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 $x\in F$


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.


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