Two norms $||-||_1 $, $||-||_2$are equivalent if:
for two constants $a,b$ and $x$ from $V$ a vector space over a field it holds that : $$a||x||_1\leqslant ||x||_2\leqslant b||x||_1.$$
This is a equivalence relation because:
$$a||x||_1\leqslant ||x||_2\leqslant b||x||_1$$ and $$c||x||_2\leqslant ||x||_3 \leqslant d||x||_2$$
it follows that there are constants such that (transitivity): $$e||x||_1 \leqslant ||x||_3 \leqslant f||x||_1$$
Reflexivity: $$a||x||_1\leqslant ||x||_1\leqslant b||x||_1$$ with $a,b = 1$; this is true.
Symmetry: $$a||x||_1 \leqslant ||x||_2 \leqslant b||x||_1$$
if we take: $$-\frac{1}{b}||x||_2\leqslant x_1 \leqslant \frac{-1}{a} ||x||_2.$$
Is this a valid proof that the equivalence of two norms is truly a equivalence relationship?
Attempt 2:
Symmetry:
$$a||x||_1 \leqslant ||x||_2 \leqslant b||x||_1$$
$\Rightarrow $: $$\frac{1}{b} ||x||_2 \leqslant ||x||_1 \leqslant \frac{1}{a}||x||_2.$$
Transitivity: given
$$a||x||_1\leqslant ||x||_2\leqslant b||x||_1$$ and $$c||x||_2\leqslant ||x||_3 \leqslant d||x||_2$$
$\Rightarrow$ : $$ac ||x||_1 \leqslant c||x||_2\leqslant ||x||_3 \leqslant d||x||_2\leqslant db||x||_1.$$