# Show that $T^2\le 1$ whenever $0\le T\le 1$ in a Hilbert space

Let $$H$$ be a Hilbert space and $$T$$ a bounded operator on $$H$$ such that $$0\le T\le 1$$. I want to show that $$0\le T^2\le 1.$$
My approach: I know, it is not true that $$0 \le a\le b \implies a^2\le b^2$$ in arbitrary C$$^*$$-algebra. So we can not just take the square on the both side and preserve the order. Thus, let $$S$$ be the square root for $$T$$, that is, $$S^2=T,$$ this is possible since $$T$$ is positive $$(T\ge 0).$$ Now notice that, for $$x \in H$$ we have: \begin{aligned} \langle T^2x,x\rangle &=\langle STSx,x\rangle\\ &=\langle TSx,Sx\rangle, \text{ since S is positive, so } S^*=S\\ &\le \langle Sx,Sx\rangle, \text{ since } T\le 1\\ & =\langle Tx,x\rangle, \text{ since } S^2=T~\text{and } S^*=S\\ &\le \langle x,x\rangle, \text{ since } T\le 1\\ \implies 0\le T^2&\le 1. \end{aligned} Is my solution correct? Or there is any easy proof or any theorem from which this is easy, please suggest me. Thank you for your time.

• That is how I would have done it too. The only thing that needs justification is the existence of a square root. Jul 16 at 22:14
• I think you are also using the fact that $S$ commutes with $T$. Jul 16 at 23:16

There is another explanation which does not make use of the square root of $$T.$$
The Cauchy-Schwarz inequality gives $$|\langle Tx,y\rangle |\le\langle Tx,x\rangle^{1/2}\langle Ty,y\rangle^{1/2}\le \|x\|\,\|y\|,\quad y\in \mathcal{H}$$ Thus $$|Tx\|\le \|x\|$$ and $$\langle T^2x,x\rangle =\|Tx\|^2\le \|x\|^2=\langle x,x\rangle$$
Since you point out that $$0\leq a\leq b$$ doesn't necessarily imply $$a^2\leq b^2$$, it's worth mentioning that in an arbitrary unital $$C^*$$-algebra $$A$$, if $$a\in A$$ and $$0\leq a\leq 1$$, then a functional calculus argument shows that $$a^2\leq 1$$.