# 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. – Martin Argerami Nov 26 '12 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 '15 at 9:02

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

• @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
• How about in infinite dimensional vector spaces? – Alan Wang Apr 7 '17 at 1:05
• @dineshdileep Can you please justify this step: $\left(\sum_{i=1}^{N}|x_i|^2+2*\sum_{i,j,i\neq j}|x_i||x_j|\right)=||x||_1^2$? – johnny09 Apr 21 '19 at 0:13
• $RHS = ||x||_1^2=(|x_1| + /dots + |x_n|)*((|x_1| + /dots + |x_n|) = LHS$ – dineshdileep Apr 21 '19 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 '15 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. – Nobody Mar 14 '18 at 13:46
• Any hint on how to approach the generalization (but only a hint please)? – Nobody Mar 14 '18 at 13:56
• are you sure this is minkowski inequality? – Mohsen_Fatemi Apr 19 '18 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 $\Vert f + g + h \Vert \le \Vert f + g \Vert + \Vert h \Vert \le \Vert f \Vert + \Vert g \Vert + \Vert h \Vert$. – Squirtle Apr 20 '18 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,
$$\||f+g||_p<=||f||_p+||g||_p$$

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

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