Let $X$ be a measurable space, $\mu$ be a $\sigma$-finite measure on $X$, and $H$ be a separable reproducing kernel Hilbert space over $X$ with a measurable kernel $k$.

At a certain part in a proof I am reading there is the condition

$S_k : L_q (\mu) \to H $ has a dense image if and only if $id: H \to L_p(\mu)$ is injective.

We know (it is shown in the proof) $id$ is the inclusion operator and is continuous. $S_k$ is the adjoint of the inclusion operator. And $H$ consists if $p$-integrable functions.

My question is how can the inclusion mapping $id$ always be injective? $H$ is a Hilbert space of function and $L_p(\mu)$ is a space of equivalence classes of functions. I am really confused on the intuition here.

Consider the following:

  • I take the function $f \in H$. I know $f$ belongs to some element of $L_p$ and is measurable. Thus the inclusion makes sense.
  • The inclusion maps $f$ to its equivalence class $[f]$
  • Say I take a measurable $p$-integrable function $g$ which equals $f$ almost everywhere. (I am obviously assuming such a function exists in $H$).
  • We see the inclusion maps $g$ to $[f]$.
  • Therefore the inclusion map can't be injective

I could see that this can fail if no such function $g$ exists in $H$. (For example if the RKHS $H$ has a continuous kernel then every $f \in H$ is continuous and the $g$ in question could not be found.) But I have no reason to believe that such a $g$ can't exist. I am obviously suffering from a knowledge gap. Where am I going wrong in how I am thinking about this.

PS: The proof can be found on page 126 here: here

  • $\begingroup$ Haven't you answered your own question? If, for instance, all the functions in $H$ are continuous, the inclusion map will be injective. For a concrete example, take $X = [0,1]$ and let $H = H^1([0,1])$ be the Hilbert space of all absolutely continuous $f : [0,1] \to \mathbb{R}$ having $f' \in L^2([0,1],m)$, equipped with the inner product $\langle f,g\rangle_H = \int_0^1 (fg + f'g')\,dm$. $\endgroup$ Mar 17 '15 at 5:35
  • $\begingroup$ So I guess I am confused as to what you are really asking. Certainly the inclusion map can be injective. There doesn't seem to be any assertion that it is always injective. $\endgroup$ Mar 17 '15 at 5:37
  • $\begingroup$ @NateEldredge The assertion is the the inclusion map is always injective. If the kernel is continuous (which means all functions in $H$ are continuous, then yes we have injectivity.) If the kernel is * measurable* then we can only say that each $f \in H$ is measurable. $\endgroup$
    – Eric
    Mar 17 '15 at 5:41
  • $\begingroup$ I don't see that assertion in what you've written; is it elsewhere in the text? The line you quote after "At a certain point" is an if and only if statement. It doesn't assert that the map is injective. If there is an assertion, somewhere, that the inclusion map is always injective, then it must follow from other conditions you have not mentioned. $\endgroup$ Mar 17 '15 at 5:44
  • $\begingroup$ @NateEldredge, thanks for clarifying! I wasn't familiar with this class of spaces. $\endgroup$ Mar 17 '15 at 5:44

It seems, from comments, that the question is based on a misreading. The text does not assert that the inclusion map is always injective; it only gives a necessary and sufficient condition for the map to be injective. It is easy to find examples where it is, and examples where it isn't.

As an incredibly trivial example, let $H = \mathbb{R}$ with its usual inner product, considered as the RKHS of real-valued functions on a one-point set $X = \{x\}$. (The reproducing kernel is $k(x,x) = 1$.) If $\mu$ is a measure on $X$ assigning positive mass to $x$, then the inclusion $H \to L^2(X,\mu)$ is injective. If $\nu$ is the zero measure, then the inclusion $H \to L^2(X,\nu)$ is not injective.


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