If $||.||_1$ and $||.||_2$ are two equivalent norms such that $$||a||_1 \le ||b||_1$$ then do we have $$||a||_2 \le ||b||_2$$ ?
$\begingroup$
$\endgroup$
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
Not at all for example in $\mathbb R^2$ take $a = (1,0)$ $b = \left(\frac12, \frac12\right)$. You have $\|a\|_1 = 1 = \|b\|_1$ but $\|a\|_2 = 1 > \frac1{\sqrt{2}} = \|b\|_2$
-
$\begingroup$ This is not a counter-example to what was stated. $\endgroup$– S. DewarCommented Apr 19, 2018 at 8:54
-
-
$\begingroup$ As $\|a\|_1 \geq \|b\|_1$ (since $\|a_1\|=\|b\|_1$ and $\|a\|_2 \geq \|b\|_2$ (as $\|a\|_2> \|b\|_2$) $\endgroup$– S. DewarCommented Apr 19, 2018 at 8:58
-
2$\begingroup$ But what was stated $\|a\|_1 \le \|b\|_1 \implies \|a\|_2 \le \|b\|_2$ which negation is $\|a\|_1 \le \|b_1\| \text{ and } \|a\|_2 > \|b\|_2$. Which is verified by the example. $\endgroup$– KrokiCommented Apr 19, 2018 at 9:01
-