Reproducing kernel Hilbert space

Remarks

For each $x$ there is then a unique $k_x \in H$ with $f(x) = \ip{f}{k_x}$ for all $f \in H$. The function $K(x, y) = k_y(x) = \ip{k_y}{k_x}$ on $X \times X$ is the reproducing kernel of $H$.

Example: $H = \ell^2(\N)$, seen as functions on $X = \N$. Evaluation at $n$ is $f \mapsto f(n)$, and $\abs{f(n)} \le \norm{f}_2$, so it is continuous. Here $k_n$ is the unit vector $e_n$ and $K(m, n) = 1$ if $m = n$ and $0$ otherwise.

Non-example: $L^2([0, 1])$ is not a reproducing kernel Hilbert space. Its elements are classes of functions equal almost everywhere, so $f \mapsto f(x)$ is not even defined on it. The continuous functions on $[0, 1]$ with the $L^2$ inner product are not one either, because this space is not complete.

The Hardy space of the unit disc is a reproducing kernel Hilbert space with kernel $K(z, w) = 1/(1 - z \bar{w})$.