If $A$ positive, does $\|A\|\leq 1$ imply $\| Id-A\|\leq 1$?

Question

Let $A$ a bounded, positive (ie: $\langle Au,u\rangle\geq 0$) operator on $H$ a Hilbert.

Is the following statement true ?

$$\|A\|\leq 1 \text{ implies }\| Id-A\|\leq 1.$$

Thanks.

Answer 1

Since $A$ is positive and $\| A\| \leq 1$ we understand the spectrum is in $[0,1]$. $A-\mathrm{id}$ Will shift the spectrum to the left by ${1}$ which implies it will be in $[-1,0]$. Since $A-\mathrm{id}$ is still self adjoint - it is clear its operator norm is less than or equal to $1$.

Answer 2

Yes. You have $$ \langle Au,u\rangle \leq\|Au\|\,\|u\|\leq\|A\|\,\|u\|^2\leq \langle u,u\rangle. $$ Then $$ \langle (Id-A)u,u\rangle =\langle u,u\rangle -\langle Au,u\rangle \geq0. $$