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

Question

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:

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

Answer 1

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

Answer 2

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.

Answer 3

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$

Answer 4

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*}

Answer 5

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