27 Poisson Integral Formulas

Recall Cauchy Integral Theorem:

1 f(ζ) 1 2π f(a)= dζ = f(a + reiθ)dθ. (109) 2πi I∂∆(a;r) ζ − a 2π Z0 which is called the mean value theorem. It means that the central value of f is equal to the average value of f along the boundary.

Question: Instead of a, for any point z ∈ ∆(a; r), can the value of f(z) be equal to the “ average value” of f along ∂∆(a; R) ?

Poisson Integral Formula for holomorphic functions

Theorem 27.1 Let f be a continuous function on the closed ball ∆(a; R) and be holomorphic in ∆(a; R). Then 81

1 2π R2 −|z − a|2 f z f a Reiθ dθ, z a R . ( )= ( + ) iθ 2 ∀ ∈ ∆( ; ) 2π Z0 |Re − (z − a)|

Proof: Assume a = 0 and R = 1. To show:

1 2π 1 −|z|2 f z f eiθ dθ, z . ( )= ( ) iθ 2 ∀ ∈ ∆(1) 2π Z0 |e − z|

ζ−z Recall that Φz(ζ)= 1−zζ is a one-to-one and onto analytic function

Φz : ∆(1) → ∆(1)

−1 −1 satisfying Φz(z) = 0. Then its inverse Φz exists and it satisﬁes Φz (0) = z so that −1 f ◦ Φz (0) = f(z). −1 Applying (109) to f ◦ Φz :

2π −1 −1 1 −1 iθ 1 f ◦ Φz (w) f(z)= f ◦ Φz (0) = f ◦ Φz (e )dθ = dw. (110) 2π Z0 2πi I∂∆(1) w

81 1 2π iθ In particular, when z = a, this formula becomes f(a)= 2π 0 f(a + Re )dθ, the mean value theorem. R

170 Recall the formula of changing integral variable in Proposition 23.1: C g(w)dw = ψ−1(C) g ◦ ′ −1 ψ(z)ψ (z)dz. Let ψ = Φz which satisﬁes Φz (∂∆(1)) = ∂∆(1). ThenR we get R 1 f ◦ Φ−1(w) 1 (f ◦ Φ−1) ◦ Φ (ζ) · Φ′ (ζ)dζ z dw = z z z −1 2πi I∂∆(1) w 2πi IΦz (∂∆(1)) Φz(ζ)

1 f(ζ) · Φ′ (ζ)dζ 1 2π f(eiθ)Φ′ (eiθ)ieiθdθ z z . = = iθ (111) 2πi I∂∆(1) Φz(ζ) 2πi Z0 Φz(e ) In the last equality we have used the standard parametric function [0, 2π] → ∂∆(1), θ 7→ eiθ. Now we continue to calculate:

ζ − z eiθ − z Φ (ζ)= , =⇒ Φ (eiθ)= z 1 − zζ z 1 − zeiθ (1 − zζ)+(ζ − z)z 1 −|z|2 =⇒ Φ′ (ζ)= = , z (1 − zζ)2 (1 − zζ)2 so that Φ′ (eiθ)ieiθ 1 −|z|2 1 − zeiθ i(1 −|z|2) i(1 −|z|2) z ieiθ iθ = iθ 2 · iθ · = −iθ iθ = Φz(e ) (1 − ze ) e − z (−z + e )(e − z) (eiθ − z)(eiθ − z) i(1 −|z|2) = . |eiθ − z|2

Substituting this into (110) and (111), we are done.

Remarks

|ζ|2−|z|2 1. Write K(ζ, z)= |ζ−z|2 . Then the formula in Theorem 27.1 can be written as

1 2π f(z)= f(ζ)K(ζ − a, z − a)dθ, ∀z ∈ ∆(a; R) 2π Z0 where ζ = Reiθ, which is called the Poisson integral formula.

2. When a = 0, we have

1 2π f(z)= f(ζ)K(ζ, z)dθ, ∀z ∈ ∆(R). 2π Z0

171 3. If we take f = 1, we ﬁnd the formula

1 2π 1= K(ζ, z)dθ. 2π Z0 Therefore, the Poisson integral formula means that a function valued can be obtained by an average of the values of the function.

4. Let us compare Cauchy integral formula

1 f(ζ) f(z)= dζ, 2πi I∂∆(R) ζ − z

and the Poisson integral formula

1 2π f(z)= f(ζ)MattewK(ζ, z)dθ. 2π Z0

f(ζ) In Cauchy integral formula the integral kernel is ζ−z which is holomorphic in z. While |ζ|2−|z|2 in the Poisson integral formula, the integral kernel is K(ζ, z) = |ζ−z|2 which is not holomorphic in z.

In some sense, Cauchy integral formula is important to study holomorphic functions , for example, it implies that every holomorphic function must be analytic; but we cannot use Poisson integral formula to get the same conclusion.

On the other hand, Poisson integral formula is important to study an important class of real-valued functions, which are called harmonic functions, see below. Another equivalence for harmonic function is: a locally deﬁned function is harmonic if and only if it is the real part of some holomorphic function. We’ll discuss more on harmonic function in next lecture.

Poisson integral formula for harmonic functions Writing f = u + iv and separating the real part, we get

1 2π u(z)= u(Reiθ)K(Reiθ, z)dθ, ∀z ∈ ∆(R). 2π Z0 Here we used the fact that K(ζ, z) is real-valued. Notice that we cannot get this formula from Cauchy integral formula.

172 Take ζ = Reiθ and z = reiψ. Then R2 − r2 K(ζ, z)= |Reiθ − reiψ|2 R2 − r2 = [R(cosθ + isinθ) − r(cosψ + isinψ)][R(cosθ − isinθ) − r(cosψ − isinψ)] R2 − r2 = [(Rcosθ − rcosψ)+ i(Rsinθ − rsinψ)][(Rcosθ − rcosψ) − i(Rsinθ − rsinψ)] R2 − r2 = [Rcosθ − rcosψ]2 + [Rsinθ − rsinψ]2 R2 − r2 = R2 − Rrcos(ψ − θ)+ r2

Then we have

Theorem 27.2 Let f = u + iv be a continuous function on the closed ball ∆(R) and be holomorphic in ∆(R). Then

1 2π R2 − r2 u z u Reiθ dθ, z reiψ R . ( )= ( ) 2 2 ∀ = ∈ ∆( ) 2π Z0 R − Rrcos(ψ − θ)+ r

Reproducing holomorphic functions We notice that

|ζ|2 −|z|2 1 ζ + z ζ + z ζ + z K(ζ, z)= = + = Re . |ζ − z|2 2ζ − z ζ − z ζ − z If u is the real part of a holomorphic function f which is deﬁned on ∆(R) and continuous on ∆(R), by removing the real part from the formula

1 2π ζ + z u(z) := u(ζ)Re dθ ∀z ∈ ∆(R), (ζ = Reiθ), 2π Z0 ζ − z we can deﬁne another function 1 2π ζ + z uˆ(z) := u(ζ) dθ ∀z ∈ ∆(R), (ζ = Reiθ) 2π Z0 ζ − z which is a holomorphic function deﬁned in ∆(R) such that its real part

Re(ˆu)(z)= u(z).

173 In other words, from a holomorphic function f, we obtain a real-valued function u; from the real-valued function u, we recovered a holomorphic functionu ˆ.

f = u + iv → u k? k uˆ ← u

Because f andu ˆ have the same real part, by Cauchy-Riemann Equations, the diﬀerence between the imaginary parts of f andu ˆ is a constant. If we write c = f − uˆ, then c = 1 2π iθ f(0) − uˆ(0) = f(0) − 2π 0 u(Re )dθ = f(0) − u(0) = iIm(f(0)). Hence R f(z)=ˆu(z)+ i Im(f(0)).

174