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.
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2$\begingroup$ I see no reason for $K$ even to be welldefined 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 $(S(s)f)(s)$) $\endgroup$ – fedja Jun 9 at 0:46

$\begingroup$ 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?) $\endgroup$ – Nate Eldredge Jun 9 at 2:56

$\begingroup$ In fact, $S(t)$ is C0_semogroup of operators on $L^2$ so it is continuous wth respect to its topology. $\endgroup$ – Gustave Jun 9 at 3:28
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Your question must be reformulated to take into account the comments. Let us just say here that HilbertSchmidt 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.