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I would have thought that this is a good candidate for an inductive proof, but I have searched for one and failed. Is there such a proof, and if not why not ? Here's how far I got.

  1. It's easy to establish that equivalence is transitive so it's only necessary to prove that any norm is equivalent to a norm of choice, e.g. $ ||.||_1$
  2. In any finite space any norm satisfies $||.|| \le M ||.||_1$. Prove this by picking a basis and applying the triangle inequality.
  3. That leaves to prove the other half of the equivalence inequality that $||.||_1 \le m ||.||$
  4. Base case for induction: $Dim(V) = 1$. For a basis vector $e_1$ in $V$ any vector $v = \nu_1 e_1$ and $||v|| = ||\nu_1 e_1|| = |\nu_1|.||e_1|| = ||v||_1.||e_1||/||e_1||_1$ So, $||v||_1 \le m||v||$ where $m = ||e_1||_1/||e_1||$

But, I can't prove the inductive step that if true for $Dim(V) = n$ then true for $Dim(V) = n+1$.

(I have some references for the topological proof - it's specifically an inductive proof that I'm looking for),

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  1. I am not sure induction is the best way to go.

  2. There are plenty of resources on MSE about the equivalence of norms in finite dimensional spaces. For example, here.

  3. One possible way to go is to observe that the unit ball of any norm in finite dimensional space is compact with respect to the usual Euclidean topology. That is, the set $B=\{x\in V: ||x||\leq 1\}$ is compact. Since $||\cdot||_1$ is continuous in the Euclidean topology, it is bounded on $B$, so there exists some constant $M>0$ so that

$$||x||_1\leq M\enspace\forall x\in B$$

Hence, since $||x/||x||||=1$ for all $x$,

$$||x||_1 \leq M||x|| \enspace \forall x\in V$$

whence you get the other direction you seek.

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  • $\begingroup$ Thanks, but it's specifically an inductive proof that I'm interested in. $\endgroup$ – Tom Collinge Oct 28 '15 at 9:12

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