# Reproducing kernel Hilbert spaces and the isomorphism theorem

A reproducing kernel Hilbert space is a Hilbert space in which the evaluation functional

$L_x : f \rightarrow f(x)$ is continuous. By continuity, the Riesz representation theorem says that this functional can be represented as an inner product.

I have a feeling there's something fundamental I've misunderstood here. If any two real Hilbert spaces of the same cardinality are isomorphic, then why is it that $l_2$ is a RKHS, but $L^2[0,1]$ is not?

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It is of fundamental importance that the elements of a reproducing kernel Hilbert space are functions on a set $X$. The reproducing kernel itself is a function on $X \times X$. The isomorphism between $\ell^2$ and $L^2[0,1]$ won't respect this interpretation of an element as a function on $\mathbb{N}$ or $[0,1]$. Jonas Meyer's answer to this question might clarify things a little. – t.b. May 21 '11 at 23:11
Beautiful, thank you. – Simon May 21 '11 at 23:12
More to the point, of course the functions can be interpreted on the respective sets, but the evaluation functionals won't necessarily be continuous anymore. – t.b. May 21 '11 at 23:27
To make part of Theo's first comment more explicit: Elements of $L^2[0,1]$ are not functions. The problem isn't that evaluation at $x\in[0,1]$ isn't continuous, but that the evaluation doesn't exist. Also, a minor nitpick: 2 Hilbert spaces are isomorphic if they have orthonormal bases with the same cardinality (as is the case for $\ell_2$ and $L^2[0,1]$), not if the spaces themselves have the same cardinality. – Jonas Meyer May 22 '11 at 1:33

To wrap up what Theo and Jonas already said: Two (complex or real) Hilbert spaces are isomorphic if they have orthonormal bases with the same cardinality, as Hilbert spaces. So, every statement that makes use of the Hilbert space structure only, and is true or false for one space, will be true or false for the other.

But a concrete Hilbert space may have more structure than just the Hilbert space structure. When you look at the statement "A reproducing kernel Hilbert space is a Hilbert space in which the evaluation functional..." then the "evaluation functional"-part presupposes that the Hilbert space under consideration has (real or complex valued) functions as elements. This is an additional property that some Hilbert spaces have and some have not.

The space $L^2[0, 1]$ for example consists of equivalence classes instead of functions, and the "evaluation functional" cannot be well defined because it depends on the representative of an equivalence class $[f]$. In fact, for any $x \in \mathbb{R}$ and for every real number $y$ including $\infty$ and $-\infty$, every equivalence class has an element $f$ such that $f(x) = y$. And I can also define an abstract complex Hilbert space by saying that it is the space spanned by an orthonormal basis $(e_n)_{n \in \mathbb{N}}$. Now one cannot make sense of the term "evaluation functional", because the elements of this Hilbert space are not functions.

On the other hand, the Hardy space of the unit disk (Wikipedia) consists of holomorphic functions, therefore the evaluation functional is well defined. It is possible to prove that it is also continuous, but the proof makes use of the fact that the elements of this Hilbert space are holomorphic functions on the unit disk, which, as I said before, is additional structure that happens to exist for the Hardy space.

All of these examples are isomorphic as separable complex Hilbert spaces, but this isomorphism does not say anything about any structure that may exist beyond the Hilbert space structure.

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What you have said makes sense, but it raises another question. If evaluation depends on the choice of representative from $[f]$, then in what sense is the Dirac delta function a functional on $L^2[0,1]$? – Simon May 23 '11 at 13:28
Strictly speaking, the delta function is not a functional on $L^2$, but is an distribution, an element of the dual of a test function space, like the space of Schwartz test functions. The Schwartz test functions, the Hilbert space and the distributions form a Gelfand triple, see <a href="en.wikipedia.org/wiki/Rigged_Hilbert_space">Rigged Hilbert space</a> (Wikipedia). – Tim van Beek May 23 '11 at 13:56