# Math Help - Complex Schwarz's Inequality

1. ## Complex Schwarz's Inequality

Prove Schwarz's for complex numbers: (x,y)^2 <= |x|^2 |y|^2
Under what conditions does equality hold?

Hint:
| lx + my |^2 = l^2 (x,x) + 2lm (x,y) + m^2 (y,y)

is positive definite wrt l and m

The hint is really confusing me...

2. ## Re: Complex Schwarz's Inequality

Let $z = y - \frac{(y,x)}{|x|^2} \cdot x$. Now, we have $(z,x) = 0$ and all you have to do is compute $0 \leq |z|^2$.

3. ## Re: Complex Schwarz's Inequality

Can you explain how (z,x)=0?

4. ## Re: Complex Schwarz's Inequality

Originally Posted by monomoco
Can you explain how (z,x)=0?

$(z,x)=\left(y-\dfrac{(y,x)}{|x|^2}x,x\right)=(y,x)-\dfrac{(y,x)}{|x|^2}(x,x)=$

$(y,x)-\dfrac{(y,x)}{|x|^2}|x|^2=(y,x)-(y,x)=0\quad (\forall x\neq 0)$

5. ## Re: Complex Schwarz's Inequality

I don't understand how you take the inner product like that. Is x a vector and its elements are also being called x? And can you explain how it distributes that way?

6. ## Re: Complex Schwarz's Inequality

I believe the problem is a notation one. Allow me to change inner products from $(y,x)$ to $\langle y \, , \,x \rangle$. So, we have:

$\langle z \, , \, x \rangle = \left\langle y - \frac{\langle y \, , \, x \rangle}{|x|^2} \, , \, x \right\rangle = \langle y \, , \, x \rangle - \frac{\langle y \, , \, x \rangle}{|x|^2} \langle x \, , \, x \rangle =$

$= \langle y \, , \, x \rangle - \frac{\langle y \, , \, x \rangle}{|x|^2} \cdot |x|^2 = \langle y \, , \, x \rangle - \langle y \, , \, x \rangle = 0$

I hope it's clearer now.

7. ## Re: Complex Schwarz's Inequality

I get it now. Thanks very much!