The adjoint of the Cesaro operator on $H^p$

Question

For each $f=\sum_{k=0}^\infty b_kz^k\in H^p(\mathbb{D})$, $1<p<\infty$, we consider the following operators; $$ C(f)(z)=\sum_{n=0}^\infty \left(\frac{1}{n+1}\sum_{k=0}^n b_k\right)z^n $$ and $$ A(f)(z)=\sum_{n=0}^\infty \left(\sum_{k=n}^\infty \frac{b_k}{k+1}\right)z^n. $$ The operators $C$ and $A$ are bounded on $H^p(\mathbb{D})$ (see for instance A. G. Siskakis, "Composition semigroups and the Cesàro operator on $H^p$, J. London Math. Soc. 36, 1987, 153-164.).

For $p=2$ the adjoint of $C$ is $A$.

Question: Can we identify the adjoint of $C$ acting on $H^p$ with $A$ acting on $H^q$; $\frac{1}{p}+\frac{1}{q}=1$?

Answer 1

The operator $A$ on $H^q$ is similar to $C^*$ for $C$ acting on $H^p$.

Let us consider the functions $e_n(t)= e^{i2\pi nt}$ ($n\in\mathbb{N}\cup \{0\}$). For $1<p<\infty$ and $1/p+1/q=1$, the map that takes $e_n\in H^q$ to the functional $e^*_n\in {H^p}^*$ defined by $e^*_n(\sum_{k=0}^\infty b_k z^k)=b_n$ extends to a bijective isomorphim $U$ from $H^q$ onto ${H^p}^*$, and is is not difficult to check that $A= U^{-1}C^*U$.