While looking over my notes, my lecturer stated the following inequality; $$\|x\|_2 \leq \|x\|_1 \leq \sqrt{n}\|x\|_2$$ where $x \in \mathbb{R^n}.$ There was no proof given, and I've been trying to prove it for a while now. I know the definitions of the $1$ and $2$ norm, and, numerically the inequality seems obvious, although I don't know where to start rigorously.

Thank you.


5 Answers 5


We will show the more general case, i.e.:

$\|\ \cdot \|_1$ , $\|\ \cdot \|_2$, and $\|\ \cdot \|_{\infty}$ are all equivalent on $\mathbb{R}^{n}$. And we have $$\|\ x \|_{\infty} \leq \|\ x \|_{2} \leq \|\ x \|_{1} \leq n \|\ x \|_{\infty}\ $$

Every $x \in \mathbb{R}^{n}$ has the representation $x = ( x_1 , x_2 , \dots , x_n )$. Using the canonical basis of $\mathbb{R}^{n}$, namely $e_{i}$, where $e_i = (0, \dots , 0 , 1 , 0 , \dots , 0 )$ for $1$ in the $i^{\text{th}}$ position and otherwise $0$, we have that $$\|\ x \|_{\infty} = \max_{1\leq i \leq n} | x_i | = \max_{1\leq i \leq n} \sqrt{ | x_i |^{2} } \leq \sqrt{ \sum_{i=1}^{n} | x_ i |^{2} } = \|\ x \|_2 $$ Additionally, $$ \|\ x \|_2 = \sqrt{ \sum_{i=1}^{n} | x_i |^{2} } \leq \sum_{i=1}^{n} \sqrt{ | x_ i |^{2} } = \sum_{i=1}^{n} |x_i| = \|\ x \|_1$$ Finally, $$ \|\ x \|_1\ = \sum_{i=1}^{n} |x_i| \leq \sum_{i=1}^{n} | \max_{1 \leq j \leq n} x_j | = n \max_{i \leq j \leq n} | x_j | = n \|\ x \|_{\infty}$$ showing the chain of inequalities as desired. Moreover, for any norm on $\mathbb{R}^{n}$ we have that: $$\|\ x - x_{n} \|\ \to 0 \hspace{1cm} \text{as} \space\ \space\ n \to \infty $$ so that they are equivalent, as this holds for any $x \in \mathbb{R}^{n}$ under any norm actually.

  • 2
    $\begingroup$ If I wanted to more explicitly show that $\|\cdot\|_1$ was equivalent to $\|\cdot\|_2$ could I use the fact that $\|x\|_1\le\ n|x\|_\infty\iff\frac1n \|x\|_1\le\|x\|_\infty$ to deduce that $\frac1n \|x\|_1\le \|x\|_2\le \|x\|_1$? Or would I need to do something more explicit? $\endgroup$ Mar 13, 2018 at 22:05
  • 17
    $\begingroup$ @JeremyJeffreyJames This should work: $$ \|\ x \|_2 = \sqrt{ \sum_{i=1}^{n} | x_i |^{2} } \leq \sum_{i=1}^{n} \sqrt{ | x_ i |^{2} } = \sum_{i=1}^{n} |x_i| = \|\ x \|_1.$$ Then, using the Cauchy–Schwarz inequality we get for all $x\in\mathbb{R}^n$: $$ \Vert x\Vert_1= \sum\limits_{i=1}^n|x_i|= \sum\limits_{i=1}^n|x_i|\cdot 1\leq \left(\sum\limits_{i=1}^n|x_i|^2\right)^{1/2}\left(\sum\limits_{i=1}^n 1^2\right)^{1/2}= \sqrt{n}\Vert x\Vert_2, $$ $\endgroup$
    – Dragonite
    Mar 19, 2018 at 12:50
  • 3
    $\begingroup$ @Dragonite it is not clear how you get $\sqrt{ \sum_{i=1}^{n} | x_i |^{2} } \leq \sum_{i=1}^{n} \sqrt{ | x_ i |^{2} }$. Could you please offer more details? $\endgroup$
    – johnny09
    Apr 20, 2019 at 23:42
  • 4
    $\begingroup$ @johnny09 see clark's answer below, the square root is subadditive. $\endgroup$
    – dasWesen
    May 10, 2020 at 9:05

The inequality $ ||x||_1 \leq \sqrt{n} ||x||_2 $ is a consequence of Cauchy-Schwarz. To see this

$$\sqrt{n} ||x||_2 =\sqrt{1+1+\cdots+1}\sqrt{\sum_{i} x_i^2 }\geq ||x||_1$$

For the first, the function $f(x)=\sqrt{x}$ is concave and $f(0)=0$, hence $f$ is subadditive

Therefore $ f(\sum_{i} x_i^2 )\leq \sum_{i} f(x_i^2) =||x||_1 $


Another way to show that $||x||_1 \leq \sqrt{n}||x||_2$ is as follows:

\begin{align*} \begin{array}{c} \displaystyle 0\leq \sum_{i\neq j}\big(|x_i|-|x_j|\big)^2 = (n-1)\sum_{i=1}^{n} |x_i|^2 - 2\sum_{i\neq j}|x_i||x_j| \\ \displaystyle\Rightarrow\quad 2\sum_{i\neq j}|x_i||x_j|\leq (n-1)\sum_{i=1}^{n} |x_i|^2 \\ \displaystyle\Rightarrow\quad \sum_{i=1}^{n} |x_i|^2 + 2\sum_{i\neq j}|x_i||x_j|\leq n\sum_{i=1}^{n} |x_i|^2 \\ \displaystyle\Rightarrow\quad \left( \sum_{i=1}^n |x_i| \right)^2 \leq n\sum_{i=1}^{n} |x_i|^2 \\ \displaystyle\Rightarrow \quad ||x||_1\leq \sqrt{n}||x||_2 \end{array} \end{align*}


$\lVert x \rVert_2 \le \lVert x \rVert_1$ is equivalent to $\lVert x \rVert_2^2 \le \lVert x \rVert_1^2$ (norms are non-negative) which can be shown in an elementary way:

$$\lVert x \rVert_2^2 = \sum_i \lvert x_i\rvert^2 \le \left( \sum_i \lvert x_i \rvert \right)^2 = \lVert x \rVert_1^2$$

By expanding the product $\left(\sum_i z_i \right)^2 = \sum_i z_i^2 + \sum_{i \neq j} z_i z_j$ where the second sum of cross-terms is $\ge 0$ since all $z_i$'s are $\ge 0$.

Intuition for inequalities: if $x$ has one component $x_0$ much larger (in magnitude) than the rest, the other components become negligible and $\lVert x \rVert_2 \approx (\sqrt{x_0})^2 = \lvert x_0 \rvert \approx \lVert x \rVert_1 $. On the other hand, if the components of $x$ are about equal (in magnitude), $\lVert x \rVert_2 \approx \sqrt{n x_i^2} = \sqrt n \lvert x_i \rvert$, while $\lVert x \rVert_1 \approx n \lvert x_i \rvert$.


Hint: Bound $\left \lVert x\right \rVert_2$ by $\left \lVert x'\right \rVert_2$ where $x'$ is the vector with all its components equal to the maximum of the components of $x$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .