Compactness of a special kind of Integral operators

Let $$(S(t))_{t>0}$$ be a continuous operator from $$L^2(0,1)$$ to its self and Let $$K$$ be the operator \eqalign{ & K:{L^2}(0,1) \to {L^2}(0,1) \cr & f: \to (Kf)(x) = \int\limits_0^1 {k(s,x)S(s)f(s)ds} \cr} where $$k \in {L^2}(0,1) \times {L^2}(0,1)$$ It is well known that if $$S=I$$ then $$K$$ is compact operator. What can I say about the compactness of $$K$$ is this case? Thank you.

• I see no reason for $K$ even to be well-defined in general, forget about boundedness or compactness. You need to assume something else to make this construction meaningful (I interpret $S(s)f(s)$ as $(S(s)f)(s)$) – fedja Jun 9 at 0:46
• I think you also need to be explicit as to exactly in what sense $S(t)$ is "continuous". Continuous for each $t$? A continuous curve in the space of (continuous?) operators (with what topology?) – Nate Eldredge Jun 9 at 2:56
• In fact, $S(t)$ is C0_semogroup of operators on $L^2$ so it is continuous wth respect to its topology. – Gustave Jun 9 at 3:28

Your question must be reformulated to take into account the comments. Let us just say here that Hilbert-Schmidt operators (operators whose kernels are in $$L^2((0,1)^2)$$) make an ideal of the bounded operators so that composing a HS operator with a bounded operator gives a HS operator.