In our lecture on functional analysis, we were given the following statement:
Let $X$ be a $\mathbb{K}$-vector space ($\mathbb{K}$ being either $\mathbb{R}$ or $\mathbb{C}$) and $<.,.>: X \times X \rightarrow \mathbb{K}$ be a positive semi-definite Hermitian form and $\Vert x \Vert = \sqrt{ <x,x>}$ for all $x \in X$. Then we have $$ \vert <x,y> \vert \leq \Vert x \Vert \cdot \Vert y \Vert$$ for all $x,y \in X$.
I tried to understand its corresponding proof in order to get familiar with the definiton of this seminorm, but I have a question regarding one step. The proof goes as follows:
Let $x,y \in X$, $\alpha, \beta \in \mathbb{K}$ and $ \alpha, \beta \neq 0$. We have $$ 0 \leq \Vert \frac{x}{\alpha} - \frac{y}{\beta} \Vert ^{2} = \frac{\Vert x\Vert^2}{\vert \alpha\vert ^2} + \frac{\Vert y\Vert^2}{\vert \beta\vert ^2} -2 Re(\frac{<x,y>}{\alpha \cdot \bar \beta}).$$ Let $\epsilon > 0$ be arbitrary and put $\alpha = \Vert x \Vert + \epsilon$ and $\beta = \Vert y \Vert + \epsilon$. We then get $$ 2Re (<x,y>) \leq \frac{\beta}{\alpha} \cdot \Vert x\Vert ^2 + \frac{\alpha}{\beta} \Vert y \Vert^2.$$
Could some please explain how I get the factors $\frac{\beta}{\alpha} $ reps. $ \frac{\alpha}{\beta}$? I have some trouble figuring that out, in particular with respect to the complex conjugate and the real part.
Thank you for your answer.
Your choice of $\alpha,\beta$ guarantee that $\alpha, \beta$ are both real and positive. Thus $$ \frac{2}{\alpha \beta} \Re\langle x,y\rangle = 2 \Re\left(\frac{\langle x,y \rangle}{\alpha \bar{\beta}} \right) \le \frac{\Vert x\Vert^2}{|\alpha|^2} + \frac{\Vert y\Vert^2}{|\beta|^2} = \frac{\Vert x\Vert^2}{\alpha^2} + \frac{\Vert y\Vert^2}{\beta^2}. $$ Now multiply both sides by $\alpha \beta$ to get
$$ 2\Re\langle x,y\rangle \le \frac{\alpha \beta \Vert x\Vert^2}{\alpha^2} + \frac{\alpha \beta \Vert y\Vert^2}{\beta^2} = \frac{\beta}{\alpha} \Vert x \Vert^2 + \frac{\alpha}{\beta} \Vert y \Vert^2. $$