# Invertibility of Hankel operators?

Let $D$ be the unit disc in the complex plane and $P$ the Bergman projection mapping $L^2(D)$ onto the closed subspace $A^2(D)$ of holomorphic square-integrable functions (w.r.t. Lebesgue measure). Let $g$ be a bounded function on $D$. Is it possible for the operator $Tf:=(1-P)(gf)$, mapping $A^2(D)$ into its orthogonal, to be invertible? If this can happen, are there reasonable sufficient condition on $g$? Since $P(gf)=gf$ if $g$ is holomorphic, this can happen only for non-holomorphic functions. Notice that an easy calculation with $g(z)=\overline{z}$ reveals that $T$ is not invertible in this case.

The question mark in the title is due to the fact that I am not sure about the terminology "Hankel" operators.

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