Cauchy-Schwarz inequality

  • Thread starter Lajka
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  • #1
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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:

Answers and Replies

  • #2
chiro
Science Advisor
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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.
 
  • #3
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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
 
  • #4
I like Serena
Homework Helper
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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:)
 
  • #5
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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
 
  • #6
I like Serena
Homework Helper
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Yes, that's what I meant exactly! :)
Nice drawing!

See you next time. Don't be a stranger!
 
  • #7
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Okay, it's a deal :D
I gotta think of a way to repay you for your help thus far :)
 
  • #8
I like Serena
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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:
 
  • #9
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Haha, not at all, I'm flattered actually. :D
 

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