Weak-* Limit in Normed Dual Space is Unique

Theorem

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

Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space for $\struct {X, \norm \cdot}$.

Let $f, g \in X^\ast$.

Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $X^\ast$ such that:

$f_n \weakstarconv f$

and:

$f_n \weakstarconv g$

where $\weakstarconv$ denotes weak-$\ast$ convergence.

Then:

$f = g$

Proof

By the definition of weak-$\ast$ convergence, we have:

$\map {f_n} x \to \map f x$ for each $x \in X$

and:

$\map {f_n} x \to \map g x$ for each $x \in X$.

Then, from Convergent Complex Sequence has Unique Limit we have:

$\map f x = \map g x$ for each $x \in X$.

That is:

$f = g$

$\blacksquare$

Sources

• 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $27.4$: Weak-$\ast$ Convergence