Consider the integral kernel $K(x,y)$ continuous on $[0,1]^2$, acting on $C([0,1]),| \cdot |_\infty $, defining an operator $T$. Given a sequence in the unit ball in this space $f_n$ , we can find weakly convergent subsequence still denoted $f_n$. Therefore for the special case of the, measure $\mu_x(y)dy=K(x,y)dy$, we have $T(f_n)(x) \to T(f)(x)$ for all $x \in [0,1]$.
One way to finish the proof is to show that $T(f_n)$ is an equicontinuous family and therefore, (without using ascoli arzela) we have that the pointwise convergence of $T(f_n)(x) \to T(f)(x)$ implies convergence in the norm.
I want to use what I have always used when trying to show that weak convergence implies strong convergence.
Question: Can you show that for any subsequence of $T(f_n)$, there is a subsequence that converges to $T(f)$ in norm?
(for reference... one can show for example Reillich Kondrakov compactness of the Sobolev embeddings using this method.)