inequality for positive contraction operator

Question

Let $H$ be a Hilbert space, let $A\in B(H)$ satify $\|A\|\le 1$. If $A$ is positive, i.e. $A$ is a self-adjoint operator and for all $x\in H$, $\langle A(x),x\rangle\ge 0$, proof that $${\|x-A(x)\|}^2\le {\|x\|}^2-{\|A(x)\|}^2, \forall x\in H.$$
This is a exercise. For $A$ is positive \begin{align*} {\|x-A(x)\|}^2 &={\|x\|}^2+{\|A(x)\|}^2-\langle A(x),x\rangle-\langle x,A(x)\rangle\\ &={\|x\|}^2-{\|A(x)\|}^2+2{\|A(x)\|}^2-2\langle x,A(x)\rangle \end{align*} so we should prove that $${\|A(x)\|}^2\le \langle x,A(x)\rangle$$ or $${\|A(x)\|}^2=\langle A(x),A(x)\rangle = \langle x,A(A(x))\rangle \le \langle x,A(x)\rangle$$ but i don't know how to use the condition "$\|A\|\le 1$" and "$\langle A(x),x\rangle\ge 0$".

Answer 1

Since $A$ is a positive operator, you can take it square root, which is also a positive operator, and $$ \|A(x)\|^2 = \langle A^{1/2}A^{1/2}x,A x\rangle = \langle A^{1/2}x,A (A^{1/2}x)\rangle $$ now by Cauchy-Schwarz and the fact that $\|A\|\leq 1$ $$ \langle A^{1/2}x,A (A^{1/2}x)\rangle ≤ \|A^{1/2}(x)\|\|A(A^{1/2}(x))\| ≤ \|A^{1/2}(x)\|^2 = \langle x, A(x)\rangle $$ which is what you wanted to prove.

Answer 2

For an alternative solution which doesn't involve the square root, consider this. From $\|A\| \le 1$ it follows that for all $x \in H$ holds $$\langle Ax,x\rangle \le \|Ax\|\|x\| \le \|x\|^2.$$ Therefore \begin{align} 0 &\le \langle A(Ax-x),Ax-x\rangle \\ &= \langle A^2x - Ax, Ax - x\rangle\\ &= \underbrace{\langle A^2x,Ax\rangle}_{\le\|Ax\|^2} - \underbrace{\langle A^2x,x\rangle}_{=\|Ax\|^2} - \underbrace{\langle Ax,Ax\rangle}_{=\|Ax\|^2} + \langle Ax,x\rangle\\ &\le \langle Ax,x\rangle - \|Ax\|^2 \end{align}

so $\langle Ax,x\rangle \ge \|Ax\|^2$.