If $||.||$ and $||.||_0$ are equivalent then $||x_{n}-x||\rightarrow 0$ implies $||x_{n}-x||_{0}\rightarrow 0$

Question

How to show that if two norms $||.||$ and $||.||_0$ on a vector space $X$ are equivalent then $||x_{n}-x||\rightarrow 0$ implies $||x_{n}-x||_{0}\rightarrow 0$ (and vice versa)?

Answer 1

if $||||_1$ and $||||_2 $ are equivalent than you have $a,b \gt 0$ s.t for each $ x \in X $ you have : $$a||x||_1 \le ||x||_2 \le b||x||_1 $$ so, if $||x_n -x ||_1 \to 0 $

you get $$||x_n -x ||_2 \le b ||x_n-x ||_1 $$

since $ ||x_n -x||_1 \to 0$ you have $||x_n -x||_2 \to 0 $

and same works if $||x_n -x||_2 \to 0$

Answer 2

Recall that two norms $\|\cdot \|$ and $\|\cdot\|_0$ are equivalent iff there are two positive constants $C_1$ and $C_2$ such that for all $v\in X$, $$C_1\|v\|\leq \|v\|_0\leq C_2\|v\|\tag{*}.$$ Now take $v=x_n-x_0$.

i) If $\|x_n-x_0\|\to 0$ then by (*), $$0\leq \|x_n-x_0\|_0\leq C_2\|x_n-x_0\|\quad\implies \|x_n-x_0\|_0\to 0.$$

ii) If $\|x_n-x_0\|_0\to 0$ then by (*) (note that $C_1>0$), $$0\leq \|x_n-x_0\|\leq \frac{1}{C_1}\|x_n-x_0\|_0\quad\implies \|x_n-x_0\|\to 0.$$