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:

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

  • 7
    $\begingroup$ 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. $\endgroup$ Nov 26, 2012 at 17:06
  • 3
    $\begingroup$ @user844541. I think you have reversed the inequality in the first part. It should be $||x||_2 \leq ||x||_1$. $\endgroup$
    – CKM
    May 14, 2015 at 9:02

5 Answers 5


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

  • $\begingroup$ @JonasMeyer ha ha,, careless me. Thanks for it $\endgroup$ Nov 27, 2012 at 7:03
  • $\begingroup$ Wow thanks!i spent hours on this thing... $\endgroup$
    – user844541
    Nov 27, 2012 at 8:42
  • $\begingroup$ How about in infinite dimensional vector spaces? $\endgroup$
    – Alan Wang
    Apr 7, 2017 at 1:05
  • $\begingroup$ @dineshdileep Can you please justify this step: $\left(\sum_{i=1}^{N}|x_i|^2+2*\sum_{i,j,i\neq j}|x_i||x_j|\right)=||x||_1^2$? $\endgroup$
    – johnny09
    Apr 21, 2019 at 0:13
  • 1
    $\begingroup$ $RHS = ||x||_1^2=(|x_1| + /dots + |x_n|)*((|x_1| + /dots + |x_n|) = LHS$ $\endgroup$ 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.

  • $\begingroup$ Can you please explain how did you use Miskowski inequality? $\endgroup$
    – CKM
    Nov 4, 2015 at 7:04
  • $\begingroup$ @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. $\endgroup$
    – Nobody
    Mar 14, 2018 at 13:46
  • $\begingroup$ Any hint on how to approach the generalization (but only a hint please)? $\endgroup$
    – Nobody
    Mar 14, 2018 at 13:56
  • $\begingroup$ are you sure this is minkowski inequality? $\endgroup$ Apr 19, 2018 at 8:10
  • $\begingroup$ 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 $\Vert f + g + h \Vert \le \Vert f + g \Vert + \Vert h \Vert \le \Vert f \Vert + \Vert g \Vert + \Vert h \Vert$. $\endgroup$
    – Squirtle
    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,

$\|a\|_p=$($\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$


Here is a proof that works for any norm on a finite dimensional vector space, not just $\ell^{N}_{p}$ norms for $p \in [1, \infty)$.

Assume wlog that $\{ \mathbf{e}_i\}_{i=1}^N$ is a basis such that $\| \mathbf{e}_i\| = 1$. (if not you can choose a new basis by normalizing the basis vectors.) In $\mathbb R^N$, you can choose the unit vectors along coordinate axes.

Now, \begin{align*} \| x \| &= \left\| \sum_{i=1}^{N} x_i \mathbf{e}_i \right\| \\ &\le \sum_{i=1}^{N} |x_i| \quad \left( \text{by triangle inequality and } \|\mathbf{e_i}\| = 1 \right) \\ &= \|x\|_1 \end{align*}


Hint for Squirtle's generalization: calculate derivative of $f(x)=(a_1^{x}+a_2^x+...+a_n^x)^{1/x}$ and prove $f'(x)<0,\forall x\geq 1$


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