2
$\begingroup$

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 $x\in F$

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.