Cfrgtkky

Let's consider a weakly compact set $Ssubset W^{1,1}(Omega)$, 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 $uin S$ can be identified with $(u,nabla u)in L^1(Omega;Bbb Rtimes Bbb R^m)$ 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.)

The answer is NO.

Take $m=1$ and $Omega=(0,2pi)$. Consider
$$
u_n(x)=n^{-1/2}sin nx,quad ninmathbb N.
$$
Then $u_nto 0$, 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 $u_nto u$ weakly, then $u_n$ bounded. (Uniform boundedness principle.)