Image of Weakly Convergent Sequence under Compact Linear Transformation is Convergent

Theorem

Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be normed vector spaces.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a weakly convergent sequence with:

$x_n \weakconv x$

Then:

$T x_n \to T x$

Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively.

Let $f \in Y^\ast$.

Then $f \circ T \in X^\ast$.

Since $x_n \weakconv x$, we have:

$\map f {T x_n} \to \map f {T x}$ as $n \to \infty$.

So we have:

$T x_n \weakconv T x$ as $n \to \infty$.

Aiming for a contradiction, suppose that $\sequence {T x_n}_{n \in \N}$ does not converge to $T x$.

Then there exists $\epsilon > 0$ such that:

$\norm {T x_{n_k} - T x}_Y \ge \epsilon$

for all $k \in \N$, for a subsequence $\sequence {x_{n_k} }_{k \in \N}$.

From Weakly Convergent Sequence in Normed Vector Space is Bounded, $\sequence {x_n}_{n \in \N}$ and hence $\sequence {x_{n_k} }_{k \in \N}$ is bounded.

Since $T$ is compact, there exists a subsequence $\sequence {T x_{n_{k_j} } }_{j \in \N}$ such that:

$T x_{n_{k_j} } \to z \in Y$

$T x_{n_{k_j} } \weakconv z$

Since:

$T x_n \weakconv T x$

But then there exists $N \in \N$ such that:

$\norm {T x_{n_{k_j} } - T x}_Y < \epsilon$

for $j \ge N$.

This contradicts that:

$\norm {T x_{n_k} - T x}_Y \ge \epsilon$

for each $k \in \N$.

So, we have $T x_n \to T x$.

$\blacksquare$