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

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. :-)

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, $$k=\tk$$.
• First of all, thanks for the quick answer! But sadly I still do not understand, why I can rewrite $\langle k_x, \hat{k}_x \rangle$ as $k_x(x)$ (and $\langle \hat{k}_x, k_x \rangle$ as $\hat{k}_x(x)$)... Oct 10 at 18:18
• Apply (1) replacing $k_x$ with $\hat{k}_x$ (legitimate since $\hat{k}$ and $k$ have the same properties) and $f$ with $k_x$. Oct 10 at 18:19