Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

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.
share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.