Exercise 2, chapter 3 of Barry Simon. A comprehensive course in analysis part 1.

Question

2. For $x, y$ in $V$, an inner product space, and for $\lambda=re^{i\theta}\in\mathbb{C}$, use $|x+\lambda y|\geq 0$ for $\theta$ fixed to get a quadratic equation in $r$ whose roots must be either equal or nonreal. Show that this implies $|Re[<re^{i\theta}y, x>]|\leq |x|^2|y|^2$ and so conclude the Schwarz inequality.

I have this:

$|x+\lambda y|^2\geq 0$

$<x+\lambda y,x+\lambda y>\geq 0$

$(e^{2i\theta}|y|^2)r^2+(e^{-i\theta}<x,y>+e^{i\theta}<y,x>)r+|x|^2\geq 0$

then discriminant $\Delta$ is $\Delta\leq 0$.

Therefore, $ (e^{-i\theta}<x,y>+e^{i\theta}<y,x>)^2-4(e^{2i\theta}|y|^2)|x|^2\leq 0$

$e^{-4i \theta}<x,y>^2+2<x,y><y,x>+<y,x>^2\leq 4|x|^2|y|^2$

And until here I arrived, I do not know how to continue to prove that $|Re[<re^{i\theta}y, x>]|\leq |x|^2|y|^2$

Answer 1

Ignoring the hints you can prove the inequality as follows: $|x+\lambda y|^{2} \geq 0$ gives $|x|^{2}+2|\lambda|^{2}|y|^{2}+2 \Re \lambda \langle x, y \rangle \geq 0$.Put $\lambda =-\frac {\langle y, x \rangle } {|y|^{2}}$ and you will get C-S inequality.

Answer 2

A straightforward proof of the CS inequality is obtained from the orthogonal (right triangle) decomposition where $x$ is the hypotenuse: $$ x = \left(x-\frac{\langle x,y\rangle}{\langle y,y\rangle}y\right)+\frac{\langle x,y\rangle}{\langle y,y\rangle}y. $$ From this it follows that $$ \|x\| \ge \frac{|\langle x,y\rangle|}{\|y\|} \\ |\langle x,y\rangle| \le \|x\|\|y\| $$

with equality iff $x=\alpha y$ for some $\alpha$. The special case where $y=0$ is handled by reversing the roles of $x,y$. The case where $x=y=0$ is trivial.