Weak-* Limit in Normed Dual Space is Unique
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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