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Iosif Pinelis
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$\newcommand\tk{\tilde k}\newcommand\ip[2]{\langle #1,#2\rangle}$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 $\ip f{k_x}=f(x)$ and $k(x,y)=\ip{k_x}{k_y}=k_x(y)$ 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 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.)

Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229