Linear Combination of Weakly Convergent Sequences is Weakly Convergent

Theorem

Let $\Bbb F \in \set {\R, \C}$.

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

Let $x, y \in X$.

Let $\alpha, \beta \in \Bbb F$.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ converging weakly to $x$.

Let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence in $X$ converging weakly to $y$.

Then:

$\alpha x_n + \beta y_n \weakconv \alpha x + \beta y$

where $\weakconv$ denotes weak convergence.

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 $f \in X^\ast$.

Similarly, since:

$y_n \weakconv y$

we have:

$\map f {y_n} \to \map f y$

$\alpha \map f {x_n} + \beta \map f {y_n} \to \alpha \map f x + \beta \map f {y_n}$

Since $f$ is linear, we have:

$\alpha \map f {x_n} + \beta \map f {y_n} = \map f {\alpha x_n + \beta y_n}$ for each $n \in \N$.

and:

$\alpha \map f x + \beta \map f y = \map f {\alpha x + \beta y}$

So we have:

$\map f {\alpha x_n + \beta y_n} \to \map f {\alpha x + \beta y}$

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

So, from the definition of weak convergence, we have:

$\alpha x_n + \beta y_n \weakconv \alpha x + \beta y$

$\blacksquare$