Let's define the operator $K$ as $$ \begin{eqnarray*} K :&H&\longrightarrow &\;\:Z&\\ &y\left( x;\sigma \right) &\mapsto &\!\!\!\!\int_{0}^{1}C\left( \sigma \right) y\left( x;\sigma \right) d\sigma& \end{eqnarray*}\: $$ where $H$, $Z$ are Hilbert spaces, $C\left( \sigma \right) $ is a linear bounded operator from $H$ to $Z$, $x$ and $\sigma$ are real parameters. My question is: how can I prove that $K$ is bounded, and what does its adjoint operator look like?
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2$\begingroup$ I don't think the question can be answered in the generality in which you've posed it, which relies at least on being able to interpret elements of $H$ as functions of a real parameter $\sigma$ (as well as possibly of some other unspecified parameter $x$). $\endgroup$– LSpiceCommented May 6, 2019 at 16:59
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I may assume that $H=Z=L^2(\mathbb R)$ and the mapping $K$ to be given by a distribution kernel $k(s,t)$ via a formula $$ Ku(s)=\int k(s,t) u(t) dt, $$ meaning that for $u,v\in C^\infty_c(\mathbb R)$, we have $\langle Ku, v\rangle_{\mathscr D'(\mathbb R),\mathscr D(\mathbb R)}=\langle k, v\otimes u\rangle_{\mathscr D'(\mathbb R^2),\mathscr D(\mathbb R^2)}$. Now your question could be reformulated as follows: are there general criteria for extending $K$ to a bounded operator on $L^2(\mathbb R)$? The answer is yes, some general criteria are available.
Schur's criterion, for a locally integrable kernel, $$ \sup_s\int \vert{k(s,t)}\vert dt,\sup_t\int \vert{k(s,t)}\vert ds\quad \text{ both finite.} $$
Singular integrals criteria, which can be applied for instance to $k=pv(1/(s-t))$, which is a Fourier multiplier by a bounded function.
Operators which are products of type 1,2. In particular pseudo-differential operators of order 0, but it is less elementary.
