : N(x) ≤ 1 and thus

a

a+ ba−1x+

b

a+ bb−1y =

x+ y

a+ b∈ x : N(x) ≤ 1

by convexity. Thus

N(x+ y) = (a+ b)N(x+ y

a+ b) ≤ (a+ b)1 = a+ b = N(x) +N(y)

Q.E.D.

We shall consider a number of “standard” spaces.

1. lp the space of sequences (xi) with finite norm ||(xi|| = (∑∞

i=1 |xi|p)1/p for 1 ≤ p <∞and l∞ is the space of bounded sequences with the norm ||(xi)|| = supi |xi|.

2. Consider the finite dimensional versions lnp , the space of sequences of length n with

norm ||(xi)|| = (∑n

i=1 |xi|p)1/p.

3. Also c0 the space of sequences converging to 0 with norm ||(xi)|| = max |xi| =sup |xi|. Note c0 ⊂ l∞ in the obvious way.

1 of 29


4. C[0, 1] the space of continuous functions on [0, 1] with norm ||f || = sup |f(x)| =max |f(x)|

5. Lp[0, 1] the space of measurable “functions” with ||f || =(∫ 1

0 |f(x)|pdx)1/p

< ∞.

We actually mean equivalence classes of functions, i.e. f ∼ g if f = g a.e.

6. L∞[0, 1] is the space of measurable functions such that ||f || = sup |f(x)| <∞.

The triangle inequality in Lp is often called the Minkowski inequality(∫|f + g|p

)1/p

≤(∫|f |p

)1/p

+

(∫|g|p)1/p

We can see that the unit ball is convex as follows. If∫|f |p,

∫|g|p ≤ 1 and λ ∈ [0, 1] then

for each x,|(1− λ)f(x) + λg(x)|p ≤ (1− λ)|f(x)|p + λ|g(x)|p

since the map t 7→ |t|p is convex on R and so∫|(1− λ)f(x) + λg(x)|p ≤ (1− λ)

∫|f(x)p + λ

∫|g(x)|p ≤ 1

Lemma 1.3 (Holder Inequality) If f, g are measurable, and 1p + 1

q = 1 and∫|f |p <∞,

∫|g|q <∞

then ∫fg ≤

(∫|f |p

)1/p(∫|g|q)1/q

The Cauchy-Schwartz inequality is Holder with p = 2 = q.

1.2 Separability

Definition 1.4 A subset U of a metric space X is said to be dense in X if for everyx ∈ X and ε > 0 there is a u ∈ U with d(x, u) < ε. A normed space is called separableif it has a countable dense subset.

Lemma 1.5 For a normed space X, the following are equivalent

1. X is separable

2. SX = x ∈ X : ||x|| = 1 is separable

3. X contains a sequence (x1, x2, ...) whose linear span is dense.

We prove the following before the proof of the above

Lemma 1.6 A subset of a separable metric space is separable

2 of 29


Proof Let (x1, x2, ...) be a dense subset in X and Y ⊂ X. We need to find a densesubset of Y .

For each n and k, if Y contains a point whose distance from xn is at most 1k call this

un,k ∈ Y . The set un,k is countable. I claim that it is dense in Y . If y ∈ Y and ε > 0choose xn with d(xn, y) < 1

k where 1k <

ε2 . There is a point un,k since y would have been

a candidate. Then

d(y, un,k) ≤ d(y, xn) + d(xn, un,k) <1

k+

1

k< ε

Q.E.D.

Proof [of lemma 1.5] 1) =⇒ 2) is clear from the previous lemma since SX ⊂ X.2) =⇒ 3) Choose (ui) dense in SX . This sequence suffices for 3). Even the multiples

(λui) are dense. If x ∈ X and x 6= 0 then x||x|| ∈ SX so we can find a ui with || x||x|| −ui|| <

ε||x|| and hence

||x− ||x||ui|| < ε

3) =⇒ 1). Let (x1, x2, ...) be a sequence whose span is dense. The span is muchbigger than countable and so 3) is weaker than 1). We want to build a countable denseset in X. We take the rational linear combination

∑n1 rixi for ri ∈ Q. Given x ∈ X, ε > 0

we can find some n ∈ N and θ1, ..., θn ∈ R with ||x−∑θixi|| < ε

2 . Now choose rationalsri with |θi − ri| < ε

2nM where M = 1 + max||x1||, ..., ||xn|| and then∣∣∣∣∣∣∑ θixi −∑

rixi

∣∣∣∣∣∣ =∣∣∣∣∣∣∑(θi − ri)xi

∣∣∣∣∣∣≤∑||(θi − ri)xi||

=∑|θi − ri|||xi||

≤∑ ε

2nMM

≤ ε

2

and combining above inequalities gives the desired result. Q.E.D.

The homework includes proving lp is separable, and this uses the above.

1.3 Linear Maps

A linear map T : X → Y between normed spaces may or may not be continuous. It iscontinuous if and only if it is bounded, i.e. there exists K > 0 with

||Tx|| ≤ K||x||

For such a map we define its norm as infK : ||Tx|| ≤ K||x||∀x = ||T || = sup||Tx|| :||x|| ≤ 1. This is in some sense the maximum scale factor.

The space of bounded linear maps form a normed vector space under this norm. Thisspace B(X,Y ) is complete if and only if Y is complete. A complete normed space is calleda Banach space. In finite dimensions, T : U → V then dim(kerT ) + dim(ImT ) = dimU .

Two Banach spaces X and Y are called isomorphic if there is an invertible linear mapT : X → Y with ||T || and ||T−1|| bounded. They are isometric if ||T || = 1 = ||T−1||.

There is a natural (algebraic) isomorphism between U/ kerT and ImT , by “ cuttingthe electric cable”.

3 of 29


We can define the quotient in a Banach space. If Y is a closed subspace of X wedefine X/Y to be the space of translates x+Y : x ∈ X with x+Y = z+Y if x−z ∈ Y .As in the finite dimensional case, addition and scalar multiplication make sense. Wedefine the norm

||x+ Y || = inf||x+ y|| : y ∈ Y

and in some sense this is the closest point of the translate to 0, but the closest point maynot exist.

1. The norm is well defined; i.e. ||x+ Y || = ||z + Y || if x− z ∈ Y . This is because wehave seen a form of the norm without reference to a representative.

2. ||λ(x+ Y )|| = |λ|||x+ Y || is easy to show

3. We show the triangle inequality explicitly. Given ε > 0 choose yx ∈ Y and yz ∈ Ywith ||x+ yx|| < ||x+ Y ||+ ε and ||x+ yz|| < ||z + Y ||+ ε and then

||x+ z + Y || ≤ ||x+ z + yx + yz|| ≤ ||x+ yx||+ ||z + yz|| ≤ ||x+ Y ||+ ||z + Y ||+ 2ε

and taking the infimum over ε gives us our result.

4. Show ||x+ Y || = 0 if and only if x+ Y = 0, i.e. x ∈ Y .

If x ∈ Y then choose y = −x and so x + y = 0 and so ||x + Y || = 0. Converselysuppose ||x + Y || = 0. Then we can find a sequence (yk) in Y with ||x + yk|| → 0,i.e. x+ yk → 0 or yk → −x, and since Y is closed, −x ∈ Y and so x ∈ Y .

5. X/Y is complete. We take a Cauchy sequence and then construct a subsequenceand use the completeness of X. We might not find a convergent sequence ofrepresentatives. It suffices to find a convergent subsequence. Choose nk so that||xnk + Y − xm + Y || ≤ 1

2kif m ≥ nk. Hence ||xnk + Y − xnk+1

+ Y || ≤ 12k

. Chooseynk one by one so that

||xnk + ynk − (xnk+1+ ynk+1

)|| ≤ 2

2k

and so xnk + ynk is Cauchy and has a limit u. Then xnk + Y → u+ Y as required.We have used:

Lemma 1.7 If (xk) is a sequence in a complete metric space with d(xk, xk+1) ≤ εk and∑∞1 εk <∞ then (xk) is Cauchy.

2 Duality and the separation theorems

If X is a normed space then a linear functional on X is a linear map X → R (or theunderlying field). The space of bounded linear functionals is denoted X?; it is a Banachspace with the operator norm ||φ|| = sup|φ(x)| : ||x|| ≤ 1. If X is a Hilbert space thenevery bounded linear functional is of the form

x 7→ 〈x, y〉

for some y ∈ X and these maps are all in X?. Thus we can identify X? with X. Whatabout lp and Lp?

4 of 29


Theorem 2.1 (Duality of lp) If X =

c0

lp 1 < p <∞l1

, then X? is isometric to

l1

lq

l∞

with 1p + 1

q = 1.

Remark If p = 2 then q = 2 and l?p = lp.Proof For each y of the supposed dual (in each case) we can define φy : X → R by

φy(x) =

∞∑1

yixi

where x = (x1, x2, ...) and this makes sense by Holder:

∞∑1

|yixi| ≤

||x||∞||y||1||x||p||y||q||x||1||y||∞

and so the series converges absolutely.φy is clearly linear and the map y 7→ φy (lq → X?) is linear

φy+z(x) =∑

(yi + zi)xi =∑

yixi +∑

zixi = φy(x) + φz(x)

We also have |φy(x)| ≤ ||x||p||y||q and so on so the map has ||φy|| ≤ ||y||. Note thatthe map y 7→ φy is obviously 1-1 but we shall get this automatically when we prove||y|| ≤ ||φy||

It remains to show that each φ ∈ X? is of the form φy for some y (in the supposeddual) with ||y|| ≤ ||φy||.

Let (ei) be the standard unit vector basis of X. For each i set yi = φ(ei). Then φ andφy agree on the (ei). The problem is the norm. We show that ||y|| ≤ ||φ|| as then φ is ofthe form φy for that y. If 1 < p <∞ then

m∑1

|yi|q =m∑1

yi|yi|q−1sgn(yi)

=

m∑1

φ(ei)|yi|q−1sgn(yi)

= φ(m∑1

|yi|q−1sgn(yi)ei)

≤ ||φ||||m∑1

|yi|q−1sgn(yi)ei||p

= ||φ||

(m∑1

|yi|(q−1)p) 1

p

= ||φ||

(m∑1

|yi|q) 1

p

We have shown thatm∑1

|yi|q ≤ ||φ||

(m∑1

|yi|q) 1

p

5 of 29


and since 1p < 1 the estimate has some content. If y = 0 then φ = 0 and there was nothing

to prove. Thus we may assume that∑|yi| > 0 and then(

m∑1

|yi|q) 1

q

≤ ||φ||

and so taking limits as m→∞ we have(m∑1

|yi|q) 1

q

= ||y|| ≤ ||φ||

we know that φ and φy agree on (ei) and are continuous linear functionals. In lp for1 < p <∞ or c0 or l1 the closed span of the (ei) is X. Thus the functionals agree on X.

If X = c0 then

m∑1

|yi| =m∑1

yisgn(yi)

=

m∑1

φ(ei)sgn(yi)

= φ

(m∑1

sgn(yi)ei

)

≤ ||φ||||m∑1

sgn(yi)ei||

= ||φ||

and if X = l1 then|yi| = |φ(ei)| ≤ ||φ||||ei|| = ||φ||

Q.E.D.

1. In each case we show that if y is in the suggested dual then it gives a φy ∈ X?, byusing Holder. In particular each y ∈ l∞ acts on l1 and each y ∈ l1 acts on (l∞ andhence) c0.

2. For each case, given φ ∈ X? there is a y in the supposed dual with φ(ei) = φy(ei)for each i. Since the closed span of the ei is the space required this gives φ = φy onX.

However, since l∞ is not separable, we cannot conclude that l?∞ = l1. Moreover l?∞is not isomorphic to l1. Note that the above proof uses separability in the constructionofthe basis.

The word duality suggests some kind of symmetry between X and X?. In finitedimensions, indeed (X?)? ∼= X in a canonical way: if x ∈ X and φ ∈ X? then the actionof x on φ is the action of φ on x.

In Banach spaces, X still acts on X?, so X ⊂ X?? as for example c0 acts on l1 andc0 ⊂ l∞ but X may not fill up X ⊂ X??, eg c0 6= l∞ (they are massively different).

Things in Lp are somewhat similar. There is no analogue of c0 and L?∞ is every bit asnasty as l?∞. C[0, 1] is a little bit like c0. We have L?1

∼= L∞ and L?p∼= Lq if 1

p + 1q = 1 for

1 < p <∞.

6 of 29


The proof of L?p∼= Lq is very similar to the proof that l?p

∼= lq. If g ∈ Lq we can

define φg on Lp by φg(f) =∫ 10 fg and Holder still works. On the other hand, if we have

a functional φ on Lp we have no “basis” vectors to apply φ to.

We can apply φ instead to 1A(x) =

1 x ∈ A0 x 6∈ A

so we get numbers φ(1A) and the map

A 7→ φ(1A) is a signed measure. So by the Radon-Nikodym theorem there is a function gwith φ(1A) =

∫g1A for all A. So φ is given by f 7→

∫fg for f = 1A and by the standard

machine φ = φg on Lp.We motivate the next steps we take. Suppose X is a Banach space, how might we

define a functional on X? Choose a basis eαα∈A and then φ(∑λαeα) = λ1 is a linear

functional. This though will not in general be continuous, or how do we show so? Weneed, ironically, the axiom of choice to prove that a continuous functional exists.

First though we consider C[0, 1] and its dual space. If µ is a finite Borel measureon [0, 1] then the map f 7→

∫ 10 fdµ is a bounded linear functional on C[0, 1]. It turns

out that C[0, 1]? ∼= finite Borel signed measures but this is a subtle theorem, the Rieszrepresentation theorem. Newton showed that the map f 7→

∫ 10 fdx is a bounded linear

functional. Thus Riesz says that the ordinary integral is given by a measure, so Rieszimplies the existence of the Lebesgue measure. Every linear functional on C[0, 1] is givenby a signed measure.

2.1 1-Codimensional Subspaces

Definition 2.2 A subspace Y of a Banach space X is called 1-codimensional if

1. Y 6= X

2. If Z is a subspace such that Y ⊂ Z ⊂ X then Z = Y or Z = X

Theorem 2.3 If Y is a subspace of a Banach space X then the following are equivalent.

1. Y is 1-codimensional

2. Y 6= X but if x ∈ X \ Y then spanY ∪ x = X

3. Y = kerφ for some non-zero linear functional φ on X. (note φ is not necessarilybounded)

Proof 1 ⇐⇒ 2 is easy. If Y is 1-codimensional and x ∈ X \Y then Y ⊂ spanY ∪xbut Y 6= spanY ∪ x and so spanY ∪ x = X. On the other hand if Y satisfies 2and Y ⊂ Z ⊂ X and Z 6= Y then choose z ∈ Z \ Y . Then Z ⊃ spanY ∪ z = X.

1, 2 =⇒ 3. Assume Y is 1-codimensional and choose x ∈ X \ Y . Each elementz ∈ X can be written as y + λx for some y ∈ Y and λ ∈ R. The representation is uniquebecause if y1 + λ1x = y2 + λ2x then (λ1 − λ2)x = y2 − y1 ∈ Y and so λ1 = λ2 and hencey1 = y2. Now define φ(y + λx) = λ. Clearly Y ⊂ kerφ and φ 6= 0. kerφ = Y becauseY ⊂ kerφ ( X and so must be Y .

3 =⇒ 2. Suppose Y = kerφ and φ 6= 0. Take any x such that φ(x) 6= 0. We want toshow that Y ∪ x spans X. Given z ∈ X we want to choose y, λ such that z = y + λx.

We do this as follows. Set λ = φ(z)φ(x) and y = z − λx. Then φ(y) = φ(z − λx) = 0 and so

y ∈ Y as required. Q.E.D.

Lemma 2.4 If Y = kerφ is 1-codimensional in X then

7 of 29


1. Y is closed if and only if φ is continuous

2. Y is dense if and only if φ is not continuous.

Proof Since Y 6= X it cannot be closed and dense. So it suffices to prove that Y isclosed if φ is continous and Y is dense if not.

If φ is continuous, then kerφ = φ−1(0) and so is closed. We need to show if φ is notcontinuous then kerφ is dense.

φ is unbounded (i.e. not continuous) so choose a sequence (xn) in X with ||xn|| ≤ 1

with |φ(xn)| → ∞. If x ∈ X choose yn = x − φ(x)φ(xn)

xn and then φ(yn) = 0 so yn ∈ kerφ.On the other hand

||yn − x|| = || −φ(x)

φ(xn)xn|| =

∣∣∣∣ φ(x)

φ(xn)

∣∣∣∣ ||xn|| → 0

and so x ∈ closure(Y ). Q.E.D.

2.2 Separation Theorems

We state the important theorem, give some applications and then prove it, as this wayyou are motivated why the theorem is so important.

Theorem 2.5 (Hahn-Banach) If Y is a subspace of a real normed linear space X andφ : Y → R is a continuous linear functional, then there is a continuous linear functionalφ : X → R which extends φ, i.e. φ(y) = φ(y) if y ∈ Y and ||φ|| = ||φ||.

Corollary 2.6 If x ∈ X is a unit vector then there is a functional φ ∈ X? with ||φ|| ≤ 1and φ(x) = 1.

The corollary thus says that there is a supporting hyperplane to the unit ball B at eachpoint on the boundary.Proof Let Y = spanx ⊂ X and define φ : Y → R by φ(λx) = λ. Then φ(x) = 1 and|φ(λx)| = |λ|||x|| = ||λx|| and so ||φ|| = 1. Then by Hahn-Banach we can extend to φwith ||φ|| = 1 and φ(x) = 1. Q.E.D.

Corollary 2.7 (Closest point witnesses) Let Y be a closed subspace of X and x ∈ X.Let d = inf||x− y|| : y ∈ Y . Then there is a linear functional φ on X with

1. φ(y) = 0 for all y ∈ Y

2. ||φ|| = 1

3. φ(x) = d

Thus φ witnesses the fact that x is at least distance d from Y , since if y ∈ Y then

||x− y|| = ||φ||||x− y|| ≥ φ(x− y) = φ(x)− φ(y) = d

Proof Let Z = spanY ∪x and define φ : Z → R by φ(y+λx) = dλ. It satisfies 1 and3 and we want to show that |φ(y+λx)| ≤ ||y+λx||. But how do we understand the righthand side of this. We know that ||x − y|| ≥ d for all y ∈ Y and so ||λx − λ(−yλ )|| ≥ |λ|dand so we have this bounded. Then we have |φ(y + λx)| = |λd| = |λ|d ≤ |λ|||x − −yλ || =||y+λx||.By the Hahn-Banach theorem there is a functional onX with the same properties.

Q.E.D.

8 of 29


2.2.1 Reflexivity

We saw that if x ∈ X we can define an element of X??, x by x(φ) = φ(x) for φ ∈ X?. Themap x 7→ x is linear and ||x|| ≤ ||x|| since |x(φ)| = |φ(x)| ≤ ||φ||||x||. In fact ||x|| = ||x||and so the map x 7→ x is an isometric embedding of X into X??, because, if ||x|| = 1 thenby the Hahn-Banach there is a φ ∈ X? with ||φ|| = 1 and φ(x) = 1 and so x(φ) = φ(x) = 1and ||φ|| = 1 and so ||x|| ≤ 1. If ||x|| 6= 1 we scale.

Definition 2.8 If the map x 7→ x is surjective then we say that X is reflexive.

2.2.2 Adjoints

Definition 2.9 If T : X → Y is linear and bounded then the adjoint T ? : Y ? → X? isdefined by

T ?(ψ)(x) = ψ(Tx)

for ψ ∈ Y ? and x ∈ X.

Lemma 2.10 The following are properties of the adjoint

1. for each ψ, T ?ψ is linear

2. T ? is linear

3. ||T ?|| = ||T ||

Proof

1. clear since T is linear.

2.T ?(ψ + φ)(x) = (φ+ ψ)(Tx) = φ(Tx) + ψ(Tx) = T ?(φ)(x) + T ?(ψ)(x)

and so T ?(ψ + φ) = T ?(φ) + T ?(ψ).

3. We have that

|T ?(ψ)(x)| = |ψ(Tx)| ≤ ||ψ||||Tx|| ≤ ||ψ||||T ||||x||

and so ||T ?(ψ)|| ≤ ||T ||||ψ|| and thus ||T ?|| ≤ ||T ||.We claim that ||T ?|| ≥ ||T ||. We can choose x ∈ X with ||x|| = 1 and ||Tx|| ≥ ||T ||−εand then by Hahn-Banach there is a ψ ∈ Y ? with ||ψ|| = 1 and |ψ(Tx)| ≥ ||T || − ε.Then |T ?(ψ)(x)| = |ψ(Tx)| ≥ ||T || − ε but ||x|| = 1 = ||ψ|| and so ||T ?|| ≥ ||T || − εand let ε→ 0.

Q.E.D.

In a Hilbert space, if we identify H? with H via the inner product then this adjointis the usual one in the real case.

In finite dimensions we can represent T as a matrix with respect to bases in X andY . Then the matrix of T ? is the transpose of that of T , if we use the dual bases for Y ?

and X?.

Theorem 2.11 ( 1 -step extension) If Y is a 1-codimensional subspace of a normedspace X and φ : Y → R is a bounded linear functional then there is an extension φ : X →R of φ with ||φ|| ≤ ||φ||

9 of 29


The inequality of the norms is crucial if we want to apply this infinitely often.To extend into 1 dimension higher we will have a single degree of freedom.

ProofBy scaling we may assume that ||φ|| = 1. Choose x ∈ X \ Y . We can write every

point of X uniquely as y + λx for y ∈ Y and λ ∈ R. For each α define φα : X → R by

φα(y + λx) = φ(y) + λα

Our aim is to show that for an appropriate α, ||φα|| ≤ 1, since each φα is clearly linear.We need to check that

|φ(y) + λα| ≤ ||y + λx||

for all y ∈ Y and λ ∈ R. This is automatic if λ = 0 so by considering λ−1y instead of yit suffices to establish this for λ = 1, and all y ∈ Y .

We thus want|φ(y) + α| ≤ ||y + x||

for all y ∈ Y . This is the same as

φ(y) + α ≤ ||y + x|| ∀y ∈ Y and φ(z) + α ≥ −||z + x|| ∀z ∈ Y

This means we need to choose α such that

−||z + x|| − φ(z) ≤ α ≤ ||y + x|| − φ(y) ∀y, z ∈ Y

such an α exists provided

−||z + x|| − φ(z) ≤ ||y + x|| − φ(y) ∀y, z ∈ Y

in other words we needφ(y)− φ(z) ≤ ||y + x||+ ||z + x||

but

φ(y)− φ(z) = φ(y − z) ≤ ||y − z|| = ||y + x− (z + x)|| ≤ ||y + x||+ ||z + x||

as required.Q.E.D.

Definition 2.12 A partially ordered set (Ω,≤) is a set Ω equipped with a relation ≤satisfying

1. x ≤ x for all x ∈ Ω

2. If x ≤ y and y ≤ z then x ≤ z.

3. For every x, y ∈ Ω, if x ≤ y and y ≤ x then x = y

An example is to take Ω to be the set of all subsets of 1, 2, 3, 4 and A ≤ B meansA ⊆ B. Then 1 ≤ 1, 2 and ∅ ≤ 3 but 1, 2 is not related to 3.

Definition 2.13 A chain in a partially ordered set is a subset in which every pair ofelements is related.

An upper bound for a chain C ⊂ Ω is an element u ∈ Ω with c ≤ u for all c ∈ CA maximal element m of Ω is an element for which there is no v with m ≤ v but

v 6= m.

10 of 29


Note that this doesnt mean that m ≥ u for all u ∈ Ω.An example of such is Ω = ∅, 1, 2, 3, 1, 2 ordered by inclusion. Then the

maximal elements are 1, 2 and 3.

Theorem 2.14 (Zorn’s Lemma) If (Ω,≤) is a non-empty partially ordered set in whichevery chain has an upper bound, then Ω contains at least one maximal element

If you have a 1-step extension then a maximal element must be the top, i.e. in our contextthe Banach space.Proof (Hahn Banach) Let Ω be the partially ordered set of pairs (Z, φZ) where Z isa subspace of X containing Y and φZ : Z → R is a linear functional extending φ andsatisfying ||φZ || ≤ ||φ||, with the ordering (Z1, φZ1) ≤ (Z2, φZ2) if Z1 ⊂ Z2 and φZ1 andφZ2 agree on Z1. It is clear that this is a partial order and Ω is non empty since (Y, φ) ∈ Ω.

Suppose we have a chain (Zα, φZα)α∈A in Ω. Let W = ∪AZα and define φW : W →R as follows: If w ∈ W then w ∈ Zα for some α and we set φW (w) = φZα(w) for thatα. This is well defined since if w ∈ Zα and w ∈ Zβ we have Zα ⊂ Zβ, say, and then φZβextends φZα because we are in a chain. W is a subspace since if u, v ∈ W then u ∈ Zαand v ∈ Zβ say and then Zα ⊂ Zβ so u, v ∈ Zβ so u + v ∈ Zβ ⊂ W , and similarly ifZα ⊂ Zβ. In the same way φW is linear.

Moreover, if w ∈W with w ∈ Zα then

|φW (w)| = |φZα(w)| ≤ ||φ||||w||

and since this works for all w, ||φW || ≤ ||φ||. Thus (W,φW ) ∈ Ω and clearly this is anupper bound for the chain. By Zorn’s lemma, Ω contains a maximal element (Z, φZ). IfZ 6= X then we can find an x ∈ X \Z and by the 1-step extension theorem we can extendφZ to spanZ ∪ x, contradicting the maximality of (Z, φZ). Q.E.D.

Homework 4 asked to use Hahn-Banach to build a functional φ on l∞ which is boundedsuch that if (xi) is convergent then φ((xi)) = limn→∞ xn. This is not unique though, andis rather strange.

2.2.3 Complex Hahn Banach

If X is a complex normed space then we can regard it as a real normed space, using onlythe real numbers as the scalars.

Lemma 2.15 If X is a complex normed space and φ : X → C is linear then there is areal linear functional u : X → R with

φ(x) = u(x)− iu(ix)

If φ is bounded then ||φ|| = ||u||. Conversely, given u we can build φ and ||φ|| = ||u||.

This is saying that for a linear functional, the real and imaginary parts are related.An example of this is with X = C and φ(z) = z. Thus φ(x+ iy) = x+ iy and so we

have u(x + iy) = x. Then the image under u of the unit disc is very small compared tothe image of it under φ. However, it still has norm 1.Proof Let u and v be the real and imaginary parts of φ so that

φ(x) = u(x) + iv(x) ∀x ∈ X

11 of 29


u, v : X → R and are real linear since for λ ∈ R,

u(λx) + iv(λx) = φ(λx) = λφ(x) = λu(x) + iλv(x)

Also we have that u(ix) + iv(ix) = φ(ix) = iφ(x) = iu(x) − v(x) and equating real andimaginary parts gives

u(ix) = −v(x) u(x) = v(ix)

and so we have φ(x) = u(x)− iu(ix).

|φ(x)| =√u(x)2 + u(ix)2 ≥ |u(x)|

and so ||φ|| ≥ ||u||. On the other hand for any x we can write |φ(x)| = eiθφ(x) = φ(eiθx) =u(eiθx)− iu(ieiθx) = u(eiθx) for some θ ∈ R. Since |φ(x)| is real we have

|φ(x)| = u(eiθx) ≤ ||u||||x||

and so ||φ|| ≤ ||u||.Conversely, if we build φ as φ(x) = u(x)− iu(ix) then φ is linear and its real part is

u so ||φ|| = ||u||. Q.E.D.

Theorem 2.16 (Complex Hahn-Banach) If Y is a subspace of a complex Banachspace X and φ : Y → C is linear then there is an extension φ of φ with ||φ|| ≤ ||φ||.

Proof Write φ(x) = u(x) − iu(ix) for a real linear functional u on Y . Extend u tou : X → R with ||u|| ≤ ||u|| by the real Hahn-Banach theorem. Set φ(x) = u(x)− iu(ix)for x ∈ X. Then φ is complex linear and extends φ. Furthermore

||φ|| ≤ ||u|| ≤ ||u|| = ||φ||

Q.E.D.

Corollary 2.17 If X is a real Banach space then X is reflexive if and only if X? isreflexive.

Without the Hahn-Banach theorem, we cannot prove the ⇐= direction.Proof We first prove the =⇒ direction. We want to show that each element F of X???

is φ for some φ ∈ X?. Define φ : X → R by φ(x) = F (x). If x ∈ X, then

φ(x) := x(φ) := φ(x) := F (x)

and so φ = F on all elements of X?? of the form x. If X is reflexive, then this is all ofX??.

Now the other direction. Suppose that X is not reflexive. Then the image of X under∧ map is a closed subspace of X?? since it is complete. Since we are assuming that itis not the whole of X?? we can find F ∈ X??? which is not 0 but with F (x) = 0 for allx ∈ X. By assumption F = φ for some φ ∈ X?. But if x ∈ X then

φ(x) = x(φ) = φ(x) = 0

This contradicts φ 6= 0 Q.E.D.

Corollary 2.18 l1 and l∞ are not reflexive.

Proof c0 is not reflexive, and l1 ∼= c?0 and l∞ ∼= l?1 Q.E.D.

12 of 29


2.3 Weak and Weak-? Topologies

Definition 2.19 Let X be a Banach space. The weak topology on X consists of allpossible unions of sets of the form

x ∈ X : φ1(x) < α1, ..., φn(x) < αn

for some finite sequence φ1, ..., φn ∈ X? and αi ∈ R.In the same way, the weak-? topology on X? consists of unions of sets of the form

φ ∈ X? : φ(x1) < α1, ..., φ(xn) < αn

for x1, ..., xn ∈ X and αi ∈ R.

The sets of the above form can be called a “ basis of neighbourhoods” for the opensets.

If X is reflexive, then the weak and weak-? topologies on X? are the same.The weak topology on X makes each φ ∈ X? continuous, because φ−1(U) is a union

of neighbourhoods if U is open in R.

U = ∪α(aα, bα) = ∪α(xα < bα ∩ −xα < −aα)

and thusφ−1(U) = ∪αx : φ(x) < bα,−φ(x) < −aα

The weak topology is the weakest topology which makes each element of X? continuous.If X is infinite dimensional then the weak and weak-? topologies are not metrisable, socompactness is not the same as sequentially compactness.

Theorem 2.20 (Banach-Alaoglu) The unit ball BX? is compact in the weak-? topology

The proof uses Tychonov’s theorem. If (Xα) are topological spaces then the product∏αXα is the Cartesian product (xα) : xα ∈ Xα∀α with the topology consisting of

unions of sets (xα) : xα1 ∈ Uα1 , ..., xαn ∈ Uαn where Uαi is open in Xαi for each i. ThenTychonov’s theorem says the product of compact spaces is compact.Proof We consider the product

∏x∈BX [−1, 1] with the topology from R.

We can embed BX? into this by φ 7→ (φ(x)) so φ : BX → [−1, 1]. This is a 1-1embedding. If φ 6= ψ there is an x ∈ BX where φ(x) 6= ψ(x). The topology inherited byBX? from the product topology is exactly the weak-? topology. Neighbourhoods restrictonly finitely many values φ(x1), ..., φ(xn). The product is compact by Tychonov’s theorem.

It suffices to prove that the image of BX? is closed. Suppose f is in the closure of theimage. We want that if x, y ∈ BX , and λ, µ ∈ R with |λ|+ |µ| ≤ 1 then

f(λx+ µy) = λf(x) + µf(y)

as then this will imply that f is the restriction to BX of some linear functional.Given ε > 0, the functions g satisfying

1. |g(x)− f(x)| < ε

2. |g(y)− f(y)| < ε

3. |g(λx+ µy)− f(λx+ µy)| < ε

13 of 29


form an open set in the product, since we are restricting three values of g. So this setmeet sthe image of BX? , say φ is a linear functional whose image satisfies the inequalities,hence

|f(λx+ µy)− λf(x)− µf(y)| = |g(x)− f(x) + g(y)− f(y) + g(λx+ µy)− f(λx+ µy)|< (1 + |µ|+ |λ|)ε< 2ε

and letting ε→ 0 we have f(λx+ µy) = λf(x) + µf(y) Q.E.D.

3 The Category Theorems

If C1 ⊃ C2 ⊃ ... is a decreasing sequence of non empty closed subsets of a compact spacethen ∩∞1 Ci 6= ∅. This is not the case in R, since [0,∞) ⊃ [1,∞) ⊃ .. and ∩∞k=1[k,∞) = ∅.

Theorem 3.1 (Cantor) If C1 ⊃ C2 ⊃ ... is a decreasing sequence of non empty closedsubsets of a complete metric space X and diam(Cn)→∞ as n→ 0 then ∩∞1 Cn 6= ∅

Proof For each n, choose xn ∈ Cn. Since xn ∈ Cm if n ≥ m then d(xn, xm) ≤diam(Cm)→ 0 and so the sequence (xn) is Cauchy, and let x be the limit. Since xn ∈ Cmfor all n ≥ m and Cm is closed then x ∈ Cm for all m and so x ∈ ∩Cm. Q.E.D.

Definition 3.2 A subset E of a metric space X is called nowhere dense if its closurecontains no non-empty open subset of X.

Theorem 3.3 (Baire) A complete metric space X is not the countable union of nowheredense subsets of itself.

Corollary 3.4 R is not countable.

This is since it is the union of all singletons x.Proof Let E1, E2, ... be nowhere dense in X. We want to find x 6∈ ∪Ei. We can assume,by taking closures, that the Eis are closed and each contains no non empty open set.

Choose x1 6∈ E1. There is a ball centred at x1 which does not meet E1, and hence aclosed ball B(x1, r1) ∩ E1 = ∅. We may assume that r1 < 1.

The open ball B(x1, r1) is not contained in E2 since E2 is nowhere dense. So choosex2 ∈ B(x1, r1) but x2 6∈ E2. Since E2 is closed there is a ball B(x2, r2) which is containedin B(x1, r1) with B(X2, r2) ∩ E2 = ∅. We may assume that r2 <

12 .

Continuing in this way we build B(x1, r1) ⊃ B(x1, r1) ⊃ B(x2, r2) ⊃ B(x2, r2) ⊃ ...with ri <

1i and B(xi, ri) ∩ Ei = ∅.

By Cantor’s theorem, there is a point in all the B(xi, ri) and hence in none of the Eis.Q.E.D.

We shall use this to prove the “analogue” of the 1-1 implies onto theorem for a finitedimension vector space. This is the Open mapping theorem.

Baire’s theorem allows you to construct nasty things. There are too few nice thingsso there must be some nasty ones. At the end of the section we shall show that there is acontinuous 2π periodic function on [−π, π] whose Fourier series diverges at 0. This usesthe Uniform Boundedness Principle.

14 of 29


Theorem 3.5 (UBP) Suppose φ1, φ2, ... are bounded linear functionals on a Banachspace X. If for each x ∈ X the values φn(x) form a bounded set, then there is some Mwith ||φn|| ≤M for all n.

This at first seems strange, if not simply wrong. Take for example the following.Suppose that X = l1. Then take φ1 big in direction e1, and zero in the rest, φ2 huge

in e2 and zero in the rest, etc. Then at e7 φ7 is gigantic, but everything else is zero.However the functionals are not uniformly bounded. This is not a contradiction as wecan find some clever point x which picks up some bad behaviour from lots of the φ s.

Suppose that φi : l1 → R is the coordinate functional φi(x) = xi. Then if x ∈ l1, wehave

∑|φi(x)| = ||x||1 < ∞ which is essentially the opposite to the above. Thus clever

means very clever.This example is as bad as it gets.

Theorem 3.6 (Sharp UBP/Plank) If (φi) are unit functionals (operator norm 1) and(wi) are positive numbers with

∑wi < 1 then there is a unit vector x ∈ X with |φn(x)| >

wn for all n.

Proof (Plank =⇒ UBP) If (||φn||) is unbounded we can choose a subsequence φnkwith ||φnk || ≥ 4k. Let ψk = φnk/||φnk || which is a unit functional. Choose x with|ψk(x)| ≥ 1

2k+1 and ||x|| ≥ 1. Then |φnk(x)| ≥ 2k−1 →∞. Q.E.D.

Proof (UBP) Suppose (φn(x)) is bounded for each x ∈ X. For each k ∈ N, let Ck =x ∈ X : |φn(x)| ≤ k,∀n. These sets are closed and they cover X. Thus by Baire’stheorem at least one is not nowhere dense, and hence includes a ball of positive radius.In fact Ck = kC1, so C1 includes such a ball, say the ball of radius r > 0 around y.

C1 is convex and symmetric so if u ∈ C1 then −u ∈ C1. If ||x|| ≤ r we have y+ x and−y + x in C1 and so the average x = 1

2(y + x) + 12(−y + x) ∈ C1 by convexity. Thus if

||x|| ≤ r then |φn(x)| ≤ 1 for all n. Hence ||φn|| ≤ 1r for all n. Q.E.D.

3.1 Open Mapping Theorem

Recall that if T : Rn → Rn is linear then it is one to one if and only if it is onto.

Theorem 3.7 (Open mapping) If T : X → Y is a linear map from a Banach space Xto a Banach space Y which is onto then T is an open map, i.e. T (BX) ⊃ rBY for somer > 0.

Corollary 3.8 (Inverse mapping theorem) If X and Y are Banach spaces and thelinear map T : X → Y is one to one and onto and bounded then T−1 is bounded.

This looks like a cheat since since we assume one to one and onto, but it isnt becausewe conclude that T is invertible in the sense that it has a bounded inverse (in the senseof analysis).

The proof of the OMT has two steps. The first uses Baire and completeness of Y andthe second uses an iterative argument and completeness of X. They are linked by a cleverremark.Proof (OMT theorem 3.7) We shall start by showing that for some positive r, theclosure T (Bα) includes rBY .

For each k let Ck = T (kBX) for k = 1, 2, ... These closed sets cover Y since T is ontoand hence by Baire’s theorem at least one has non empty interior. Since Ck = kC1 in

15 of 29


fact C1 has non empty interior. Suppose C1 includes the ball of radius r > 0 around y.Since C1 is convex and symmetric, it includes the ball of radius r around 0, rBY . ThusT (Bα) ⊃ rBY .

In particular, if ||y|| ≤ r there is an x ∈ X with ||x|| ≤ 1 so that

||y − Tx|| ≤ r

2(3.1)

We shall now prove that T (2BX) ⊃ rBY , and so T (BX) ⊃ r2BY .

Suppose that y ∈ Y and ||y|| ≤ r. Choose x1 ∈ BX with ||y − Tx1|| ≤ r2 by (3.1).

Now by applying (3.1) to the vector y − Tx1 we can find x2 with norm ||x2|| ≤ 12 with

||y − Tx1 − Tx2|| ≤r

4

Continuing we obtain (xj) with ||xi|| ≤ 12i−1 and ||y − T (x1 + ... + xi)|| ≤ r

2i. Since X

is complete, the sum∑∞

1 xj converges to x and ||y − Tx|| = lim ||y − T (∑i

1 xj)|| = 0 soy = Tx and ||x|| ≤

∑∞1 ||xj || ≤

∑∞1

12i−1 = 2 Q.E.D.

We could try to find ui such that ||y − Tui|| → 0. If the ui have a convergentsubsequence then its limit is in BX and has image Y . This though requires compactness.Proof (Inverse mapping theorem 3.8) By the OMT, we know that if ||y|| ≤ r thereis an x with ||x|| ≤ 1 and y = Tx for some r > 0. Thus ||T−1y|| = ||x|| ≤ 1 and so||T−1|| ≤ 1

r . Q.E.D.

3.2 Closed Graph Theorem

If T : X → Y is a linear map, its graph is the set (x, Tx) : x ∈ X ⊂ X × Y .If X and Y are Banach spaces then X × Y is a Banach space with

||(x, y)|| = ||x||+ ||y||

but note that this isn’t the only norm we can use. We can thus ask whether the graph isclosed in X × Y . The answer is yes if whenever xn → x and Txn → y then y = Tx.

If T is continuous, the graph is closed since in this case xn → x =⇒ Txn → Tx, evenif we do not know that Txn converges.

Theorem 3.9 (Closed Graph) If T is a linear map between Banach spaces X and Ythen T is bounded if and only if its graph is closed in X × Y .

Proof Let G = (x, Tx) : x ∈ X be the graph of T . It is a closed subspace of X × Yso is a Banach space in the inherited norm. Define

π1 : G→ X by π1(x, Tx) = x

π2 : X × Y → Y by π2(x, y) = y

These are obviously linear. π2 is continuous because

||y|| ≤ ||x||+ ||y|| = ||(x, y)||

π1 is continuous because

||x|| ≤ ||x||+ ||Tx|| = ||(x, Tx)||

π1 is one to one and onto, since if u = x we have (u, Tu) = (x, Tx). By the IMT (orOMT), π1 has a bounded inverse. Since T = π2 π−11 , it is bounded. Q.E.D.

16 of 29


3.3 Application to Fourier Analysis

There is a continuous 2π periodic function on [−π, π] whose Fourier series diverges at 0.The Fourier series is

∑∞−∞ ane

int where

an =1

2π

∫ π

−πf(x)e−inxdx

The Nth partial sum is

N∑−N

aneint =

1

2π

N∑−N

eint∫ π

−πf(x)e−inxdx =

1

2π

∫ π

−πf(x)

N∑−N

ein(t−x)dx

The sum inside is

N∑−N

ein(t−x) =e−N(t−x) − ei(N+1)(t−x)

1− ei(t−x)

=e−(N+ 1

2)(t−x) − ei(N+ 1

2)(t−x)

e−i(t−x)

2 − ei(t−x)

2

=sin((N + 1

2)(t− x))

sin(12(t− x))

and thusN∑−N

aneint =

1

2π

∫ π

−πf(x)

sin((N + 12)(t− x))

sin(12(t− x))dx

In particular, the value at t = 0 is

N∑−N

an =1

2π

∫ π

−πf(x)

sin((N + 12)(x))

sin(12(x))dx

The bit on the left hand side is supposed to converge to f(0), and so you would expectthe graph of the kernel DN to spike at 0. We would like this Fourier series to approximatef , meaning that

1

2π

∫ π

−πf(x)DN (x)dx ≈ f(0)

We would like the kernel to reproduce f . Thus the kernel should have integral 1 with itsmass concentrated near t, it is an “approximation to the identity”

The map f 7→ 12π

∫ π−π f(x)DN (x)dx = SN (f) is a bounded linear functional on the

space of continuous 2π periodic functions, with norm ||f || = maxx∈[−π,π] |f(x)|. By UBP,we can deduce that there is an f where (SN (f)) is unbounded if (||SN ||) is unbounded.

We claim that

||SN || =1

2π

∫ π

−π

∣∣∣∣∣sin(N + 12)x

sin 12x

∣∣∣∣∣ dx = ||DN ||1

SN acts on L∞[−π, π] by the same integral formula and its norm is at most ||DN ||1.On L∞ the norm clearly is ||DN ||1 because we can apply it to sgn(DN ) in L∞ and||sgn(DN )||∞ = 1 but SN (sgn(DN )) = 1

2π

∫ π−π(sgn(DN ))DN = ||DN ||1. However sgn(DN )

17 of 29


is not continuous but we can approximate by continuous functions of L∞ norm 1 and whichare 2π periodic.

Now DN vanishes at ± πN+ 1

2

,± 2πN+ 1

2

, ...,± NπN+ 1

2

and

||DN ||1 =1

π

∫ π

0

∣∣∣∣∣sin(N + 12)x

sin 12x

∣∣∣∣∣ dx ≥ 1

π

N∑k=1

kπ

N+ 12∫

(k−1)π

N+ 12

∣∣∣∣∣sin(N + 12)x

sin 12x

∣∣∣∣∣ dx

≥ 1

π

N∑k=1

kπ

N+ 12∫

(k−1)π

N+ 12

∣∣sin(N + 12)x∣∣∣∣∣sin kπ

2N+1

∣∣∣ dx

because x 7→ sinx is increasing on [0, π]. Now if x > 0 we have sinx ≤ x so for each k,1

sin kπ2(N+1)

≤ 2(N+1)kπ and so we get that

||DN ||1 ≥2(N + 1)

π2

N∑k=1

1

k

kπ

N+ 12∫

(k−1)π

N+ 12

∣∣∣∣sin(N +1

2)x

∣∣∣∣ dx

=2(N + 1)

π2

N∑k=1

1

k

∫ π

N+ 12

0sin(N +

1

2)xdx

=4

π2

N∑k=1

1

k

>4

π2logN →∞

and so we have the result that we want

4 Unbounded Operators

We motivate the following with a classical example. We have a taut string fixed at 0 and1 and at time t = 0 it is plucked into shape x 7→ f(x, 0). Thereafter its shape is f(x, t).Then f obeys the wave equation

c2∂2f

∂x2=∂2f

∂t2

with the boundary conditions f(0, t) = 0 = f(1, t). We now expand f(x, t) as a Fourierseries to get it as

f(x, t) =∞∑1

an(t) sin(nπx)

Note that sinnπx vanishes at 0 and 1 so we immediately satisfy the boundary conditions,and this is also the reason why we do not consider the cosine terms. By extending to anodd function on [−1, 1] we have an expression in L2. If we can differentiate term by termthen

∂2f

∂t2=∞∑1

a′′n(t) sin(nπx)∂2f

∂x2= −

∞∑1

an(t)n2π2 sin(nπx)

18 of 29


By the wave equation we have

a′′n(t) = −c2n2π2an(t)

and so an(t) = An cos(cnπt) +Bn sin(cnπt). At t = 0 we have f(x, 0) =∑∞

1 Cn sin(nπx)

and if we furthermore assume that ∂f∂t |t=0 = 0, i.e. we have a slope but no speed, then we

get that Bn = 0 for all n and thus

an(t) = An cos(cnπt)

and An = Cn and so

f(x, t) =∞∑1

An cos(cnπt) sin(nπx)

The method above worked because the functions sinnπx are eigenfunctions of themap f 7→ f ′′ satisfying the boundary conditions f(0) = 0 = f(1) and are orthogonal inL2[0, 1]. This is what we expect from a self adjoint operator. The Laplacian f 7→ f ′′ isnot a bounded operator on any useful space, so we need a theory of unbounded operators.The Laplacian is “ like ” a self adjoint operator though, since using integration by partstwice we get:

〈f ′′, g〉 =

∫ 1

0f ′′(x)g(x)dx

= f ′(x)g(x)|10 −∫ 1

0f ′(x)g′(x)dx

= f ′(x)g(x)|10 − f(x)g′(x)|10 +

∫ 1

0f(x)g′′(x)dx

so if f and g are C2 and f(0) = 0 = f(1) and g(0) = 0 = g(1) then we get 〈f ′′, g〉 = 〈f, g〉.So f ′′ looks self adjoint on smooth functions satisfying the Dirichlet boundary conditions.The boundary conditions are thus part of the definition of the operator. Dirichlet andNeumann boundary conditions make it “like” a self adjoint operator.

The apparent self adjointness is not enough though for spectral theory.

Definition 4.1 A linear operator T is called densely defined in the Hilbert space H ifit is defined and linear on a dense subspace T : D(T )→ H.

Note that f ′′ is defined and linear on C2 which is dense in L2

Definition 4.2 We say that S extends T if D(S) ⊃ D(T ) and S(x) = T (x) for x ∈D(T ).

Definition 4.3 We say that T is symmetric if

〈Tx, y〉 = 〈x, Ty〉

if x, y ∈ D(T ).

Definition 4.4 We define the adjoint of a densely defined T as follows. We defineD(T ?) to be all y for which x 7→ 〈Tx, y〉 is a bounded linear functional on D(T ). In thiscase the functional extends uniquely to H. It can be represented uniquely by an innerproduct x 7→ 〈x, z〉 for some z ∈ H. We call this z by T ?y, so

〈Tx, y〉 = 〈x, T ?y〉

if x ∈ D(T ) and y ∈ D(T ?).

19 of 29


Definition 4.5 T is called self adjoint if D(T ) = D(T ?) and T = T ?

Note that T is symmetric is the same as saying that T ? extends T .Warning though that D(T ?) might not even be dense.The map f 7→ f ′′ on C2[0, 1] with f(0) = 0 = f(1) is not self adjoint, but it is

symmetric. There are functions g for which∫ 10 f′′g =

∫ 10 fg

′′ but with g not in the givendomain, because g′′ is not continuous.

The integration by parts repeated works fine if f(0) = 0 = f(1) = g(1) = g(0) and f ′

and g′ are absolutely continuous, meaning

f ′(x) =

∫ x

0u(t)dt g′(x) =

∫ x

0v(t)dt

and then∫f ′g′ makes sense by Holder and f ′′ = u and g′′ = v. To find the correct domain

on which to study f ′′ we set H1[0, 1] to be the space of absolutely continuous functionsf(x) = c +

∫ x0 u(t)dt with u ∈ L2[0, 1] and H2[0, 1] to be the space of differentiable

functions f on [0, 1] with f ′(x) ∈ H1.f ′′ may not exist everywhere but f(x) = a + bx +

∫ x0

∫ t0 u(s)dsdt and u ∈ L2. Let D

be the space of functions

D = f ∈ H2[0, 1] : f(0) = 0 = f(1)

Theorem 4.6 The map f 7→ f ′′ is self adjoint on D.

Proof We need to show that if the map f 7→∫f ′′g is a bounded linear functional on D

then g ∈ D and g(x) = a+ bx+∫ x0

∫ t0 u(s)dsdt and

∫f ′′g =

∫fu, as then we would have

D(T ) = D(T ?) and 〈Tx, y〉 = 〈x, T ?y〉 as required.If the map f 7→

∫f ′′g is a bounded linear functional then we can write it as

∫fh

for some h ∈ L2 by Riesz-Frechet. Then let G(x) =∫ x0

∫ t0 h(s)dsdt. Then G ∈ H2 and

G′′(x) = h.We want to check that G is almost g. Then∫ 1

0f ′′(x)G(x)dx = f ′(x)G(x)|10 −

∫ 1

0f ′(x)G′(x)dx

= f ′(x)G(x)|10 − f(x)G′(x)|10 +

∫ 1

0f(x)G′′(x)dx

= f ′(x)G(x)|10 +

∫ 1

0f(x)G′′(x)dx

Recall that∫f ′′g =

∫fh but also G′′ = h so we have∫ 1

0f ′′(x)G(x)dx = f ′(x)G(x)|10 +

∫ 1

0f(x)h(x)dx = f ′(x)G(x)|10 +

∫ 1

0f ′′g

We apply to f which not only satisfies f(0) = 0 = f(1) but to those which satisfy as wellf ′(0) = 0 = f ′(1).

Thus G − g is orthogonal to all f ′′ = u ∈ L2 for functions f ∈ D satisfying alsof ′(0) = 0 = f ′(1).

Since f ′(x) =∫ x0 u(t)dt the condition f ′(1) = 0 is equivalent to the statement∫ 1

0u(t)dt = 0

20 of 29


and since f(x) =∫ x0 f′(t)dt so we have

0 =

∫ 1

0f ′(t)dt = xf ′|10 −

∫ 1

0xf ′′(x)dx = −

∫ 1

0xf ′′(x)dx = −

∫ 1

0xu(x)dx

and so G− g is orthogonal to all elements of L2 satisfying∫ 1

0u = 0

∫ 1

0xu(x)dx = 0

thus G−g is orthogonal to all u ∈ L2 which are themselves orthogonal to linear functions.Thus G− g must be linear. This means

g(x) = a+ bx+G(x) ∈ H2

and g′′ = G′′ = h. This means∫ 10 f′′g =

∫ 10 fh =

∫ 10 fg

′′ but also, integrating by partstwice gives ∫ 1

0f ′′g = f ′(x)g(x)|10 +

∫ 1

0f(x)g′′(x)dx

and so in fact f ′(x)g(x)|10 = 0 for all f ∈ D. We can find f ∈ D with arbitrary values off ′(0), f ′(1) and so we must have g(0) = 0 = g(1).

We have proved that the domain of the adjoint is not too big. Conversely, if f, g ∈ Dthen integration by parts shows ∫ 1

0f ′′g =

∫ 1

0fg′′

Q.E.D.

4.1 Closed Operators

Definition 4.7 A densely defined operator T in a Hilbert space H with domain D(T ) iscalled closed if its graph is closed, i.e. if xn ∈ D(T ) and xn → x and T (xn) → y in Hthen x ∈ D(T ) and T (x) = y.

The closed graph theorem can thus be reworded as if T is closed and D(T ) = H thenT is bounded. In other words, closed is almost bounded, if you are not defined on thewhole space.

Example 4.1 Suppose H = l2 and define T : C00 → l2 by

T (x1, x2, ..., xn, 0, 0, ...) = (x1, 2x2, ..., nxn, 0, ...)

where C00 is the set of all finitely non zero sequences. This operator is not closed. Let

u1 = (1, 0, ...), u2 = (1, 1/4, 0, ...), ... un = (1, 1/4, 1/9, ..., 1/n2, 0, ...), ...

and then Tu1 = u1, Tun = (1, 1/2, ..., 1/n, 0, ...) and un → (1, 1/4, 1/9, ...) = x andTun → (1, 1/2, 1/3, ...) = y but x 6∈ C00 = D(T ).

However, we can extend this operator to be closed. Choose

D = (xi)∞1 :∑

n2x2n <∞

21 of 29


and observe that∑n2x2n <∞ enforces Tx ∈ l2. You might call this the natural domain.

Then T is defined on D and we claim that it is closed.Suppose ui → x and Tui → y = (y1, y2, ...) in l2. The map

(θ1, θ2, ...) 7→ (θ1, θ2/2, θ3/3, ...)

is bounded on l2, and we denote it by S. Then if u ∈ D we have S(Tu) = u. So

ui = S(Tui)→ S(y)

since S is bounded and so x = Sy and also∑n2x2n =

∑n2(ynn

)2=∑

y2n <∞

and so x ∈ D and so Tx = y since xi = 1i yi.

In other words what we have done in the above is to define the domain of definitionso that the “inverse” S works as it should.

Consider the map f 7→ f ′ in L2[0, 1], initially defined on C1[0, 1] the space of contin-uously differentiable functions. The operator is not symmetric on this space, because ofthe change of sign when you perform integration by parts.

However, f 7→ if ′ is symmetric with the right boundary conditions.The map is not closed on C1; we find fn → f in L2 with f ′n → g in L2 but f 6∈ C1.Choose f to be the function 0 at 0 and 1, to be 1/2 at 1/2 and piecewise linear. Then

choose g to be the function 1 up to 1/2 and -1 from 1/2 to 1, as in the pictures (YETTO DRAW).

Then choose continuous approximations gn converging to g in L2, for example

gn(x) =

1 0 ≤ x ≤ 1/2− 1/n

linear 1/2− 1/n ≤ x ≤ 1/2 + 1/n

−1 1/2 + 1/n ≤ x ≤ 1

and so gn are indeed continuous and gn → g in L2. Set fn(x) =∫ x0 gn(t)dt and so fn ∈ C1

and f ′n = gn by construction. Note that f 6∈ C1. It is easy to see that fn → f in L2

because the map which takes u to the function x 7→∫ x0 u is a bounded linear operator on

L2. Thus since f ′n → g in L2 we have fn →∫g in L2 and

∫g = f .

We now extend the domain so that the function is closed.

Theorem 4.8 The map f 7→ f ′ is closed on the domain

f ∈ H1[0, 1] : f(0) = f(1)

which is called periodic H1

Note that this is the natural space for f ′.Proof Suppose fn is in periodic H1 and fn → f in L2 and f ′n → g in L2. Let G(x) =∫ x0 g(t)dt and note that G ∈ H1 and G′ = g. We now want to show that G is almost f .

The map that takes u to x 7→∫ x0 u(t)dt is bounded on L2 and so∫ x

0f ′n(t)dt→

∫ x

0g(t)dt = G(x)

22 of 29


in L2 and also

fn(x) = fn(0) +

∫ x

0f ′n(t)dt

and these converge to f . Thus fn(0) → f − G in L2 and so f − G is constant. We canthus write

f(x) = c+G(x)

and so f ∈ H1 and f ′ = G′ = g. Also note that 0 = fn(1) − fn(0) =∫ 10 f′n for all n and

since f ′n → g in L2 then∫ 10 g = 0 and so f(1) − f(0) =

∫ 10 f′ =

∫ 10 g = 0 and so f is in

periodic H1 Q.E.D.

For if ′, we have ∫ 1

0if ′g = if g|10 −

∫ 1

0if g′ =

∫ 1

0f ¯ig′

We now show that the integral as an operator is bounded. We are showing

u 7→(x 7→

∫ x

0u(t)dt

)is bounded on L2, i.e. ∫ 1

0

(∫ x

0u(t)dt

)2

dx ≤ K∫ 1

0u(t)2dt

Now ∫ 1

0

(∫ x

0u(t)dt

)2

dx =

∫ 1

0

(∫ 1

0u(t)χ[0,x]dt

)2

dx

≤∫ 1

0

(∫ 1

0u2)(∫ 1

0χ[0,x]

)dx

=

∫ 1

0u2∫ 1

0xdx

=1

2

∫ 1

0u2

Theorem 4.9 If T is densely defined in H then

1. T ? is closed

2. If D(T ?) is dense then T has a closed extension, T ??.

3. If T is symmetric then it has closed extension T ?.

In particular, if T is self adjoint then it is closed.Proof

1. Suppose xn → x in H and T ?xn → y and xn ∈ D(T ?). The map u 7→ 〈Tu, xn〉 is abounded linear functional for each n. Then

〈Tu, xn〉 → 〈Tu, x〉

since xn → x and also〈Tu, xn〉 = 〈u, T ?xn〉 → 〈u, y〉

since T ?xn → y. Then 〈Tu, x〉 = 〈u, y〉 and because the map u 7→ 〈u, y〉 is a boundedlinear functional we conclude that x ∈ D(T ?) and T ?x = y.

23 of 29


2. Since D(T ?) is dense, T ?? does exist. By 1 it is closed, but does it extend T .

Suppose x ∈ D(T ). The map u 7→ 〈T ?u, x〉 = 〈u, Tx〉 is defined on D(T ?) and is abounded linear functional since Tx ∈ H. So x ∈ D(T ??) and 〈u, Tx〉 = 〈T ?u, x〉 =〈u, T ??x〉 for all u ∈ D(T ?). Since this is dense, T ??x = Tx.

3. This is immediate from 1, since T symmetric means T ? extends T , and T ? is closed.

Q.E.D.

4.2 The Spectrum

Definition 4.10 For bounded operator T : H → H on a Hilbert space, we define

1. The resolvent consists of λ ∈ C for which T − λI is invertible.

2. The point spectrum consists of λ ∈ C for which T − λI is not injective

3. The continuous spectrum consists of λ ∈ C for which T − λI is one to one, isnot onto but for which Im(T − λI) is dense.

4. The residual spectrum consists of λ ∈ C with T −λI being one to one but Im(T −λI) is not dense.

Definition 4.11 For an operator T : H → H (not necessarily bounded) on a Hilbertspace, we define

1. The resolvent consists of λ ∈ C for which T − λI is one to one and Im(T − λI) isdense in H and (T − λI)−1 is bounded on the image of D(T ).

2. The point spectrum consists of λ ∈ C for which T −λI is not one to one on D(T ).

3. The continuous spectrum consists of λ ∈ C for which T − λI is one to one,Im(T − λI) is dense but (T − λI)−1 is not bounded on Im(T − λI).

4. The residual spectrum consists of λ ∈ C with T −λI is one to one but Im(T −λI)is not dense.

Theorem 4.12 If T is a self adjoint operator in a Hilbert space then its residual spectrumis empty.

Proof Suppose Im(T − λI) is not dense. Lets use S to denote T − λI. There is anon-zero y ∈ H with 〈Sx, y〉 = 0 for all x ∈ D(T ) = D(S). So the map x 7→ 〈Sx, y〉 is abounded linear functional, and hence y ∈ D(S?) = D(T ?) = D(T ) as T is self adjoint. So0 = 〈x, Sy〉 for all x ∈ D(T ) and so Sy = 0. Thus S = T − λI is not one to one. Q.E.D.

Note that if T is bounded then the two definitions agree. It is enough to check theresolvent since 2 and 4 are identical. We need to check that if S is bounded and denselydefined then its extension to H is invertible if and only if S is one to one, Im(S) is denseand S−1 is bounded on Im(S).

If the extension is invertible then clearly S is one to one and S−1 is bounded on Im(S).Initially S is defined on a dense subspace D(S). If y ∈ H the extension S is onto so thereis an x with Sx = y. Choose xn ∈ D(S) with xn → x. Then Sxn → Sx = y because S isbounded, but Sxn = Sxn so y can be approximated from Im(S) so Im(S) is dense.

In the other direction, S−1 has a continuous extension, call it T . We shall show thatS T and T S are the identity so S is invertible. Suppose y ∈ H and choose yn ∈ Im(S)with yn → y. Then S−1yn → Ty so S(S−1yn) = yn → S(Ty) so y = S(Ty) so 4S T isthe identity. Similarly T S is the identity.

24 of 29


5 The Laplacian and Quantum Harmonic Oscillator.

We have already seen that the map f 7→ f ′′ is self-adjoint on Dirichlet H2[0, 1]. Thus ithas no residual spectrum. We also observed that there is a complete orthonormal basisof L2[0, 1] consisting of the eigenfunctions x 7→ sinnπx with eigenvalue −n2π2.

We would like an analogue of the spectral theorem

1. The spectrum consists only of the eigenvalues (the operator has purely point spec-trum).

2. The corresponding eigenvectors form a complete orthonormal basis.

3. The map f 7→ f ′′ on H2 can be expressed as follows. If f ′′ = g =∑θn sinnπx then

f(x) =∑ θn−n2π2 sinn pix.

In other words the operator multiplies the coefficient by the eigenvalue −n2π2 orfor each f ∈ Dirichlet H2, f(x) =

∑γn sinnπx where

∑n2π2γ2n < ∞ and f ′′(x) =

−∑n2π2γn sinnπx.

This is also true for the Laplacian on a nice domain Ω ⊂ R2 but we shall only proveit in one dimension, by directly writing down the resolvent.

We need that if λ 6= −n2π2 then (T − λI)−1 is bounded on Im(T ), i.e. everythingthat is one to one is in the resolvent, and so not in the spectrum. We know that iff(x) = sinnπx then f ′′(x) − λf(x) = (−n2π2 − λ) sinnπx so the resolvent should be∑θn sinnπx→

∑ θn−n2π2−λ sinπnx

This is clearly a bounded operator since we divide the sequence of coefficients with re-spect to an orthonormal basis by non-zero numbers which tend to infinity. The operator iseven compact. We need to check that if f ∈ Dirichlet H2 and f ′′−λf = g =

∑θn sinnπx

then f =∑ θn−n2π2−λ sinnπx, i.e. if g ∈ Im(T−λI) then we have (T−λI)−1g well defined.

We shall use “boot strapping”. We have

f ′′ = λf + g (5.1)

and g ∈ L2 is nice and f ∈ H2 is very nice. The fact that f satisfies a differential equationmeans that smoothness of f and g passes to f ′′ and so that f is much smoother.

To do this we start by observing that f ∈ L2 so we may write

f(x) =∑

γn sinnπx

Our aim is to prove that γn = θn−n2π2−λ for each n.

Note that we can’t state that f ′′(x) =∑−n2π2γn sinnπx because this might not even

make sense.By the differential equation (5.1) above we have f ′′(x) =

∑(λγn + θn) sinnπx. Inte-

gration is a bounded operator on L2 so we can integrate term by term, so as to get

f(x) = cx+ d+∑ (λγn + θn)

−n2π2sinnπx

Then if c = d = 0 we have∑ (λγn + θn)

−n2π2sinnπx =

∑γn sinπnx

in L2 and so we get −γnn2π2 = λγn + θn as required.

25 of 29


To eliminate the linear term we use the boundary conditions f(0) = 0 = f(1). Onthe face of it this sum at 0 and 1 vanishes since we chose sinnπx to do this, but an L2

convergent series might not even have a meaning at a particular point. We need somethingstronger. However,

∑(λγn + θn) <∞ because f, g ∈ L2 and so∑∣∣∣∣λγn + θn−n2π2

∣∣∣∣ ≤ (∑(λγn + θn)2)(∑ 1

n4π4

)<∞

so the series∑ (λγn+θn)

−n2π2 sinnπx converges uniformly.

Since f ∈ H2, not just L2 it should come as no surprise that its Fourier series convergesmuch better than in L2. Thus the sum does vanish at 0 and 1 and f(0) = d and f(1) = c+dso c = d = 0.

We have that (T − λI)−1 is bounded on Im(T − λI) if λ 6= −n2π2. Setting λ = 0 weget that the spectral representation for T−1 and hence T .

We now consider higher dimensions. The laplacian is

∆f(x, y) =∂2f

∂x2+∂2f

∂y2

and we consider the Dirichlet Laplacian. The eigenfunctions are (x, y) 7→ sinmπx sinnπyand the eigenvalue is −(n2 +m2)π2. In k dimensions we replace

∑ 1n4π4 with∑ 1

(n21 + ...+ n2k)2π4≈∫Rk

1

|x|4dx =∞

If the dimension k is large enough we don’t get the boundary conditions automatically.There are no tricks, on Ω a random domain we dont know the eigenfunctions. You needa big theorem.

On R the map f 7→ f ′′ has many more eigenvalues eitx 7→ −t2eitx so −t2 is aneigenvalue. On [0, 1] we have discrete eigenvalues, but on R we have the whole of (−∞, 0]in the spectrum.

A Schrodinger operator is a map L : f 7→ −f ′′+V (x)f where V is a potential. To solvethe Schrodinger equation in the same way as the vibrating string we need the spectral

properties of L. The Dirichlet Laplacian corresponds to V =

0 x ∈ [0, 1]

∞ x ∈ (−∞, 0) ∪ (1,∞).

The particle is bound in [0, 1] and this forces the discrete energy levels.

5.1 Quantum Harmonic Oscillator

The simplest example of a Schrodinger operator on R is

Tf(x) = −f ′′(x) +x2

4f(x)

where we have taken our potential V to be x2/4. This potential tends to∞ fairly rapidly.Particles can wander over all of R but typically they are at distance about 1 from 0. Thestates are bound. If we put f(x) = p(x)e−x

2/4 then

−f ′′(x) +x2

4f(x)− λf(x) = e−x

2/4(p′′(x) + xp′(x)− (λ− 1

2)p(x))

and we claim that for n = 0, 1, 2, ... the map p 7→ −p′′(x) + xp′(x) has eigenvalue n witheigenfunction a polynomial of degree n. This gives rise to the eigenvalues 1

2 ,32 ,

52 , ... for

the original T .

26 of 29


The map L which is p 7→ −p′′(x) + xp′(x) acts on polynomials of degree n as follows:

L(1) = 0, L(x) = x, L(x2) = −2 + 2x, L(x3) = −6x+ 3x3

and can be represented by 0 0 −2 0 . . .0 1 0 −6 . . .0 0 2 0 . . .0 0 0 3 . . .

and this matrix is upper triangular and so its eigenvalues are 0, 1, 2, ... and the eigenfunc-tion corresponding to n is a polynomial of degree at most n. These are related to theHermite polynomials. Let

F (x, t) = ext−t2/2 =

∞∑0

pn(x)

n!tn

We can show that pn is a polynomial of degree n in x. This is left as an exercise. Wewant that −p′′n(x) + xp′(x) = npn(x) or equivalently∑ (−p′′n(x) + xp′(x))

n!=∑ npn(x)

n!

or alternatively

−∂2F

∂x2+ x

∂F

∂x= t

∂F

∂t

which is true upon differentiating F . We call such an F a generating function.T is supposed to be self adjoint so we would like the eigenfunctions pn(x)e−x

2/4 to beorthogonal. We would like to know that if m 6= n then∫ ∞

−∞pn(x)pm(x)

e−x2/2

√2π

dx = 0

∫ ∞−∞

p2n(x)e−x

2/2

√2π

dx = ||pn||2

or in other formulation∫ ∞−∞

∑m≥0

pm(x)

m!sm∑n≥0

pn(x)

n!tne−x

2/2

√2π

dx =

∫ ∞−∞

∑n≥0

pn(x)

n!(st)n

e−x2/2

√2π

dx = W (st)

Lets see if this holds:∫ ∞−∞

F (x, s)F (x, t)e−x

2/2

√2π

dx =

∫ ∞−∞

exs−s2/2ext−t

2/2 e−x2/2

√2π

dx

=

∫ ∞−∞

e12(x−s−t)2est

dx√2π

= est∫ ∞−∞

e−12(x−s−t)2 dx√

2π

= est =∑ (st)n

n!

and so the pn are orthogonal and ||pn||2 = n!. In order to show that T has pure pointspectrum 1

2 ,32 ,

52 , ... the main thing we need is that the pn(x)e−x

2/4 form a completeorthonormal basis in L2(R). We want that the pn form a complete orthonormal basis in

L2(R, e−x2/2√2π

).

27 of 29


We know that polynomials are dense in L2[a, b] because they are dense in C[a, b] with

the much stronger uniform norm. On R with weight e−x2/2

√2π

we can approximate in L2 by

continuous functions with bounded support. We can approximate these on the boundedsupport by polynomials but the polynomials will explode elsewhere and not approximateon the line.

The usual proof of the density in L2(R, e−x2/2√2π

) uses the invertibility of the Fourier

transform. We shall use a different approach.

If q ∈ L2(R, e−x2/2√2π

) then its coefficients with respect to the pn are

an =

∫ ∞−∞

q(x)pn(x)

n!

e−x2/4

√2π

dx = 〈q, pnn!〉

and so the Nth partial sum of the expansion is

y 7→N∑0

anpn(y)√n!

=

∫ ∞−∞

q(x)

N∑0

pn(x)pn(y)

n!

e−x2/2

√2π︸︷︷︸

KN (y,x)

dx

and so KN is like the Dirichlet kernel.It gives the best L2 approximations, but potentially bad with uniform approximations.

We shall use a better kernel built with a smooth cut of. The kernel given above cuts offsharply at N , and so we take

Kt(y, x) =∞∑0

pn(y)pn(x)

n!tne−x

2/4

√2π

for 0 ≤ t < 1 and we will let t→ 1. We try to rewrite this kernel in a better manner. Weknow that

F (y, tu) =∞∑0

pn(y)

n!tnun

but instead of u we have pn(x) and so the idea is to make pn like u. We do this as follows:

Lemma 5.1

inpn(x)e−x2/2 =

∫ ∞−∞

uneixue−u2/2 du√

2π

Proof We first multiply by tn/n! and sum. On the left we get

∞∑0

inpn(x)e−x2/2 t

n

n!= e−x

2/2F (x, it) = e−x2/2exit+t

2/2 = e−12(x−it)2

On the right we get∫ ∞−∞

∞∑0

tn

n!uneixu

e−u2/2

√2π

du =

∫ ∞−∞

etu+ixu−u2/2 du√

2π

=

∫ ∞−∞

e−12(u−ix−t)2e

12(ix+t)2 du√

2π

= e12(ix+t)2

∫ ∞−∞

e−12(u−ix−t)2 du√

2π

28 of 29


The integral is a contour integral along v 7→ v − ix of e−12 (z−t)2√2π

. By Cauchy’s integral

theorem, the integral around a rectangle is 0. We thus need the integral on the verticalsides to tend to zero. This does as we are integrating e−z

2. Q.E.D.

We thus have

Kt(y, x) =∞∑0

pn(y)

n!tn∫ ∞−∞

(−iu)neixue−u2/2 du√

2π

=1

2π

∫ ∞−∞

F (y,−itu)eixue−u2/2du

=1

2π

∫ ∞−∞

e−iytu+t2u2/2eixu−u

2/2du

=1

2π

∫ ∞−∞

e− 1

2(1−t2)

(u− ix−iyt

1−t2

)2

e− 1

2(x−yt)2

1−t2 du

=1√2πe− 1

2(x−yt)2

1−t2

∫ ∞−∞

e−12

(u−fish)2

sharkdu√2π

=1√2πe− 1

2(x−yt)2

1−t21√

1− t2

This kernel as a function of v is the density of a Gaussian with mean yt and variance1− t2 and so ∫

q(x)Kt(y, x)dx ≈ q(yt) ≈ q(y)

29 of 29

ma3g8 functional analysis ii - warwick

Documents