Prove that for the defined $\langle .,. \rangle$ there exist $0 < a \le b$ such that $a\|x\| \le \|x\|_\ast ≤ b\|x\|$ for all $x \in H$.

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 $$\|x\|_\ast = \sqrt{\langle x, x \rangle} = \sqrt{(Ax,x)} = \sqrt{(x,Ax)} \le\sqrt{ \|A\| \|x\|^2} = \sqrt{\alpha} \|x\|\equiv a \|x\|$$