Compactness of the closed unit ball of ${\rm Lip}_0(X)$ in the topology of pointwise convergence implies that ${\rm Lip}_0(X)$ is a dual space

Question

It is proven that a closed unit ball in the set of real-valued Lipschitz functions ${\rm Lip}_0(X)$ defined on a Banach space $X$ is compact for the topology of pointwise convergence.

However, I fail to understand that this implies that ${\rm Lip}_0(X)$ is a dual space. In other words, the space ${\rm Lip}_0(X)$ is a predual.

Can anyone explain to me?

score 2 · Accepted Answer · answered Nov 10 '15 at 09:17

2

This follows from a general fact that if $B$ is a Banach space and $E\subset B^*$ is a set of funtionals that separate points in $E$ and the closed unit ball of $B$ is compact with respect to the weak topology introduced by $E$, then $B$ is isometric to $({\rm span}\, E)^*$. See this paper.

In your case $E=\{\delta_x\colon x\in X\}$.

answered Nov 10 '15 at 09:17

Tomasz Kania

16,361

I think it should be '$B$ is isometric to $(\overline{span(E)})^*$'. – Idonknow Nov 10 '15 at 12:55
Sure, however a normed space and its completion have the same dual space. – Tomasz Kania Nov 10 '15 at 13:03
I am wondering why weak topology and topology of pointwise convergence of $Lip(X)$ coincide. – Idonknow Nov 23 '15 at 15:05

Compactness of the closed unit ball of ${\rm Lip}_0(X)$ in the topology of pointwise convergence implies that ${\rm Lip}_0(X)$ is a dual space

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