# Any two norms on a finite-dimensional vector space are Lipschitz equivalent

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Does the following work as a proof that any two norms on a finite-dimensional vector space are Lipschitz equivalent? I can't see a mistake in it, but it's simpler than the one in my notes.

Let be the Euclidean norm, and another norm. If these two norms are Lipschitz equivalent, we are done by transitivity.

Now, , but is continuous, and only outputs onto the unit sphere-shell (which is compact) if we provide it with input on the unit circle, so must have compact image (being continuous image of compact set). In particular, by Heine-Borel, it is closed and bounded; hence there are such that , and so *on the unit circle*. Then that extends easily to all of because and we multiply both sides of the inequality by a constant.

Also, with equality at some (by "continuous function on compact set attains its bounds"); but by continuity: the term is , and since is a norm, it is only zero at one point (which is isolated from this circle).

Hence .

This proof seems simpler to me than the one which appears in my notes, which involves actually expanding the Euclidean norm in terms of its components - it's doing a different thing in each direction, rather than this one which has just the one idea.

Let be the Euclidean norm, and another norm. If these two norms are Lipschitz equivalent, we are done by transitivity.

Now, , but is continuous, and only outputs onto the unit sphere-shell (which is compact) if we provide it with input on the unit circle, so must have compact image (being continuous image of compact set). In particular, by Heine-Borel, it is closed and bounded; hence there are such that , and so *on the unit circle*. Then that extends easily to all of because and we multiply both sides of the inequality by a constant.

Also, with equality at some (by "continuous function on compact set attains its bounds"); but by continuity: the term is , and since is a norm, it is only zero at one point (which is isolated from this circle).

Hence .

This proof seems simpler to me than the one which appears in my notes, which involves actually expanding the Euclidean norm in terms of its components - it's doing a different thing in each direction, rather than this one which has just the one idea.

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**Smaug123**)..

To my mind, a more major issue is that it's very unclear throughout your proof what topology (norm) you're referring to when you say things like compact, continuous etc. I think you're assuming things that aren't necessarily true, but it is, as I say, unclear.

Note that the result you are claiming is NOT true for infinite dimensional vector spaces, so it's worrying that your proof doesn't appear to need the dimension to be finite.

Concrete counterexample you might want to consider:

Take our vector space to be the set of cts (real valued) functions on [0,1].

Define f_n(x) = 1/sqrt(x) (x > 1/n), sqrt(n) (x <= 1/n) (in other words, f_n(x) = min(1/sqrt(x), sqrt(n)).

Define two norms

Then .

, so .

Clearly there's no C s.t.

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There's one "sort of trivial" issue, which is that you seem to think your vector space has to be embedded in for some n, which need not be the case (and is one of the reasons most proofs start by choosing a basis and defining a norm in terms of that basis).

**DFranklin**)There's one "sort of trivial" issue, which is that you seem to think your vector space has to be embedded in for some n, which need not be the case (and is one of the reasons most proofs start by choosing a basis and defining a norm in terms of that basis).

To my mind, a more major issue is that it's very unclear throughout your proof what topology (norm) you're referring to when you say things like compact, continuous etc. I think you're assuming things that aren't necessarily true, but it is, as I say, unclear.

Note that the result you are claiming is NOT true for infinite dimensional vector spaces, so it's worrying that your proof doesn't appear to need the dimension to be finite.

Note that the result you are claiming is NOT true for infinite dimensional vector spaces, so it's worrying that your proof doesn't appear to need the dimension to be finite.

I've gone through it and I'm pretty sure that when I say "compact" (under either norm) I'm right. "Continuous image of compact set is compact" is true; unit circle under Euclidean topology is compact; so its image under the (continuous) M-norm is compact in R^n (whatever topology it induces) and hence is closed and bounded by Heine-Borel. (Or does Heine-Borel assume Euclidean topology? I think my standard proof of Heine-Borel works, but it's lunchtime now so I can't check it.)

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OK - so if my proof is correct then it has shown that all norms on a *real* f.d. space are Lipschitz equivalent?

**Smaug123**)OK - so if my proof is correct then it has shown that all norms on a *real* f.d. space are Lipschitz equivalent?

My proof requires finiteness for "closed and bounded iff compact" - Heine-Borel only holds in finite-dimensional spaces, right?

*image*of your map , and the image is a subset of R. So this works even for infinite dimensional spaces.

[On the other hand, having checked a bit on infinite dimensional vector spaces, I think that the actual point of failure is that the unit sphere/ball isn't compact in such a space, and you could say that claiming it without proof is fine for a f.d. real vector space is fine. Again, I don't recall whether that proof is obvious without relying on being able to identify the vectorspace with R^n).

I've gone through it and I'm pretty sure that when I say "compact" (under either norm) I'm right. "Continuous image of compact set is compact" is true

unit circle under Euclidean topology is compact; so its image under the (continuous) M-norm is compact in R^n (whatever topology it induces) and hence is closed and bounded by Heine-Borel. (Or does Heine-Borel assume Euclidean topology? I think my standard proof of Heine-Borel works, but it's lunchtime now so I can't check it.)

What you need to be aware of is that the result you are trying to prove is "foundational". Many results are easily proved by using this result to say "it's equivalent to R^n under a particular norm, and we already proved that back in Analysis I". So you've got to be very careful about circular reasoning here. That you're "not sure" how Heine-Borel is defined certainly doesn't convince me you're being careful enough. It's not that you can't "make it work", it's that you should be providing me with a proof that leaves me with no questions or doubts.

Of course, if you want to be sure, ask your lecturer/examiner what they think.

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That you're "not sure" how Heine-Borel is defined certainly doesn't convince me you're being careful enough. It's not that you can't "make it work", it's that you should be providing me with a proof that leaves me with no questions or doubts.

Of course, if you want to be sure, ask your lecturer/examiner what they think.

**DFranklin**)…

That you're "not sure" how Heine-Borel is defined certainly doesn't convince me you're being careful enough. It's not that you can't "make it work", it's that you should be providing me with a proof that leaves me with no questions or doubts.

Of course, if you want to be sure, ask your lecturer/examiner what they think.

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