Completion of some $C[0,1]$ functions with some inner product is a reproducing kernel Hilbert space

Question

The problem

Let $X$ be the space of $C^1[0,1]$ functions with the special property $f(0) = 0$. Consider the inner product defined in the following way: $$\langle f, g \rangle = \int_0^1 f'(x) \overline{g'(x)} dx$$ I want to show that the space $H$ corresponding to the completion of $X$ under the induced norm metric is a reproducing kernel Hilbert space, i.e. $\forall x \in [0,1]$ that we have $$|L_x(f)| = |f(x)| \leq C_x||f||_H, \forall f \in H$$ for some $C_x > 0$.

My progress

It is not particularly difficult to show that $L_x$ is bounded on the dense subspace $X$, as we can use the fundamental theorem of calculus effectively, i.e. for $f \in X$ we have \begin{aligned}|L_x(f)| = |f(x)| = \Big | \int_0^x f'(y) dy \Big | \leq \int_0^x |f'(y)| dy \\ \leq \sqrt{\int_0^x 1^2dy}\sqrt{\int_0^x|f'(y)|^2dy} \leq \sqrt{x}||f'||_2 = \sqrt{x}||f||_H, \end{aligned}

and so $C_x = \sqrt{x}$ suffices. My difficulty is enforcing this on the completion $X$.

In my mind, there seems to be two potential ways to approach this; we use that $X$ is a dense subspace (i.e. realize every $f \in H$ by some sequence $\{f_n\}_{n = 1}^\infty \subseteq X$ where $f_n \xrightarrow{H} f$) and either exploit some inequalities, or we attempt to say something meaningful about the limit $f$.

In particular, it would be nice if we could recover the fundamental theorem of calculus for the limit function $f$. This does not seem entirely unreasonable, as they are lots of known exercises regarding the limits of $C_1$ functions (in particular, that we can sometimes get uniform limits to be absolutely continuous functions, subject to some addition conditions, and so we still have FTC).

My question

Is there any result here I am missing which could tether this together? Or am I missing some sort of obvious forcing of limits? Any help would be appreciated.

Answer 1

This is a standard density argument, which is used for instance in Sobolev space theory.

1. The functional $L_x$ is continuous on the dense subspace. Hence, there is a unique extension to its closure, and this extension $L_x: X \to \mathbb R$ is linear and continuous. (In fact, this is true for every uniformly continuous function.)

2. The space $X$ is a closed subspace of the usual Sobolev space $H^1(0,1)$. Any function in $X \subset H^1(0,1)$ can be changed on a set of measure zero to obtain a continuous function (this is a Sobolev embedding theorem). For such functions, the extended functional $L_x$ indeed returns the evalution at $x$.