If $A$ is an hyponormal operator on a complex Hilbert space $F$ (i.e. $AA^\* \leq A^\*A$). 



> 
> Is $A^2$ also hyponormal? i.e. is
>  $$A^2(A^2)^\* \leq (A^2)^\*A^2?$$
> 
> 
> 


Recall that $AA^\* \leq A^\*A$ iff $\langle (AA^\*-A^\*A)x,x\rangle\leq 0$.


Note also that $A$ is hyponormal iff 
$$\|Ax\|\geq \|A^\*x\|,$$
for all $x\in F$



The square of an hyponormal operator
In general no; if $S$ is the unilateral shift then e.g. the operator $A=S^\*+2S$ is hyponormal but $A^2$ is not. This example appears as an exercise in Chapter II.4 of Conway's "Subnormal Operators" book, and can be checked by hand via a mildly tedious calculation. It may also follow from more general results of Ito and Wong ("Subnormality and Quasinormality of Toeplitz operators, Proc. Amer. Math. Soc., 34 (1972), pp157-164, MR0303334) --Conway gives this as a reference but I haven't read through it. 


