# Are two norms equivalent if they induce the same topology on a vector space?

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?

• 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).
– user2093
Commented Feb 15, 2012 at 20:11
• 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... Commented Feb 15, 2012 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
Commented Feb 15, 2012 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
Commented Feb 15, 2012 at 20:27
• @WNY: It is not possible to have two nonequivalent norms on a vector space which induce the same topology. See Matthias's answer. Commented Feb 15, 2012 at 21:06

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 $$n > c$$.

• 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. Commented Feb 15, 2012 at 21:03
• Can you clarify the inclusions? Commented Jan 19, 2019 at 17:39
• @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. Commented Jan 25, 2019 at 6:17

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 $$1/c_1 := \sup_{x\in E} [x] / |x| < \infty$$.