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Let $V$ be a $\mathbb C-$vector space and let $\Phi$ be a hermitian quadratic form on $V$. Assume that $\Phi$ is positive definite, i. e., $\Phi(x)>0$ for all nonzero vectors $x\in V$. Let $\varphi$ be the polar form of $\Phi$. Prove the Cauchy-Schwarz inequality: for all $x,y\in V$ we have $$|\varphi(x,y)|^2\le\Phi(x)\Phi(y).$$

This is my attempt: Let $x,y\in V, t\in\mathbb C$ and $F:\mathbb C\to\mathbb C$ such that $F(t)=\Phi(tx+y)$. Then, \begin{align*} F(t)&=\Phi(tx+y)\\ &=\varphi(tx+y,tx+y)\\ &=\varphi(tx,tx)+\varphi(tx,y)+\varphi(y,tx)+\varphi(y,y)\\ &=\overline{t}t\varphi(x,x)+\overline{t}\varphi(x,y)+t\varphi(y,x)+\varphi(y,y)\\ &=|t|^2\Phi(x)+\overline{t}\varphi(x,y)+t\varphi(y,x)+\Phi(y). \end{align*}

I think that the proof of this inequality is similar to the case of bilinear forms, but I don't know what to do. :( Help

Selena J
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  • If you can show that $\varphi(x, y)$ is an inner product, then you can use the usual Cauchy-Schwarz to prove this. You could also adapt my answer in the linked question; try expanding $\Phi\left(x - \frac{\varphi(x, y)}{\Phi(y)} y\right)$ and see what you get. – Theo Bendit Jul 27 '21 at 01:21

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