# Necessary and sufficient condition for two complete norms to be equivalent

Let $$E$$ be a normed vector space with two complete norms $$\left \|\cdot \right \|_1$$ and $$\left \|\cdot \right \|_2$$. Prove that those norms are equivalent if and only if every Cauchy sequence with respect to one norm is also Cauchy with respect to the other.

We only have to prove that $$\text{id}:\left (E,\left \|\cdot \right \|_1\right )\to \left (E,\left \|\cdot \right \|_2\right )$$ is continuous. Since both spaces are Banach, the open mapping theorem will guarantee that $$\text{id}$$ is a homeomorphism, and therefore the norms will be equivalent.

So, I thought of closed graph. Suppose that there is a sequence $$x_n$$ which tends to $$x$$ in $$\left \|\cdot \right \|_1$$, and $$x_n=\text{id}(x_n)$$ tends to $$y$$ in $$\left \|\cdot \right \|_2$$. We want $$y=\text{id}(x)=x$$.

But this is not true in general: Banach space with respect to two norms must be Banach wrt the sum of the norms?

Presumably the problem is that the example given previously does not satisfy the strong statement: "for every Cauchy sequence with respect to one norm, is also Cauchy with respect to the other".

There is a suggestion:

If $$(a_n)_n\subset \mathbb{R}$$ tends to $$0$$, then there exists another sequence $$(\varepsilon_n)_n\subset \mathbb{R}_{>0}$$ such that $$\varepsilon_n\to +\infty$$ but $$\varepsilon_na_n\to 0$$.

But I cannot see how to make use of it. How would you solve the exercise?

EDIT: This post has been marked as a duplicate, but I think that in an unfair way. The question equivalence of two definitions of norm equivalence: "$|\cdot|_1=|\cdot|_2^\alpha$" vs. "being a Cauchy sequence is the same for both norms" makes use of a knowledge that is more advanced with respect to the simpler knowledge that I am using here. I did not understand what they were talking about because I do not know anything about p-adic Analysis. Moreover, they use definitions of norm-equivalence that I am not familiarized with. This is an exercise given on a more basic subject, and our professors are no way interested on a solution like the one given in that post (providing that it really solves my problem, which I cannot assert since I do not understand it).

• May 6, 2018 at 16:55
• Due to completeness, Cauchy is the same as convergent. If you state the result for convergent sequences, then there is a similar result for arbitrary normed vector spaces over $\mathbf R$ or $\mathbf C$ (or other fields equipped with a nontrivial absolute value). See Theorems 2.1 and A.1, and Definition 1.3, in kconrad.math.uconn.edu/blurbs/gradnumthy/equivnorms.pdf.
– KCd
Nov 18, 2020 at 5:19

Assume that $\text{id}:\left (E,\left \|\cdot \right \|_1\right )\to \left (E,\left \|\cdot \right \|_2\right )$ is not bounded, i.e. there does not exists $M > 0$ such that $\|x\|_2 \le M\|x\|_1, \forall x \in X$. Hence there exists a sequence $(x_n)_n$ in $X$ such that $\|x_n\|_1 = 1$ and $\|x_n\|_2 \ge n^2$ for all $n \in \mathbb{N}$.

Consider $\left(\frac1n x_n\right)_n$. We have

$$\left\|\frac1n x_n\right\|_1 = \frac1n \xrightarrow{n\to\infty} 0$$

so $\frac1n x_n \xrightarrow{\|\cdot\|_1} 0$. In particular, $\left(\frac1n x_n\right)_n$ is Cauchy w.r.t. $\|\cdot\|_1$ so the assumption implies that $\left(\frac1n x_n\right)_n$ is Cauchy w.r.t. $\|\cdot\|_2$. In particular, it is bounded with respect to $\|\cdot\|_2$. On the other hand, we have

$$\left\|\frac1n x_n\right\|_2 \ge \frac1n \cdot n^2 = n$$

which is a contradiction.

Hence there exists $M > 0$ such that $\|x\|_2 \le M\|x\|_1, \forall x \in X$.