Poisson extension is an Lp contraction and converges in finite Lp

By [L5] (Hölder against the constant function $1$, whose conjugate norm is $m(\mathbb{T}{)}^{1-1\mathrm{/}p}=1$ for finite $p$ and $\mathrm{\parallel }1{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ for $p=\mathrm{\infty }$), every $f\in L^p(\mathbb{T},m;\mathbb{C})$ satisfies ${\int }_{\mathbb{T}}\mathrm{\mid }f\mathrm{\mid }dm\le \mathrm{\parallel }f{\mathrm{\parallel }}_p<\mathrm{\infty }$; hence $f\in L^1(\mathbb{T},m)$ and the Poisson integral $P[f]$ of [L1] is defined.

For every $0\le r<1$ and every $\zeta \in \mathbb{T}$, translation invariance of $m$ and [L3] give ${\int }_{\mathbb{T}}{P}_r(\zeta -\eta )dm(\eta )={\int }_{\mathbb{T}}{P}_{r}dm=1$; since the kernel is positive, the elementary estimate $\mathrm{\mid }\int gdm\mathrm{\mid }\le \int \mathrm{\mid }g\mathrm{\mid }dm$ of [L5] applied to $g=P_r(\zeta -\cdot )f$ gives $\mathrm{\mid }(P_r\ast f)(\zeta )\mathrm{\mid }\le {\int }_{\mathbb{T}}{P}_r(\zeta -\eta )\mathrm{\mid }f(\eta )\mathrm{\mid }dm(\eta )\mathrm{.}$

If $g\in C(\mathbb{T},\mathbb{C})$, write $g=u+iv$ with real continuous $u,v$. By the agreement clause of [L1] the integrals $P[u]$ and $P[v]$ coincide with the published Poisson integrals of the real continuous boundary data $u\circ {\phi }^{-1}$ and $v\circ {\phi }^{-1}$; [L7] makes both real harmonic, and linearity of the integral gives $P[g]=P[u]+iP[v]$, so $P[g]$ is complex harmonic.

For $p=\mathrm{\infty }$: step 1.2 and the total mass from [L3] give $\mathrm{\mid }(P_r\ast f)(\zeta )\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\mathbb{T}}{P}_r(\zeta -\eta )dm(\eta )=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $\zeta$, so $\mathrm{\parallel }{P}_r\ast f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and ${\sup }_{0\le r<1}\mathrm{\parallel }(P[f]{)}_r{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$.

For $p=1$: integrating the display of step 1.2 over $\zeta$ and applying Tonelli's theorem [L2] to the nonnegative product-measurable integrand $(\zeta ,\eta )\mapsto P_r(\zeta -\eta )\mathrm{\mid }f(\eta )\mathrm{\mid }$, then translation invariance of $m$, gives $\mathrm{\parallel }{P}_r\ast f{\mathrm{\parallel }}_1\le {\int }_{\mathbb{T}}\text{\hspace{0.17em}⁣}{\int }_{\mathbb{T}}{P}_r(\zeta -\eta )\mathrm{\mid }f(\eta )\mathrm{\mid }dm(\eta )dm(\zeta )={\int }_{\mathbb{T}}\mathrm{\mid }f(\eta )\mathrm{\mid }({\int }_{\mathbb{T}}{P}_r(\zeta -\eta )dm(\zeta ))dm(\eta )=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{.}$

For $1<p<\mathrm{\infty }$: fix $\zeta$ and put $w_{\zeta }(\eta ):=P_r(\zeta -\eta )$, a nonnegative measurable density with $\int w_{\zeta }dm=1$ by step 1.2, so [L4] makes ${\nu }_{\zeta }:=w_{\zeta }dm$ a probability measure on $\mathbb{T}$ with $\int \mathrm{\mid }f\mathrm{\mid }d{\nu }_{\zeta }=\int P_r(\zeta -\eta )\mathrm{\mid }f(\eta )\mathrm{\mid }dm(\eta )\le {\sup }_{\eta }{P}_r\mathrm{\parallel }f{\mathrm{\parallel }}_1<\mathrm{\infty }$, so $\mathrm{\mid }f\mathrm{\mid }\in L^1({\nu }_{\zeta })$ and $\mathrm{\mid }f{\mathrm{\mid }}^p\in L^1({\nu }_{\zeta })$ because $\int \mathrm{\mid }f{\mathrm{\mid }}^{p}d{\nu }_{\zeta }\le {\sup }_{\eta }{P}_r\mathrm{\parallel }f{\mathrm{\parallel }}_p^p$. Jensen's inequality [L4] applied to the convex function $t\mapsto t^p$ and $\mathrm{\mid }f\mathrm{\mid }$ gives, using step 1.2, $\mathrm{\mid }(P_r\ast f)(\zeta ){\mathrm{\mid }}^p\le ({\int }_{\mathbb{T}}\mathrm{\mid }f\mathrm{\mid }d{\nu }_{\zeta }{)}^p\le {\int }_{\mathbb{T}}\mathrm{\mid }f{\mathrm{\mid }}^{p}d{\nu }_{\zeta }={\int }_{\mathbb{T}}{P}_r(\zeta -\eta )\mathrm{\mid }f(\eta ){\mathrm{\mid }}^{p}dm(\eta );$ integrating this display over $\zeta$ and applying the same Tonelli and translation-invariance argument yields $\mathrm{\parallel }{P}_r\ast f{\mathrm{\parallel }}_p^p\le \mathrm{\parallel }f{\mathrm{\parallel }}_p^p$, hence $\mathrm{\parallel }{P}_r\ast f{\mathrm{\parallel }}_p\le \mathrm{\parallel }f{\mathrm{\parallel }}_p$.

$P[f]$ is complex harmonic. Indeed $f\in L^1$ by step 1.1; choose continuous $g_n$ with $\mathrm{\parallel }f-g_n{\mathrm{\parallel }}_1\to 0$ [L6] at $p=1$. For $z\in \mathbb{D}$ the bound of [L3] and step 1.1 give $\mathrm{\mid }P[f](z)-P[g_n](z)\mathrm{\mid }=\mathrm{\mid }P[f-g_n](z)\mathrm{\mid }\le \frac{1+\mathrm{\mid }z\mathrm{\mid }}{1-\mathrm{\mid }z\mathrm{\mid }}\mathrm{\parallel }f-g_n{\mathrm{\parallel }}_1,$ and the factor $(1+\mathrm{\mid }z\mathrm{\mid })\mathrm{/}(1-\mathrm{\mid }z\mathrm{\mid })$ is bounded on every compact subset of $\mathbb{D}$; hence $P[g_n]\to P[f]$ locally uniformly on $\mathbb{D}$. Each $P[g_n]$ is complex harmonic by step 1.3, so the locally uniform limit $P[f]$ is complex harmonic by [L7].

Convergence for continuous data: let $g\in C(\mathbb{T},\mathbb{C})$. Applying [L8] to the real and imaginary parts and using the identification of [L1] gives $\mathrm{\mid }P[g](r{e}^{i\alpha })-g(e^{i\alpha })\mathrm{\mid }\to 0$ uniformly in $\alpha$ as $r\uparrow 1$; consequently, for every $1\le p<\mathrm{\infty }$, $\mathrm{\parallel }{P}_r\ast g-g{\mathrm{\parallel }}_p\le \mathrm{\parallel }{P}_r\ast g-g{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$ because $m$ is a probability measure [L2].

Combining steps 2.1, 2.2 and 2.3, for every $1\le p\le \mathrm{\infty }$ and every $0\le r<1$ the contraction $\mathrm{\parallel }{P}_r\ast f{\mathrm{\parallel }}_p\le \mathrm{\parallel }f{\mathrm{\parallel }}_p$ holds. With the harmonicity proved in step 2.4 this yields ${\sup }_{0\le r<1}\mathrm{\parallel }(P[f]{)}_r{\mathrm{\parallel }}_p\le \mathrm{\parallel }f{\mathrm{\parallel }}_p<\mathrm{\infty }$, so $P[f]\in h^p(\mathbb{D})$ with $\mathrm{\parallel }P[f]{\mathrm{\parallel }}_{h^p}\le \mathrm{\parallel }f{\mathrm{\parallel }}_p$.

Let $p<\mathrm{\infty }$ and $\epsilon >0$. By [L6] choose continuous $g$ with $\mathrm{\parallel }f-g{\mathrm{\parallel }}_p<\epsilon \mathrm{/}3$. For every $0\le r<1$, step 3.1 and the triangle inequality for the $L^p$ norm give $\mathrm{\parallel }{P}_r\ast f-f{\mathrm{\parallel }}_p\le \mathrm{\parallel }{P}_r\ast (f-g){\mathrm{\parallel }}_p+\mathrm{\parallel }{P}_r\ast g-g{\mathrm{\parallel }}_p+\mathrm{\parallel }g-f{\mathrm{\parallel }}_p\le 2\mathrm{\parallel }f-g{\mathrm{\parallel }}_p+\mathrm{\parallel }{P}_r\ast g-g{\mathrm{\parallel }}_p,$ and step 2.5 makes the last term smaller than $\epsilon \mathrm{/}3$ for all $r$ sufficiently close to $1$; hence $\mathrm{\parallel }{P}_r\ast f-f{\mathrm{\parallel }}_p<2\epsilon \mathrm{/}3+\epsilon \mathrm{/}3=\epsilon$ for those $r$, so $\mathrm{\parallel }{P}_r\ast f-f{\mathrm{\parallel }}_p\to 0$ as $r\uparrow 1$. This proves the finite-$p$ norm limit, while at $p=\mathrm{\infty }$ no norm-convergence claim is made; the contraction and the $h^p$ membership are step 3.1, completing the proof.