Is a contraction idempotent operator self-adjoint?
Solution 1:
Here is a proof.
Note that $\|T^*\|=\|T\|\leq1$. For an $x\in H$, $\langle Tx,T^*x\rangle=\langle T^2x,x\rangle=\langle Tx,x\rangle$. Then \begin{align} \|Tx-T^*x\|^2&=\|Tx\|^2+\|T^*x\|^2-2\text{Re}\,\langle Tx,T^*x\rangle\\ &\leq\|Tx\|^2+\|x\|^2-2\text{Re}\,\langle Tx,x\rangle\\ &=\|Tx-x\|^2. \end{align} If we now take $x=Ty$ for some $y\in H$, we have $Tx=x$, and so $T^*x=Tx$, which translates to $$T^*Ty=TTy=Ty.$$ This shows that $T^*T=T$. Then $T^*=(T^*T)^*=T^*T=T$. So $T$ is selfadjoint (actually, it is positive).