# Cauchy Schwarz's proof

Gold Member

## Homework Statement

I'm trying to follow the demonstration of the Cauchy-Schwarz's inequality proof given in http://mathworld.wolfram.com/SchwarzsInequality.html.
I am stuck at the last step, namely that $\langle \bar g , f \rangle \langle f , \bar g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle \Rightarrow |\langle f , g \rangle |^2 \leq \langle f , f \rangle \langle g , g \rangle$.

I don't know.

## The Attempt at a Solution

$\langle \bar g , f \rangle \langle f , \bar g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle \Rightarrow \langle \bar f , g \rangle \langle \bar f , g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle$. I'm stuck here.
I know that $||f||=\sqrt {\langle f , f \rangle}$ but I don't even know if I can use this fact. Any tip is appreciated.

Gold Member
I've made some progress I think.
Mathworld didn't specify it explicitely but I think that f and g are real functions.
So that I reach $\langle \bar g , f \rangle \langle f , \bar g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle \Rightarrow \langle \bar f , g \rangle \langle \bar f , g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle \Rightarrow \langle f , g \rangle ^2 \leq \langle f , f \rangle \langle g , g \rangle$. So I "almost" reach the proof. I have a missing absolute value though. Any idea why?

Gold Member
Ok I got it wrong, f and g aren't real valued function because lambda (which is complex) is defined by some inner products involving f and g and their complex conjugate only.
If someone could tell me how to understand the last step I'd be grateful.

HallsofIvy
Homework Helper
I've made some progress I think.
Mathworld didn't specify it explicitely but I think that f and g are real functions.
So that I reach $\langle \bar g , f \rangle \langle f , \bar g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle \Rightarrow \langle \bar f , g \rangle \langle \bar f , g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle \Rightarrow \langle f , g \rangle ^2$
This is incorrect. $\langle f, g\rangle^2$ is a complex number. You want $\left|\langle f, g\rangle\right|^2$.

$\leq \langle f , f \rangle \langle g , g \rangle$. So I "almost" reach the proof. I have a missing absolute value though. Any idea why?
Again, those are wrong. You want $\langle f, \bar f\rangle= |f|^2$ and $\langle g, \bar g\rangle= |g|^2$.

Gold Member
Thanks HallsofIvy!
This is incorrect. $\langle f, g\rangle^2$ is a complex number. You want $\left|\langle f, g\rangle\right|^2$.
Yeah you are right, I realized this in my previous post.

You want $\langle f, \bar f\rangle= |f|^2$ and $\langle g, \bar g\rangle= |g|^2$.
Ok... How do I prove these, for any inner product? I'm looking at the properties of Hermitian inner product given there: http://mathworld.wolfram.com/HermitianInnerProduct.html but I've no clue how to relate it with the $|.|^2$ (same as norm squared?)