Brezis' exercise 6.2.3: is it true that $\|Tu_{n_k} - v_1\|_\infty \xrightarrow{k \to \infty} 0$?

Question

I'm trying to solve an exercise in Brezis' Functional Analysis

Let $E:=C([0, 1])$ be the Banach space with the supremum norm $\|\cdot\|_\infty$. We define a bounded linear operator $T:E\to E$ by $$ Tu(t) := \int_0^t u(s) \, \mathrm d s \quad \forall u \in E, t \in [0, 1]. $$ Then $T$ is a compact operator.

In below attempt, I'm able to prove that $\|Tu_{n_k} - v_1\|_p \xrightarrow{k \to \infty} 0$ for $p \in \mathbb N^*$.

Is it true that $\|Tu_{n_k} - v_1\|_\infty \xrightarrow{k \to \infty} 0$?

Thank you so much for your elaboration!

Let $B$ be the closed unit ball of $E$. Then $T(B) \subset B$. Let $(u_n) \subset B$. We need to prove that $(Tu_n)$ has compact closure in $E$.

For $u\in E$, we extend its domain to the whole $\mathbb R$ by defining $u(t) :=0$ for $t \in \mathbb R \setminus [0, 1]$.

Let $p \in \mathbb N^*$. Then $T(B)$ is bounded in $L^p (\mathbb R)$. Also, for every $\varepsilon>0$ there is a bounded measurable subset $\Omega$ of $\mathbb R$ such that $\sup_{u\in B} \|Tu\|_{L^p (\mathbb{R} \setminus \Omega)}<\varepsilon$. By Corollary 4.27 (in the same book), $T(B)$ has compact closure in $L^p (\mathbb R)$.

By diagonal extraction, there is a subsequence $(Tu_{n_k})_k$ and a sequence $(v_p)$ such that $v_p \in L^p (\mathbb R),\|v_p\|_p \le 1$ and $\|Tu_{n_k} - v_p\|_p \xrightarrow{k \to \infty} 0$ for $p \in \mathbb N^*$. Convergence in $L^p (\mathbb R)$ implies a.e. convergence of a subsequence, so $v_1=v_p$ a.e. for $p \in \mathbb N^*$. Then $\|Tu_{n_k} - v_1\|_p \xrightarrow{k \to \infty} 0$ for $p \in \mathbb N^*$.

This is a duplicate and it is an easy application of Arzela-Ascoli Theorem. — geetha290krm, Jun 16 '23 at 09:39
@geetha290krm I asked for a (possible) fix of my attempt, so it's not a duplicate. — Akira, Jun 16 '23 at 09:40
You are going way off the track. It makes no sense to bring in $L^{p}(\mathbb R)$. — geetha290krm, Jun 16 '23 at 09:49
@geetha290krm I got it. I have just edited my thread and my updated question is whether $|Tu_{n_k} - v_1|_\infty \xrightarrow{k \to \infty} 0$ or not. — Akira, Jun 16 '23 at 09:53
@geetha290krm I meant I have figured out how to solve the exercise by Arzela-Ascoli theorem. Now I'm interested in whether $|Tu_{n_k} - v_1|_\infty \xrightarrow{k \to \infty} 0$ or not. — Akira, Jun 16 '23 at 10:19

score 0 · Accepted Answer · answered Jun 16 '23 at 20:50

As @geetha290krm mentioned in a comment, it's not possible to have $\|Tu_{n_k} - v_1\|_\infty \xrightarrow{k \to \infty} 0$. The appropriate proof would be by Arzela-Ascoli theorem as follows.

Clearly, $T(B)$ is pointwise bounded. It remains to prove that $T(B)$ is pointwise equi-continuous. Let $u\in B$ and $0 < t_1<t_2<1$. Indeed, $$ \begin{align*} |Tu(t_2)-Tu(t_1)| &= \left | \int_0^{t_2} u(s) \, \mathrm d s - \int_0^{t_1} u(s) \, \mathrm d s \right | \\ &= \left | \int_{t_1}^{t_2} u(s) \, \mathrm d s \right |\\ &\le \int_{t_1}^{t_2} |u(s)| \, \mathrm d s \\ &\le (t_2-t_1) \|u\|_\infty \\ &\le (t_2-t_1). \end{align*} $$

The claim then follows.

Brezis' exercise 6.2.3: is it true that $\|Tu_{n_k} - v_1\|_\infty \xrightarrow{k \to \infty} 0$?

1 Answers1