# showing that l2 norm is smaller than l1

How can I show that L2<=L1

$||x||_1\le \cdot ||x||_2$

and also

$\|x\|_2\leq \sqrt m\|x\|_{\infty}$

regarding the first part, can I say that:

$$\sqrt{\sum\limits_{i=1}^n x^2 } \leq {\sum\limits_{i=1}^n {\sqrt x}^2 }$$

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You really have to specify your context here. What is $x$? Ell-$p$ spaces can be considered in different settings ($n$-tuples, infinite sequences, measurable functions over finite measure spaces, measurable functions over infinite measure spaces, say) and the answer to your question varies depending on each of those contexts. – Martin Argerami Nov 26 '12 at 17:06

This implies $||x||_2\leq ||x||_1$. Now \begin{align} ||x||_2^{2}=\sum_{i=1}^{N}|x_i|^2\leq N*\max_{i}(|x_i|^2)=N||x||_{\infty}^{2} \end{align} This implies $||x||_2\leq \sqrt{N}||x||_{\infty}$