This is an exercise from Brezis (8.12).

Exercise Question

Let $I=(0,1)$ and $1\leq p\leq\infty$. Set, \begin{align} B_{p}=\{u\in W^{1,\,p}(I)|\,\|u\|_{W^{1,\,p}(I)}\leq 1\}. \end{align} 1. Prove that $B_{p}$ is a closed subset of $L^{p}(I)$ when $1<p\leq\infty$; more precisely, $B_{p}$ is compact in $L^{p}(I)$.

My Solution

Let $1<p\leq\infty$. Consider a sequence $(u_{n})\in B_{p}$ which converges to $u$ in $L^{p}(I)$. Since $(u_{n})$ is bounded in $W^{1,\,p}(I)$ we have that $u_{n}'\rightharpoonup v$ in $B_{p}$. That is, \begin{align} \langle g,u_{n}'\rangle\rightarrow\langle g,v\rangle\quad\forall g\in L^{p^{*}}(I). \end{align} Furthermore, we have, \begin{align} \int\varphi' u_{n}=-\int\varphi u_{n}'\quad\forall\varphi\in C_{c}^{\infty}(I). \end{align} Since $C_{c}^{\infty}(I)\subset L^{p'}(I)$ we have, by Riesz representation, that $\int\varphi u_{n}'$ is the unique representation of the dual map for some $g\in L^{p^{*}}(I)$. Hence we have, \begin{align} \int\varphi v=\langle g,v\rangle\leftarrow\langle g,u_{n}'\rangle=\int\varphi u_{n}'=-\int\varphi' u_{n}\rightarrow-\int\varphi' u. \end{align} Therefore we have $v=u'$ and so $u\in W^{1,\,p}(I)$. Finally, by uniform boundedness, since $u_{n}\rightharpoonup u$ in $W^{1,\,p}(I)$ we have, \begin{align} \|u\|_{W^{1,\,p}(I)}\leq\liminf_{n\rightarrow\infty}\|u_{n}\|_{W^{1,\,p}(I)}\leq 1. \end{align} Therefore $\|u\|_{W^{1,\,p}(I)}\leq 1$ and hence $u\in B_{p}$. So $B_{p}$ is closed for $1<p\leq\infty$.

My Issues

I feel something is wrong with my proof. I can't account why this does not work for $p=1$ and I also don't see why it does work for $p=\infty$. When I initially look at the weak convergence, should I be treating the $u_{n}'$ as the functional? and the $g$ as the element of $L^{p}(I)$?

Second Solution

I first prove that every bounded $W^{1,\,p}(I)$ sequence has a convergent subsequence whos limit belongs to $W^{1,\,p}(I)$:

Let $1<p\leq\infty$. Consider a bounded sequence $(u_{n})$ in $W^{1,\,p}(I)$. Since $W^{1,\,p}(I)$ is compactly embedded in $C(\overline{I})$ for $1<p\leq\infty$. Since the $u_{n}$ are continuous the sequence $(u_{n})$ has at least one accumulation point $u$. Hence there exists a convergent subsequence $(u_{n_{k}})$ such that $u_{n_{k}}\rightarrow u$ in $C(\overline{I})$, i.e., $\|u_{n_{k}}-u\|_{\infty}\rightarrow 0$. We can now show that by Proposition 8.3 (Brezis) that $u\in W^{1,\,p}(I)$. \begin{align} \bigg|\int u\varphi'\bigg|=\lim_{k\rightarrow\infty}\bigg|\int u_{n_{k}}\varphi'\bigg|\leq\lim_{k\rightarrow\infty}\|u'_{n_{k}}\|_{p}\|\varphi\|_{p'}\leq C\|\varphi\|_{p'}. \end{align} This implies $u\in W^{1,\,p}(I)$. This shows that given a bounded sequence $(u_{n})$ in $W^{1,\,p}(I)$ there is a subsequence $(u_{n_{k}})$ which has a strong (and weak) limit $u\in W^{1,\,p}(I)$ (more precisely $u_{n_{k}}'\rightharpoonup u'$ in $L^{p}(I)$ for $1<p<\infty$).

Now, suppose $(u_{n})\in B_{p}$ such that $u_{n}\rightarrow u\in L^{p}(I)$. Then $u_{n}'\rightharpoonup v$ in $L^{p}(I)$. So by the above $u_{n}\rightarrow u\in W^{1,\,p}(I)$.

Finally we check that $\|u\|_{W^{1,\,p}(I)}\leq 1$. By the uniform boundedness principle we have $\|u\|_{W^{1,\,p}(I)}\leq\liminf_{n\rightarrow\infty}\|u_{n}\|_{W^{1,\,p}(I)}\leq 1$. Therefore, given a convergent sequence $(u_{n})\in B_{p}$ we have $u_{n}\rightarrow u\in B_{p}$.

This proof does not account for $p=\infty$. I do not know how to do this. I am not so proficient with topologies and weak* convergence.

  • $\begingroup$ I believe what I am trying to say is, there is something wrong with my proof because if right, it proves the case $p=1$ and not the case $p=\infty$. Unless I am misunderstanding something here. $\endgroup$ – Zeta-Squared Jul 10 '19 at 5:18

Your argument "Since $(u_{n})$ is bounded in $W^{1,\,p}(I)$ we have that $u_{n}'\rightharpoonup v$ in $B_{p}$" only works when $W^{1,\,p}(I)$ is a reflexive Banach space, which excludes the cases $p=1$ and $p=\infty$.

You can stil show the assertion for $p=\infty$, but you have to use a different or modified argument.

Additional comment:

$p=\infty$: Consider $(u_n)\subset B_\infty$ with $u_n\rightarrow u$ in $L^\infty$. Consider $W^{1,\infty}\subset (L^1\times L^1)^*$ via the identification $(u,u')\in L^\infty\times L^\infty = (L^1\times L^1)^*$. By Banach-Alaoglu's Theorem we have $u_n\rightharpoonup^* v$ (for some sub-sequence which we WLOG still denote $(u_n)$). The norm is weak-star lower semi-continuous, so $v\in B_\infty$. To verify that $v=u$, it suffices to observe that $$ \int_I u \phi\, dx = \int_I v \phi\, dx $$ for all $\phi\in C_c^\infty(I)$.

  • $\begingroup$ I believe I have another solution, however this depends on the fact that a bounded sequence of continuous functions has an accumulation point. Is this true? $\endgroup$ – Zeta-Squared Jul 10 '19 at 7:45
  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – nmasanta Jul 10 '19 at 9:47
  • $\begingroup$ @nmasanta: I disagree. The OP asked specifically why his proof did not exclude the case $p=1$ (a good question). I gave a specific answer to this question. $\endgroup$ – StarBug Jul 10 '19 at 12:53
  • $\begingroup$ @Jack: Your argument that a bounded sequence in $W^{1,p}$ has an accumulation point in $C(\overline{I})$ is correct due to the compact embedding of $W^{1,p}$ into $C(\overline{I})$. $\endgroup$ – StarBug Jul 10 '19 at 15:40

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