Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent/Proof 1

Theorem

Let $\mathbb F$ be a subfield of $\C$.

Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\mathbb F$.

Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$.

Let $f \in X^\ast$.

Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $X^\ast$ converging weakly to $f$.

Then $\sequence {f_n}_{n \mathop \in \N}$ converges weakly-$\ast$ to $f$.

Let $x \in X$.

We aim to show that:

$\map {f_n} x \to \map f x$

Then, since $x \in X$ was arbitrary, we will obtain that $\sequence {f_n}_{n \mathop \in \N}$ converges weakly-$\ast$ to $f$.

Since $\sequence {f_n}_{n \mathop \in \N}$ converges weakly to $f$, we have:

$\map F {f_n} \to \map F f$ for each $F \in \paren {X^\ast}^\ast$.

Define $x^\wedge : X^\ast \to \mathbb F$ by:

$\map {x^\wedge} f = \map f x$

for each $f \in X^\ast$.

$x^\wedge \in \paren {X^\ast}^\ast$

So, setting $F = x^\wedge$, we obtain:

$\map {x^\wedge} {f_n} \to \map {x^\wedge} f$

That is:

$\map {f_n} x \to \map f x$

$\blacksquare$