Proof: If a reproducing kernel exists for a Hilbert space, then it is unique

Question

I really want to prove the statement in the title but I'm struggling with it. Here my current state:

Proof via contradiction. Let $\mathcal{H}$ be a RKHS with two reproducing kernels $k$ and $\hat{k}$ and let $x \in \mathcal{H}$. Then:

\begin{align} \|{k_x - \hat{k}_x}\|^2 &= \langle k_x - \hat{k}_x, k_x - \hat{k}_x \rangle \\ &= \langle k_x - \hat{k}_x , k_x \rangle - \langle k_x - \hat{k}_x , \hat{k}_x \rangle \\ &= \color{orange}{\langle k_x, k_x \rangle + \langle \hat{k}_x, \hat{k}_x \rangle} - \color{blue}{\langle \hat{k}_x, k_x \rangle - \langle k_x, \hat{k}_x \rangle}\\ &= ~... \\ &= \color{orange}{k(x,x) - \hat{k}(x,x)} - \color{blue}{k(x,x) + \hat{k}(x,x)} \\ &= 0. \end{align}

And this would be a contradiction since $\|x-y\| = 0 \Longleftrightarrow x = y$.

So the orange terms look fine but I don't know how to get the blue terms from the third to the fifth line. Please help.

Cheers. :-)

Answer 1

$\newcommand\tk{\tilde k}\newcommand\ip[2]{\langle #1,#2\rangle}$Let $k$ be a reproducing kernel of a reproducing kernel Hilbert space (RKHS) $H:=\mathcal H$ of real-valued functions on a set $X$. Then $$\ip f{k_x}=f(x)\tag{1}$$ and $$k(x,y)=\ip{k_x}{k_y}=k_x(y)\tag{2}$$ for all $f\in H$ and all $x$ and $y$ in $X$, where $\ip\cdot\cdot$ is the inner product on $H$.

Let now $\tk$ be another reproducing kernel of $H$. Then, by (1) and (2), for all $x\in X$ $$\begin{aligned} \|k_x-\tk_x\|^2& =\ip{k_x}{k_x}+\ip{\tk_x}{\tk_x}-\ip{k_x}{\tk_x}-\ip{\tk_x}{k_x} \\ & =k(x,x)+\tk(x,x)-k_x(x)-\tk_x(x) \\ & =k(x,x)+\tk(x,x)-k(x,x)-\tk(x,x)=0, \end{aligned}$$ whence $k_x=\tk_x$ for all $x$, that is, $k=\tk$.

(You mismatched colors.)