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Show that $\forall x\in \mathbb R^n$ : $|x|_{M}\leq|x|\leq|x|_{s}\leq n|x|_{M}$

where,

$|x|_{M}$ $=$ $max$ {${|x_{1}|,...,|x_{n}|}$} $=$ maximum of absolute values of the components $=$ sup norm of $x$ $=$ infinity norm of $x$

$|x|_{s}$ $=$ $|x_{1}|$ $+$ $|x_{2}|$ $+$ $...$ $+$ $|x_{n}|$ $=$ sum of absolute values of the components $=$ one norm of $x$

$|x|$ $=$ $\sqrt{<x,x>}$ $=$ $\sqrt{\sum x_{i}^2}$ $=$ Euclidean norm of $x$ $=$ two norm of $x$

I have the following definition as a help: two norms are equivalent if exist $K,M \in \mathbb R$ such that $K|x|^{*}$ $\leq$ $|x|^{+}$ $\leq$ $M|x|^{*}$

this is what i wear so far:

$|x|_{M}^2\leq \sum|x_{i}|^2=|x|^2$. Thus, $|x|_{M} \leq |x|$ But could you give me an idea of how the others would be?

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  • $\begingroup$ This post math.stackexchange.com/questions/218046/… gives a generalization of the answer you are looking for. It applies to all "proper" norms - which include the 1-norm and $\infty$-norm. $\endgroup$ – firdaus Oct 25 '20 at 21:27
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Let $e_i \in \mathbb R^n$ be the vector whose $j$-th entry is a $1$ if $j=i$ and $0$ if $j \neq i$. Then: \begin{align}\|x\| &= \|x_1e_1 + \cdots + x_ne_n\| \\ &\leq \|x_1e_1\| + \cdots + \|x_ne_n\| \\ &= |x_1|\|e_1\| + \cdots + |x_n|\|e_n\| = |x_1| + \cdots + |x_n|. \end{align} Also, since $|x_i| \leq \|x\|_M$ for all $i$, we have $|x_1| + \cdots + |x_n| \leq \|x\|_M + \cdots + \|x\|_M = n\|x\|_M$.

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