Weak Limit in Normed Vector Space is Unique

Theorem

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

Let $x, y \in X$.

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

$x_n \weakconv x$

and:

$x_n \weakconv y$

where $\weakconv$ denotes weak convergence.

Then:

$x = y$

Proof

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

Since:

$x_n \weakconv x$

we have:

$\map f {x_n} \to \map f x$ for each $f \in X^\ast$.

Since:

$x_n \weakconv x$

we have:

$\map f {x_n} \to \map f y$ for each $f \in X^\ast$.

From Convergent Complex Sequence has Unique Limit, we have:

$\map f x = \map f y$ for each $f \in X^\ast$.

From Normed Dual Space Separates Points, we therefore have:

$x = y$

$\blacksquare$

Sources

• 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $27.1$: Weak Convergence