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Demonstration for Lp norm

  • Thread starter Telemachus
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  • #1
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Homework Statement


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.

Thanks in advance.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


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.

Thanks in advance.
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
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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
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I think I see what you meant. I'll try and tell you. Thanks.
 
  • #5
Ray Vickson
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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##
So, in general, you need to show that if ##I_n = \{ (i,j): 1 \leq i < j \leq n \}## then
[tex] 2 \sum_{I_n} |a_i| | a_j| \leq (n-1) \sum a_i^2 [/tex]
The case for ##n = 2## is easy: ##0 \leq (|a|-|b|)^2 = a^2 +b^2 - 2 |a| |b|##.
 
  • #6
LCKurtz
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@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
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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:
  • #8
Samy_A
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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.
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
LCKurtz
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What if your ##\vec x## vector is ##\langle |x_1|,|x_2|,...|x_n|\rangle## in the first place?
 
  • #10
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Done :D thanks.
 

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