Show $T: C([0,1]) \rightarrow C([0,1])$ is compact

Question

Consider $T: C([0,1]) \rightarrow C([0,1])$ defined by $$(Tf)(t) := \int_0^1 \kappa_t(s)f(s)ds,$$ where $\kappa:[0,1]^2 \rightarrow \mathbb{R}$ satisfies the following properties:

1. for all $t\in [0,1]$, the function $\kappa_t(s)$ is integrable in $s$;
2. $t\rightarrow \kappa_t(s) \in L^1([0,1])$ is continuous.

Show that $T$ is a compact operator.

My proof: Let $f_n$ be a bounded sequence in $C([0,1])$ with the $\sup$ norm, we show that the sequence satisfies the assumption of Arzelà–Ascoli theorem, which would imply that $T$ is a compact operator.

1. To show $(Tf_n)(t)$ is uniformly bounded. By Schwartz inequality, we have $$|(Tf_n)(t)| \leq ||\kappa_t||_1 ||f_n ||_\infty,$$ the quantity $||f_n||_\infty$ is bounded for each $n$ by assumption, and $||\kappa_t||_1$ is bounded over $t\in [0,1]$ because Property 2.

2. To show $(Tf_n)(t)$ is equicontinuous, let $t_i \rightarrow t$ $$\lim_{i\rightarrow \infty}\bigg|(Tf_n)(t_i) - (Tf_n)(t)\bigg|\leq \lim_{i\rightarrow \infty}\int_0^1 |\kappa_{t_i}(s) - \kappa_t(s)|\cdot |f_n(s)|ds \leq \lim_{i\rightarrow \infty}||\kappa_{t_i} - \kappa_t||_1 ||f_n ||_\infty,$$ which goes to zero (because of Property 2) independent of $n$ since $||f_n ||_\infty$ is bounded.

This is a problem from a Ph.D. entrance exam, I am a little unsure because my proof seemed very short compared to other problems in the set. Is this correct? Thank you very much for reading.

Answer 1

There is no problem for boundedness.

However, for equi-continuity, you have to be more careful. First of all, we should show that $\lim_{i\to\infty}\sup_n\dots\to 0$ (there won't be any problem because the estimates are uniform in $n$, but we have to take the supremum). Second, this proves the equi-continuity at $t$ and we want a uniform equi-continuity. This can be done using the suggested estimate, namely, $$\sup_n|T(f_n)(s)-T(f_n)(t)|\leqslant \lVert\kappa_s-\kappa_t\rVert_1$$ and using the unifom continuity of $s\mapsto \kappa_s$ in $L^1$.