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?
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.)