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I have a homework problem which I feel should be simple but is actually surprisingly tricky. This is why I love math sometimes....

Let $X$ be a normed linear space. Suppose $\|\cdot\|_1$ and $\|\cdot\|_2$ are two norms on $X$ such that $x_n\to x_0$ in $\|\cdot\|_1$ if and only if $x_n\to x_0$ in $\|\cdot\|_2$.

Show that $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent.

What I have done so far is show that if the topologies induced by $\|\cdot\|_1$ and $\|\cdot\|_2$ are equal, then the norms are equivalent. So I feel that this is only a partial solution because I have failed to show that the given assumption about convergence implies that the induced topologies by each norm are equal. I think this is a well known fact but I should still prove it.

I start by taking a $\|\cdot\|_1$-neighborhood $U$ of $0$ in $X$. Now I would need to show that for any $x\in U$, there is a $\|\cdot\|_2$-neighborhood $V$ of $x$ such that $V\subset U$. This would establish that every $\|\cdot\|_1$-open set is $\|\cdot\|_2$-open. The other direction would be exactly the same. Now I feel pretty dense here (not in the topological sense either) but I can't see how to apply the sequence condition. Any hints or advice on how to do this?

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2 Answers 2

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Let $B_i(x,r)$ be the $\|\cdot\|_i$-nbhd of radius $r$ centred at $x$. Suppose that no $\|\cdot\|_2$-nbhd $V$ of $x$ is a subset of $U$. For $n\in\mathbb{Z}^+$ choose a point $$x_n \in B_2\left(x,\frac1n\right)\setminus U.$$ What can you say about the sequence of $x_n$’s?

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We would then have a sequence that converges in the $||\cdot||_{2}$ norm, but not in the $||\cdot||_{1}$ norm. I see the argument now. Thank you very much. –  Kyle Schlitt Oct 2 '11 at 2:41
    
@Kyle: You’ve got it. You’re welcome. –  Brian M. Scott Oct 2 '11 at 2:44
    
@ Brian M. Scott: It turns out that my argument showing that the norms are equivalent if they induce the same topologies is broken (and not easily fixable). My new argument goes like this: Since the topologies induced by the norms are the same, the identity map from $T:(X, ||\cdot||_{1})\rightarrow (X,||\cdot||_{2})$ is continuous. In particular it is continuous at $0$. I can use the definition of continuity of a linear operator to finally get equivalence of the norms. I feel like I'm going the long way around to get to this conclusion. Is this normal? –  Kyle Schlitt Oct 2 '11 at 4:25
    
@Kyle: I've added a proof of this fact in an edit to my answer. Hope this helps. –  Giuseppe Negro Oct 2 '11 at 9:41
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@Kyle: Are you asking whether it’s normal to feel that way about this argument, or more generally? Offhand I don’t see any significantly simpler argument. In general I find that I often start with an unnecessarily complicated argument and then (eventually) end up simplifying it. (My Doktormutter was moderately notorious for this.) –  Brian M. Scott Oct 2 '11 at 9:48

A topological approach.

The fact that the topology of a space is completely determined by its convergent sequences is true in all metrizable spaces. For in those spaces the closure of a set coincides with the sequential closure, so if two metrizable topologies have the same convergent sequences their closure operator is the same and so they share the same closed sets. Then the topologies are equal.

Now clearly the topologies induced by $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ are metrizable.

EDIT Since the OP is asking I'd like to provide a proof of the fact that two norms inducing the same topology are equivalent. I like to start from the following.

Fact Let $p_1, p_2$ be norms ($\tiny{\text{see note below}}$) on a vector space $X$. Call $B_1=\{x \in X \mid p_1(x) <1\}$ and $B_2=\{x \in X \mid p_2(x)\}$. The following statements are equivalent:

  1. $p_1(x)\le p_2(x)$ for all $x\in X$;
  2. $B_2 \subset B_1$. (Mnemonically: the bigger the norm, the smaller the unit ball).

Proof The only non trivial implication is $2 \Rightarrow 1$. The sought inequality is obviously true if $x=0$, so take $x\ne 0$ and put $y=x /cp_2(x)$ with $c>1$. We have

$$p_1(y)=\frac{p_1(x)}{cp_2(x)}<1$$

so $p_1(x) < cp_2(x)$ for all $c>1$. As $c \to 1$ we get inequality 2. $\square$

We proceed now to the main point. Suppose $\lVert \cdot \rVert_1, \lVert \cdot \rVert_2$ are two norms on $X$ inducing the same topology. In particular, the $\lVert \cdot \rVert_1$-unit ball $B_1$ is a neighborhood of the origin in both topologies, meaning that there exists $r > 0$ such that the $\lVert \cdot \rVert_2$-ball $rB_2$ is contained in $B_1$. Define $p_1(x)=\lVert x \rVert_1, p_2(x)=\frac{\lVert x \rVert_2}{r}$. From the preceding Fact we infer that $p_1(x) \le p_2(x)$ for all $x$, that is

$$\lVert x \rVert_1 \le \frac{1}{r}\lVert x \rVert_2,\qquad \forall x \in X.$$

Proceeding in an analogous fashion we can prove an inequality going the other way round. So the two norms are equivalent. $\square$


Note: Seminorms is enough, actually. But we do not need this generality here.

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See Pete's answer here for an easy example showing that it fails horribly without metrics (cocountable topology on an uncountable set). –  t.b. Oct 2 '11 at 2:28
    
@Guiseppe: Thanks for the details, this does provide an argument that doesn't involve continuity, which I was curious about. I'm grateful for the assistance but don't forget this was a homework problem! :) –  Kyle Schlitt Oct 2 '11 at 17:44

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