# Is the unit ball closed in $W^{1,\,p}(I)$ for $1<p\leq\infty$

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 $$1; more precisely, $$B_{p}$$ is compact in $$L^{p}(I)$$.

My Solution

Let $$1. Consider a sequence $$(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.

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 $$1. Consider a bounded sequence $$(u_{n})$$ in $$W^{1,\,p}(I)$$. Since $$W^{1,\,p}(I)$$ is compactly embedded in $$C(\overline{I})$$ for $$1. 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).

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.

• 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. – 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.
$$p=\infty$$: 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 $$\phi\in C_c^\infty(I)$$.
• @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. – StarBug Jul 10 '19 at 12:53
• @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 $C(\overline{I})$. – StarBug Jul 10 '19 at 15:40