Let $\Omega \subset \mathbb{R}^n$ be open and bounded and let $S \subset \Omega$ be a hypersurface in $\mathbb{R}^n$. Let further be $C_0(\Omega)$ the space of all continuous functions with compact support. In addition to that $B:\mathbb{R}^n \times C_0(\Omega) \rightarrow \mathbb{R}$ shall be a continuous linear operator in its second argument with respect to the $L_\infty$-norm. So $B(x)(h)$ should be continous and linear in $h$, but right now - aside from usual integrability assumptions - nothing else is specified regarding the first argument $x$.

I do have the following functional: $$\Phi(h):=\int\limits _{S}B(x)(h)\, dA \mspace{2in} h \in C_0(\Omega)$$

My question now is which additional properties - to the ones already mentioned above - the operator $B$ has to fulfill so that the operator $\Phi(.)$ is a continous linear functional with respect to the $L_\infty$-norm on $C_0(\Omega)$, i.e. $|\Phi(h)| \leq C \|h\|_\infty$ for all $h \in C_0(\Omega)$.

note: I am pretty sure that I can prove continuity in the above sense if I force the operator $B(.)(.)$ to be continuous in its first argument. So I am especially interested in any weaker assumptions. If it helps one might also restrict the space of the arguments of $\Phi$ to $C_0^\infty(\Omega)$ instead of using $C_0(\Omega)$.

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Flip coordinates and consider $B$ as a linear operator from $C_0(\Omega)$ into a suitable space $E$ of functions on $\mathbb R^n$. Now $f\mapsto \int_S f(x)dA$ must be defined as a linear functional on $E$, thus each function in $E$ should have a restriction to $S$ which is in $L^1(S,dA)$. So, if $B$ induces a bounded linear mapping $C_0(\Omega)\to L^1(S,dA)$, your desire is fulfilled.

If you restrict to $C^\infty_0(\Omega)$, then by the Schwartz kernel theorem $B$ is given by a distribution again called $B$ in $\mathcal D'(\Omega\times \mathbb R^n)$, which again has to be regular enough along $S\subset \mathbb R^n$ that the integral makes sense.

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  • $\begingroup$ Thanks for the answer, I honestly did not think that I would still get one after quite some time had passed after posting the question. I will accept the answer now, as I know it steers me in a useful direction, though I did not have time to check any details. $\endgroup$ – Wizard Sep 25 '14 at 9:31

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