Positive contraction operator on Hilbert space

Solution 1:

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.