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