proof of RKHS for a particular kernel is unique Suppose that I have a kernel $K$. Then show that the RKHS $H_1$ and $H_2$ of $K$ are the same.
So I need to prove the above statement. To begin with, as an exercise, I proved the reverse statement "If $H$ is a RKHS then it has a unique kernel $K$" with basic tools using inner products and reproducing kernel property. However, I even could not start to prove the other direction. So can you provide a roadmap to prove this statement?
 A: Firstly, some missing context: There is some set $X$ such that $K:X\times X\to\mathbb C$.  Each of $H_1$ and $H_2$ is a Hilbert space whose elements are functions from $X$ to $\mathbb C$, and the vector space operations are the usual pointwise addition and scalar multiplication of functions.  For each $y\in X$, the function $k_y:X\to\mathbb C$ defined by $k_y(x)=K(x,y)$ is in each $H_m$, and if $f$ is in $H_m$ ($m=1,2$), then for all $x\in X$, $f(x)=\langle f,k_x\rangle_m$.  (Due to this last fact, it was redundant of me to point out that the operations are pointwise.)
Here is one approach.  


*

*Show that the span $V$ of $\{k_x\}_{x\in X}$ is dense in each $H_m$.  This follows from the fact that if $f(x)=0$ for each $x\in X$, then $f=0$.  

*Considering $V$ as a subspace of each $H_m$, define the map $T:(V,\langle\cdot,\cdot\rangle_1)\to (V,\langle\cdot,\cdot\rangle_2)$ by $Tv=v$.  That is, $T$ is the identity map on the subspace $V$, but with domain and range given the (potentially) different inner products.

*Show that $T$ is an isometry, and therefore extends uniquely to an isometry $S:H_1\to H_2$.  Note why $S$ is also surjective.

*Show that if $f\in H_1$, then for all $x\in X$, $(Sf)(x)=f(x)$, using the fact that $S^*k_x=k_x$.  Thus, $Sf=f$ as functions, and conclude that $H_1=H_2$ as sets of functions.

*Finally, note that the inner products are identical because the identity map is an isometry.

