Positive Operator on Hilbert Space I am stuck with this problem for a long time.
If $T\geq0$ then show that $T^2\leq$||T||T.
Any help will be appreciated.
 A: A positive semidefinite hermitian operator has a positive semidefinite hermitian square root, say $T = S^2$, with $S \geqslant 0$. For positive semidefinite hermitian operators, the norm is the largest eigenvalue, hence $\lVert S\rVert^2 = \lVert T\rVert$, and
$$\begin{align}
\langle T^2x,x\rangle &= \langle Tx,Tx\rangle\\
&= \langle S^2 x, S^2x\rangle\\
&= \lVert S(Sx)\rVert^2\\
&\leqslant (\lVert S\rVert\cdot \lVert Sx\rVert)^2\\
&= \lVert S\rVert^2 \langle Sx,Sx\rangle\\
&= \lVert T\rVert \langle S^2x,x\rangle\\
&= \langle \lVert T\rVert Tx,x\rangle,
\end{align}$$
which means $T^2 \leqslant \lVert T\rVert T$.
A: You can avoid the square root by noting that $(x,y)_{T}=(Tx,y)$ acts as an inner product, except that it may not be positive definite. Because of this, the Cauchy-Schwarz inequality holds:
$$
           |(x,y)_{T}|^{2} \le (x,x)_{T}(y,y)_{T} = (x,x)_{T}(Ty,y) \le (x,x)_{T}\|Ty\|\|y\| \le \|T\|(x,x)_{T}\|y\|^{2}
$$
That is,
$$
          |(Tx,y)|^{2} \le \|T\|(Tx,x)\|y\|^{2}.
$$
Letting $y=Tx$,
$$
   \|Tx\|^{4} \le \|T\|(Tx,x)\|Tx\|^{2}
$$
Whether or not $Tx=0$,
$$
     \|Tx\|^{2} \le \|T\|(Tx,x)
$$
The final result holds by noticing that $(T^{2}x,x)=\|Tx\|^{2} \le (\|T\|Tx,x)$. So $T^{2} \le \|T\|T$
