Using Arzela-Ascoli in order to show that 1-lipschitz and weak convergence implies strong convergence in $L^3$.

Question

In this question 1-Lipschitz and weak convergence implies strong convergence in $L^3$, there is a comment by FourierWho that one can apply Arzela-Ascoli in order to obtain the conclusion, but I don't see how one can use Arzela-Ascoli to obtain such a conclusion.

Thanks.

Answer 1

This follows from the fact given in the answer I gave here.

To see this first note that since $(f_n)$ converges weakly in $L^3$, we have $\| f_n\|_3\leq M$. Together with the 1-Lipschitz condition, this means that $\| f_n\|_\infty+ \|f_n'\|_\infty\leq M'$ uniformly in $n$ (1), and so by Arzela-Ascoli, there is a convergent subsequence, but since the limit has to be 0 (by the weak convergence assumption), the whole sequence has to go to 0 (in fact the limit is not only in $L^3$ norm, but uniform).

To see why (1) is true we use the fundamental theorem of calculus: $$ |f_n(y)| \leq |f_n(x)| + \int_0^1 |f_n'(z)|\, dz, \qquad x,y \in [0,1]. $$ Now integrate with respect to $x$ to get
$$ |f_n(y)|\leq \| f_n\|_1 + \| f_n'\|_1\leq \| f_n\|_3 + \| f_n'\|_\infty \leq M+1, $$ owing to Hölder's inequality. This gives (1).

Answer 2

Claim 1: If $(f_{n})_{n \in \mathbb{N}}$ are $1$-Lipschitz in $[a,b]$, then $\|f_{n}\|_{L^{\infty}([a,b])}$ is unbounded if and only if $\inf \{|f_{n}(x)| \, \mid \, x \in [a,b]\} \to \infty$ as $n \to \infty$.

Claim 2: If $\inf \{|f_{n}(x)| \, \mid \, x \in [a,b]\} \to \infty$, then $\|f_{n}\|_{L^{p}([a,b])} \to \infty$ for any given $p \in (0,\infty]$.

To see Claim 1, observe that $|f_{n}(y)| \geq \|f_{n}\|_{\infty} - |b - a|$.