Let $H$ be a Hilbert space over $\mathbb{R}$ with an inner product $(·, ·)$ and the norm $\|x\| = \sqrt {(x, x)}$. Let $A$ be a bounded strictly positive definite linear operator on $H$ with $A^\ast = A$. For $x, y \in H$, let $\langle x, y \rangle = (Ax, y)$ and $\|x\|_\ast = \sqrt{\langle x, x \rangle}$.
Prove that $\langle .,. \rangle$ is an inner product on the vector space $H$, and that there exist constants $0 < a \le b$ such that $a\|x\| \le \|x\|_\ast ≤ b\|x\|$ for all $x \in H$.
$\text{My Attempt}$:
- for the first part to show that it is inner product:
1- $\langle x, x \rangle = (Ax, x) \ge 0$ and $(Ax, x)=0$ iff $x=0$
2- $\langle x, y \rangle = (Ax, y) = (x, Ay)= \langle y , x \rangle$
3- $\langle x+z, y \rangle = (Ax+Az, y) = (Ax, y) +(Az, y) =\langle x, y \rangle +\langle z, y \rangle$
4- $\langle \alpha x, y \rangle = (\alpha Ax, y)= \alpha (Ax, y)=\alpha \langle x, y \rangle$
- For the second part
1- Since $A$ is a positive definite linear operator $\|x\|_\ast = \sqrt{\langle x, x \rangle} = \sqrt{(Ax,x)} \ge\sqrt{\beta \|x\|^2} = \sqrt{\beta} \|x\|\equiv b \|x\|$
2- Let $\|A\| = \alpha$. Since $A$ is symmetric $\|x\|_\ast = \sqrt{\langle x, x \rangle} = \sqrt{(Ax,x)} = \sqrt{(x,Ax)} \le\sqrt{ \|A\| \|x\|^2} = \sqrt{\alpha} \|x\|\equiv a \|x\|$
