Prove that $lip_0(K)$ is separating if and only if it is weak$^*$-dense in $Lip_0(K)$.

Question

Please refer here for the definition of little Lipschitz functions.

The following statement is the first sentence in the proof of Lemma $2.4$.

Using Hahn-Banach Theorem, one can prove that $lip_0(K)$ is separating if and only if it is weak$^*$-dense in $Lip_0(K)$.

Question: How to prove the statement above?

Remark: Let $X$ be a Banach space. A subspace $S$ of $X^*$ is called separating if $x^*(x)=0$ for all $x^* \in S$ implies $x=0$.

Answer 1

More generally, let $X$ be a Banach space and $S\subseteq X^*$ be a linear subspace. Then $S$ is separating iff it is weak*-dense.

First, suppose $S$ is separating and let $U\subseteq X^*$ be a nonempty open set. Shrinking $U$, we may assume it has the following form: for some finite-dimensional subspace $Y\subseteq X$, some $g\in Y^*$, and some $\epsilon>0$, $$U=\{f\in X^*:\|f|_Y-g\|<\epsilon\}.$$

Let $T$ be the subset of $Y^*$ obtained by restricting each element of $S$ to $Y$. Then $T$ is a separating subspace of $Y^*$. Choose a basis $B$ for $T$ and extend it to a basis $B'$ of $Y^*$, and consider the dual basis on $Y$. Then every element of $T$ vanishes on the dual basis elements corresponding to elements of $B'\setminus B$. Since $T$ is separating, there can be no such elements, so $B=B'$ and $T=Y^*$. In particular, $g\in T$. This means there exists $f\in S$ such that $f|_Y=g$, and hence this $f$ is in $U$. Thus $S\cap U\not=\emptyset$.

Conversely, suppose $S$ is weak*-dense and let $x\in X$. By Hahn-Banach, there is some $f\in X^*$ such that $f(x)\neq 0$. The set $U=\{f\in X^*:f(x)\neq 0\}$ is weak*-open, and we just saw it is nonempty. Thus by density of $S$, $U\cap S\neq \emptyset$, i.e. there exists $f\in S$ such that $f(x)\neq 0$. Thus $S$ is separating.