Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was working on a homework problem which stated:

Let $X$ be a normed linear space with two norms $\|\cdot\|_1$ and $\|\cdot\|_2$. Assume that $\|x_n - x^1\|_1\to 0$ for some $x^1\in X$ if and only if $\|x_n - x^2\|_2\to 0$ for some $x^2\in X$. Show that $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent.

At first I errantly misinterpreted the question as $x_n\to x$ in $\|\cdot\|_1$ if and only if $x_n\to x$ in $\|\cdot\|_2$.

UPDATE: I found a way to go directly from the convergence condition to equivalence of norms. So I don't need to prove that the sequences converge to the same limit in this way (since it now follows from equivalence of norms). I won't erase anything though, as the answer from Andres below may be helpful to someone else.

Now that I realize my mistake I'm trying to fix the proof. Since the conclusion I'm trying to prove (that the norms are equivalent) implies that the limits in either norm must agree, I know that must also be true, even though it's not given in the question. After failing to prove this extra condition, I am getting suspicious that it might be needed as a hypothesis after all. Can anyone confirm or refute this?

My attempt:

Taking $\alpha = \dfrac12\min\left(\|x^1 - x^2\|_1, \|x^1 - x^2\|_2\right)$,

I let $B_{1} = \{x\in X : ||x - x^1||_{1} < \alpha\}$ and let $B_{2} = \{x\in X : ||x - x^{2}||_{2} < \alpha\}$.

Now suppose for a contradiction that there is some $x\in B_{1}\cap B_{2}$. Then $||x - x^{1}||_{1} < \alpha$. And $||x - x^{2}||_{2} < \alpha$. I want to use this to prove $||x^{1} - x^{2}||$ (in either norm) is less than $2\alpha$ (which would give me a contradiction). But I keep ending up with these floating $||x - x^{2}||_{1}$ or $||x - x^{1}||_{2}$ terms that I can't do anything with:

$ \begin{eqnarray*} ||x^{1} - x^{2}||_{1} &\leq& ||x^{1} - x||_{1} + ||x - x^{2}||_{1}\\ &<& \alpha + ||x - x^{2}||_{1} \end{eqnarray*} $

and

$ \begin{eqnarray*} ||x^{1} - x^{2}||_{2} &\leq& ||x^{1} - x||_{2} + ||x - x^{2}||_{2}\\ &<& ||x^{1} - x||_{2} + \alpha \end{eqnarray*} $

share|improve this question
    
Are you missing something like "$ \to 0$" after the norms of those differences in the homework problem? –  Ragib Zaman Oct 5 '11 at 3:12
    
Shouldn’t those be $\|x_n-x^2\|_2\to 0$ and $\|x_n-x^1\|_1\to 0$? –  Brian M. Scott Oct 5 '11 at 3:16
    
Absolutely. Thank you. –  Kyle Schlitt Oct 5 '11 at 3:17
    
Is it valid for EVERY sequence $x_n$? Then take the constant sequence $x_n = x^1$. –  André Caldas Oct 6 '11 at 6:11
    
Kyle: It would be great if you posted your solution as an answer if you have the time. –  t.b. Oct 7 '11 at 1:46
add comment

2 Answers

up vote 1 down vote accepted

Take $x_n = x^1$. Then, since $x_n \rightarrow x^1$ in any norm, we have that $\|x^1 - x^2\|_2 = \|x_n - x^2\|_2 \rightarrow 0$. That is, $x^1 = x^2$.

The above means that the identity $i: (X, \|\cdot\|_1) \rightarrow (X, \|\cdot\|_2)$ is continuous with continuous inverse at $x^1$. But this means that the identity is a homeomorphism. Now, this is "my definition" of equivalent norms: when the topology induced by both is the same. And this is the same as saying that the identity is a homeomorphism.

share|improve this answer
    
Our definition is that each norm is uniformly bounded by a constant multiple of the other. Of course both definitions are equivalent, but our homework problem was essentially to justify that. This definitely works. So thanks for this. I already submitted my argument which actually went directly from the convergence condition to the definition (my definition) of equivalence of norms. –  Kyle Schlitt Oct 9 '11 at 17:55
add comment

Edit: As Kyle pointed out, the answer below is not correct.


Take $\alpha = \dfrac12\min\left(\|x^1 - x^2\|_1, \|x^1 - x^2\|_2\right)$. Now, notice that the ball of radius $\alpha$ around $x^1$ in the first norm does not intersect the ball of radius $\alpha$ around $x^2$ in the second norm.

This means that the sequence cannot be in both balls at the same time.

share|improve this answer
    
Very elegant. Thank you times a million - I can modify my argument with this trick now. –  Kyle Schlitt Oct 5 '11 at 17:14
    
@ André Caldas: For some reason I am still stuck - although I geometrically see (if one can visualize two norms simultaneously...) why this should work. (See edits above.) –  Kyle Schlitt Oct 6 '11 at 0:53
    
@Kyle: It was very elegant... but completely wrong, it seems... :-) Sorry if I mislead you! –  André Caldas Oct 6 '11 at 5:31
    
No problem whatsoever! I got where I needed to go in the end. :) Thanks for helping all the same. –  Kyle Schlitt Oct 6 '11 at 14:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.