Norm equivalence

Theorem: If is a finite-dimensional vector space over , then any two norms on are equivalent.

Proof: Let be any basis for . Then, for any element of , e.g., , we have with . Let , and introduce

Assume be some norm on , then we just need to illustrate the equivalence of and . Obviously, we have the following estimate

For the reversed inequality, we decompose the proof into three steps.
Step 1: Prove is compact with respect to the norm .
Let be a sequence in , then we have . That is to say, for each , there must be some such that . Considering the index can only take finitely many values, there must be infinitely many such that for some fixed index . Since are contained in (closed set), we can select a converged subsequence. For the sequence selected just before, we can further select a converged subsequence when . Iteratively, we finally find a converged sequence. Because , the subsequences are always contained in , which illustrates the compactness of .
Step 2: Prove is compact with respect to the norm .
Let be a sequence in , then, according to Step 1, we can find a subsequence and in such that . Hence, we have

It means that is a subsequence converges to with respect to , which shows the compactness of .
Step 3: The norm is a continuous real valued function and is compact with respect to . A real valued continuous function on a compact set must achieve a maximum and minimum on that set. Hence, there must exist two constants and such that for all . Since if and only if , this implies that

Step 4: Show . If we had , then this would imply that there is an such that . So , and , which contradicting the fact that . Hence, we have .
Combining the statements from Step 1 to 4, we finally conclude the proof.

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2020