Exercise 1.11.3

Answers

$| v |_{2} = \sqrt{| v_{1} |^{2} + \dots + | v_{n} |^{2}} \leq \sqrt{| v |_{\infty}^{2} + \dots + | v |_{\infty}^{2}} = \sqrt{n} | v |_{\infty}$ .

Let $w = (sign (v_{1}), \dots, sign (v_{n}))$ , so we have $ |v|_1 = |v^Tw| $\leq | v |_2 | w |_2 = \sqrt{n} | v |_2 $ .$

The inequality comes from Cachy-Schwarz inequality

niuers

2020-03-20 00:00