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How can I show that $L_2\le L_1$

$||x||_1\ge ||x||_2$

and also we have that

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

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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    @user844541. I think you have reversed the inequality in the first part. It should be $||x||_2 \leq ||x||_1$.2015-05-14

3 Answers 3

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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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In fact, we can do something stronger than this

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

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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    Yes. And thanks for reaching out and asking. It is the Minkowshi inequality under repeated use, where the measure employed is the counting measure. Here is the idea: The M inequality says that $\Vert f+g \Vert \le \Vert f \Vert + \Vert g \Vert$. So if we apply it multiple times we can see that $\Vert f + g + h \Vert \le \Vert f + g \Vert + \Vert h \Vert \le \Vert f \Vert + \Vert g \Vert + \Vert h \Vert$.2018-04-20
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This is a short proof to show that any vector norm is less than or equal to 1-norm.

According to Minkowski Inequality,
$\||f+g||_p<=||f||_p+||g||_p$

$||a||_p=$($\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-2} |a_i|^p)^{1/p} +(|a_{n-1}|^p)^{1/p} + (|a_n|^p)^{1/p} \\ ...\\...\\ \le(|a_1|^p)^{1/p} +(|a_2|^p)^{1/p} +....+ (|a_n|^p)^{1/p}\\=|a_1| +|a_2| +....+ |a_n| = ||a||_1 \\ $

$\implies||a||_p\le ||a||_1$ for any $ p\gt1$