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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.

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

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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 $$\langle T^2x,x\rangle =\|Tx\|^2\le \|x\|^2=\langle x,x\rangle $$

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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 $a^2\leq 1$.

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