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  1. Two norms $\def\norm#1{\lVert#1\rVert}\norm\cdot_1$ and $\norm\cdot_2$ are equivalent iff $\;\exists\;c_1,c_2>0$ such that $c_1\norm x_1\le \norm x_2\le c_2\norm x_1$

Show that $\norm x_1=\sum_{i=1}^n \lvert x_i \rvert$ and $\norm x_2=\left ( \sum_{i=1}^n x_i^2 \right )^{1/2}$ are equivalent.

It looks like $c_2$ is $1$, and that this can be proven with induction. But what could $c_1$ be?

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I mean what I don't understand is the following: If I square both terms, and expand $(\sum_{i=1}^n \lvert x_i \rvert)^2 = \left ( \sum_{i=1}^n x_i^2 \right ) + \sum_{i=1}^n x_i(x_k+\dotsb+x_n)_{x_k\neq x_i}$. However, the second term grows with $n$, so how can $\frac{\norm x_1}{\norm x_2} \leq C$ at all?

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  • $\begingroup$ Well, I think that $n$ is kept fixed. $\endgroup$ – Siminore Nov 5 '13 at 8:49
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You want to show that for a fixed $n$ the norms $\|\|_1,\|\|_2$ are equivalent in $\mathbb R^n$. The constant $c_1$ will depend on $n$.

Hint:
Cauchy Schwarz inequality.

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Let $x=x_i$ for all $i$. Then $||x||_1=x\dim x$, and $||x||_2=\sqrt{\sum_{i=1}^{\dim x}x^2}=\sqrt{x^2\dim x}=x\sqrt{\dim x}$. Therefore $$ \lim_{\dim x\to\infty} \frac{||x||_1}{||x||_2}=\sqrt{\dim x}=\infty $$ So the two cannot be equivalent by your definition.

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