Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm doing practice physics qualifying exam problems and came across this one I didn't know how to solve:

Show that if $f(x)$ is bounded and analytic for $|z|=|x+iy|<1$, then $$f(\zeta)=\frac{1}{\pi}\int_{|z|<1}\frac{f(z)dxdy}{(1-\bar{z}\zeta)^2}$$ Hint: First express the area integral in polar coordinates, then transform one of the integrals to a suitable line integral of a rational function that can be evaluated using the calculus of residues.

I tried using $z=re^{i\theta}$ and messing around with the integral, but after a long writeout I am left with a puddle of muddled thoughts. Could someone explain the next steps?

share|improve this question
add comment

2 Answers 2

up vote 1 down vote accepted

You can prove this by Green formula: $$\int_{|z|=1}F(z)dz=2i\int_{|z|<1}\frac {\partial F}{\partial \bar{z}}dxdy.$$ By the Cauchy Int Formula $$f(\zeta)=\frac{1}{2\pi i }\int_{|z|=1} \frac{f(z)}{z-\zeta}dz.$$

Let $F(z)=\frac{f(z)}{z-\zeta}$, on the circle $|z|=1$, we have $z=\frac{1}{\bar z}$, so $F(z)=\frac{f(z)}{z-\zeta}=\frac{\bar zf(z)}{1-\bar z\zeta}$, by easy computation , we can get $\frac {\partial F}{\partial \bar{z}}=\frac{f(z)}{(1-\bar z\zeta)^2}$. Finally, using the Green Formula, we get $$f(\zeta)=\frac{1}{\pi }\int_{|z|<1}\frac{f(z)dxdy}{(1-\bar z\zeta)^2}.$$

share|improve this answer
    
Great, thanks a lot for your help. –  Peter May 4 '12 at 17:38
    
@Peter My pleasure. –  Riemann May 5 '12 at 1:53
add comment

The function $$K(\zeta,z)=\frac{1}{\pi}\frac{1}{(1-\bar{z}\zeta)^2}$$ is known as the Bergman reproducing kernel.

Hint:

  1. Compute the series expansion of $g(x)=\frac{1}{(1-x)^2}$, e.g. by using that $$\frac{1}{(1-x)^2} =\frac{d}{dx}\frac{1}{1-x}$$

  2. First prove the statement for $f_n(z)=z^n$, e.g. by using the fact $$\int_0^{2\pi}e^{ikt}dt=0,\text{ for all integers $k\ne0$}$$


Hopefully leaving the fun parts to you...

share|improve this answer
    
Thanks for the tip. I'll try working it out, and hopefully won't run into any more problems. –  Peter May 4 '12 at 17:38
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.