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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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4  
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

2 Answers 2

up vote 4 down vote accepted

I assume you are using finite dimensional vector spaces (looks like a familiar question from golub and loan).

\begin{align} ||x||_2^{2}=\sum_{i=1}^{N}|x_i|^2\leq\left(\sum_{i=1}^{N}|x_i|^2+2*\sum_{i,j,i\neq j}|x_i||x_j|\right)=||x||_1^2 \end{align}

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

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@JonasMeyer ha ha,, careless me. Thanks for it –  dineshdileep Nov 27 '12 at 7:03
    
Wow thanks!i spent hours on this thing... –  user844541 Nov 27 '12 at 8:42

In fact, we can do something stronger than this

$${\Vert a \Vert}_p = (\sum_{i=0}^n a_i^p)^{1/p} \le (\sum_{i=0}^{n-1} a_i^p)^{1/p} + (a_n^p)^{1/p} \le .... \sum_{i=0}^n a_i = {\Vert a \Vert}_1$$

Where each inequality is using the Minkowski inequality. Moreover, we can generalize this idea further to show

$${\Vert * \Vert}_q \le {\Vert * \Vert}_p \text{ whenever } p\le q$$

It is a good exercise.

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Why the down vote? The math is correct. –  Squirtle Nov 14 '13 at 3:05

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