# Showing that $l_2$ norm is smaller than $l_1$

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:

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

• 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. Nov 26, 2012 at 17:06
• @user844541. I think you have reversed the inequality in the first part. It should be $||x||_2 \leq ||x||_1$.
– CKM
May 14, 2015 at 9:02

## 3 Answers

I assume you are using finite dimensional vector spaces (looks like a familiar question from Golub and Van Loan).

\begin{align} ||x||_2^{2}=\sum_{i=1}^{N}|x_i|^2\leq\left(\sum_{i=1}^{N}|x_i|^2+2\cdot\sum_{i,j,i< 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\cdot\max_{i}(|x_i|^2)=N||x||_{\infty}^{2} \end{align} This implies $$||x||_2\leq \sqrt{N}||x||_{\infty}$$

• @JonasMeyer ha ha,, careless me. Thanks for it Nov 27, 2012 at 7:03
• Wow thanks!i spent hours on this thing... Nov 27, 2012 at 8:42
• How about in infinite dimensional vector spaces? Apr 7, 2017 at 1:05
• @dineshdileep Can you please justify this step: $\left(\sum_{i=1}^{N}|x_i|^2+2*\sum_{i,j,i\neq j}|x_i||x_j|\right)=||x||_1^2$? Apr 21, 2019 at 0:13
• $RHS = ||x||_1^2=(|x_1| + /dots + |x_n|)*((|x_1| + /dots + |x_n|) = LHS$ Apr 21, 2019 at 14:14

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.

• Can you please explain how did you use Miskowski inequality?
– CKM
Nov 4, 2015 at 7:04
• @chandresh consider one vector which consists of $(a_0,a_1,\cdots,a_{n-1},0)$ and one $(0,0,\cdots,0,a_n)$. However I don't see yet how the generalization works. Mar 14, 2018 at 13:46
• Any hint on how to approach the generalization (but only a hint please)? Mar 14, 2018 at 13:56
• are you sure this is minkowski inequality? Apr 19, 2018 at 8:10
• 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 $\Vert f + g + h \Vert \le \Vert f + g \Vert + \Vert h \Vert \le \Vert f \Vert + \Vert g \Vert + \Vert h \Vert$. Apr 20, 2018 at 22:52

This is a short proof to show that any vector norm is less than or equal to 1-norm.

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

$$\|a\|_p=$$($$\sum_{i=0}^n |a_i|^p)^{1/p}\le (\sum_{i=0}^{n-1} |a_i|^p)^{1/p} +(|a_n|^p)^{1/p}\\\le (\sum_{i=0}^{n-2} |a_i|^p)^{1/p} +(|a_{n-1}|^p)^{1/p} + (|a_n|^p)^{1/p} \\ ...\\...\\ \le(|a_1|^p)^{1/p} +(|a_2|^p)^{1/p} +....+ (|a_n|^p)^{1/p}\\=|a_1| +|a_2| +....+ |a_n| = \|a\|_1 \\$$

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