Metric Functionals and Weak Convergence
Given a metric space \((X,d)\), pick a point \(o\in X\) and define the set
\[ X^\vee := \{x \mapsto d(x,w)-d(o,w) \mid w\in X\}. \]Denote by \(X^\diamondsuit\) the closure of \(X^\vee\) in the topology of pointwise convergence. We call each element in \(X^\diamondsuit\) a metric functional. Note that every metric functional is a \(1\)-Lipschitz map \(X\to\mathbb{R}\) that vanishes at the point \(o\).
Definition Let \((X,d)\) be a metric space and \(X^\diamondsuit\) the space of all metric functionals on \(X\). We say that a sequence \((a_n)\) in \(X\) converges \(d\)-weakly to \(a\in X\) if for every \(\mathbf{h}\in X^\diamondsuit\) we have
\[ \liminf_{n\to\infty}\, \mathbf{h}(a_n) \geq \mathbf{h}(a).\]Theorem A bounded sequence in a normed linear space converges \(d\)-weakly if and only if it converges in the classical weak sense.
Yes. For example, if \(X\) is the real line equipped with the metric
\[ d(x,y)=\sqrt{|x-y|},\]we have \(X^\diamondsuit = X^\vee\cup\{0\}\) and the unbounded sequence \(a_n=n\) converges \(d\)-weakly to every point in \(X\).
Theorem If \(X\) is the space \(\ell_1\) or is a normed linear space whose dual space is strictly convex, then \(d\)-weakly convergent sequences are bounded.
There is an error in Theorem 1.4 in the printed version. Theorem 1.4 is not true for the space \(C[0,1]\) as pointed out here.
