# 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?

-

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.)
• 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.
I didn't understand why we showed the span $V$ is dense in $H_m$ in first point. Where did we use that property in the proof later on? Also why do we need to show the surjectivity in third point? – neticin Jan 29 '13 at 15:31
@neticin: Density of $V$ in $H_1$ is used to conclude that $T$ has a (unique) isometric extension on $H_1$. Density of $V$ in $H_2$ is used to conclude that $S$ is surjective. (Note also where completeness of $H_1$ and $H_2$ is used.) Survectivity is useful because when it is shown that $Sf=f$ for all $f\in H_1$, we want to conclude that $S$ is the identity map, not just inclusion. (One could also use symmetry, however; if you show that $H_1\subseteq H_2$, it automatically follows that $H_2\subseteq H_1$. So there are alternative approaches to being explicit about surjectivity at the start.) – Jonas Meyer Jan 29 '13 at 18:10