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 "$\langle A(x),x\rangle\ge 0$".
from Hot Weekly Questions - Mathematics Stack Exchange
Post a Comment