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Given a sequence $u_k\in W^{1,p}(B_1)\cap C^{\alpha}(B_1)$ such that $\|u_k\|_{C^{\alpha}(B_1)}\le 1$ for all $k\in \mathbb N$. Suppose we have $$ u_k \rightharpoonup u\;\;\mbox{weakly in $W^{1,p}(B_1)$} $$ and we also have $$ u_k \rightarrow u_0\;\;\mbox{in $C^{\alpha}(B_1)$}. $$

Can I imply from the above two informations that $u_k\rightarrow u_0$ strongly in $W^{1,p}(B_1)$?

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  • $\begingroup$ I don't think so. Let $f_n: [0,1] \to \mathbf{R}$ be a zig-zag function with step size $1/n$ and $\lvert f \rvert_\infty = 1/n$. Then every $f_n$ is bounded as a Lipschitz function, and the sequence converges to zero weakly in $W^{1,p}$ and strongly in $C^{0,\alpha}$ for all $\alpha \in (0,1)$. However $\lvert f_n \rvert_{1,p} \geq 1$ for all $n$ because $f_n' = 1$ almost everywhere. $\endgroup$
    – Leo Moos
    Commented Jul 2, 2021 at 13:22
  • $\begingroup$ @LeoMoos Thanks for the comment, but the example you mentioned does not have a uniform bound on Holder norm. In fact the Holder norm of $f_n$ blows up as $n\rightarrow \infty$ $\endgroup$
    – Harish
    Commented Jul 2, 2021 at 13:36
  • $\begingroup$ Are you sure? The Lipschitz norm is $1 + 1/n$, so that seems strange. I have the Holder norm decaying like $n^{\alpha-1}$. $\endgroup$
    – Leo Moos
    Commented Jul 2, 2021 at 13:38
  • $\begingroup$ @LeoMoos Please let me know the definition of zig zag function you are considering. $\endgroup$
    – Harish
    Commented Jul 2, 2021 at 13:39
  • $\begingroup$ I defined them in my answer below. $\endgroup$
    – Leo Moos
    Commented Jul 2, 2021 at 13:54

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I don't think so. Define a sequence of 'zig-zag' functions $(f_n)$ on $[0,1]$ as follows. Let $f_n$ be $\frac{2}{n}$-periodic with $f_n(x) = x$ when $x \in [0,\frac{1}{n}]$ and $f_n(x) = \frac{2}{n}- x$ when $x \in [\frac{1}{n},\frac{2}{n}]$. Every function is Lipschitz with $\lvert f_n \rvert_\infty \leq 1/n$ and $\lvert f_n' \rvert_\infty = 1$.

Moreover the sequence converges to zero weakly in $W^{1,p}$ for every $p \in (1,\infty)$ and strongly in $C^{0,\alpha}$ for every $\alpha \in (0,1)$, while $\lvert f_n \rvert_{1,p} \geq 1$ for every $n$.

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