Theorem: Evaluation Mapping of Infinite-Dimensional Normed Vector Space is not Jointly Continuous with Normed Dual Space given Weak-* Topology

Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\R$.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $w^\ast$ be the weak-$\ast$ topology on $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.
Let $\struct {X, \norm {\, \cdot \,}_X} \times \struct {X^\ast, w^\ast}$ be the direct product of $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {X^\ast, w^\ast}$ with the product topology.
Define $\innerprod \cdot \cdot : \struct {X, \norm {\, \cdot \,}_X} \times \struct {X^\ast, w^\ast} \to \R$ by:
:$\innerprod x f = \map f x$ for each $\tuple {x, f} \in X \times X^\ast$.
Then $\innerprod \cdot \cdot$ is not continuous at $\struct { {\mathbf 0}_X, {\mathbf 0}_{X^\ast} }$.

Proof:
Let $\struct {\FF, \subseteq}$ be the set of non-empty finite subsets of $X$ equipped with set inclusion.
From Set of Finite Subsets is Directed Set with Set Inclusion, $\struct {\FF, \subseteq}$ is directed.
For each $x \in X$, define $\hat x : X^\ast \to \GF$ by:
:$\map {\hat x} f = \map f x$ for each $f \in X^\ast$.
From Intersection of Finitely Many Kernels of Linear Functionals in Infinite-Dimensional Vector Space is Infinite-Dimensional with Finite Codimension:
:$\ds \bigcap_{x \in F} \map \ker {\hat x} \ne \set { {\mathbf 0}_{X^\ast} }$ for each $F \in \FF$.
For each $F \in \FF$, take:
:$\ds f_F \in \bigcap_{x \in F} \map \ker {\hat x} \ne \set { {\mathbf 0}_{X^\ast} }$
with $\norm {f_F}_{X^\ast} = \size F$.
We have:
:$\size F = \norm {f_F}_{X^\ast} = \sup \set {\size {\map {f_F} x} : \norm x_X = 1}$
Hence we can take $y_F \in X$ with $\norm {y_F}_X = 1$ and:
:$\ds \size {\map {f_F} {y_F} } \ge \frac 1 2 \size F$
For each $F \in \FF$, take:
:$\ds x_F = \frac 2 {\size F} y_F$
so that:
:$\map {f_F} {x_F} = 1$ 
From {{NormAxiomVector|2}}, we have:
:$\ds \norm {x_F}_X = \frac 2 {\size F}$
Since $X$ is infinite, for each $n \in \N$ there exists a finite set $F_n \subseteq X$ with $\size {F_n} = n$.
Then for each $F \in \FF$ with $F \supseteq F_n$, we have:
:$\ds \norm {x_F}_X \le \frac 1 n$
Hence we have $x_F \to {\mathbf 0}_X$ in $\struct {X, \norm {\, \cdot \,}_X}$.
We show that $f_F \to {\mathbf 0}_{X^\ast}$ in $\struct {X^\ast, w^\ast}$.
Let $U$ be an open neighborhood of ${\mathbf 0}_{X^\ast}$ in $\struct {X^\ast, w^\ast}$.
Then there exists $\epsilon > 0$ and $F_0 \in \FF$ such that:
:$\set {g \in X^\ast : \size {\map g x} < \epsilon \text { for each } x \in F_0} \subseteq U$ 
Then for $F \in \FF$ with $F \supseteq F_0$, we have:
:$\map {f_F} x = 0$ for each $x \in F$
since:
:$\ds f_F \in \bigcap_{x \in F} \map \ker {\hat x}$ 
Hence:
:$f_F \in \set {g \in X^\ast : \size {\map g x} < \epsilon \text { for each } x \in F_0} \subseteq U$
for $F \supseteq F_0$. 
Since $U$ was an arbitrary open neighborhood of ${\mathbf 0}_{X^\ast}$, it follows that $f_F \to {\mathbf 0}_{X^\ast}$ in $\struct {X^\ast, w^\ast}$.
From Characterization of Convergence of Net in Product Topology, we have:
:$\tuple {x_F, f_F} \to \tuple { {\mathbf 0}_X, {\mathbf 0}_{X^\ast} }$
We have:
:$\innerprod {x_F} {f_F} = \map {f_F} {x_F} = 1$ 
while:
:$\innerprod { {\mathbf 0}_X} { {\mathbf 0}_{X^\ast} } = 0$
Hence $\innerprod \cdot \cdot$ is not continuous at $\struct { {\mathbf 0}_X, {\mathbf 0}_{X^\ast} }$, by Characterization of Continuity in terms of Nets.
{{qed}}