Positive contraction operator on Hilbert space

Question

This is an exercise from a functional analysis textbook. I am stuck in it.

Let $H$ be a Hilbert space, $A\in\mathcal{B}(H)$. We call $A$ a contraction operator if $\|A\|\leq 1$ and call $A$ a positive operator if for any $x\in H$, $\langle A(x),x\rangle\geq 0$.

We can prove that for any positive contraction operator $A$ on $H$, $A$ is self-adjoint and the following inequality $$\|x-A(x)\|^2\leq \|x\|^2-\|A(x)\|^2,\quad\forall x\in H$$ holds. (Proof omitted.)

Now give $r$ positive contraction operators $A_1,A_2,\cdots,A_r$ on $H$, let $T=A_1A_2\cdots A_r$, the goal is to prove $$\|x-T(x)\|^2\leq r\left(\|x\|^2-\|T(x)\|^2\right),\quad\forall x\in H.$$

My attempt is to write the LHS of the above inequality as $$\|x-A_1(x)+A_1(x)-A_1A_2(x)+\cdots+A_1A_2\cdots A_{r-1}(x)-A_1A_2\cdots A_r(x)\|^2$$ and expand it, but it doesn’t seem to work.

Answer 1

You were on the right track, you just have to to do it in the opposite order. We have \begin{align*} \|x-Tx\|^2&=\|x-A_rx+A_rx-A_{r-1}A_rx+\dots+A_2\dots A_rx-A_1\dots A_rx\|^2\\ &\leq (\|x-A_rx\|+\|A_rx-A_{r-1}(A_rx)\|+\dots+\|A_2\dots A_rx-A_1(A_2\dots A_r)x\|)^2\\ &\leq r(\|x-A_r x\|^2+\|A_r x-A_{r-1}(A_r x)\|^2+\dots+\|A_2\dots A_rx-A_1(A_2\dots A_r)x\|^2)\\ &\leq r(\|x\|^2-\|A_r x\|^2+\dots+\|A_2\dots A_r x\|^2-\|A_1\dots A_r x\|^2)\\ &=r(\|x\|^2-\|Tx\|^2). \end{align*} The first inequality is simply the triangle inequality, the second one is the Cauchy-Schwarz inequality, and the third the inequality for the individual positive contractions from the question.

Answer 2

Suppose the inequality is true for some $r\ge 1$. Let $T_r(x) = A_1A_2\dots A_rx$. Note that \begin{align*} \|x - T_{r + 1}(x)\|^2 & = \|x - A_{r + 1}x + A_{r + 1}x - T_r(A_{r + 1}x)\|^2 \\ & \le 2\|x - A_{r + 1}x\|^2 + 2\|A_{r + 1}x - T_r(A_{r + 1}x)\|^2 \\ & \le 2\Big[\|x\|^2 - \|A_{r + 1}x\|^2\Big] + 2r\Big[\|A_{r + 1}x\|^2 - \|T_r(A_{r + 1}x)\|^2\Big]. \end{align*} Note that $\|A_{r + 1}x\|^2\le \|A_{r + 1}\|^2\|x\|^2\le \|x\|^2$ and the same argument shows that $\|T_r(A_{r + 1}x)\|^2\le \|A_{r + 1}x\|^2$. It follows that \begin{align*} \|x - T_{r + 1}(x)\|^2 & \le 2\|x\|^2 - 2\|T_r(A_{r + 1}x)\|^2 + 2r\|x\|^2 - 2r\|T_r(A_{r + 1}x)\|^2 \\ & = 2(r + 1)\Big[\|x\|^2 - \|T_r(A_{r + 1}x)\|^2\Big] \\ & = 2(r + 1)\Big[\|x\|^2 - \|T_{r + 1}x\|^2\Big]. \end{align*}

Edit: This is an attempt using induction which doesn't quite work out (thanks to OP for pointing it out), but I'm leaving it here in case anyone sees some way of making induction work.