$L^1$ norm equivalent to weak topology of $W^{1,1}$? Let's consider a weakly compact set $S\subset W^{1,1}(\Omega)$, where $\Omega$ is a domain in $\Bbb R^m$ with smooth boundary. It turns out that $(S,w)$ is metrizable.

Is the topology induced by the metric $d(u,v):=\int_{\Omega} |u(x)-v(x)|\, dx$ on $S$ equivalent to the weak topology $(S,w)$ that $S$ inherits from $W^{1,1}$?

We know that the map $T:(S,w)\to L^1(\Omega)$ defined by
$$
Tu:=u
$$
is a bijection from a compact space into a Hausdorff space, hence if we manage to show continuity of $T$ then $T$ is a homeomorphism onto its image.
An element $u\in S$ can be identified with $(u,\nabla u)\in L^1(\Omega;\Bbb R\times \Bbb R^m)$ so we can view $T$ as a projection of the first coordinate. It is clearly continuous but I don't know if it is weakly continuous or not. Am I missing something obvious or is the statement simply not true?
 A: By Rellich-Kondrachov the inclusion map from $W^{1,1}$ to $L^1$ is compact (with respect to the norm topologies).  And a compact operator is weak-to-norm sequentially continuous (standard exercise, use a double subsequence trick and Hahn-Banach).  So if we are right about $S$ being weakly metrizable, then it is true that the inclusion map from $(S,w)$ into $L^1$ is continuous and hence a homeomorphism onto its image.
(I got a little worried about the weak metrizability when Google turned up https://people.math.gatech.edu/~heil/6338/summer08/section9f.pdf, where Heil says on page 363 that weakly compact does not imply weakly metrizable, even in a separable space.  But his proposed counterexample is the closed unit ball of $\ell^1$, which isn't actually weakly compact.  And Daniel Fischer's proof looks good to me.)   (Added: I had an email conversation with Professor Heil, who acknowledges that this is a mistake in the notes.)
A: The answer is NO.
Take $m=1$ and $\Omega=(0,2\pi)$. Consider
$$
u_n(x)=n^{-1/2}\sin nx,\quad n\in\mathbb N.
$$
Then $u_n\to 0$, in the $L^1-$sense, 
but $\|u_n\|_{W^{1,1}}>\pi n^{1/2}$, i.e., $\{u_n\}$ is unbounded, and hence $u_n$ does not tend to zero in the weak $W^{1,1}-$sense.
Note. If $u_n\to u$ weakly, then $u_n$ bounded. (Uniform boundedness principle.)
