The following is extracted from the book 'Lipschitz Algebras' by N. Weaver. (page $36$)

It is standard that a bounded linear functional $\phi \in Lip_0(X)^*$ belongs to the predual of $Lip_0(X)$ if and only if it is continuous with respect to the weak$^*$ topology.

How to prove the above statement?

Note that $X$ is a pointed metric space (We denote this point as $0$). $Lip_0(X)$ is the set of all real-valued Lipschitz functions which satisfy $f(0)=0$.

  • $\begingroup$ This is true for general Banach spaces. In fact, if $\phi \in X^{**}$ is continuous w.r.t. the weak-$*$ topology on $X^{*}$, there is $x \in X$ with $J(x) = \phi$, $J : X \to X^{**}$ being the canonical embedding. This should be contained in many books on functional analysis. $\endgroup$ – gerw Nov 23 '15 at 10:05

This is a general fact. Let $Y$ be any Banach space, $Y^*$ its dual (this is ${\rm Lip}_0(X)$ in your question) and $j_Y \colon Y \to Y^{**}$ the canonical inclusion map. Suppose $y^{**} \colon Y^{*} \to \mathbf K$ is a linear, weak$^*$ continuous functional (note that $y^{**} \in Y^{**}$ as weak$^*$ continuity implies norm continuity). We will show that $y^{**} \in j_Y[Y]$. As, $y^{**}$ is weak$^*$ continuous, there are $y_1, \ldots, y_k \in Y$ and $C> 0$, such that $$ \def\abs#1{\left|#1\right|}\abs{y^{**}(y^*)} \le C \max_i \abs{y^*(y_i)} $$ This implies $\bigcap_i \ker j_Y(y_i) \subseteq \ker y^{**}$, hence, there are $a_i \in \mathbf K$ with $y^{**} = \sum_i a_i j_Y(y_i)$ or $$ y^{**} = j_Y\left(\sum_i a_i y_i \right) \in j_Y[Y]. $$

  • $\begingroup$ Just want clarify something, your $[Y]$ means closure of $Y$? $\endgroup$ – Idonknow Nov 23 '15 at 13:27
  • $\begingroup$ No. $j_Y[Y]$ means the image of $Y$ in $Y^{**}$ under the canonical embedding $j_Y \colon Y \hookrightarrow Y^{**}$ $\endgroup$ – martini Nov 23 '15 at 13:28
  • $\begingroup$ For the converse, suppose $y^{**} \in j_Y[Y]$ and $y^*_n$ converges to $y^*$ in weak$^*$ topology. Then we have $j_Y(y)=y^{**}$ for some $y \in Y$. By the definition of convergence in weak$^*$ topology, we have $y^*_n(y)$ converges to $y^*(y)$. Hence, we have $y^{**}(y^*_n)$ converges to $y^{**}(y^*)$. $\endgroup$ – Idonknow Nov 24 '15 at 4:14

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