We say that two norms, and , on the same vector space are equivalent if there exist such that for every ,

.

In a finite dimensional normed space, all norms are equivalent.

**Proof**: Suppose is an n-dimensional space with basis elements and that and are norms on . We know (from an earlier lemma) there exists a such that for any we choose

(Where .)

Now consider and apply the triangle inequality.

Where . If we then apply our earlier inequality we’re left with

.

If we repeat this process but with and reversed, we achieve the other inequality. Since our choice of was arbitrary, and since our and don’t depend on our choice of , we see that and are equivalent. Since these were arbitrary norms, all norms on are equivalent.

Sadly, there is fatal flaw in this argument here. The quantifiers for x and c have been switched. the statement of the theorem says that there must exist a beta independent of x, however your beta depends on c which depends on x.

This was a subtle mistake that I fell for at first. I’ve been searching for more elementary proofs of this statement (we have a functional test coming up), and sadly I feel like the essence of this statement requires “work”. So anyone reading this comment, the best I’ve come up with is equating w/ sup norm. Then negating the “hard” direction.

Comment by Kevin Joyce — November 23, 2008 @ 2:51 pm |

I’m not 100% sure that I understand your concern. I agree that the quantifiers are (or were … fixed now) backwards, but I don’t see what’s wrong aside from that.

Comment by cjohnson — November 25, 2008 @ 9:24 pm |