Weak Convergence in Normed Dual Space of Reflexive Normed Vector Space is Equivalent to Weak-* Convergence/Proof 1

Theorem

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

Let $\struct {X, \norm {\, \cdot \,}_X}$ be a reflexive 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 $f \in X^\ast$.

Then:

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

Proof

Sufficient Condition

Suppose that $\sequence {f_n}_{n \mathop \in \N}$ converges weakly to $f$.

From Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent, we have:

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

$\Box$

Necessary Condition

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

Then:

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

for each $x \in X$.

We aim to show that:

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

Let $F \in X^{\ast \ast}$.

Since $X$ is reflexive, we have:

for each $F \in X^{\ast \ast}$ there exists $x_F \in X$ such that $F = x_F^\wedge$

where $x_F^\wedge : X^\ast \to \mathbb F$ is defined by:

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

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

Then, we have:

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

That is:

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

Since $F = x_F^\wedge$, we therefore have:

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

Since $F \in X^{\ast \ast}$ was arbitrary, we have:

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

$\blacksquare$