Let's consider a compact metric space X, $K:X\times X \to \mathbb{R}$ is a continuous kernel, $H_K$ is the induced RKHS and $\nu$ is a finite measure on X. We define the integral operator associated to K as:
\begin{align*} L_K \colon & L_2(\nu) \to H_K\\ & f \mapsto \int_X K_x f(x) \nu(dx) \end{align*}
I want to prove the following (which I am not sure is true): $$\| i(f) \|_2^2 = \langle f, L_K \circ i(f) \rangle_{H_K}$$ where $f$ is in $H_K$ and $i$ is the continuous embedding from $H_K$ to $L_2(\nu)$ . My proof is as follows, since $f$ is in $H_K$, using the reproducing property:
$$\| i(f) \|_2^2 = \int f(x)^2 \nu(dx) = \int_X f(x)\langle f, K(x,.) \rangle_{H_K} \nu(dx) = \langle f, \int_X f(x) K(x,.) \nu(dx) \rangle_{H_K} = \langle f, L_K \circ i(f) \rangle_{H_K}$$
I don't know how to justify the step with the exchange of the integral with the scalar product. Is it true that if $H$ is a Hilbert space, $X$ is a metric space, $\phi:X \to H$ and $A$ is a continuous linear operator from $H$ to $\mathbb{R}$ then $$A\left( \int_X \phi(x)\nu(dx) \right) = \int_X A( \phi(x) ) \nu(dx)$$
And if true, what would be a proof?
Yes. An integral is a limit of linear combinations; those two are addressed precisely by the continuity and the linearity of $A$.