I would like to know if the following proof is fine. I haven't filled in all the detail but please let me know what you think about the basic outline.(I am aware that there are posts which have dealt with this at some level, but I would like to know if this proof is valid.)

I will show that any weakly convergent sequence $\{u_{k}\}_{k}$ in $W^{1,\infty}(I)$ converges strongly in $L^{\infty}(I)$, for bounded open interval $I \subset \mathbb{R}^{n}$.

Proof: From Morrey's inequality we have the continuous embedding into Holder space: $W^{1,\infty}(I) \hookrightarrow C^{0,1}(I)$. This follows since we can show that $$\Vert u \Vert_{C^{0,1}(I)}\leq C\Vert u \Vert_{W^{1,\infty}(I)}$$ for some $C > 0$. To show that $C^{0,1}(I) \Subset C(I)$ we note that since $\{u_{k}\}_{k}$ is weakly convergent, it is bounded in $W^{1,\infty}(I)$. From the continuous embedding it follows that it is also bounded in $C^{0,1}(I)$. This gives: $$\Vert u_{n} \Vert_{C^{0,1}} = \sup\limits_{x \in I}|u_{n}| + \sup\limits_{x,y \in I, x \neq y}\frac{|u_{n}(x)-u_{n}(y)|}{|x-y|} \leq M\text{ } \text{ for some } M > 0$$ Therefore $\sup\limits_{x \in I}|u_{n}| \leq M_{1}$ (for some constant $M_{1}$) and $\sup\limits_{x,y \in I, x\neq y}\frac{|u_{n}(x)-u_{n}(y)|}{|x-y|} \leq M_{2}$ (for some constant $M_{2}$), this shows that $\{u_{n}\}_{n}$ is uniformly bounded and equicontinuous. It follows then from Arzela-Ascoli Theorem that $\{u_{n}\}_{n}$ has a uniformly convergent subsequence, say $\{{u_{n}}_{j}\}_{j}$. Since this subsequence is measurable, uniformly bounded and uniformly convergent, it follows that $\{{u_{n}}_{j}\}_{j}$ converges in $L^{\infty}(I)$. We can then use a well known Lemma to show that the whole sequence converges strongly in $L^{\infty}(I)$. $\square$


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    $\begingroup$ For me the proof is fine. $\endgroup$ – Tomás Mar 12 '14 at 19:23
  • $\begingroup$ Okay thanks. Does the equivalence of the limit follow simply from the continuity of the mappings involved and the lemma that I made reference to? So that I can state that $u_{n} \rightharpoonup u$ in $W^{1,\infty}$ implies $u_{n} \rightarrow u$ in $L^{\infty}$. $\endgroup$ – user100431 Mar 12 '14 at 19:29
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    $\begingroup$ Yes, its is a topological lemma and this lemma was proved in this site plenty of times $\endgroup$ – Tomás Mar 12 '14 at 19:55

The proof is correct.

The first embedding $W^{1,\infty}(I) \hookrightarrow C^{0,1}(I)$ is actually an isomorphism; depending on how you norm $W^{1,\infty}(I)$, it may be an isometric isomorphism. I would not call this (easy) fact an application of Morrey's lemma myself, but on the other hand it makes sense to classify it as an endpoint case of Morrey's lemma.

The well known Lemma you alluded to is "if every subsequence has a subsubsequence converging to $u$, then the sequence converges to $u$". To ensure that the limit of every $L^\infty$-convergent subsubsequence is indeed $u$, we can use some notion of convergence that is

  • weaker than weak $W^{1,\infty}$
  • weaker than strong $L^\infty$
  • not absurdly weak, i.e., uniqueness of a limit holds

E.g., weak $L^\infty$ convergence does the job. A subsubsequence converging in $L^\infty$ to some $w$ will also converge weakly in $L^\infty$ to the same limit $w$; on the other hand, the whole sequence converges weakly in $L^\infty$ to $u$. So $w=u$.

Everywhere here weak means weak${}^*$.


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