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.

  • 2
    $\begingroup$ Technically I think you should be a little bit more careful in your last step, referencing exactly what is meant by "$t \mapsto \kappa_t$ is continuous" (i.e. if $t \to s$ then $\kappa_t \to \kappa_s$ in the sense of $L^1$, as opposed to, say, pointwise). Otherwise it looks fine. $\endgroup$ – Ian Aug 6 '14 at 16:18
  • $\begingroup$ Check this related problem. $\endgroup$ – Mhenni Benghorbal Aug 6 '14 at 16:41
  • $\begingroup$ @MhenniBenghorbal Could you explain your hint in that problem please, I did not get it :( I know that in general $||T||_{op} \geq \sup_i ||T e_i||_H$. Also I couldn't see the connection between the two problems. Thank you very much! $\endgroup$ – Xiao Aug 7 '14 at 10:54
  • $\begingroup$ @Xiao: Basically what you did in your proof is applying the definition I referred you to it by using Ascoli theorem. $\endgroup$ – Mhenni Benghorbal Aug 7 '14 at 18:44

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 $L^1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.