RKHS of functions vanishing at single point

Question

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.

Answer 1

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 $$ K_0(x,y)=\langle Pk_y,k_x\rangle=\langle k_y,k_x\rangle-\frac{\langle k_y,k_{x_0}\rangle}{\|k_{x_0}\|^2}\,\langle k_{x_0},k_x\rangle =K(x,y)-\frac{K(x_0,y)K(x,x_0)}{K(x_0,x_0)}. $$