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 $C\_0(\Omega)$.



Continuity of Functional Represented by Surface Integral
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. 


