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Can we say that; if the $l_1$-norm of an arbitrary vector $a$ is smaller that $l_1$-norm of $b$ ($||a||_1 \le ||b||_1$) then the $l_2$-norm of $a$ is smaller than $l_2$-norm of $b$ ($||a||_2 \le ||b||_2$) as well?


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Take, e.g. in $\Bbb R^4$, $b=(1/3,1/3,1/3,1/3)$ and $a=(1,0,0,0)$. – David Mitra Nov 29 '12 at 18:45

Consider $\Bbb R^2$, $a=(1,3)$ and $b=(2,2)$. They have the same $l^1$ norm, but their $l^2$-norm is different.

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Have you computed the $l^2$ norms? – Davide Giraudo Nov 29 '12 at 18:25

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