1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Demonstration for Lp norm

  1. Feb 29, 2016 #1
    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.

    Thanks in advance.
     
  2. jcsd
  3. Feb 29, 2016 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  4. Feb 29, 2016 #3
    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##
     
  5. Feb 29, 2016 #4
    I think I see what you meant. I'll try and tell you. Thanks.
     
  6. Feb 29, 2016 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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|##.
     
  7. Mar 1, 2016 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    @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.
     
  8. Mar 1, 2016 #7
    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
  9. Mar 1, 2016 #8

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    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##.)
     
  10. Mar 1, 2016 #9

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What if your ##\vec x## vector is ##\langle |x_1|,|x_2|,...|x_n|\rangle## in the first place?
     
  11. Mar 1, 2016 #10
    Done :D thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Demonstration for Lp norm
  1. Lp Subspaces (Replies: 1)

  2. Is this a norm? (Replies: 13)

  3. Irrational demonstration (Replies: 10)

Loading...