harmonic functions and brownian motion
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Harmonic functions and Brownian Motion
Oleg Ivrii
October 19, 2020
Oleg Ivrii Harmonic functions and Brownian Motion
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What is potential theory?
Potential theory lies in the intersection of
complex analysis,
probability,
partial differential equations.
The main objects we will study are
harmonic and subharmonic functions,
Brownian motion,
Sobolev spaces.
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Harmonic functions
Let Ω ⊂ Rn. A function u : Ω→ R is harmonic if it continuous
and satisfies the mean-value property (MVP)
u(x) =
∂B(x ,r)
u(y)dS(y),
provided B(x , r) ⊂ Ω.
Alternatively, a function is harmonic if it is C 2 and satisfies
∆u =∂2u
∂x21
+∂2u
∂x22
+ · · ·+ ∂2u
∂x2n
= 0.
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Harmonic functions
Remark. There are 6 different versions of the MVP. One can
require the MVP on
1 Spheres,
2 Small spheres,
3 A sequence of arbitrarily small spheres,
4 Balls,
5 Small balls,
6 A sequence of arbitrarily small balls.
In fact, all 6 versions of the MVP are equivalent.
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Harmonic functions
Lemma. The first definition requires u to be continuous, while the
second definition requires u to be C 2. In fact, any harmonic
function is C∞.
Proof. Given ε > 0, construct a bump function
φ ∈ C∞, φ(x) ≥ 0, φ(x) only depends on |x |,
suppφ ⊂ B(0, ε),
ˆRn
φ(x) |dx |n = 1.
If dist(x , ∂Ω) > ε,
u(x) =
ˆRn
u(y)φ(x − y) |dy |n ∈ C∞.
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Harmonic functions
Lemma. The first definition requires u to be continuous, while the
second definition requires u to be C 2. In fact, any harmonic
function is C∞.
Proof. Given ε > 0, construct a bump function
φ ∈ C∞, φ(x) ≥ 0, φ(x) only depends on |x |,
suppφ ⊂ B(0, ε),
ˆRn
φ(x) |dx |n = 1.
If dist(x , ∂Ω) > ε,
u(x) =
ˆRn
u(y)φ(x − y) |dy |n ∈ C∞.
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Harmonic functions
Lemma. The first definition requires u to be continuous, while the
second definition requires u to be C 2. In fact, any harmonic
function is C∞.
Proof. Given ε > 0, construct a bump function
φ ∈ C∞, φ(x) ≥ 0, φ(x) only depends on |x |,
suppφ ⊂ B(0, ε),
ˆRn
φ(x) |dx |n = 1.
If dist(x , ∂Ω) > ε,
u(x) =
ˆRn
u(y)φ(x − y) |dy |n ∈ C∞.
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Harmonic functions
Lemma. For C 2 functions, MVP ⇐⇒ ∆u = 0.
Green’s formula. If u ∈ C 2(B(x , r)
), then
∂B(x ,r)
u(y)dS(y)− u(x) =1
2π
ˆB(x ,r)
∆u(y) logr
|y − x ||dy |2
(A similar formula holds in higher dimensions.)
Exercise. ∆u(y) = limr→0(4/r2) ·ffl
∂B(x ,r) u(y)dS(y)− u(x)
.
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Harmonic functions
Lemma. For C 2 functions, MVP ⇐⇒ ∆u = 0.
Green’s formula. If u ∈ C 2(B(x , r)
), then
∂B(x ,r)
u(y)dS(y)− u(x) =1
2π
ˆB(x ,r)
∆u(y) logr
|y − x ||dy |2
(A similar formula holds in higher dimensions.)
Exercise. ∆u(y) = limr→0(4/r2) ·ffl
∂B(x ,r) u(y)dS(y)− u(x)
.
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Harmonic functions
Lemma. For C 2 functions, MVP ⇐⇒ ∆u = 0.
Green’s formula. If u ∈ C 2(B(x , r)
), then
∂B(x ,r)
u(y)dS(y)− u(x) =1
2π
ˆB(x ,r)
∆u(y) logr
|y − x ||dy |2
(A similar formula holds in higher dimensions.)
Exercise. ∆u(y) = limr→0(4/r2) ·ffl
∂B(x ,r) u(y)dS(y)− u(x)
.
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Subharmonic functions
The same argument shows that for C 2 functions, the sub-mean
value property ∂B(x ,r)
u(y)dS(y) ≥ u(x)
is equivalent to ∆u ≥ 0.
Warning! Typically when discussing subharmonic functions, one
asks that they are merely upper semi-continuous. Thus, the
equation ∆u ≥ 0 needs to be understood weakly in the sense of
distributions.
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Subharmonic functions
The same argument shows that for C 2 functions, the sub-mean
value property ∂B(x ,r)
u(y)dS(y) ≥ u(x)
is equivalent to ∆u ≥ 0.
Warning! Typically when discussing subharmonic functions, one
asks that they are merely upper semi-continuous. Thus, the
equation ∆u ≥ 0 needs to be understood weakly in the sense of
distributions.
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Maximum modulus principle
Maximum modulus principle. Suppose u : Ω→ R is a harmonic
function. If it achieves its maximum value in the interior, then u is
constant.
Dirichlet’s problem. Suppose Ω is a bounded domain. Given a
continuous function f : ∂Ω→ R, find a harmonic function
u : Ω→ R which continuously to ∂Ω and agrees with f .
The maximum-modulus principle shows that if the solution exists,
then it is unique.
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Maximum modulus principle
Maximum modulus principle. Suppose u : Ω→ R is a harmonic
function. If it achieves its maximum value in the interior, then u is
constant.
Dirichlet’s problem. Suppose Ω is a bounded domain. Given a
continuous function f : ∂Ω→ R, find a harmonic function
u : Ω→ R which continuously to ∂Ω and agrees with f .
The maximum-modulus principle shows that if the solution exists,
then it is unique.
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Maximum modulus principle
Maximum modulus principle. Suppose u : Ω→ R is a harmonic
function. If it achieves its maximum value in the interior, then u is
constant.
Dirichlet’s problem. Suppose Ω is a bounded domain. Given a
continuous function f : ∂Ω→ R, find a harmonic function
u : Ω→ R which continuously to ∂Ω and agrees with f .
The maximum-modulus principle shows that if the solution exists,
then it is unique.
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Poisson Integral Formula
Poisson Integral Formula. If u : B(0, 1)→ R is harmonic and
continuous up to the boundary, then
u(x) =
∂B(0,1)
u(ζ)1− |x |2
|x − ζ|ndS(ζ).
Dirichlet’s problem on the ball. Since the Poisson kernel is
harmonic, for any cts. function f : ∂B(0, 1)→ R,
u(x) := P[f ](x) =
∂B(0,1)
f (ζ)1− |x |2
|x − ζ|ndS(ζ)
defines a harmonic function.
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Poisson Integral Formula
Poisson Integral Formula. If u : B(0, 1)→ R is harmonic and
continuous up to the boundary, then
u(x) =
∂B(0,1)
u(ζ)1− |x |2
|x − ζ|ndS(ζ).
Dirichlet’s problem on the ball. Since the Poisson kernel is
harmonic, for any cts. function f : ∂B(0, 1)→ R,
u(x) := P[f ](x) =
∂B(0,1)
f (ζ)1− |x |2
|x − ζ|ndS(ζ)
defines a harmonic function.
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Properties of the Poisson kernel
To see that P[f ] extends f , notice that the Poisson kernel
Px(ζ) =1− |x |2
|x − ζ|n
satisfies the following properties:
Px(ζ) > 0,ffl∂B(0,1) Px(rζ)dS(ζ) = 1 for 0 < r < 1,
∀ε > 0 ∃δ > 0 s.t.
Px(ζ) < ε,
provided 1− |x | < δ and |x − ζ| > δ.
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Harnack’s inequality
Suppose u is a positive harmonic function defined on a
neighbourhood of B(0, 1). Since
C1 <1− |x |2
|x − ζ|n< C2, x ∈ B(0, 1), ζ ∈ ∂B(0, 1),
there exists universal constants c,C > 0 such that
c <u(x1)
u(x2)< C , x1, x2 ∈ B(0, 1/2).
If u is harmonic on a domain Ω and K is a compact subset, then
c(K ,Ω) <u(x1)
u(x2)< C (K ,Ω), x1, x2 ∈ K .
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Harnack’s inequality
Suppose u is a positive harmonic function defined on a
neighbourhood of B(0, 1). Since
C1 <1− |x |2
|x − ζ|n< C2, x ∈ B(0, 1), ζ ∈ ∂B(0, 1),
there exists universal constants c,C > 0 such that
c <u(x1)
u(x2)< C , x1, x2 ∈ B(0, 1/2).
If u is harmonic on a domain Ω and K is a compact subset, then
c(K ,Ω) <u(x1)
u(x2)< C (K ,Ω), x1, x2 ∈ K .
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Harnack’s inequality
Suppose u is a positive harmonic function defined on a
neighbourhood of B(0, 1). Since
C1 <1− |x |2
|x − ζ|n< C2, x ∈ B(0, 1), ζ ∈ ∂B(0, 1),
there exists universal constants c,C > 0 such that
c <u(x1)
u(x2)< C , x1, x2 ∈ B(0, 1/2).
If u is harmonic on a domain Ω and K is a compact subset, then
c(K ,Ω) <u(x1)
u(x2)< C (K ,Ω), x1, x2 ∈ K .
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Harnack’s Theorem
Theorem. Suppose
u1 ≤ u2 ≤ · · · ≤ un ≤ . . .
is an increasing sequence of positive harmonic functions defined on
a domain Ω. Either they converge uniformly on compact subsets of
Ω or they converge to +∞.
The proof is left as an exercise.
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Green’s identities
Suppose Ω ⊂ Rn is a bounded domain with C 1 boundary. Let η be
the outward pointing normal. If u ∈ C 1(Ω), then
ˆΩuxidV =
ˆ∂Ω
uηidS , i = 1, 2, . . . n.
To see this, just apply Stokes theorem
ˆΩdω =
ˆ∂Ωω
to the differential form
ω = u(x) dx1 dx2 . . . dxi . . . dxn.
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Green’s identities
Suppose Ω ⊂ Rn is a bounded domain with C 1 boundary. Let η be
the outward pointing normal. If u ∈ C 1(Ω), then
ˆΩuxidV =
ˆ∂Ω
uηidS , i = 1, 2, . . . n.
To see this, just apply Stokes theorem
ˆΩdω =
ˆ∂Ωω
to the differential form
ω = u(x) dx1 dx2 . . . dxi . . . dxn.
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Green’s identities
Substituting uv instead of u into
ˆΩuxidV =
ˆ∂Ω
uηidS ,
we get the integration-by-parts formula:
ˆΩ
(uxi v + vxiu)dV =
ˆ∂Ω
uvηidS .
If u ∈ C 2(Ω), we can plug in uxi instead of u:
ˆΩ
(uxixi v + vxiuxi )dV =
ˆ∂Ω
uxi vηidS .
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Green’s identities
Substituting uv instead of u into
ˆΩuxidV =
ˆ∂Ω
uηidS ,
we get the integration-by-parts formula:
ˆΩ
(uxi v + vxiu)dV =
ˆ∂Ω
uvηidS .
If u ∈ C 2(Ω), we can plug in uxi instead of u:
ˆΩ
(uxixi v + vxiuxi )dV =
ˆ∂Ω
uxi vηidS .
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Green’s identities
Summing over i = 1, 2, . . . , n, we get
ˆΩ
(∆u · v +∇u · ∇v)dV =
ˆ∂Ω
Dηu · v dS .
Switching the roles of u and v , we obtain
ˆΩ
(∆v · u +∇u · ∇v)dV =
ˆ∂Ω
Dηv · u dS .
Subtracting, we arrive at:
ˆΩ
(∆u · v −∆v · u)dV =
ˆ∂Ω
(Dηu · v − Dηv · u) dS .
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Green’s identities
Summing over i = 1, 2, . . . , n, we get
ˆΩ
(∆u · v +∇u · ∇v)dV =
ˆ∂Ω
Dηu · v dS .
Switching the roles of u and v , we obtain
ˆΩ
(∆v · u +∇u · ∇v)dV =
ˆ∂Ω
Dηv · u dS .
Subtracting, we arrive at:
ˆΩ
(∆u · v −∆v · u)dV =
ˆ∂Ω
(Dηu · v − Dηv · u) dS .
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Green’s identities
Summing over i = 1, 2, . . . , n, we get
ˆΩ
(∆u · v +∇u · ∇v)dV =
ˆ∂Ω
Dηu · v dS .
Switching the roles of u and v , we obtain
ˆΩ
(∆v · u +∇u · ∇v)dV =
ˆ∂Ω
Dηv · u dS .
Subtracting, we arrive at:
ˆΩ
(∆u · v −∆v · u)dV =
ˆ∂Ω
(Dηu · v − Dηv · u) dS .
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Green’s formula
We now apply Green’s identity in n = 2, Ω = D \ B(0, ε),
v(z) = log 1|z| and take ε→ 0+:
ˆε<|z|<1
(∆u · v −∆v · u)|dz |2 =
ˆε<|z|<1
∆u · log1
|z ||dz |2. (1)
As ε→ 0+, this tends toˆD
∆u · log1
|z ||dz |2.
ˆ∂D
(Dηu · v − Dηv · u) |dz | =
ˆ∂D
u(z) |dz |. (2)
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Green’s formula
We now apply Green’s identity in n = 2, Ω = D \ B(0, ε),
v(z) = log 1|z| and take ε→ 0+:
ˆε<|z|<1
(∆u · v −∆v · u)|dz |2 =
ˆε<|z|<1
∆u · log1
|z ||dz |2. (1)
As ε→ 0+, this tends toˆD
∆u · log1
|z ||dz |2.
ˆ∂D
(Dηu · v − Dηv · u) |dz | =
ˆ∂D
u(z) |dz |. (2)
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Green’s formula
Finally,
ˆ∂B(0,ε)
(Dηu · v − Dηv · u) |dz | = o(1)− 1
ε
ˆ∂B(0,ε)
u(z) |dz | (3)
tends to −2π u(0) as ε→ 0+.
Putting everything together, we arrive at:
ˆD
∆u · log1
|z ||dz |2 =
ˆ∂D
u(z) |dz | − 2π u(0).
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Green’s formula
Finally,
ˆ∂B(0,ε)
(Dηu · v − Dηv · u) |dz | = o(1)− 1
ε
ˆ∂B(0,ε)
u(z) |dz | (3)
tends to −2π u(0) as ε→ 0+.
Putting everything together, we arrive at:
ˆD
∆u · log1
|z ||dz |2 =
ˆ∂D
u(z) |dz | − 2π u(0).
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∆ log |z | = 2πδ0
If Ω is an arbitrary domain in the plane with C 1 boundary, the
above reasoning shows:
ˆΩ
∆u log1
|z |dV =
ˆ∂Ω
Dηu · log
1
|z |−Dη log
1
|z |·udS−2π u(0).
If u = φ ∈ C∞c (C), then
ˆC
∆u log1
|z |dV = −2π u(0).
One summarizes this as ∆ log |z | = 2πδ0 in the sense of
distributions.
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∆ log |z | = 2πδ0
If Ω is an arbitrary domain in the plane with C 1 boundary, the
above reasoning shows:
ˆΩ
∆u log1
|z |dV =
ˆ∂Ω
Dηu · log
1
|z |−Dη log
1
|z |·udS−2π u(0).
If u = φ ∈ C∞c (C), then
ˆC
∆u log1
|z |dV = −2π u(0).
One summarizes this as ∆ log |z | = 2πδ0 in the sense of
distributions.
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∆ log |z | = 2πδ0
If Ω is an arbitrary domain in the plane with C 1 boundary, the
above reasoning shows:
ˆΩ
∆u log1
|z |dV =
ˆ∂Ω
Dηu · log
1
|z |−Dη log
1
|z |·udS−2π u(0).
If u = φ ∈ C∞c (C), then
ˆC
∆u log1
|z |dV = −2π u(0).
One summarizes this as ∆ log |z | = 2πδ0 in the sense of
distributions.
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Green’s formula via Brownian Motion
Run Brownian motion starting from the origin, stopped at time τ
when it hits the unit circle.
The Green’s function has the probabilistic interpretation as the
occupation density of Brownian motion.
That is, for any measurable set E ⊂ D,
E0(time BM spends in E ) =
ˆE
1
πlog
1
|z ||dz |2.
Remark.´D
1π log 1
|z| |dz |2 = 1
2 .
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Green’s formula via Brownian Motion
If f were harmonic then f (Bt) would be a martingale with respect
to the Brownian filtration, so that
1
2π
ˆf (ζ)|dζ| − f (0) = Ef (Bτ )− f (0) = 0.
Of course, if f is not harmonic, then f (Bt) isn’t martingale.
The correct formula is:
1
2π
ˆf (ζ)|dζ| − f (0) = Ef (Bτ )− f (0) = E
ˆ t
0
1
2∆f (Bt)dt
=1
2
ˆD
∆f · 1
πlog
1
|z ||dz |2 =
1
2π
ˆD
∆f · log1
|z ||dz |2.
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Greens functions for C 1 domains
In the analytic proof of Green’s formula, we used that the Green’s
function with singularity at z0 satisfies the following properties:
1 G (z) is zero on ∂Ω.
2 G (z) on Ω \ z0 is harmonic.
3 G (z)− log 1|z−z0| is harmonic in a neighbourhood of z0.
By the maximum-modulus principle, if such a function exists, then
it is unique.
Conversely, let u(z) be the solution of Dirichlet’s problem with
boundary values log |z − z0|. Then, G (z) = u(z) + log 1|z−z0|
satisfies the above properties.
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Greens functions for C 1 domains
In the analytic proof of Green’s formula, we used that the Green’s
function with singularity at z0 satisfies the following properties:
1 G (z) is zero on ∂Ω.
2 G (z) on Ω \ z0 is harmonic.
3 G (z)− log 1|z−z0| is harmonic in a neighbourhood of z0.
By the maximum-modulus principle, if such a function exists, then
it is unique.
Conversely, let u(z) be the solution of Dirichlet’s problem with
boundary values log |z − z0|. Then, G (z) = u(z) + log 1|z−z0|
satisfies the above properties.
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Greens functions for C 1 domains
In the analytic proof of Green’s formula, we used that the Green’s
function with singularity at z0 satisfies the following properties:
1 G (z) is zero on ∂Ω.
2 G (z) on Ω \ z0 is harmonic.
3 G (z)− log 1|z−z0| is harmonic in a neighbourhood of z0.
By the maximum-modulus principle, if such a function exists, then
it is unique.
Conversely, let u(z) be the solution of Dirichlet’s problem with
boundary values log |z − z0|. Then, G (z) = u(z) + log 1|z−z0|
satisfies the above properties.
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Conformal invariance of Green’s function
Lemma. If f : Ω→ Ω′ is holomorphic and u : Ω′ → R is harmonic,
then u f : Ω→ R is also harmonic.
Lemma. If ϕ : Ω→ Ω′ is conformal, then
GΩ(z ,w) = GΩ′(ϕ(z), ϕ(w)
).
Examples.
GD(z ,w) = log
∣∣∣∣1− zw
z − w
∣∣∣∣, GH(z ,w) = log
∣∣∣∣w − z
w − z
∣∣∣∣ = log
∣∣∣∣z − w
z − w
∣∣∣∣.
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Conformal invariance of Green’s function
Lemma. If f : Ω→ Ω′ is holomorphic and u : Ω′ → R is harmonic,
then u f : Ω→ R is also harmonic.
Lemma. If ϕ : Ω→ Ω′ is conformal, then
GΩ(z ,w) = GΩ′(ϕ(z), ϕ(w)
).
Examples.
GD(z ,w) = log
∣∣∣∣1− zw
z − w
∣∣∣∣, GH(z ,w) = log
∣∣∣∣w − z
w − z
∣∣∣∣ = log
∣∣∣∣z − w
z − w
∣∣∣∣.
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Conformal invariance of Green’s function
Lemma. If f : Ω→ Ω′ is holomorphic and u : Ω′ → R is harmonic,
then u f : Ω→ R is also harmonic.
Lemma. If ϕ : Ω→ Ω′ is conformal, then
GΩ(z ,w) = GΩ′(ϕ(z), ϕ(w)
).
Examples.
GD(z ,w) = log
∣∣∣∣1− zw
z − w
∣∣∣∣, GH(z ,w) = log
∣∣∣∣w − z
w − z
∣∣∣∣ = log
∣∣∣∣z − w
z − w
∣∣∣∣.
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Subordination principle
Lemma. If f : Ω→ Ω′ is a holomorphic function, then
GΩ(z ,w) ≤ GΩ′(f (z), f (w)
).
Proof. For a fixed z ∈ Ω, the difference
u(w) = GΩ(z ,w)− GΩ′(f (z), f (w)
).
is a harmonic function whose boundary values are negative in
lim sup sense.
Corollary. If Ω ⊂ Ω′, then GΩ(z ,w) ≤ GΩ′(z ,w
).
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Subordination principle
Lemma. If f : Ω→ Ω′ is a holomorphic function, then
GΩ(z ,w) ≤ GΩ′(f (z), f (w)
).
Proof. For a fixed z ∈ Ω, the difference
u(w) = GΩ(z ,w)− GΩ′(f (z), f (w)
).
is a harmonic function whose boundary values are negative in
lim sup sense.
Corollary. If Ω ⊂ Ω′, then GΩ(z ,w) ≤ GΩ′(z ,w
).
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Subordination principle
Lemma. If f : Ω→ Ω′ is a holomorphic function, then
GΩ(z ,w) ≤ GΩ′(f (z), f (w)
).
Proof. For a fixed z ∈ Ω, the difference
u(w) = GΩ(z ,w)− GΩ′(f (z), f (w)
).
is a harmonic function whose boundary values are negative in
lim sup sense.
Corollary. If Ω ⊂ Ω′, then GΩ(z ,w) ≤ GΩ′(z ,w
).
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Symmetry of the Green’s function
Lemma. For any domain Ω ⊂ C with C 1 boundary,
GΩ(z ,w) = GΩ
(w , z
).
Proof. For a fixed z ∈ Ω, the difference
u(w) = GΩ(z ,w)− GΩ(w , z)
= GΩ(z ,w)−ˆ∂Ω
log1
|ζ − z |dωw (ζ) + log
1
w − z.
is harmonic and has negative boundary values in lim sup sense.
Hence, GΩ(z ,w) ≤ GΩ(w , z).
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Symmetry of the Green’s function
Lemma. For any domain Ω ⊂ C with C 1 boundary,
GΩ(z ,w) = GΩ
(w , z
).
Proof. For a fixed z ∈ Ω, the difference
u(w) = GΩ(z ,w)− GΩ(w , z)
= GΩ(z ,w)−ˆ∂Ω
log1
|ζ − z |dωw (ζ) + log
1
w − z.
is harmonic and has negative boundary values in lim sup sense.
Hence, GΩ(z ,w) ≤ GΩ(w , z).
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Symmetry of the Green’s function
Lemma. For any domain Ω ⊂ C with C 1 boundary,
GΩ(z ,w) = GΩ
(w , z
).
Proof. For a fixed z ∈ Ω, the difference
u(w) = GΩ(z ,w)− GΩ(w , z)
= GΩ(z ,w)−ˆ∂Ω
log1
|ζ − z |dωw (ζ) + log
1
w − z.
is harmonic and has negative boundary values in lim sup sense.
Hence, GΩ(z ,w) ≤ GΩ(w , z).
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Positive Harmonic Functions
For a function u on the disk, write ur (ζ) := u(rζ).
For f (ζ) ∈ C (∂D), form the Poisson extension
u(x) = P[f ](x) =1
2π
ˆ∂D
1− |x |2
|x − ζ|2f (ζ)|dζ|.
Define Pr [f ] := ur .
Lemma. For f ∈ C (∂D),
Pr [f ]→ f , as r → 1−, uniformly on ∂D.
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Positive Harmonic Functions
Theorem. There is a bijection between positive harmonic functions
and measures on the unit circle given by integration against the
Poisson kernel:
u(x) =1
2π
ˆ∂D
1− |x |2
|x − ζ|2dµ(ζ)
Examples. Lebesgue measure corresponds to the function 1, while
the Poisson kernel Pζ(x) corresponds to 2π δζ .
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Positive Harmonic Functions
Proof. Let u be a positive harmonic function on the disk with
u(0) = 1. We want to find a measure µ such that u = P[µ].
1 For 0 < r < 1, consider the probability measure
µr := ur (θ)(dθ/2π).
2 Since the space of probability measures is weak-∗ compact,
there exist rk → 1− such that µr → µ.
3 Weak-∗ convergence implies that for any x ∈ D,
P[µr ](x)→ P[µ](x), as r → 1.
4 As P[µr ] = u(rx), we see that P[µ](x) = u(x) as desired.
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Positive Harmonic Functions
Proof. Let u be a positive harmonic function on the disk with
u(0) = 1. We want to find a measure µ such that u = P[µ].
1 For 0 < r < 1, consider the probability measure
µr := ur (θ)(dθ/2π).
2 Since the space of probability measures is weak-∗ compact,
there exist rk → 1− such that µr → µ.
3 Weak-∗ convergence implies that for any x ∈ D,
P[µr ](x)→ P[µ](x), as r → 1.
4 As P[µr ] = u(rx), we see that P[µ](x) = u(x) as desired.
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Positive Harmonic Functions
Proof. Let u be a positive harmonic function on the disk with
u(0) = 1. We want to find a measure µ such that u = P[µ].
1 For 0 < r < 1, consider the probability measure
µr := ur (θ)(dθ/2π).
2 Since the space of probability measures is weak-∗ compact,
there exist rk → 1− such that µr → µ.
3 Weak-∗ convergence implies that for any x ∈ D,
P[µr ](x)→ P[µ](x), as r → 1.
4 As P[µr ] = u(rx), we see that P[µ](x) = u(x) as desired.
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Positive Harmonic Functions
Proof. Let u be a positive harmonic function on the disk with
u(0) = 1. We want to find a measure µ such that u = P[µ].
1 For 0 < r < 1, consider the probability measure
µr := ur (θ)(dθ/2π).
2 Since the space of probability measures is weak-∗ compact,
there exist rk → 1− such that µr → µ.
3 Weak-∗ convergence implies that for any x ∈ D,
P[µr ](x)→ P[µ](x), as r → 1.
4 As P[µr ] = u(rx), we see that P[µ](x) = u(x) as desired.
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Positive Harmonic Functions
Proof. Let u be a positive harmonic function on the disk with
u(0) = 1. We want to find a measure µ such that u = P[µ].
1 For 0 < r < 1, consider the probability measure
µr := ur (θ)(dθ/2π).
2 Since the space of probability measures is weak-∗ compact,
there exist rk → 1− such that µr → µ.
3 Weak-∗ convergence implies that for any x ∈ D,
P[µr ](x)→ P[µ](x), as r → 1.
4 As P[µr ] = u(rx), we see that P[µ](x) = u(x) as desired.
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Positive Harmonic Functions
We now show that the measures µr := ur (θ)(dθ/2π) converge to
µ, as r → 1.
Unwinding definitions, we want to show that
ˆf · Pr [µ] · dθ
2π→
ˆf dµ, f ∈ C (∂D).
By Fubini’s theorem,
ˆf · Pr [µ] · dθ
2π=
ˆPr [f ] dµ.
Remark. We have seen any positive harmonic function u can be
represented as P[µ] for some measure µ. The above convergence
result shows that the measure µ is unique.
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Positive Harmonic Functions
We now show that the measures µr := ur (θ)(dθ/2π) converge to
µ, as r → 1. Unwinding definitions, we want to show that
ˆf · Pr [µ] · dθ
2π→
ˆf dµ, f ∈ C (∂D).
By Fubini’s theorem,
ˆf · Pr [µ] · dθ
2π=
ˆPr [f ] dµ.
Remark. We have seen any positive harmonic function u can be
represented as P[µ] for some measure µ. The above convergence
result shows that the measure µ is unique.
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Positive Harmonic Functions
We now show that the measures µr := ur (θ)(dθ/2π) converge to
µ, as r → 1. Unwinding definitions, we want to show that
ˆf · Pr [µ] · dθ
2π→
ˆf dµ, f ∈ C (∂D).
By Fubini’s theorem,
ˆf · Pr [µ] · dθ
2π=
ˆPr [f ] dµ.
Remark. We have seen any positive harmonic function u can be
represented as P[µ] for some measure µ. The above convergence
result shows that the measure µ is unique.
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Weyl’s lemma
Lemma. A function u : Ω→ R is harmonic if and only if u ∈ L1loc
and ˆΩu∆φ = 0
for all φ ∈ C∞c (Ω).
Proof. If u was C 2, we could integrate by parts to see that
ˆΩ
∆u · φ = 0, for all φ ∈ C∞c (Ω),
which would imply that ∆u = 0.
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Weyl’s lemma
Lemma. A function u : Ω→ R is harmonic if and only if u ∈ L1loc
and ˆΩu∆φ = 0
for all φ ∈ C∞c (Ω).
Proof. If u was C 2, we could integrate by parts to see that
ˆΩ
∆u · φ = 0, for all φ ∈ C∞c (Ω),
which would imply that ∆u = 0.
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Weyl’s lemma
For the general case, form the functions uε = u ∗ ηε. The functions
uε are smooth and converge to u in L1loc.
Since ˆΩuε∆ϕ =
ˆΩu ∗ ηε ·∆ϕ =
ˆΩu · ((∆ϕ) ∗ ηε) =
=
ˆΩu ·∆(ϕ ∗ ηε) = 0,
each uε satisfies the MVP on balls.
Therefore, u itself satisfies the MVP on balls.
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Weyl’s lemma
For the general case, form the functions uε = u ∗ ηε. The functions
uε are smooth and converge to u in L1loc.
Since ˆΩuε∆ϕ =
ˆΩu ∗ ηε ·∆ϕ =
ˆΩu · ((∆ϕ) ∗ ηε) =
=
ˆΩu ·∆(ϕ ∗ ηε) = 0,
each uε satisfies the MVP on balls.
Therefore, u itself satisfies the MVP on balls.
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Weyl’s lemma
For the general case, form the functions uε = u ∗ ηε. The functions
uε are smooth and converge to u in L1loc.
Since ˆΩuε∆ϕ =
ˆΩu ∗ ηε ·∆ϕ =
ˆΩu · ((∆ϕ) ∗ ηε) =
=
ˆΩu ·∆(ϕ ∗ ηε) = 0,
each uε satisfies the MVP on balls.
Therefore, u itself satisfies the MVP on balls.
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Thank you for your attention!
Oleg Ivrii Harmonic functions and Brownian Motion