Sesquilinear Forms: Cauchy-Schwarz

Question

This thread is related: Parallelogram

Given a Hilbert space $\mathcal{H}$.

Consider a quadratic form: $$q:\mathcal{H}\to\mathbb{C}:\quad q[\lambda\varphi]=|\lambda|^2q[\varphi]$$

Suppose it satisfies: $$q[\varphi+\psi]+q[\varphi-\psi]=2q[\varphi]+2q[\psi]$$

Define a sesquilinear form: $$s:\mathcal{H}\times\mathcal{H}\to\mathbb{C}:\quad s(\varphi,\psi):=\frac{1}{4}\sum_{\alpha=0\ldots3}i^\alpha q[\varphi+i^\alpha\psi]$$

Then for positive forms: $$q\geq0:\quad|s(\varphi,\psi)|\leq q[\varphi]q[\psi]$$

How can I check this?

Answer 1

For this, you need only assume $Q$ is real and non-negative on a complex space with $$ Q(x+y)+Q(x-y)=2Q(x)+2Q(y),\\ Q(ix)=Q(x). $$ Then it is true that $B(x,y)=\frac{1}{4}\sum_{n=0}^{3}i^{n}Q(x+i^{n}y)$ satisfies $$ |B(x,y)| \le Q(x)^{1/2}Q(y)^{1/2}. $$ Therefore $|x|_{Q}=Q(x)^{1/2}$ satisfies the triangle inequality. Then, if $Q(\lambda x)=|\lambda|^{2}Q(x)$ holds for all scalars $\lambda$, it is also true that $B$ is sesquilinear.

Outline

1. $B(x+x',y)=B(x,y)+B(x',y)$. So $B(\alpha x,y)=\alpha B(x,y)$ for all complex scalars with rational real and imaginary parts.

2. $4|B(x,y)|=\lim_{n}4B(\alpha_n x,y)= \lim_{n}Q(\alpha_n x + y)-Q(\alpha_n x-y) \le 2Q(x)+2Q(y)$ if $Q(x) \ge 0$ for all $x$.

3. Use rational positive real scalars to obtain $|B(rx,sy)| \le 1$ and $|B(x,y)| \le Q(x)^{1/2}Q(y)^{1/2}$.

4. Continuity: $|\lambda B(x,y)-B(\lambda x,y)|=\lim_{n}|B((\lambda_n-\lambda)x,y)|\le\lim_{n}|\lambda_{n}-\lambda|Q(x)^{1/2}Q(y)^{1/2}=0$ if $Q(\alpha x)=|\alpha|^{2}Q(x)$. So $\lambda B(x,y)=B(\lambda x,y)$ for general $\lambda\in\mathbb{C}$. That makes $B$ fully sesquilinear.