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

I am seeking a proof of the following claim.

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

Thank you.

• What are your thoughts on the matter? Apr 15, 2016 at 16:09
• All norms on a finite dimensional space are equivalent. I think this has to play into the proof. Finite dimensional spaces are complete. I cannot find a solid definition of weak convergence and I'm not sure how to piece it together. Apr 15, 2016 at 16:11
• Are you familiar with the fact that a sequence is weakly convergent if and only if it converges in the weak topology?
– user228113
Apr 15, 2016 at 16:13
• I am vaguely familiar. I am approaching this from more of a functional analysis perspective and not a topological one, if that makes sense. Apr 15, 2016 at 16:14
• Could you not just pick the standard basis and go from there? That will tell you that weak convergence implies coordinate-wise convergence which implies strong convergence? Apr 18, 2016 at 1:39

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 $$v$$.

• I don't think equivalence of norms holds for finite-dimensional vector spaces over an arbitrary subfield of $\mathbb{C}$. The proofs I know all use some kind of completeness and/or compactness argument. Apr 23, 2016 at 18:51
• @JossevanDobbendeBruyn I believe you're right. Edited! Apr 26, 2023 at 20:23

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.

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.