Radial p-means of a holomorphic function are nondecreasing

Reduction and subharmonicity. If $f\equiv 0$, then both sides of the asserted inequality vanish and the further claims are immediate, so assume $f$ is not identically zero. Then $u=\mathrm{\mid }f{\mathrm{\mid }}^p$ is subharmonic on $\mathbb{D}$ and continuous, by [L1].

The $L^p$ comparison. Let $0<p<q<\mathrm{\infty }$ and let $g$ be measurable on $\mathbb{T}$. By [L5], $\mathrm{\parallel }g{\mathrm{\parallel }}_{L^p}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^q}$; and if $q=\mathrm{\infty }$, $\mathrm{\parallel }g{\mathrm{\parallel }}_{L^p}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$.

The modification is the Poisson extension of the boundary data. Fix $0<R<1$. By [L2] and continuity of $u$ on $\overline{D(0,R)}$ (a compact subset of $\mathbb{D}$), the Poisson modification $h:=P_{D(0,R)}u$ is harmonic on $D(0,R)$, satisfies $h\ge u$ on $\mathbb{D}$, and is the Poisson extension of $u{\mathrm{\mid }}_{\mathrm{\partial }D(0,R)}$; hence by [L3], $h(z)={\int }_{\mathbb{T}}P(z\mathrm{/}R,\eta )u(R\eta )dm(\eta )(\mathrm{\mid }z\mathrm{\mid }<R),h(0)={\int }_{\mathbb{T}}u(R\eta )dm(\eta )\mathrm{.}$

Containment of the classes. Let $0<p<q\le \mathrm{\infty }$ and $f\in H^q(\mathbb{D})$; every radius $r\in [0,1)$ satisfies $\mathrm{\parallel }{f}_r{\mathrm{\parallel }}_{L^p}\le \mathrm{\parallel }{f}_r{\mathrm{\parallel }}_{L^q}$ by step 1.2, and $\mathrm{\parallel }{f}_r{\mathrm{\parallel }}_{L^q}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^q}$ by the definition of the supremum when $q<\mathrm{\infty }$, while $\mathrm{\parallel }{f}_r{\mathrm{\parallel }}_{L^q}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^q}$ for $q=\mathrm{\infty }$ as well because $\mathrm{\mid }{f}_r\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^{\mathrm{\infty }}}$ pointwise. Taking suprema over $r$ gives $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^q}<+\mathrm{\infty }$, so $f\in H^p(\mathbb{D})$ and the containment $H^q(\mathbb{D})\subseteq H^p(\mathbb{D})$ holds with the asserted norm comparison.

Monotonicity of the means. Let $0<r\le R<1$. The case $r=R$ is an equality of the two integrals, so assume $r<R$. Step 2.1 gives $h\ge u$ on $D(0,r)$ and $h(0)={\int }_{\mathbb{T}}u(R\eta )dm(\eta )$; since $h$ is harmonic on $D(0,R)$, hence on a neighbourhood of the closed disc $\overline{D(0,r)}$ for $r<R$, integrating the inequality $u(r\zeta )\le h(r\zeta )$ over $\mathbb{T}$ against $m$ and applying the mean value property [L4] gives ${\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta )={\int }_{\mathbb{T}}u(r\zeta )dm(\zeta )\le {\int }_{\mathbb{T}}h(r\zeta )dm(\zeta )=h(0)={\int }_{\mathbb{T}}\mathrm{\mid }f(R\eta ){\mathrm{\mid }}^{p}dm(\eta )\mathrm{.}$

The supremum is the limit. Let $f\in H^p(\mathbb{D})$. Continuity of $u=\mathrm{\mid }f{\mathrm{\mid }}^p$ at $0$ gives ${\sup }_{\zeta \in \mathbb{T}}\mathrm{\mid }u(r\zeta )-u(0)\mathrm{\mid }\to 0$ as $r\downarrow 0$. Thus the mean inequality of step 3.1 also holds when the smaller radius is $0$. The map $r\mapsto {\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta )$ is therefore nondecreasing on $[0,1)$ and bounded by $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p<+\mathrm{\infty }$ by [L6]. A nondecreasing bounded real function on $[0,1)$ has supremum equal to its limit as $r\uparrow 1$, so ${\sup }_{0\le r<1}({\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta ){)}^{1\mathrm{/}p}={\lim }_{r\uparrow 1}({\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta ){)}^{1\mathrm{/}p},$ that is, $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}={\lim }_{r\uparrow 1}\mathrm{\parallel }{f}_r{\mathrm{\parallel }}_{L^p}$.

Assembly. The mean inequality is step 3.1, the limit description of the norm is step 4.1, and the containment together with the norm comparison is step 2.2; all were proved under the given hypotheses.