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

canbe injective. There doesn't seem to be any assertion that it isalwaysinjective. $\endgroup$alwaysinjective. 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. $\endgroup$isinjective. 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$1more comment