Feel Reproduce Kernel

Yesterday when I was processing with Ahlfors’ Complex Analysis, I encountered the problem about Residue.

Problem:

If $|z|\textless 1$, $f(z)$ is analytic and bounded, assuming $|\zeta|\textless 1$, then

$f(\zeta)=\displaystyle\frac{1}{\pi}\int\int_{|z|\textless 1}\frac{f(z)\mathrm{d}x\mathrm{d}y}{(1-\overline{z}\zeta)^2}$

It is an equation, but more than an equation. It is the meaning of Reproduce Kernel.

When learning Functional Analysis, I found some interesting schemes for proofs, esp. with Riesz Theorem and its corollaries. One of them is the Reproduce Kernel, which is more of an operator in Hilbert Space.

Definition:

Assuming $H$ is a Hilbert Space of complex-valiued functions on set $X$. We will say $H$ is a Reproduce Kernel Hilbert Space if every linear map:

$\displaystyle L_{x}:f\rightarrow f(x)$ is continuous at every $x$.

By Riesz Theorem, there exists a function $K_{x}$ of H s.t.

$\displaystyle f(x)=\langle f, K_{x}\rangle$.

We can define the function $K_{x}$ this way:

$\displaystyle K(x,y):=\overline{K_{x}(y)}$.

This is called reproduce kernel.

Let’s see what is its good properties:

•  $K(x,y)=\displaystyle\overline{K_{x}(y)}=\langle K_{y}, K_{x}\rangle$, which is the inner product;
• $K(x,x)=\displaystyle \langle K_{x},K_{x}\rangle \ge 0$;
• $K_{x}=0$ iff every $f(x)=0$.

However if there is an orthonormal basis for H, $\{\phi_n\}_{n=1}^\infty$, then

$\displaystyle K(x,y)=\sum_{n=1}^\infty \phi_n(x)\overline{\phi_n(y)}$.

Let’s prove the problem on Ahlfors’ book, we will use Residue Theorem.

Proof:

$z=\rho e^{i\theta}$,

$\displaystyle\frac{1}{\pi}\int\int_{|z|\textless 1}\frac{f(z)\mathrm{d}x\mathrm{d}y}{(1-\overline{z}\zeta)^2}=\displaystyle\frac{1}{\pi}\int_{\rho\textless 1}\int_{0}^{2\pi}\frac{f(z)\mathrm{d}x\mathrm{d}y}{(1-\overline{z}\zeta)^2}=\displaystyle\frac{1}{\pi}\int_{\rho\textless 1}\int_{0}^{2\pi}\frac{f(\rho e^{i\theta})\rho\mathrm{d}\rho\mathrm{d}\theta}{(1-\rho e^{-i\theta}\zeta)^2}=\displaystyle\frac{1}{i\pi}\int_{\rho\textless 1}\int_{C_{\rho}}\frac{f(z)z\mathrm{d}z}{(z-\rho^2\zeta)^2}\rho\mathrm{d}\rho=\displaystyle\frac{1}{2i\pi}\int_{\rho\textless 1}\int_{C_{\rho}}\frac{f(z)z\mathrm{d}z}{(z-\rho^2\zeta)^2}\mathrm{d}\rho^2=\displaystyle\int_{\rho^2\textless 1}(f(z)z)'|_{z=\rho^2\zeta}\mathrm{d}\rho^2=\displaystyle \int_{0}^1(\rho^2 f(\rho^2\zeta))'\mathrm{d}\rho^2=f(\zeta)$

$\Box$