This document discusses harmonic and subharmonic functions. It begins by defining harmonic functions and providing examples to show properties like the maximum/minimum principle and mean-value property. It then introduces the Dirichlet problem for the disk and proves there is always a solution. Harnack's inequality relating values of positive harmonic functions is presented. Finally, it defines subharmonic functions and provides examples to show properties like the submean-value property.

1. Subharmonic Functions and the Solution to the Dirichlet
Problem
Author: Bryan R. Clark
Advisor: Dr. Edoh Amiran
Western Washington University
brclark9@gmail.com
Thursday 26th September, 2013
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2. Overview
1 Harmonic Functions
Properties of Harmonic Functions
Dirichlet Problem for the Disk
Harnack’s Principle
2 Subharmonic Functions
Properties
Barrier Function
Perr´on Method
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3. Harmonic Functions
Definition
A real-valued function u : U → R on an open set U ⊆ C is harmonic if it
is C2 on U and
∆u ≡ 0
there where the Laplacian ∆u is defined by
∆u =
∂2
∂x2
+
∂2
∂y2
u.
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4. Harmonic Functions
Example
Is u(x, y) = e3x cos(3y) harmonic?
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5. Harmonic Functions
Example
Is u(x, y) = e3x cos(3y) harmonic?
∂u
∂x
= 3e3x cos(3y)
∂2u
∂x2
= 9e3x cos(3y)
∂u
∂y
= −3e3x sin(3y)
∂2u
∂y2
= −9e3x cos(3y)
Therefore ∆u = ∂2
∂x2 + ∂2
∂y2 u = 9e3x cos(3y) − 9e3x cos(3y) = 0 thus u
is harmonic.
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6. Maximum/Minimum Principle
Theorem
If u : U → R is harmonic on a connected open set U and if there is a point
P0 ∈ U with the property that u(P0) = supQ∈U u(Q) or
u(P0) = infQ∈U u(Q), then u is constant on U.
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7. Maximum/Minimum Principle
Theorem
If u : U → R is harmonic on a connected open set U and if there is a point
P0 ∈ U with the property that u(P0) = supQ∈U u(Q) or
u(P0) = infQ∈U u(Q), then u is constant on U.
An immediate consequence of this theorem is the following corollary.
Corollary
Let U ⊆ C be a bounded, connected open set and let u be a continuous,
real valued function on the closure U of U that is harmonic on U. Then
max u
U
= max u
∂U
min u
U
= min u
∂U
.
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8. Maximum/Minimum Principle
Lets prove this corollary.
Proof.
A continuous function on a closed bounded set attains its maximum. Thus
there exists a P ∈ U such that u(P) is maximal. If P ∈ U then by the
maximum principle u is constant and thus attains its maximum. If P ∈ ∂U
then the result is certainly true.
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9. Mean-value Property
Theorem
Suppose that u : U → R is a harmonic function on an open set U ⊆ C and
that D(P, r) ⊆ U for some r > 0 then
u(P) =
1
2π
2π
0
u(P + reiθ
)dθ.
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10. Mean-value Property
Theorem
Suppose that u : U → R is a harmonic function on an open set U ⊆ C and
that D(P, r) ⊆ U for some r > 0 then
u(P) =
1
2π
2π
0
u(P + reiθ
)dθ.
Q: What does this mean?
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11. Mean-value Property
A:
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12. Mean-value Property
A:
Q: What does this remind you of?
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13. Mean-value Property
A:
Q: What does this remind you of?
A: The Cauchy integral formula.
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14. Mean-value Property
Example
Let u(x, y) = x2 − y2, harmonic, and the disk D(P, 1) where P = 1 + i.
Then
u(1, 1) =
1
2π
2π
0
u(1 + cos(θ), 1 + sin(θ)) dθ
=
1
2π
2π
0
[1 + cos(θ)]2
− [1 + sin(θ)]2
dθ
=
1
2π
2π
0
2 cos(θ) − 2 sin(θ) + cos(2θ) dθ
=
1
2π
2π
0
2 cos(θ) dθ −
1
2π
2π
0
2 sin(θ) dθ +
1
2π
2π
0
cos(2θ) dθ
= 0
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15. Poisson Integral Formula
Theorem
Let u : U → R be a harmonic function on a neighborhood of D(0, 1).
Then, for any point a ∈ D(0, 1),
u(a) =
1
2π
2π
0
u(eiψ
)
1 − |a|2
|a − eiψ|2
dψ.
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16. Poisson Integral Formula
Theorem
Let u : U → R be a harmonic function on a neighborhood of D(0, 1).
Then, for any point a ∈ D(0, 1),
u(a) =
1
2π
2π
0
u(eiψ
)
1 − |a|2
|a − eiψ|2
dψ.
The Poisson integral formula says that the value u(reiψ) for some interior
point reiψ is a weighted average of the values of u on the boundary of the
circle where the weight
P(θ − ψ) =
1
2π
·
1 − r2
1 + r2 − 2r cos(θ − ψ)
> 0
and has total weight 1.
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17. Dirichlet Problem for the Disk
Definition
Let D denote the unit disk. Given a bounded, piecewise continuous
function on the boundary of the disk,
u : ∂D → R
is there a harmonic function on the disk,
u : D → R
that extends to match the original u on the boundary of the disk, the
extension begin continuous other than the original discontinuities?
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18. Dirichlet problem for the Disk
Does there always exist a solution to the Dirichlet problem on the disk?
Yes or No
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19. Solution to the Dirichlet Problem for the Disk
Theorem
Let f be a continuous function on ∂D(0, 1). Define
u(z) =



1
2π
2π
0 f (eiψ) · 1−|z|2
|z−eiψ|2 dψ if z ∈ D(0, 1)
f (z) if z ∈ ∂D(0, 1)
Then u is continuous on D(0, 1) and harmonic on D(0, 1).
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20. Solution to the Dirichlet Problem for the Disk
Proof.
It can be shown that
1 − |z|2
|z − eiθ|2
=
eiθ
eiθ − z
+
e−iθ
e−iθ − ¯z
− 1.
When z ∈ D(0, 1),
u(z) =
1
2π
2π
0
f (eiθ
)
eiθ
eiθ − z
dθ +
1
2π
2π
0
f (eiθ
)
e−iθ
e−iθ − ¯z
dθ
−
1
2π
2π
0
f (eiθ
)dθ
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21. Solution to the Dirichlet Problem for the Disk
Continued.
Using the fact that ∆ = 4 ∂2
∂z∂¯z we conclude that the first integral is
harmonic.
∂
∂¯z
1
2π
2π
0
f (eiθ
)
eiθ
eiθ − z
dθ =
1
2π
2π
0
f (eiθ
)
∂
∂¯z
eiθ
eiθ − z
dθ
= 0
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22. Solution to the Dirichlet Problem for the Disk
Continued.
Using the fact that ∆ = 4 ∂2
∂z∂¯z we conclude that the first integral is
harmonic.
∂
∂¯z
1
2π
2π
0
f (eiθ
)
eiθ
eiθ − z
dθ =
1
2π
2π
0
f (eiθ
)
∂
∂¯z
eiθ
eiθ − z
dθ
= 0
Since eiθ
eiθ−z
is holomorphic. Thus it satisfies the criteria for being
harmonic. The second and third integrals are harmonic by the same
argument. So we have a sum of harmonic functions which is harmonic.
Therefore if z ∈ D(0, 1) then u is harmonic.
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23. Solution to the Dirichlet Problem for the Disk
Continued.
Now I will convince you that the function u(z) is continuous.
When z ∈ D(0, 1) we want to show that u is continuous. The Poisson
integral formula says that the value u(reiθ) for some interior point reiθ is a
weighted average of the values of f on the boundary of the circle where
the weight P(θ − ψ) > 0 and has total weight 1. We see that if we fix θ
and take the limit limr→1− Pr (θ − ψ) we get 0 given θ = ψ. If r is very
close to 1, i.e. eiθ very close to the boundary, then Pr (θ − ψ) is close to 0
except when ψ is close to θ. We are taking a weighted average with most
of the weight concentrated near the point eiθ. Thus it makes sense that
u(reiθ) converges to f (eiθ).
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24. Solution to the Dirichlet Problem for the Disk
Example
Find the function u harmonic on the unit disk, D(0, 1) = {z : |z| < 1} and
takes on the boundary vales f (t) =



π for π
2 < t < π
0 for − π
2 < t < π
2
−π for − π < t < −π
2
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25. Solution to the Dirichlet Problem for the Disk
Example (Continued)
Solution:
u(z) =
− tan−1 1+r
1−r tan θ+π
2 + tan−1 1+r
1−r tan 2θ+π
4
π
+
tan−1 1+r
1−r tan 2θ−π
4 − tan−1 1+r
1−r tan θ−π
2
π
if z ∈ D(0, 1), and
u(z) = f (z)
if z ∈ ∂D(0, 1)
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26. The Harnack Inequality
Proposition
Let u be a nonnegative, harmonic function on a neighborhood of D(P, R).
Then for any z ∈ D(P, R),
R − |z − P|
R + |z − p|
· u(P) ≤ u(z) ≤
R + |z − P|
R − |z − P|
· u(P).
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27. The Harnack Inequality
Proposition
Let u be a nonnegative, harmonic function on a neighborhood of D(P, R).
Then for any z ∈ D(P, R),
R − |z − P|
R + |z − p|
· u(P) ≤ u(z) ≤
R + |z − P|
R − |z − P|
· u(P).
Harnack’s inequality is an inequality relating the values of a positive
harmonic function at two points.
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28. The Harnack Inequality
Example
Let u(x, y) = 1
2(x + 1) and which is nonnegative on D(0, 1). Our upper
and lower bounds for a point z = 1
2 +
√
3
4 i are,
R − |z|
R + |z|
· u(0) ≤ u(z) ≤
R + |z|
R − |z|
· u(0).
1 − |1
2 +
√
3
4 |
1 + |1
2 +
√
3
4 |
· 0.5 u
1
2
+
√
3
4
i
1 + |1
2 +
√
3
4 |
1 − |1
2 +
√
3
4 |
· 0.5.
0.10188831 u
1
2
+
√
3
4
i 2.45366725
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29. Harnack’s Principle
Using Harnack’s inequality we can can conclude that non-increasing
sequences of harmonic functions always converge. This is called Harnack’s
Theorem or Harnack’s Principle.
Theorem
Let u1 ≤ u2 ≤ . . . be harmonic functions on a connected open set U ⊆ C.
Then either uj → ∞ uniformly on compact sets or there is a harmonic
function u on U such that uj → u uniformly on compact sets.
Alternatively
Let u1 ≤ u2 ≤ . . . be harmonic functions on a connected open set U ⊆ C.
If for some point z0 ∈ U it happens that limn→∞ un(z0) = ∞, then
limn→∞ un(z) = ∞ uniformly on every compact set in U. However if
limn→∞ un(z0) = L ≤ ∞ then un(z) converges throughout U to a
harmonic function u uniformly on compact subsets of U.
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30. Subharmonic Functions
We now know what a harmonic function is and some of its properties so
lets define what a subharmonic function is and look at some of its
properties.
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31. Subharmonic Functions
We now know what a harmonic function is and some of its properties so
lets define what a subharmonic function is and look at some of its
properties.
Definition
Let U ⊆ C be an open set and f a real valued continuous function on U.
Suppose that for each D(P, r) ⊆ U and every real valued harmonic
function h defined on a neighborhood of D(P, r) which satisfies f ≤ h on
∂D(P, r) , it holds that f ≤ h on D(P, r). Then f is said to be
subharmonic on U.
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32. Subharmonic Functions
Example
Is the function |z|2 − 1 subharmonic on D(0, 1)?
It is hard to decide this based on the definition of a subharmonic function
thus we have a proposition to help of decide this.
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33. Subharmonic Functions
Example
Is the function |z|2 − 1 subharmonic on D(0, 1)?
It is hard to decide this based on the definition of a subharmonic function
thus we have a proposition to help of decide this.
Proposition
If f is C2 on an open set U, f is subharmonic if and only if ∆f ≥ 0 on U.
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34. Subharmonic Functions
Example
Is the function f (z) = |z|2 − 1 subharmonic on D(0, 1)?
∂f
∂x
= 2x
∂2f
∂x2
= 2
∂f
∂y
= 2y
∂2f
∂y2
= 2
Therefore ∆u = ∂2
∂x2 + ∂2
∂y2 u = 2 + 2 = 4 > 0 and f is subharmonic.
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35. Submean-value Property
Proposition
Let f : U → R be continuous. Suppose that, for each D(P, r) ⊆ U,
f (P) ≤
1
2π
2π
0
f (P + reiθ
)dθ. (∗)
Then f is subharmonic.
Conversely, if f : U → R is a (continuous) subharmonic function and if
D(P, r) ⊆ U, then the inequality (∗) holds.
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36. Submean-value Property
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37. Maximum Principle for Subharmonic Functions
Proposition
If f is subharmonic on U, and if there is a P ∈ U such that f (P) ≥ f (z)
for all z ∈ U, then f is constant.
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38. Maximum Principle for Subharmonic Functions
Proposition
If f is subharmonic on U, and if there is a P ∈ U such that f (P) ≥ f (z)
for all z ∈ U, then f is constant.
Q: Is there a minimum principle for subharmonic functions?
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39. Maximum Principle for Subharmonic Functions
Proposition
If f is subharmonic on U, and if there is a P ∈ U such that f (P) ≥ f (z)
for all z ∈ U, then f is constant.
Q: Is there a minimum principle for subharmonic functions?
A: No. By the the submean-value principle for subharmonic functions
there does not exist a minimum principle because the value of f can
be strictly smaller than the average on the corresoponding circle.
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40. Barrier Functions
Now we will define a special function, that singles out a point P in the
boundary of a set, called a barrier.
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41. Barrier Functions
Now we will define a special function, that singles out a point P in the
boundary of a set, called a barrier.
Definition
Let U ⊆ C be an open set and P ∈ ∂U. We call a function b : U → R a
barrier of U at P if
(I) b is continuous ;
(II) b is subharmonic on U;
(III) b ≤ 0;
(IV) {z ∈ ∂U : b(z) = 0} = {P}.
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42. Barrier Function
Example
Let U = D(0, 1) and P = 1 + 0i.
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43. Barrier Function
Example
Let U = D(0, 1) and P = 1 + 0i.
Then b(z) = x − 1 is a barrier for U at P, where x = Re z
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44. Barrier Function
Example
Let U = D(0, 1) and P = 1 + 0i.
Then b(z) = x − 1 is a barrier for U at P, where x = Re z
In general if U ⊆ C is open and bounded. Let P ∈ U be the point in U
that is the furthest from 0.
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45. Barrier Function
Example
Let U = D(0, 1) and P = 1 + 0i.
Then b(z) = x − 1 is a barrier for U at P, where x = Re z
In general if U ⊆ C is open and bounded. Let P ∈ U be the point in U
that is the furthest from 0.
Let r = |P| and θ0 = arg P, 0 ≤ θ0 ≤ 2π then
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46. Barrier Function
Example
Let U = D(0, 1) and P = 1 + 0i.
Then b(z) = x − 1 is a barrier for U at P, where x = Re z
In general if U ⊆ C is open and bounded. Let P ∈ U be the point in U
that is the furthest from 0.
Let r = |P| and θ0 = arg P, 0 ≤ θ0 ≤ 2π then
z −→ Re(e−iθ0
z) − r
is a barrier for U at P.
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47. Barrier Functions
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48. Perr´on Method
Q: Does every set U ⊆ C with a given boundary condition have a
solution to the Dirichlet problem?
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49. Perr´on Method
Q: Does every set U ⊆ C with a given boundary condition have a
solution to the Dirichlet problem?
A: No, for example
Example
Let U = D(0, 1) {0}. Then ∂U = {z : |z| = 1} ∪ {0}. Set
f (z) =
1 if |z| = 1
0 if z = 0
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50. Perr´on Method
Example
Let U = D(0, 1) {0}. Then ∂U = {z : |z| = 1} ∪ {0}. Set
f (z) =
1 if |z| = 1
0 if z = 0.
f is continuous on ∂U, if there is a solution to the Dirichlet problem it
must be that u(z) = u(eiθz) for some fixed θ. This is because if u(z) is a
solution then u(eiθz) is a solution for a fixed θ and because the Dirichlet
problem as a unique solution then u(z) = u(eiθz).
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51. Perr´on Method
Example (Cont.)
By using the polar form of the Laplacian, ∆ = 1
r
∂
∂r r ∂
∂r + 1
r2
∂2
∂θ2 , we see
that
0 = ∆u =
1
r
∂
∂r
r
∂u
∂r
(Independence of θ)
0 =
∂
∂r
r
∂u
∂r
(When r = 0)
Thus r ∂u
∂r = C for some constant C ∈ R.
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52. Perr´on Method
Example (Cont.)
By using the polar form of the Laplacian, ∆ = 1
r
∂
∂r r ∂
∂r + 1
r2
∂2
∂θ2 , we see
that
0 = ∆u =
1
r
∂
∂r
r
∂u
∂r
(Independence of θ)
0 =
∂
∂r
r
∂u
∂r
(When r = 0)
Thus r ∂u
∂r = C for some constant C ∈ R. This is a simple ODE and thus
u = C log(r) + D for some constant D ∈ R.
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53. Perr´on Method
Example (Cont.)
By using the polar form of the Laplacian, ∆ = 1
r
∂
∂r r ∂
∂r + 1
r2
∂2
∂θ2 , we see
that
0 = ∆u =
1
r
∂
∂r
r
∂u
∂r
(Independence of θ)
0 =
∂
∂r
r
∂u
∂r
(When r = 0)
Thus r ∂u
∂r = C for some constant C ∈ R. This is a simple ODE and thus
u = C log(r) + D for some constant D ∈ R.
There are no choices for C and D that will allow such a u to agree with
the given f on the boundary of U. Thus this Dirichlet problem cannot be
solved.
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54. Perr´on Method
Q: Can we ever guarantee that a solution to the Dirichlet problem will
exist?
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55. Perr´on Method
Q: Can we ever guarantee that a solution to the Dirichlet problem will
exist?
A: Yes!
Theorem
Let U be a bounded, connected open subset of C such that U has a
barrier bP for each P ∈ ∂U. Then the Dirichlet problem can always be
solved on U. That is, if f is continuous function on ∂U, then there is a
function u continuous on U, harmonic on U, such that u ∂U
= f . The
function u is uniquely determined by these conditions.
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56. Perr´on Method
Q: So what went wrong this our example?
Example
Let U = D(0, 1) {0}. Then ∂U = {z : |z| = 1} ∪ {0}. Set
f (z) =
1 if |z| = 1
0 if z = 0
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57. Perr´on Method
Q: So what went wrong this our example?
Example
Let U = D(0, 1) {0}. Then ∂U = {z : |z| = 1} ∪ {0}. Set
f (z) =
1 if |z| = 1
0 if z = 0
A: There does not exist a barrier function at the point z = 0 in the
boundary of U.
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58. Perr´on Method
Theorem
Let U be a bounded, connected open subset of C such that U has a
barrier bP for each P ∈ ∂U. Then the Dirichlet problem can always be
solved on U. That is, if f is continuous function on ∂U, then there is a
function u continuous on U, harmonic on U, such that u ∂U
= f . The
function u is uniquely determined by these conditions.
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59. Perr´on Method
Proof.
Without loss of generality assume that f is real valued. First set
S = {ψ : ψ is subharmonic of U and lim sup
U z→P
ψ(z) ≤ f (P), ∀P ∈ ∂U}
The set S is not empty because ∂U is compact and bounded below by
some real constant m and therefore ψ(z) ≡ m is in S
Define , for each z ∈ U
u(z) = sup
ψ∈S
ψ(z).
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60. Perr´on Method
Proof.
Without loss of generality assume that f is real valued. First set
S = {ψ : ψ is subharmonic of U and lim sup
U z→P
ψ(z) ≤ f (P), ∀P ∈ ∂U}
The set S is not empty because ∂U is compact and bounded below by
some real constant m and therefore ψ(z) ≡ m is in S
Define , for each z ∈ U
u(z) = sup
ψ∈S
ψ(z).
I claim that u solves the Dirichlet problem for f and U. We can show this
in 3 steps.
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61. Perr´on Method
Proof.
Without loss of generality assume that f is real valued. First set
S = {ψ : ψ is subharmonic of U and lim sup
U z→P
ψ(z) ≤ f (P), ∀P ∈ ∂U}
The set S is not empty because ∂U is compact and bounded below by
some real constant m and therefore ψ(z) ≡ m is in S
Define , for each z ∈ U
u(z) = sup
ψ∈S
ψ(z).
I claim that u solves the Dirichlet problem for f and U. We can show this
in 3 steps.
1 The function u is bounded above.
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62. Perr´on Method
Proof.
Without loss of generality assume that f is real valued. First set
S = {ψ : ψ is subharmonic of U and lim sup
U z→P
ψ(z) ≤ f (P), ∀P ∈ ∂U}
The set S is not empty because ∂U is compact and bounded below by
some real constant m and therefore ψ(z) ≡ m is in S
Define , for each z ∈ U
u(z) = sup
ψ∈S
ψ(z).
I claim that u solves the Dirichlet problem for f and U. We can show this
in 3 steps.
1 The function u is bounded above.
2 The function u is harmonic.
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63. Perr´on Method
Proof.
Without loss of generality assume that f is real valued. First set
S = {ψ : ψ is subharmonic of U and lim sup
U z→P
ψ(z) ≤ f (P), ∀P ∈ ∂U}
The set S is not empty because ∂U is compact and bounded below by
some real constant m and therefore ψ(z) ≡ m is in S
Define , for each z ∈ U
u(z) = sup
ψ∈S
ψ(z).
I claim that u solves the Dirichlet problem for f and U. We can show this
in 3 steps.
1 The function u is bounded above.
2 The function u is harmonic.
3 For each w ∈ ∂U, limU z→w u(z) = f (w).
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64. Perr´on Method
Cont.
1 The function u is bounded above.
Let M = maxζ∈∂U f (ζ), take ψ ∈ S and ε > 0, and let
Eε = {z ∈ U : ψ(z) ≥ M + ε}. Assume that Eε is nonempty.
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65. Perr´on Method
Cont.
1 The function u is bounded above.
Let M = maxζ∈∂U f (ζ), take ψ ∈ S and ε > 0, and let
Eε = {z ∈ U : ψ(z) ≥ M + ε}. Assume that Eε is nonempty.
It can be shown that Eε is closed.
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66. Perr´on Method
Cont.
1 The function u is bounded above.
Let M = maxζ∈∂U f (ζ), take ψ ∈ S and ε > 0, and let
Eε = {z ∈ U : ψ(z) ≥ M + ε}. Assume that Eε is nonempty.
It can be shown that Eε is closed.
Since Eε ⊂ U and U is bounded then Eε is bounded then Eε is
compact. Since ψ is continuous and Eε is compact then then ψ
attains its maximum at some point P ∈ Eε. Since the max is at least
M + ε on Eε and ψ is less than M + ε outside Eε then ψ(P) is a max
for ψ on all of U. Therefore by the maximum principle ψ is constant.
Since P ∈ Eε then the constant must be greater than or equal to
M + ε which contradicts ψ ∈ S. So Eε must be empty and thus
ψ(ζ) ≤ M for all ζ ∈ U. Therefore u is bounded above.
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67. Perr´on Method
Cont.
2 The function u is harmonic
Let D(P, r) ⊆ U and p ∈ D(P, r). By definition of u there exists a
sequence {ψj }∞
j=1 ⊆ S such that limj→∞ ψj (p) = u(p). Let
Ψn(z) = max{ψ1(z), . . . , ψn(z)}
for each z ∈ U. By properties of subharmonic functions Ψn(z) is
subharmonic on U and Ψ1 ≤ Ψ2 ≤ . . . . Let
Φn(z) =
Ψn(z) if z ∈ U D(P, r)
Poisson integral of Ψn ∂D(P,r)
if z ∈ D(P, r)
By the submean-value property the function Φn is subharmonic on U
and also Φn ∈ S. Furthermore Φ1 ≤ Φ2 ≤ . . . on U by maximum
principle.
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68. Perr´on Method
Cont.
Finally,
ψn(p) ≤ Ψn(p) ≤ Φn(p) ≤ u(p) (∗)
for each n and thus limn→∞ Φn(p) = u(p) By Harnack’s principle the
function {Φn} converse to a harmonic function Φ on the disk D(P, r).
Moreover Φ(P) = u(P) by (∗).
Take a point q ∈ D(P, r), there is a sequence {ρj }∞
j=1 ⊆ S such that
limj→∞ ρj (q) = u(q). Let ˜ρj (z) = max{ρj (z), ψj (z)} for z ∈ U and let
Λn(z) = max{˜ρ1(z), . . . , ˜ρn(z)}
for each z ∈ U. Again Λn is a subharmonic function on U with
Λ1 ≤ Λ2 ≤ . . . and Λn(q) → u(q).
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69. Perr´on Method
Cont.
Define Hn(z) in the same way we defined Φn(z) before
Hn(z) =
Λn(z) if z ∈ U D(P, r)
Poisson integral of Λn ∂D(P,r)
if z ∈ D(P, r)
{Hn} has a harmonic limit H on D(P, r) and H(q) = u(q). By design,
Φ(z) ≤ H(z) ≤ u(z) on D(P, r)
Hence H(p) = u(p) because Φ(p) = u(p). The harmonic function Φ − H
is such that Φ − H ≤ 0 on ∂D(P, r) and (Φ − H)(p) = 0. The maximum
principle says then that Φ ≡ H. Thus for any q ∈ D(P, r) the function H
agrees with the function Φ on p. Thus u(p) agrees with u(q) for each
point in D(P, r). Since P and r are chosen at random then we can
conclude that u is harmonic.
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70. Perr´on Method
Cont.
3 For each w ∈ ∂U, limU z→w u(z) = f (w).
Fix w ∈ ∂U. For ε, k > 0 we have that g(z) = f (w) − ε + kbw (z) is
subharmonic for z ∈ U. Choose δ > 0, a function of ε, so that
f (z) > f (w) − ε whenever z ∈ ∂U ∩ D(w, δ). Thus
g(z) ≤ f (z) for z ∈ ∂U ∩ D(w, δ)
Note that bp(z) has a strictly negative upper bound on ∂U D(w, δ).
Choose k, a function of ε, large enough such that
g(z) ≤ f (z) for z ∈ ∂U D(w, δ)
Thus g ∈ S, by the maximum principle for subharmonic functions, so
g ≤ u and
f (w) − ε = lim
U z→w
g(z) ≤ lim inf
U z→w
u(z)
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71. Perr´on Method
Cont.
By a similar argument lim supU z→w u(z) ≤ f (w) + ε. Thus
f (w) − ε ≤ lim inf
U z→w
u(z) ≤ lim sup
U z→w
u(z) ≤ f (w) + ε
Therefore we have the desired result limU z→w u(z) = f (w) : for w ∈ ∂U.
The final thing to show is that u is unique. This follows from the
maximum principle which says that the maximum for a harmonic function
occurs on the boundary.
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72. References
Greene, Robert E., and Krantz, Steven G. Function Theory of One Complex
Variable. 3rd ed. 40. American Mathematical Society, 2006. Print.
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73. Thank You
Thank you all for coming. I would
like to give a special thank you to
Dr. Edoh Amiran
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