Can anyone give small hint for me to solve the following:

Suppose $f_n:[0,1]\rightarrow \mathbb{R}$ are $1$-Lipschitz. If $f_n\rightarrow 0$ weakly in $L^3[0,1]$, then $f_n\rightarrow 0$ strongly in $L^3[0,1]$.

My attempt: Let $x_0\in[0,1]$. We have $|f_n(x)-f_n(x_0)|\leq |x_0-x|\leq 1$. Then

$\int_0^1 |f_n|^3=\int_0^1 |f_n-f_n(x_0)+f_n(x_0)|^3\leq 8(\int_0^1 |f_n-f_n(x_0)|^3+\int_0^1 |f_n(x_0)|^3)$.

Now $f_n\rightarrow 0$ weakly in $L^3[0,1]$ means that $\int_0^1 f_ng\rightarrow 0$ for any $g\in L^{3/2}[0,1]$.

I do not know what $x_0$ and what $g$ I should choose. I am completely lost.


Suppose $|f_n(x) | \ge 2$ for some $x,n$, then $|f_n(x')| \ge 1$ for all $x'$ and furthermore $f_n(x), f_n(x')$ have the same sign. Hence $|\int f_n| \ge 1$. Since $\int f_n \to 0$, we can assume (by renumbering if necessary) that $|f_n(x)| \le B$ for all $x$, and for all $n$.

If we can show that $\lim_n f_n(x) = 0$ for all $x$ then the dominated convergence theorem shows that $\int |f_n|^3 \to 0$.

Suppose there is some $x'$ such that $\delta=\limsup_n |f_n(x')| > 0$. By considering $-f_n$ if necessary we can assume that there is some subsequence $n_k$ such that $f_{n_k}(x') \ge {\delta \over 2}$. Hence $f_{n_k}(x) \ge f_{n_k}(x')-|x-x'|\ge {\delta \over 2} - |x-x'|$ and so $\int_{|x-x'| \le {\delta \over 2}} f_{n_k} \ge ({\delta \over 2})^2$. By choosing $g=1_{[x'-{\delta \over 2},x'+{\delta \over 2}]}$ we get $\int g f_{n_k} \ge ({\delta \over 2})^2$, a contradiction. Hence $\lim_n f_n(x) = 0$.

(Note that this proof holds for any $p \in [1,\infty)$, not just $p=3$.)

  • 1
    $\begingroup$ Nice idea. Since $(f_{n})_{n \in \mathbb{N}}$ is uniformly bounded in $L^{\infty}([0,1])$ by the first paragraph of your response, one can apply the Arzela-Ascoli Theorem and obtain a slightly simpler proof that way. $\endgroup$ – fourierwho Jul 16 '17 at 22:00
  • $\begingroup$ @fourierwho: Thanks! I think the compactness argument ends up being about the same length, so I opted for the less technical approach. $\endgroup$ – copper.hat Jul 17 '17 at 1:05

The space $\text{Lip}([0,1])$ of Lipschitz continuous functions on $[0,1]$ equals the Sobolev space $W^{1,\infty}([0,1])$. Since $[0,1]$ is a finite measure space (with Lebesgue measure), $W^{1,\infty}([0,1]) \subseteq W^{1,3}([0,1])$. One can show that since the sequence of Sobolev derivatives $(f_{n}')_{n \in \mathbb{N}}$ is bounded in $L^{\infty}([0,1])$, the derivatives $(f_{n}')_{n \in \mathbb{N}}$ converge weakly to zero in $L^{3}([0,1])$. In particular, $(f_{n},f_{n}')_{n \in \mathbb{N}}$ converges weakly to zero in $W^{1,3}([0,1])$. By the Rellich-Kondrachov Theorem, it follows that $f_{n} \to 0$ strongly in $L^{3}([0,1])$.

There may be a simpler proof, particularly since we're working in one dimension, but I don't see it yet.

  • $\begingroup$ Why does $(f_n')$ bounded in $L^\infty$ imply $f_n'\rightharpoonup0$ weakly in $L^3$? Take $f_n(x)=x$ for a counterexample. $\endgroup$ – Jason Jul 16 '17 at 21:21
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    $\begingroup$ In your example $(f_{n})_{n \in \mathbb{N}}$ does not converge weakly to zero. $\endgroup$ – fourierwho Jul 16 '17 at 21:56
  • $\begingroup$ Ah, okay. So you meant $f_n\rightharpoonup0$ weakly and $(f_n')$ bounded in $L^\infty$ implies $f_n'\rightharpoonup0$ weakly? How would one show that? $\endgroup$ – Jason Jul 16 '17 at 23:17
  • $\begingroup$ Intuitively, one tests against smooth functions and then concludes. Put more slowly, one doesn't need the bound on $(f_{n}')_{n \in \mathbb{N}}$ to see that $\int_{0}^{1} f_{n}'(x) g(x) \, dx \to 0$ when $g \in C^{\infty}_{c}([0,1])$. The bound is necessary to pass from this (by density) to arbitrary functions in $L^{3}([0,1])^{*}$. Note that it's necessary and sufficient for $\|f_{n}'\|_{L^{3}([0,1])}$ to be bounded independently of $n$. $\endgroup$ – fourierwho Jul 16 '17 at 23:33

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