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?


1 Answer 1


Yes. An integral is a limit of linear combinations; those two are addressed precisely by the continuity and the linearity of $A$.

  • $\begingroup$ The integral here is a bit strange since $\phi(x)$ is a function in $H$ for every $x \in X$. Does the argument still hold? Also when you say "an integral is a limit of linear combinations" you also include the Lebesgue integral? $\endgroup$
    – DimSum
    Feb 5, 2021 at 9:17
  • $\begingroup$ Yes and yes. How do you define the Lebesgue integral? $\endgroup$ Feb 5, 2021 at 13:42
  • $\begingroup$ As a sup of simple functions but that does not rigorously answer my question. There is still something not well defined with this "Hilbert-valued" integral. I think we need to introduce the notion of Bochner integrability to do things properly. I found exactly the result I was looking for in the book of Steinwart and Christmann "Support Vector Machines" where they state under which conditions the linear operator and the integral commute (Equation A.32). Though in my initial problem I assumed X compact so it might be directly Bochner integrable, I am not sure. $\endgroup$
    – DimSum
    Feb 6, 2021 at 14:17
  • $\begingroup$ In that book the integral is defined, as I said, as a limit of linear combinations. So of course the integral commutes with $A$ as long as it is linear and continuous. $\endgroup$ Feb 6, 2021 at 16:07
  • $\begingroup$ I agree. The answer is nevertheless incomplete without the mention of « Bochner integrale », it is not Lebesgue. $\endgroup$
    – DimSum
    Feb 6, 2021 at 18:18

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