Weak convergence $\iff$ strong convergence in finite dimensional space

Question

I am seeking a proof of the following claim.

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

Thank you.

Answer 1

This is just an elaboration on what others have already said in the comments. Let $V$ be a finite-dimensional normed linear space (over $\mathbb F \in \{\mathbb R, \mathbb C\}$). As you say, all norms on a finite-dimensional vector space are equivalent, so we may assume that we have the usual 2-norm on $\mathbb F^n$. The definition of weak convergence is

Definition. If $(v_j)_{j=1}^\infty$ is a sequence of vectors in $V$, we say $v_j$ weakly converges to $v$, i.e., $v_j \rightharpoonup v$, iff $\ell(v_j) \to \ell(v)$ in $\mathbb F$ for all linear maps $\ell : V \to \mathbb F$, i.e., $\ell \in V^*$, where $V^*$ is the dual space of $V$.

We use the following theorem to prove that weak convergence implies strong convergence. (As the name implies, strong convergence implies weak convergence, which is a standard result.)

Theorem. The functionals $(e_i^*)_{i=1}^n$ given by $e_i^*(a_1 e_1 + \cdots a_n e_n) = a_i$ are a basis of $V^*$.

Suppose that $v_j \rightharpoonup v$. Then $e_i^*(v_j) \to e_i^* (v)$ for all $i \in \{1, \dots, n\}$. That is, $| e_i^*(v_j) - e_i^*(v) | \to 0$ in $\mathbb F$. Hence $\| v_j - v \|^2 = \sum_{i=1}^n |e_i^*(v_j) - e_i^*(v)|^2 \to 0$, which is to say that $v_j$ converges strongly to $v$.

Answer 2

This is perhaps best understood from a topological perspective.

A base for the weak topology is formed by finite intersections of sets of the form, $$\{u:a < \phi(u) < b\},$$

for any continuous linear functionals $\phi$. Geoometrically, one of these sets looks like the infinite slab between two parallel hyperplanes.

In finite dimensions, you can intersect a finite number of these sets to get an open set resembling a cube (or any other convex set really). Imagine intersecting the parallel planes in the x-direction with those in the y-direction, with those in the z-direction: in the middle you are left with a cube.

Now, if you make your cube small enough, you can fit it inside any open ball that is used to generate the strong topology. Thus the weak topology is at least as fine as the strong topology in finite dimensions, so weak convergence implies strong convergence there.

This also provides some intuition about why the result fails in infinite dimensions -- you can only "control" one direction per set of parallel hyperplanes. So in infinite dimensions, whatever finite set of hyperplanes you use, there will always be a complementary infinity of directions that are uncontrolled.

Answer 3

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.