How can the Cauchy-Schwarz inequality be applied to L^2 functions in a book?

[Lajka]

Hi,

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

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

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

I suppose that both of these correct, but I don't how to justify the transition from |\int^{b}_{a}f(x)g(x)dx| to \int^{b}_{a}|f(x)g(x)|dx.
I suppose I should use the fact that
|\int^{b}_{a}f(x)g(x)dx| \leq \int^{b}_{a}|f(x)g(x)|dx
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 going to 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)

 

[chiro]

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.

 

[Lajka]

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

 

[I like Serena]

Hi Lajka!

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? )

 

[Lajka]

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

I'm guessing you meant something like this, and it makes sense, so I guess that's that.

Thanks serena! :D

 

[I like Serena]

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

See you next time. Don't be a stranger!

 

[Lajka]

Okay, it's a deal :D
I got to think of a way to repay you for your help thus far :)

 

[I like Serena]

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:

 

[Lajka]

Haha, not at all, I'm flattered actually. :D