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

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

  • 0
    Yes, correct...2012-12-05

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