# RKHS of functions vanishing at single point

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 $$\phi$$.

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