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Whenever two norms are equivalent in the sense that $\|x\|_1\le c_1\cdot \|x\|_2$ and $\|x\|_2\le c_2\cdot \|x\|_1$, they generate the same topology. Is the reverse also true, i.e. if a topology is generated by two different norms, are the norms equivalent in the above sense? We know this to be true for $\mathbb{R}^n$, but is it generally true, and if not, what are some counterexamples?

What about a Hausdorff, translationally invariant vector space topology generated by two different metrics, are the metrics equivalent in the sense $d_1(x,y)\le c_1\cdot d_2(x,y)$ and vice versa?

s.harp
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Ivan
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    The answer here is "no." Consider the following two metrics generating the same topology on the positive real halfline: $d_1(x,y) = |x - y|$, and $d_2(x,y) = |1/x - 1/y|$. Because $x\mapsto 1/x$ is not Lipschitz continuous near the origin, the metrics are not equivalent (try to prove this). –  Feb 15 '12 at 20:11
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    It's been shown that any two separable infinite dimensional Banach spaces are topologically equivalent. See here. Of course, they need not be norm equivalent as you have... – David Mitra Feb 15 '12 at 20:15
  • WNY:Sorry, I meant a translationally invariant metric. One rather trivial example is $|x-y|/(1+|x-y|)$ but that's because it is a bounded metric. – Ivan Feb 15 '12 at 20:22
  • David Mitra: I am not completely sure about why a norm on space X should induce a norm on a homeomorphic space Y? It'd be good for me to have an example of non-equivalent norms on a Banach space that induce the same topology (and no, this isn't a homework question, it's for my own elucidation:-)) – Ivan Feb 15 '12 at 20:27
  • I don't see why two norms on two different Banach spaces will induce two different norms on one of the spaces? – Ivan Feb 15 '12 at 20:33
  • Ivan: Why not take the metric along with the homeomorphism? Say $h$ is your homeomorphism. Then define $d_Y(x,y) = d_X(h^{-1}(x), h^{-1}(y))$. Then $h$ is actually an isometric homeomorphism. Do the same for norms. In fact, all you really need is a set-theoretic bijection. You're basically inducing the same metric structure on $Y$ that you have on $X$ (i.e., you are just relabeling elements of $X$). –  Feb 15 '12 at 20:34
  • The problem with norms is that $||h^{-1}(x)||$ is not linear, and it doesn't satisfy the triangle inequality since $h^{-1}$ might not be linear. Am I missing something? – Ivan Feb 15 '12 at 20:38
  • Yes, in the case of vector spaces, I was implicitly assuming that $h$ is linear. –  Feb 15 '12 at 20:42
  • Bottom line is: For a finite dimensional vector space, all norms are equivalent. For infinite dimensional spaces, it is possible to have two nonequivalent norms inducing the same topology. For general metric spaces it is possible to have nonequivalent metrics inducing the same topology. –  Feb 15 '12 at 20:44
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    @WNY: It is not possible to have two nonequivalent norms on a vector space which induce the same topology. See Matthias's answer. – Nate Eldredge Feb 15 '12 at 21:06
  • @Ivan I'm sorry, I didn't read your question carefully enough; my previous comment talks about something different. – David Mitra Feb 15 '12 at 21:16
  • @Nate, Ivan: Yes, yes, of course. I wasn't being careful. To restate: It is possible to have homeomorphic infinite dimensional Banach spaces with nonequivalent norms; however, in this case the homeomorphism in question cannot be linear. –  Feb 15 '12 at 21:22

2 Answers2

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First, let's have a look at the case of a metric space $X$ and two metrics $d_1$ and $d_2$ on it. If these are equivalent, meaning $d_1(x,y) \leq c_1 d_2(x,y)$ and $d_2(x,y)\leq c_2 d_1(x,y)$ for some real numbers $c_1,c_2 > 0$ and all $x,y \in X$, we get the following inclusions for open balls $B_{r}^i(x) = \{y \in X \:|\: d_i(x,y) < r\}$: $$ B^2_{\frac{r}{c_1}}(x) \subseteq B^1_r(x) $$ $$ B^1_{\frac{r}{c_2}}(x) \subseteq B^2_r(x) $$ for all $x \in X$ and $r > 0$. As every open subset is a union of open balls, it follows that the metrics induce the same topology. The case for norms is a corollary now.

Conversely, if the topologies induced by two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space $X$ are identical, we have an inclusion $B_r^1(0) \subset B^2_1(0)$ for some $r > 0$. Let $x \in X\setminus{\{0\}}$ and set $y = \frac{rx}{2\|x\|_1}$. Then $\|y\|_1 = \frac{r}{2} < r$, hence $\|y\|_2 < 1$ which shows $\|x\|_2 \leq \frac{2}{r} \|x\|_1$. By symmetry, you get the desired equivalence.

Now the bad news: Two metrics inducing the same topology do not have to be equivalent: Take $X = \mathbb{R}$ and $d_1$ the normal norm induced metric and set $d_2 = \frac{d_1}{d_1 + 1}$. It is not difficult to show that $d_1$ and $d_2$ induce the same topology, but there is no $c > 0$ with $ d_1(x,y) \leq c \cdot d_2(x,y)$ as the inequality $n \leq c \frac{n}{n + 1}$ does not hold for $n > c$.

Matthias Klupsch
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    You've probably seen this argument in the proof that a linear operator between normed spaces is continuous iff it is bounded. This also gives an easy proof for normed spaces: If two norms $||\cdot||_1, ||\cdot||_2$ generate the same topology on $X$ then the identity operator $I : (X, ||\cdot||_1) \to (X, ||\cdot||_2)$ is a homeomorphism. So $I$ and $I^{-1}$ are continuous, hence bounded, and if you write down the definition of bounded you see that it means that the two norms are equivalent. The same argument in reverse also gives the converse. – Nate Eldredge Feb 15 '12 at 21:03
  • Can you clarify the inclusions? – mavavilj Jan 19 '19 at 17:39
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    @mavavilj : If $y \in B^2_{\frac{r}{c_1}}(x)$, then $d_1(x,y) \leq c_1d_2(x,y) < r$, so $y \in B^1_r(x)$. The second inclusion uses the same argument. – Matthias Klupsch Jan 25 '19 at 06:17
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I present my proof below which may be more intuitive.

Assume $|\cdot|$ and $[\cdot]$ are two norms that induce the same topology on $E$. Then there exist $c_1, c_2 >0$ such that $c_1|x| \le [x] \le c_2 |x|$ for all $x\in E$.

Let $c_2 := \sup_{x\in E} |x| / [x]$. Assume the contrary that $c_2 = \infty$. Then there is a sequence $(x_n)$ such that $|x_n| > n [x_n]$. Let $y_n := \frac{x_n}{n[x_n]}$. Then $y_n \to 0$ in $[\cdot]$. On the other hand, $|y_n| >1$ and thus $y_n \not \to 0$ in $|\cdot|$. This is a contradiction. Similarly, we get $1/c_1 := \sup_{x\in E} [x] / |x| < \infty$.

Analyst
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