Self - adjoint contraction operators on Hilbert spaces

Question

Let $\mathbb{H}$ be a complex Hilbert space.

Let $T:\mathbb{H}\rightarrow \mathbb{H}$ be a self - adjoint contraction operator.

why is the following statement true ? $$0 \le T^2 \le I$$

Answer 1

Let $( \cdot, \cdot)$ denote the inner product on $H$.

1. $(T^2x,x)=(Tx,Tx) = ||Tx||^2 \ge 0 $ for all $x \in H.$ This shows that $T^2 \ge 0.$

2. Since $T$ is a contraction, we have $||Tx|| \le ||x||$ for all $x \in H.$ Hence

$(T^2x,x)=(Tx,Tx) = ||Tx||^2 \le ||x||^2 =(x,x) $ for all $x \in H.$ This gives $T^2 \le I.$

Answer 2

$\langle T^{2}x , x \rangle =\langle Tx , Tx \rangle =\|Tx\|^{2}$ so $0 \leq \langle T^{2}x , x \rangle \leq \|x\|^{2}$. This says $0 \leq T^{2} \leq I$.