# Demonstration for Lp norm

1. Feb 29, 2016

### Telemachus

1. The problem statement, all variables and given/known data
Hi there. I have to prove this inequality:

$||x||_2 \leq ||x||_1 \leq \sqrt{n} ||x||_2$

Where $||x||_2$ is the $l_p$ norm with p=2, so that:

$||x||_2=(|x_1|^2+|x_2|^2+...+|x_n|^2)^{\frac{1}{2}}$

And similarly $||x||_1=|x_1|+|x_2|+...+|x_n|$ is the $l_1$ vectorial norm.

so, the first part I think its easy (I suspect the second part is also easy, but I couldn't get through it).

I have that:

$||x||_2=(|x_1|^2+|x_2|^2+...+|x_n|^2)^{\frac{1}{2}}\leq |x_1|+|x_2|+...+|x_n|=||x||_1$ which is directly satisfied by applying the triangle inequality. So I think thats done.

Now, for the other inequality I have to show that: $||x||_1 \leq \sqrt{n} ||x||_2$

I couldn't find the way to show that, so I thought that perhaps someone here could help me.

2. Feb 29, 2016

### LCKurtz

Not sure what you mean by follows "directly" from the triangle inequality. I would suggest looking at $n=2$ for ideas. Then it says$$\sqrt{a^2+b^2}\le |a| + |b| \le \sqrt 2\sqrt{a^2+b^2}$$Square all three sides and see if that gives you any ideas.

3. Feb 29, 2016

### Telemachus

I mean that this follows by the triangle inequality: $\sqrt{a^2+b^2}\leq |a|+|b|$. But what about this: $|a|+|b|\leq \sqrt{2} \sqrt{a^2+b^2}$? how do I prove that's true? and how do I do that for arbitrary n?

By squaring all sides I get: $0 \leq 2|ab| \leq a^2+b^2$

4. Feb 29, 2016

### Telemachus

I think I see what you meant. I'll try and tell you. Thanks.

5. Feb 29, 2016

### Ray Vickson

So, in general, you need to show that if $I_n = \{ (i,j): 1 \leq i < j \leq n \}$ then
$$2 \sum_{I_n} |a_i| | a_j| \leq (n-1) \sum a_i^2$$
The case for $n = 2$ is easy: $0 \leq (|a|-|b|)^2 = a^2 +b^2 - 2 |a| |b|$.

6. Mar 1, 2016

### LCKurtz

@Telemachus: Not knowing your background, I wonder if you already have the Cauchy-Schwartz inequality or not:$$|(\vec x, \vec y)|\le \parallel \vec x \parallel \parallel \vec y \parallel$$where the left side is just the dot product in $\mathbb R^n$ and those are $l_2$ norms on the right. One proof of your right hand inequality is often demonstrated using it.

7. Mar 1, 2016

### Telemachus

Yes, I've tried to proove it by using Cauchy Schwartz inequality, but I coulnd't get it. I'll show you:

$| \sum_i x_i y_i | \leq ||x||_2 ||y||_2$

So I have: $|x_1 y_1+x_2 y_2+...+x_n y_n | \leq \sqrt{ \left ( |x_1|^2+|x_2|^2+...+|x_n|^2 \right )\left ( |y_1|^2+|y_2|^2+...+|y_n|^2 \right ) }$

What I've tried (which is probably not in the right way) was just setting all $y_i=1 \forall i$. And then I get:

$|x_1+x_2 +...+x_n| \leq \sqrt{n} \sqrt{\left ( |x_1|^2+|x_2|^2+...+|x_n|^2 \right )}=\sqrt{n} ||x||_2$

Which looks close, but I aslo have this:

$|x_1+x_2 +...+x_n| \leq |x_1|+|x_2| +...+|x_n|=||x||_1$

So its still inconclusive.

Last edited: Mar 1, 2016
8. Mar 1, 2016

### Samy_A

You are almost there.

Cauchy Schwarz is valid for any choice of $x_i, y_i$.
You chose $y_i=1$. Why not also make a specific choice for $x_i$? (Imagine for a moment that all $x_i \geq 0$.)

9. Mar 1, 2016

### LCKurtz

What if your $\vec x$ vector is $\langle |x_1|,|x_2|,...|x_n|\rangle$ in the first place?

10. Mar 1, 2016

### Telemachus

Done :D thanks.