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Cauchy-Schwarz inequality

  1. Jul 11, 2011 #1
    Hi,

    Quick question here: I know that C-S inequality in general states that
    [itex]|<x,y>| \leq \sqrt{<x,x>} \cdot \sqrt{<y,y>}[/itex]

    and, in the case of [itex]L^2(a,b)[/itex]functions (or [itex]L^2(R)[/itex] functions, for that matter), this translates to
    [itex]|\int^{b}_{a}f(x)g(x)dx| \leq \sqrt{\int^{b}_{a}|f(x)|^2dx} \cdot \sqrt{\int^{b}_{a}|g(x)|^2dx}[/itex]

    What I don't understand is, in a book I read, it says
    [itex]||fg||_1 \leq ||f||_2 \cdot ||g||_2[/itex]
    which means
    [itex]\int^{b}_{a}|f(x)g(x)|dx \leq \sqrt{\int^{b}_{a}|f(x)|^2dx} \cdot \sqrt{\int^{b}_{a}|g(x)|^2dx}[/itex]

    I suppose that both of these correct, but I don't how to justify the transition from [itex]|\int^{b}_{a}f(x)g(x)dx|[/itex] to [itex]\int^{b}_{a}|f(x)g(x)|dx[/itex].
    I suppose I should use the fact that
    [itex]|\int^{b}_{a}f(x)g(x)dx| \leq \int^{b}_{a}|f(x)g(x)|dx[/itex]
    but that can't be sufficient, e.g., if 2<5 and 2<17 doesn't mean that 5<17. Any thoughts?

    Thanks.

    EDIT: I'm just gonna get greedy and pop-in another small question from the book I use
    http://i.imgur.com/l4jD0.png
    Can anybody explain to me why is this cleary true? (I hate it when they say it like that, I feel dumb)
     
    Last edited: Jul 11, 2011
  2. jcsd
  3. Jul 11, 2011 #2

    chiro

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    Science Advisor

    Hello Lajka.

    I'm a little confused with statement:

    ∫ba|f(x)g(x)|dx≤∫ba|f(x)|2dx−−−−−−−−−√⋅∫ba|g(x)|2dx−−−−−−−−−√

    I can't see the connection between the 1-norm of (f+g) and this integral.

    One thing that pops to mind that if you need to find something about the bounds of norm of (f+g), that you should use the triangle inequality which will place bounds of (f+g) in terms (f) and (g) (norms).

    Off the top of my head, I can't remember how various norms are linked in terms of inequalities, but I'm guessing that if you find some sort of inequality relationship that categorizes the 1-norm with the 2-norm in terms of an inequality, you could use the triangle inequality to establish that identity.

    I'm sorry I can't help you any more than this though.
     
  4. Jul 11, 2011 #3
    whoops, sorry about that plus, it was just a typo (i probably was thinking of minkowsky inequality at that moment), there should be multiplication in there, of course.

    thanks for the response, in any case :D
     
  5. Jul 11, 2011 #4

    I like Serena

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    Homework Helper

    Hi Lajka! :smile:

    Why don't you try to apply C-S on <|f|,|g|>?
    That should give you your inequality.

    As for your pop-in question.
    Try to write the integral as a sum of rectangles with width 1.
    You should be able to see the inequality holds for each rectangle.
    (Should I make a drawing? I like drawings! Or perhaps you will? :wink:)
     
  6. Jul 13, 2011 #5
    Hey serena! :)

    You must be my guardian angel on this forum or something. Anyway, the solution to the first question worked out flawlessly, and as for the second one, here's the picture! :D
    NyfvB.png
    I'm guessing you meant something like this, and it makes sense, so I guess that's that.

    Thanks serena! :D
     
  7. Jul 13, 2011 #6

    I like Serena

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    Yes, that's what I meant exactly! :)
    Nice drawing!

    See you next time. Don't be a stranger!
     
  8. Jul 13, 2011 #7
    Okay, it's a deal :D
    I gotta think of a way to repay you for your help thus far :)
     
  9. Jul 13, 2011 #8

    I like Serena

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    Homework Helper

    I have to admit that it was your drawings in previous threads that drew my attention - I liked them! :D
    Since then I've been hovering around.
    I hope you don't mind. :shy:
     
  10. Jul 13, 2011 #9
    Haha, not at all, I'm flattered actually. :D
     
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