Show that the following is true for a bounded sequence in $W^{1,p}(I)$

Question

Let $I=(0,1)$. Assume that $u_n$ is a bounded sequence in $W^{1,p}(I)$ with $1 \lt p \le \infty$.

1. Show that there exist a subsequence $(u_{n_k})$ and some $u$ in $W^{1,p}(I)$ such that $||u_{n_k}-u||_{L^{\infty}} \to 0$.

2. Show that $u_{n_k}'\rightharpoonup u'$ weakly in $L^{p}(I)$ if $1 \lt p \lt \infty$

3. Show that $u_{n_k}'\rightharpoonup^{*} u'$ in $\sigma(L^{\infty},L^1)$ if $p=\infty$

For (1) $W^{1,p}(I)$ is compactly embedded in $C(\bar{I})$ for $1 \lt p \lt \infty$. Thus the identity map being compact has a convergent subsequence in $C(\bar{I})$. I am not able to show that $u \in W^{1,p}(I) $

For (2),do I have to use that $C_c^{\infty}(I)$ is dense in $L^{P}(I)$??

For (3), How do I show this??

Thanks for the help!!

Answer 1

For part (1), to not stack subscripts too deeply, let us assume that the full sequence $(u_n)$ converges uniformly to the continuous function $u\in C([0,1])$.

If $1 < p < \infty$, by the reflexivity of $W^{1,p}(I)$ we know that $(u_n)$ has a subsequence $(u_{n_m})$ converging weakly to some $v \in W^{1,p}(I)$. Then $(u_{n_m}')$ converges weakly to $v'$ in $L^p(I)$, whence

$$u(y) - u(x) = \lim_{m\to \infty} \bigl(u_{n_m}(y) - u_{n_m}(x)\bigr) = \lim_{m\to\infty} \int_x^y u_{n_m}'(t)\,dt = \int_x^y v'(t)\,dt = v(y) - v(x)$$

for all $0 < x < y < 1$. That shows $u - v$ is constant. But $(u_{n_m} - v)$ converges weakly to $0$ and uniformly to $u-v$, which implies $u = v$, and we have shown that $u = v \in W^{1,p}(I)$.

For $p = \infty$, note that $W^{1,\infty}(I)$ consists precisely of the (equivalence classes modulo equality almost everywhere of) Lipschitz-continuous functions on $I$ (the one direction, that every $w \in W^{1,\infty}(I)$ has a Lipschitz continuous representative is easy, the other direction is a consequence of Rademacher's theorem, which is non-elementary). The boundedness of the sequence $(u_n)$ implies that the sequence is equi-Lipschitz, i.e. there is a $K\in \mathbb{R}$ with $\lvert u_n(x) - u_n(y)\rvert \leqslant K\cdot\lvert x-y\rvert$ for all $n$ and all $x,y\in I$. Then it's easy to see that a pointwise limit of any subsequence of $(u_n)$ is also Lipschitz continuous with Lipschitz constant $K$. If you already know/can use that every Lipschitz continuous function on $I$ belongs to $W^{1,\infty}(I)$, this part of (1) is also complete. If you can't use Rademacher's theorem, use that $W^{1,\infty}(I)$ embeds continuously into $W^{1,p}(I)$ for all $p < \infty$, so from the previous part you know that the limit function $u$ belongs to $W^{1,2}(I)$ (to pick a specific $p$, it's not important which we pick). It remains to see that the derivative $u'$ belongs to $L^{\infty}(I)$, and that follows from the Lipschitz-continuity of $u$ (if $u'$ were unbounded, we could find an interval $(x,y)$ such that $\lvert u(y) - u(x)\rvert > (K+1)\cdot \lvert y-x\rvert$, contradicting the fact that $u$ has Lipschitz constant $K$).

For (2), you don't need to use $C_c^{\infty}(I)$. The argument above shows that $(u_{n_k}')$ has subsequences converging weakly to $u'$ in $L^p(I)$, and all that remains is to deduce that the full sequence converges weakly to $u'$. This criterion is useful for that.

(3) is similar to (2), but due to the non-reflexivity of $L^{\infty}(I)$ requires an extra step. The boundedness of $(u_n)$ in $W^{1,\infty}(I)$ implies the boundedness of $(u_n')$ in $L^{\infty}(I)$. Without loss of generality we can assume $\lVert u_n'\rVert_{L^{\infty}(I)} \leqslant 1$ for all $n$. By the Banach-Alaoglu theorem, since $L^{\infty}(I)$ is the dual of $L^1(I)$, the closed unit ball of $L^{\infty}(I)$ is $\sigma(L^{\infty}, L^1)$-compact. Since $L^1(I)$ is separable, the closed unit ball of $L^{\infty}(I)$ is metrisable in the topology $\sigma(L^{\infty}, L^1)$, and therefore it is $\sigma(L^{\infty},L^1)$-sequentially compact. So every subsequence of $(u_{n_k})$ has a further subsequence that is $\sigma(L^{\infty},L^1)$-convergent, and it remains to show that $u'$ is the only possible limit.

Answer 2

Here is another solution. Let $1 < p \le \infty$.

Since $(u_n)_{n \ge 1}$ in bounded in $W^{1,p}(I)$, which is compactly embedded in $C(\overline I)$, there exist a subsequence $(u_{n_k})_{k\ge 1}$ and $u \in C(\overline I)$ such that $\| u_{n_k} - u \|_{L^\infty} \to 0$.

The sequence $(u'_{n_k})_{k \ge 1}$ is bounded in $L^{p}(I)$, hence there exist a subsequence $(u'_{n_{k_j}})_{j \ge 1}$ and $g \in L^{p}(I)$ such that $u'_{n_{k_j}} \rightharpoonup g$ weakly in $L^{p}(I)$ if $1 < p < \infty$ or weakly-$\star$ in $L^{\infty}(I)$ if $p=\infty$. But we have $$ \int_I u_{n_{k_j}} \varphi' = -\int_I u'_{n_{k_j}} \varphi \quad \forall\varphi \in C^\infty_c(I) $$ so that in the limit we obtain $$ \int_I u \varphi' = -\int_I g \varphi \quad \forall\varphi \in C^\infty_c(I), $$ which implies $g=u'$ an $u \in W^{1,p}(I)$.

Finally, we have that any subsequence of $(u'_{n_k})_{k \ge 1}$ has a further subsequence that converges to $u'$ (for the weak topology if $1 < p < \infty$ or the weak-$\star$ topology if $p=\infty$), hence the whole sequence $(u'_{n_k})_{k \ge 1}$ converges.