2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $\phi\in C_c^\infty(I)$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.