# Equivalence of norms is a equivalence relation

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 $$c||x||_2\leqslant ||x||_3 \leqslant d||x||_2$$

$\Rightarrow$ : $$ac ||x||_1 \leqslant c||x||_2\leqslant ||x||_3 \leqslant d||x||_2\leqslant db||x||_1.$$

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There are some problems with your "proof". In particular the minus signs that crop up in proving symmetry are off-base. Your discussion of transitivity needs to be more specific about where the magic constants $e$ and $f$ come from. Reflexivity is okay! –  hardmath Dec 5 '12 at 16:45
Constants $a$ and $b$ need to be positive, something you omitted in defining equivalence of norms. –  hardmath Dec 5 '12 at 16:47
Looks fine except for the above mentioned positivity requirement, and you haven't demonstrated transitivity. You need to show that $e=ac$ and $f=db$ work. –  copper.hat Dec 5 '12 at 16:56
Is it correct now ? Thanks for all the feedback. –  bakabakabaka Dec 5 '12 at 17:26
Yes, correct... –  Berci Dec 5 '12 at 21:59