came across this exercise I haven't been able to solve. I saw a very similar exercise (to prove an integral operator with a kernel is compact), but it is just different enough, I think, to warrant its own question. The question is as follows:
Exercise: Consider an integral operator $T$, defined as: $$ (Tf)(t):=\int_{0}^{1}k(t,s)f(s)ds $$ with kernel $k(t,s):[0,1]^{2}\to \mathbb{R}$, which satisfies the following: (i) for each $s\in [0,1]$, the function $k_{s}(t)=k(t,s)$ is integrable in $t$; (2) the map $ s\mapsto k_{s}$ is a continuous map from $[0,1]$ to $L_1[0,1]$. Show that $T$ is compact in $C[0,1]$.
I know I'm supposed to use the Arzelà–Ascoli theorem here and show that the closed unit ball $B_{C[0,1]}$ gets mapped to a precompact set in $C[0,1]$. To show that $K=T(B_{C[0,1]})$ is precompact by Arzelà–Ascoli, I need to show it is uniformly bounded and equicontinuous.
To show uniform boundedness, I tried following the scheme of the question I linked, to obtain (for $f\in B_{C[0,1]}$, $||f||\leq 1$): $$|(Tf)(t)|=\lvert\int_0^1k(t,s)f(s)ds\rvert\leq|\int_0^1k(t,s)ds|\leq\int_0^1|k(t,s)|ds$$ And here is where I got stuck. I am not told that $k(t,s)$ is integrable in $s$, so I cannot say that this integral is less than $\infty$. This is contrary to the question I linked, where the question states that $k(t,s)$ is integrable in $t$. What do I do now?