0
$\begingroup$

Exercise 5.3. Let $\mathcal H$ be an RKHS on $X$ with reproducing kernel $K$, fix $x_0\in X$ and let $\mathcal H_0=\{f\in\mathcal H: f(x_0)=0\}$. Compute the kernel function for $\mathcal H_0$.

It is easy to show that $H_{0}$ is RKHS and I am aware of the fact that if $P:H\rightarrow H_{0}$ is the orthogonal projection then the kernel of $H_{0}$ is $K_{0}(x,y)=\langle Pk_{y},k_{x}\rangle$. My question is how to calculate the kernel explicitly?

I tried to find a function $\phi:X\rightarrow \mathbb{C}$ such that $\phi(x_{0})=0$ and for every function $f\in H_{0}$ there exist a function $g\in H_{0}$ such that $f(x)=\phi(x)g(x)$. Also if the multiplication operator $M_{\phi}$ on $H$ is isometry, then $f(x)=\phi(x)g(x)=\phi(x)\langle g,k_{x}\rangle=\phi(x)\langle \phi g,\phi k_{x}\rangle= \langle \phi g,\phi k_{x}\overline{\phi(x)}\rangle=\langle f,\phi k_{x}\overline{\phi(x)}\rangle$.

Then the kernel of $H_{0}$ is $K_{0}(y,x)=\phi(y) k(y,x)\overline{\phi(x)}$. But I am not able to find such $\phi$.

Please help me regarding this. Thanks in advance.

$\endgroup$
1

1 Answer 1

4
$\begingroup$

I have to admit that I cannot really follow what you are trying to do. How do you justify the existence of $g$?

Here is a way to find $K_0$. The subspace $H_0$ consists of those $f$ such that $$ 0=f(x_0)=\langle f,k_{x_0}\rangle. $$ Thus $H_0=\{k_{x_0}\}^\perp$. So $P=I-Q$, where $Q$ is the orthogonal projection onto $\mathbb C\,k_{x_0}$. That is, $$ Qg=\frac{\langle g,k_{x_0}\rangle}{\|k_{x_0}\|^2}\,k_{x_0}. $$ Then $$ Pk_y=k_y-=\frac{\langle k_y,k_{x_0}\rangle}{\|k_{x_0}\|^2}\,k_{x_0}, $$ and $$ K_0(x,y)=\langle Pk_y,k_x\rangle=\langle k_y,k_x\rangle-\frac{\langle k_y,k_{x_0}\rangle}{\|k_{x_0}\|^2}\,\langle k_{x_0},k_x\rangle =K(x,y)-\frac{K(x_0,y)K(x,x_0)}{K(x_0,x_0)}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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