# Cauchy-Schwarz inequality

1. Jul 11, 2011

### 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 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. Jul 11, 2011

### 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.

3. Jul 11, 2011

### 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

4. Jul 11, 2011

### I like Serena

Hi Lajka!

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

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? )

5. Jul 13, 2011

### 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

6. Jul 13, 2011

### I like Serena

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

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

7. Jul 13, 2011

### Lajka

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

8. Jul 13, 2011

### 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:

9. Jul 13, 2011

### Lajka

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