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. :-)



Proof: If a reproducing kernel exists for a Hilbert space, then it is unique
$\newcommand\tk{\tilde k}\newcommand\ip[2]{\langle #1,#2\rangle}$Let $k$ be a reproducing kernel of a [reproducing kernel Hilbert space (RKHS)](https://en.wikipedia.org/wiki/Reproducing_kernel_Hilbert_space#Definition) $H:=\mathcal H$ 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.)


