Suppose that we have a functional $\mathcal{I}$ defined on $L^1(\Omega)^d$, for a spatial dimension $d\geq 2$, by
\begin{align*}
\mathcal{I} (u) =
\begin{cases}
\int_\Omega |\nabla u| \,\mathrm{d}x &\quad\text{ if } u \in W^{1,1}(\Omega),
\\
\infty &\quad \text{ otherwise}.
\end{cases}
\end{align*}
It is clear by standard results that the functional $\mathcal{I}$ is weakly lower semi-continuous with respect to weak convergence in $W^{1,1}(\Omega)$.
My question is this: is the functional weakly lower semi-continuous with respect to weak convergence in $L^1(\Omega)^d$?
I begin the argument as follows. Let $(u_n)$ be a sequence in $L^1(\Omega)^d$ such that it converges weakly in $L^1(\Omega)$ to a limit $u$. Without loss of generality (because otherwise the result is trivial), and taking a subsequence in $n$ not relabelled, we have
\begin{align*}
\liminf_{n\rightarrow\infty} \mathcal{I}(u_n) = \lim_{n\rightarrow\infty} \mathcal{I}(u_n) < \infty.
\end{align*}
Hence the sequence $(u_n)$ must be bounded in $W^{1,1}(\Omega)^d$. This is where I am now stuck.
Best Answer
This functional is not lower semicontinuous on $L^1$. To make things more explicit, I will assume $\Omega=(-1,1)^d$, but a similar construction works for arbitrary domains $\Omega$.
Let $u(x)=1$ if $x_1\geq 0$ and $u(x)=0$ if $x_1<0$. It is not hard to see that the distributional derivative $\partial_1 u$ is the surface measure of the hyperplane $\{x_1=0\}$. Thus $u\notin W^{1,1}(\Omega)$. Now define $u_n$ by $$ u_n(x)=\begin{cases}1&\text{if }x_1\geq 1/n,\\nx_1&\text{if }0\leq x_1<1/n,\\0&\text{if }x_1<0.\end{cases} $$ Again it is not hard to see that $\nabla u=n1_{0\leq x_1\leq 1/n}$. Moreover, $u_n\to u$ in $L^1$ and $\mathcal I(u_n)=1$. Thus $\mathcal I(u)>\lim_{n\to\infty}\mathcal I(u_n)$.