# 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?

• 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

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.
• 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