TITLE: Weak convergence $\iff$ strong convergence in finite dimensional space
QUESTION [8 upvotes]: I am seeking a proof of the following claim.

Weak convergence $\implies$ strong convergence in a finite-dimensional normed linear space.

Thank you.

REPLY [2 votes]: It is sufficient to prove that if $\mathrm{dim}(E) < \infty$
then $\sigma(E,E') = \mathcal{T}_E$ where $\sigma(E,E')$ is weak topology and $\mathcal{T}_E$ strong topology, or topology induced by norm on $E$. Equivalently every open $\mathcal{T}_E$-neighborhood of origin $B_E(\epsilon)$ it's also an open $\sigma(E,E')$-neighborhood. Let $x=(x_1,...,x_n)$, then $\left \| x \right \|_E:=\max_{1 \leq j \leq n} |x_j|$ defines a norm on $E$ 
(I can take this norm, since in finite dimensions are all equivalent), and
\begin{align*}
\displaystyle B_E(\epsilon)&=\lbrace x \in E : \left \| x \right \|_E < \epsilon \rbrace = \lbrace x \in E : |x_i| < \epsilon , \forall i=1,...,n \rbrace = \bigcap_{i=1}^n B_{\mathbb{K}}(\epsilon)
\end{align*}
where this intersection is by definition a $\sigma(E,E')$-neighborhood open.
Note that we have weak convergence iff there is convergence with respect weak topology, substantially by definition.