Boundedness of integral operators on spaces of continuous functions Consider a standard integral operator $T$ formally defined by
$$
Tf(x):=\int_{K} k(x,y)f(y)dy,\qquad x\in K,
$$
where $K$ is a locally compact metric measure space. It is immediate to see that the operator is bounded in $\infty$-norm if $\sup_{x\in K}\int_K |k(x,y)| dy<\infty$.
Are there any conditions implying this operator to be bounded on $C_b(K)$ and/or on $BUC(K)$ and/or on $C_0(K)$?
 A: Too long for a comment. Your requirement is too stringent and it is quite likely that to get continuity from $L^\infty$ into itself, it is indeed necessary to have
$$
\text{esssup}_x\int\vert k(x,y)\vert dy<+\infty.
$$
On the other hand, if you are interested in $L^2$ or $L^p$ continuity, you have a much more interesting question. By the way, no iff condition is known for the $L^2(\mathbb R^N)$ continuity, although you have several quite refined sufficient conditions. For instance, Calder\'on-Zygmund singular integrals and zeroth order pseudo-differential operators are bounded on $L^p(\mathbb R^N)$ for $p\in (1,+\infty)$, but not in general on $L^1$ or $L^\infty$. If you relax your $L^\infty$ requirement and for instance look for $BMO(\mathbb R^N)$ continuity, I believe that singular integrals with odd kernels are indeed bounded on
$BMO(\mathbb R^N)$: to give a simple example, I think that with 
$$
k(x,y)=\text{pv}\frac{1}{x-y}=\frac{d}{dx}\bigl(\ln \vert x-y\vert\bigr)
$$ 
you get an operator which is not bounded on $L^\infty(\mathbb R)$, but is bounded on $BMO(\mathbb R)$.
You can raise also a similar question for the Sobolev continuity, say from $H^s(\mathbb R^N)$ into
$H^{s'}(\mathbb R^N)$.
