Why are weak convergence and strong convergence equivalent in finite dimensional spaces? Please tell me, Where can I find a proof for this?



2 Answers 2


Take a basis $\{v_1,...,v_k \} $ of $V$, and define projection $\pi_i \in V^* $ as $$ \pi_i(\sum_j c_j v_j) = c_i $$ Then $\{\pi_1,...,\pi_k \} $ is also a basis of $V^*$ . Now introduce the $l^1$ norm as $ \|v\| = \sum_i |\pi_i(v)| $ for all $ v \in V $. Now if $v^n \rightharpoonup v \in V $ then $ \pi_i(v^n) \rightarrow \pi_i(v) $ for all $i$, hence $ \|v^n -v\| \rightarrow 0 $. But in finite dimensional space, all norms are equivalent, hence the convergence is strong.


I think most functional analysis or topology textbooks will answer your question. But in any case, the key for a finite dimensional space $E$ is that every point $x \in E$ can be written as $x = \sum_{i=1}^k x_i e_i$, where $e_1, e_2, \ldots, e_k$ form a basis in $E$ such that $||e_i|| = 1$ for all $i$. This allows you to choose a particular radius for neighborhoods in the weak topology based on neighborhoods in the strong topology.

We only need to check that an open set in the strong topology is open in the weak topology. Consider a neighborhood of $y$ with radius $r$, denoted by $B(y, r)$. The map $f_i(x) = x_i$ is continuous and linear. Hence

$$ ||x - x_0|| \leq \sum_{i=1}^k |\langle f_i, x - y \rangle |. $$

Let $V$ be the neighborhood

$$ V = \{ x \in E : | \langle f_i, x - y \rangle | < \frac{r}{k}, i = 1, \ldots, k \}. $$

Then clearly $||x - y|| < r$ for every $x \in V$, hence $V \subset B(y, r)$. We see here that $E$ being finite dimensional allowed us to choose $\frac{r}{k}$ as our radius in our neighborhood $V$ so that $V \subset B(y, r)$.


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