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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*} $


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

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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
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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.

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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
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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.

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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
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