A theorem of Hardy and Littlewood and its extension for Abel

means says that the following are equivalent for functions $f$ in the Hardy space $ H^p$ of the disc:

(a) $\Vert f(e^{it}z)-f(z)\Vert_p=O(t^{\alpha})$ as $t\to 0$, i.e. $f\in \Lambda\left(p,\alpha\right)$, the mean Lipschitz class.

(b) $M_p(r,f^\prime) =O\left ((1-r)^{\alpha -1}\right )$ as $ r\to 1^{-}$,

(c) $\Vert f(rz)-f(z)\Vert_p=O((1-r)^{\alpha})$ as $ r\to 1^{-}$,

We will describe an "elementary" proof of the theorem, and will discuss an analogue on Bergman spaces.

Recall that in the abstract setting, the Favard class $F_{\alpha}$, $0<\alpha\leq 1$, for a strongly continuous

operator semigroup $(T_t)$ on a Banach space $X$ is

$$F_{\alpha}=\{f\in X: \Vert T_t(f)-f \Vert_X=O(t^{\alpha})\}.$$

The above conditions then, interpreted in terms of the composition semigroups $f\to f(e^{it}z)$ (rotations) and $f\to f(e^{-t}z)$ (dilations) say that $F_{\alpha}=\Lambda\left(p,\alpha\right)$ for both of these semigroups on $H^p$. The question arises, given a general composition semigroup $T_t(f)= f\circ \phi_t$ acting on Hardy space (or on other spaces of analytic functions), to identify its Favard class $F_{\alpha}$ in terms of a condition analogous to (b), involving the derivative.