Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent

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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 $f$.


Proof 1

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 $f \in X^\ast$.

From Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have:

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

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

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

That is:

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

$\blacksquare$


Proof 2

Let $J$ be the evaluation linear transformation on $X$.

By Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual:

$J : X \to X^{\ast \ast}$


Thus, for each $x \in X$:

\(\ds \map {f_n} x\) \(=\) \(\ds \map {\map J x} {f_n}\)
\(\ds \) \(\stackrel {n \to \infty} {\longrightarrow}\) \(\ds \map {\map J x} f\)
\(\ds \) \(\) \(\, \ds = \, \) \(\ds \map f x\)

$\blacksquare$


Sources

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