# Norms on a normed vector space $X$ are equivalent if and only if Cauchy sequences are preserved

This is my attempt in proving the above statement.

Suppose $\left \| . \right \|_{1}$ and $\left \| . \right \|_{2}$ are equivalent.$\\$ And suppose $\left \{ x_{n} \right \}$ is a Cauchy sequence under the first norm.$\\$ Let $\varepsilon$ be arbitrary.
So $\exists N$ such that $\forall m\geq n> N$ $\left \| x_{m}-x_{n} \right \|_{1}< C\varepsilon$ where, $C\left \| x_{m}-x_{n} \right \|_{2}\leq \left \| x_{m}-x_{n} \right \|_{1}$ (since the norms are equivalent).
so we can have $\left \| x_{m}-x_{n} \right \|_{2}< \varepsilon$ Hence the sequence is Cauchy under the second norm.

Now for the converse(that is where I got stuck) suppose that the Cauchy sequences are preserved under the two norms.

For the equivalence in norms the alternating definition that I'm going to use is, If norms are equivalent then
$C\overline{B_{1}}\subset \overline{B_{2}}\subset c\overline{B_{1}}$ Where $\overline{B_{1}}$ and $\overline{B_{2}}$ are the closed unit balls under the norm 1 and norm 2

Let $x \in \overline{B_{1}}$ So we have a sequence $\left \{ x_{n} \right \}$ in $B_{1}$ that converges to x. So that sequence is Cauchy in $\left \| . \right \|_{1}$ Hence by the supposition it is cauchy under $\left \| . \right \|_{2}$
After this could somebody please let me know how to proceed?

• You'll need to use Countable Choice to get that result. ​ ​
– user57159
Commented Sep 15, 2016 at 10:23

$\newcommand{\nrm}[1]{\left\lVert{#1}\right\rVert}\newcommand{\norm}{\nrm\bullet}$Personally, I would have gone directly with "$\neg A\implies\neg B$", instead of "$B\implies A$":

Suppose there is no constant $C$ such that $\norm_2\le C\cdot \norm_1$. Then, there is a $\norm_1$-Cauchy sequence which is not $\norm_2$-Cauchy.

Since for all $C$ there is $x$ such that $\nrm{x}_2>C\cdot\nrm{x}_1$, we can find a sequence $(x_n)$ such that $\nrm{x_n}_2>4^{n}\nrm{x_n}_1$

Now, let's consider $$v_n=\sum_{k=0}^{n} \frac{x_k}{2^k\nrm{x_k}_1}$$

We have (say, $m\le n$) $$\nrm{v_n-v_m}_1=\nrm{\sum_{k=m+1}^n \frac{x_k}{2^k\nrm{x_k}_1}}_1\le \sum_{k=m+1}^n \frac{\nrm{x_k}_1}{2^k\nrm{x_k}_1}=\\=2\left(1-2^{-n-1}\right)-2\left(1-2^{-m-1}\right)=2^{-m}-2^{-n}<2^{-m}$$

So $(v_n)$ is $\norm_1$-Cauchy.

However, $$\nrm{v_n-v_{n-1}}_2=\nrm{\frac{x_n}{2^n\nrm{x_n}_1}}_2=\frac{\nrm{x_n}_2}{2^n\nrm{x_n}_1}>\frac{4^n}{2^n}=2^n$$

So $(v_n)$ isn't $\norm_2$-Cauchy.

• ("we can find a sequence" is where Countable Choice was used.) ​ ​
– user57159
Commented Sep 15, 2016 at 12:12
• Yes. I use AC without giving much thought about it.
– user228113
Commented Sep 15, 2016 at 12:41
• Thank you so much for the responses. It helped alot. Well.. This was my first post on mathstack :D Commented Sep 15, 2016 at 14:34