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

About YiMin

This is just a nerd PhD student of Math@UT Austin.