There exists M such that $M||\textbf{x}||_{\infty} \le ||\textbf{x}||$

Question

Show there exists M such that $M||\textbf{x}||_{\infty} \le ||\textbf{x}||$.

M positive constant, $||\circ||_{\infty}$ uniform norm, and $||\circ||$ a norm on $\mathbb{R}^m$.

This is the usual equivalence of norms in finite-dimensional vector spaces. See for example https://math.stackexchange.com/questions/599824/any-two-norms-on-finite-dimensional-space-are-equivalent? — v_lentin, Jan 12 '21 at 13:34

Hermis14 · Answer 1 · 2021-01-12T14:58:22.443

0

$\newcommand{\norm}[1]{\Vert #1 \Vert}$ If $\Vert \cdot \Vert$ is an Euclidean norm $\Vert \cdot \Vert_2$, $$\norm{x} _\infty = \max_{1\leq i \leq m} |x_i| \leq \sqrt{x_1^2 + \cdots + x_m^2} = \norm{x}_2 $$ Therefore, $M = 1$ will suffice in this case.

edited Jan 12 '21 at 14:58

answered Jan 12 '21 at 13:46

Hermis14

2,597

@casper54 did not say that $\lVert . \rVert$ was the Euclidean norm on $\mathbb{R}^{m}$. – v_lentin Jan 12 '21 at 14:26
@v_lentin I edited the answer. – Hermis14 Jan 12 '21 at 14:58

There exists M such that $M||\textbf{x}||_{\infty} \le ||\textbf{x}||$

1 Answers1