Does a bounded sequence in $W^{1,1}(\Omega)$ necessarily have a weakly convergent subsequence

functional-analysissobolev-spaces

Let $\Omega \subset \mathbb{R}^3$ be a bounded region with smooth boundaries.

I am aware that the Sobolev space $W^{1,1}(\Omega)$ is compactly embedded in $L^1(\Omega)$,

Therefore, any sequence $\{ f_n \} \subset W^{1,1}(\Omega)$ bounded with respect to the $W^{1,1}$-norm must have a subsequence convergent with respect to $L^1$-norm.

Now, I am curious whether we can go further and find a subsequence "weakly convergent" in $W^{1,1}(\Omega)$.

What I mean is that: can we find a subsequence $\{ f_{n_k}\}$ and some $f \in W^{1,1}(\Omega)$ such that
\begin{equation}
\int_{\Omega} \Bigl( \sum_{i=1}^3 [\partial_i f_{n_k}] g_i+f_{n_k}g \Bigr) \to \int_\Omega \Bigl( \sum_{i=1}^3 [\partial_i f] g_i+fg \Bigr) \text{ as } n_k \to \infty
\end{equation}
for any $g_1, g_2, g_3, g \in L^\infty(\Omega)$?

I am of course aware that the Banach-Alaoglu Theorem does NOT apply to $L^1$ spaces. However, I wonder if $W^{1,1}$ would make any difference.

Could anyone please help me?

Best Answer

The $W^{k,p}$ are reflexive for $p\in (1,\infty)$ as explained here. Consider the sequence

$$f_{n}=0, x\leq 0, f_{n}(x)=nx, 0\leq x\leq 1/n, f_{n}(x)=1, x\in [1/n,1], f_{n}(x)=e^{-(x-1)^{2}},x\geq 1. $$

Then $f_{n}\in W^{1,1}$ and converges to the step function $f_{n}\to 1_{[0,1)}+1_{(1,\infty)}e^{-(x-1)^{2}}$ in $L^{1}$, but it is not in $W^{1,1}$ (see nice answer Weak Derivative Heaviside function.

Let's suppose by contradiction that $\exists w\in L^1_{loc}(\mathbb{R})$ such that $$\int_{\mathbb{R}}H(x)v'(x)dx=-\int_{\mathbb{R}}w(x)v(x)dx\qquad\forall v\in C^{\infty}_c(\mathbb{R}) $$ By the Fundamental Theorem of Calculus we have that $$\int_0^{+\infty}w(x)v(x)dx=v(0)\qquad\forall v\in C^{\infty}_c(\mathbb{R}) $$ By the Lebesgue Dominated Convergence Theorem we have that $$\lim_{r\to 0}\int_{-r}^{r}|w(x)|dx=0 $$ Then there exists $\delta>0$ such that $\int_{-\delta}^{\delta}|w(x)|dx\le 1/2$.

Let $v\in C^{\infty}_c(\mathbb{R})$ such that $\operatorname{supp}(v)\subset [-\delta,\delta]$ and $\max (v)=v(0)=1$. Then $$1=v(0)=\int_0^{+\infty}w(x)v(x)dx=\int_{-\delta}^{\delta}w(x)v(x)dx \le \max(v)\int_{-\delta}^{\delta}|w(x)|dx\le 1/2$$ Contradiction!!!

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