h^p is the Poisson image of Lp for 1<p<=infinity

Setup. Let $u\in h^p(\mathbb{D})$ with $M=\mathrm{\parallel }u{\mathrm{\parallel }}_{h^p}$ and let $r_j=1-\frac{1}{j+1}\uparrow 1$. By [L1] the function $u$ is complex harmonic, hence continuous, on $\mathbb{D}$, and $u_r(\zeta )=u(r\zeta )$ is measurable with $\mathrm{\parallel }{u}_r{\mathrm{\parallel }}_p\le M$ for every $0\le r<1$; if $p=\mathrm{\infty }$ this reads $\mathrm{\mid }{u}_r\mathrm{\mid }\le M$ everywhere, and then the probability measure $m$ gives $\mathrm{\parallel }{u}_r{\mathrm{\parallel }}_1\le \mathrm{\parallel }{u}_r{\mathrm{\parallel }}_{\mathrm{\infty }}\le M$ by [L5].

Choice bookkeeping. By [L9] the Axiom of Choice supplies ${\mathrm{A}\mathrm{C}}_{\omega }$, the ultrafilter lemma, DC and HB, so the reflexivity and weak-subsequence hypotheses recorded in [L4] are met.

The finite exponent case: a weak limit. Assume $1<p<\mathrm{\infty }$. The sequence $(u_{r_j})$ is norm-bounded by $M$ in the reflexive space $L^p(\mathbb{T},m;\mathbb{C})$ by steps 1.1 and 1.2, so [L4] provides a subsequence, relabelled $(u_{r_j})$, and an element $f\in L^p(\mathbb{T},m;\mathbb{C})$ with $u_{r_j}\rightharpoonup f$; explicitly ${\int }_{\mathbb{T}}{u}_{r_j}gdm\to {\int }_{\mathbb{T}}fgdm$ for every $g\in L^q(\mathbb{T},m;\mathbb{C})$.

The exponent infinity: a boundary measure. Assume $p=\mathrm{\infty }$. By step 1.1, $\mathrm{\parallel }{u}_r{\mathrm{\parallel }}_1\le M$ for every $r$, so $u\in h^1(\mathbb{D})$ with $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^1}\le M$; [L3] therefore gives a unique finite regular complex Borel measure $\mu$ on $\mathbb{T}$ with $u=P[\mu ]$, $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^1}=\mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})$ and ${\int }_{\mathbb{T}}gd\mu ={\lim }_{r\uparrow 1}{\int }_{\mathbb{T}}g{u}_{r}dm$ for every continuous $g$.

The finite exponent case: identification of $u$. Assume $1<p<\mathrm{\infty }$ and let $f$ be the weak limit of step 2.1. Fix $z\in \mathbb{D}$. For every $j$ with $r_j>\mathrm{\mid }z\mathrm{\mid }$ the function $u$ is harmonic on an open set containing the closed disc of radius $r_j$, so [L2] gives $u(z)={\int }_{\mathbb{T}}{P}^{(j)}(z,\eta )u_{r_j}(\eta )dm(\eta ),P^{(j)}(z,\eta ):=\frac{r_j^2-\mathrm{\mid }z{\mathrm{\mid }}^2}{\mathrm{\mid }{r}_j\eta -z{\mathrm{\mid }}^2},$ and writing $P(z,\eta )=(1-\mathrm{\mid }z{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }\eta -z{\mathrm{\mid }}^2$ this becomes $u(z)={\int }_{\mathbb{T}}P(z,\eta )u_{r_j}(\eta )dm(\eta )+{\int }_{\mathbb{T}}(P^{(j)}(z,\eta )-P(z,\eta ))u_{r_j}(\eta )dm(\eta )\mathrm{.}$ The function $P(z,\cdot )$ is continuous on $\mathbb{T}$, hence belongs to $L^q$, so step 2.1 gives ${\int }_{\mathbb{T}}P(z,\cdot )u_{r_j}dm\to {\int }_{\mathbb{T}}P(z,\cdot )fdm=P[f](z)$; the second integral is bounded in modulus by $\mathrm{\parallel }{P}^{(j)}(z,\cdot )-P(z,\cdot ){\mathrm{\parallel }}_q\mathrm{\parallel }{u}_{r_j}{\mathrm{\parallel }}_p$ by [L5], and this tends to $0$ because $r_j\uparrow 1$, the point $z$ stays at positive distance from $\mathbb{T}$, and $\mathrm{\parallel }{u}_{r_j}{\mathrm{\parallel }}_p\le M$. Hence $u(z)=P[f](z)$ for every $z\in \mathbb{D}$, that is $u=P[f]$.

The exponent infinity: extension of the boundary functional. Assume $p=\mathrm{\infty }$ and let $\mu$ be the measure of step 2.2. For continuous $g$ and $0\le r<1$, [L5] gives $\mathrm{\mid }{\int }_{\mathbb{T}}g{u}_{r}dm\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{\parallel }{u}_r{\mathrm{\parallel }}_{\mathrm{\infty }}\le M\mathrm{\parallel }g{\mathrm{\parallel }}_1$; letting $r\uparrow 1$ along the convergence of step 2.2 yields $\mathrm{\mid }{\int }_{\mathbb{T}}gd\mu \mathrm{\mid }\le M\mathrm{\parallel }g{\mathrm{\parallel }}_1$. Hence $W(g):={\int }_{\mathbb{T}}gd\mu$ is a complex linear functional on the dense subspace $C(\mathbb{T},\mathbb{C})$ of $L^1(\mathbb{T},m;\mathbb{C})$ satisfying $\mathrm{\mid }W(g)\mathrm{\mid }\le M\mathrm{\parallel }g{\mathrm{\parallel }}_1$, and the bound in particular shows that $W$ vanishes on continuous functions that are $m$-almost everywhere zero, so $W$ is well defined on the corresponding subspace of the quotient; [L6] therefore extends it uniquely to a bounded complex linear functional $\tilde{W}$ on $L^1(\mathbb{T},m;\mathbb{C})$ with $\mathrm{\parallel }\tilde{W}\mathrm{\parallel }\le M$.

The exponent infinity: the density. Keep $p=\mathrm{\infty }$, $\tilde{W}$ as in step 3.2. The functional $h\mapsto \mathrm{Re}\tilde{W}(h)$ is real linear and bounded on the real space $L^1(\mathbb{T},m;\mathbb{R})$ with norm at most $\mathrm{\parallel }\tilde{W}\mathrm{\parallel }\le M$, so [L6] (applied with $p=1$ on the sigma-finite space $(\mathbb{T},m)$) provides $f_1\in L^{\mathrm{\infty }}(\mathbb{T},m;\mathbb{R})$ with $\mathrm{Re}\tilde{W}(h)={\int }_{\mathbb{T}}h{f}_{1}dm$ for all real $h\in L^1$ and $\mathrm{\parallel }{f}_1{\mathrm{\parallel }}_{\mathrm{\infty }}\le M$; applying the same theorem to $h\mapsto \mathrm{Im}\tilde{W}(h)$ gives $f_2\in L^{\mathrm{\infty }}(\mathbb{T},m;\mathbb{R})$ with $\mathrm{\parallel }{f}_2{\mathrm{\parallel }}_{\mathrm{\infty }}\le M$. Put $f:=f_1+i{f}_2\in L^{\mathrm{\infty }}(\mathbb{T},m;\mathbb{C})$: by complex linearity of $\tilde{W}$ and of the integral, $\tilde{W}(h)={\int }_{\mathbb{T}}hfdm$ for every $h\in L^1(\mathbb{T},m;\mathbb{C})$, and in particular ${\int }_{\mathbb{T}}gd\mu ={\int }_{\mathbb{T}}gfdm$ for every continuous $g$. Both $\mu$ and the density measure $fm$ are finite Borel measures on the second-countable space $\mathbb{T}$, hence regular by [L7], and they agree on all continuous functions, so the uniqueness clause of [L7] gives $\mu =fm$; consequently $u=P[\mu ]=P[fm]=P[f]$ by [L1].

The finite exponent case: norm equality. Assume $1<p<\mathrm{\infty }$, let $f$ be the weak limit of step 2.1 and keep the identification $u=P[f]$ of step 3.1. For every complex finite simple $s$ of finite-measure support with $\mathrm{\parallel }s{\mathrm{\parallel }}_q\le 1$, step 2.1 gives ${\int }_{\mathbb{T}}fsdm={\lim }_j{\int }_{\mathbb{T}}{u}_{r_j}sdm$, and [L5] bounds $\mathrm{\mid }{\int }_{\mathbb{T}}{u}_{r_j}sdm\mathrm{\mid }\le \mathrm{\parallel }{u}_{r_j}{\mathrm{\parallel }}_p\mathrm{\parallel }s{\mathrm{\parallel }}_q\le M$; the norm identity of [L5] therefore gives $\mathrm{\parallel }f{\mathrm{\parallel }}_p\le M$. Since $u=P[f]$, the contraction in [L1] gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^p}=\mathrm{\parallel }P[f]{\mathrm{\parallel }}_{h^p}\le \mathrm{\parallel }f{\mathrm{\parallel }}_p$, and hence $\mathrm{\parallel }f{\mathrm{\parallel }}_p=M=\mathrm{\parallel }u{\mathrm{\parallel }}_{h^p}$.

The finite exponent case: uniqueness and norm convergence. Assume $1<p<\mathrm{\infty }$ and let $g\in L^p(\mathbb{T},m;\mathbb{C})$ satisfy $P[g]=u=P[f]$. Then $P[f-g]=0$, and the convergence clause of [L1] gives $\mathrm{\parallel }(P[f-g]{)}_r-(f-g){\mathrm{\parallel }}_p\to 0$, so $f-g=0$ almost everywhere and $f$ is the unique representing function. The same convergence clause applied to $f$ yields $\mathrm{\parallel }{u}_r-f{\mathrm{\parallel }}_p=\mathrm{\parallel }(P[f]{)}_r-f{\mathrm{\parallel }}_p\to 0$ as $r\uparrow 1$.

The exponent infinity: the sharp norm bound. Keep $p=\mathrm{\infty }$ and $f=f_1+i{f}_2$ from step 4.1. For every complex finite simple $s$ of finite-measure support with $\mathrm{\parallel }s{\mathrm{\parallel }}_1\le 1$ one has $s\in L^1(\mathbb{T},m;\mathbb{C})$, so step 4.1 gives ${\int }_{\mathbb{T}}fsdm=\tilde{W}(s)$ and hence $\mathrm{\mid }{\int }_{\mathbb{T}}fsdm\mathrm{\mid }\le \mathrm{\parallel }\tilde{W}\mathrm{\parallel }\mathrm{\parallel }s{\mathrm{\parallel }}_1\le M$. The norm identity of [L5] at $p=\mathrm{\infty }$, whose hypothesis $\int \mathrm{\mid }fs\mathrm{\mid }dm<\mathrm{\infty }$ holds because $f\in L^{\mathrm{\infty }}$ and $s$ is bounded with finite-measure support, gives $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le M$.

The exponent infinity: norm equality and uniqueness. Assume $p=\mathrm{\infty }$. By step 4.1, $u=P[f]$ with $f\in L^{\mathrm{\infty }}(\mathbb{T},m;\mathbb{C})$, so the contraction of [L1] at $p=\mathrm{\infty }$ gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^{\mathrm{\infty }}}=\mathrm{\parallel }P[f]{\mathrm{\parallel }}_{h^{\mathrm{\infty }}}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le M=\mathrm{\parallel }u{\mathrm{\parallel }}_{h^{\mathrm{\infty }}}$, and therefore $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }u{\mathrm{\parallel }}_{h^{\mathrm{\infty }}}$. If also $P[g]=u$ with $g\in L^{\mathrm{\infty }}$, then $f-g\in L^1$ and $P[f-g]=0$, so the finite exponent convergence clause of [L1] at $p=1$ gives $\mathrm{\parallel }(P[f-g]{)}_r-(f-g){\mathrm{\parallel }}_1\to 0$, whence $f=g$ almost everywhere.

The exponent infinity: weak-star convergence of the radial functions. Keep $p=\mathrm{\infty }$ and $f$ as in step 4.1. For $g\in L^1(\mathbb{T},m;\mathbb{C})$ and $0\le r<1$ one has $u_r=P_r\ast f$ by [L1], and the product integrand $(\zeta ,\eta )\mapsto g(\zeta )P_r(\zeta -\eta )f(\eta )$ is integrable for the product of the probability measure $m$ with itself, because $\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and ${\int }_{\mathbb{T}}{P}_r(\zeta -\eta )dm(\zeta )=1$ for every $\eta$; Fubini [L8] and the translation invariance of $m$ therefore give ${\int }_{\mathbb{T}}g{u}_{r}dm={\int }_{\mathbb{T}}({\int }_{\mathbb{T}}g(\zeta )P_r(\zeta -\eta )dm(\zeta ))f(\eta )dm(\eta )={\int }_{\mathbb{T}}(P_r\ast g)(\eta )f(\eta )dm(\eta )\mathrm{.}$ Consequently [L5] bounds $\mid {\int }_{\mathbb{T}}g{u}_{r}dm-{\int }_{\mathbb{T}}gfdm\mid \le \mathrm{\parallel }{P}_r\ast g-g{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, which tends to $0$ as $r\uparrow 1$ by the $L^1$ convergence clause of [L1]; hence in $\sigma (L^{\mathrm{\infty }}(\mathbb{T},m),L^1(\mathbb{T},m))$.

Assembly. If $1<p<\mathrm{\infty }$, steps 2.1, 3.1, 4.2 and 5.1 produce a unique $f\in L^p(\mathbb{T},m;\mathbb{C})$ with $u=P[f]$, the norm identity $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^p}=\mathrm{\parallel }f{\mathrm{\parallel }}_p$ and the $L^p$ convergence $\mathrm{\parallel }{u}_r-f{\mathrm{\parallel }}_p\to 0$. If $p=\mathrm{\infty }$, steps 2.2, 3.2, 4.1, 5.2, 6.1 and 7.1 produce a unique $f\in L^{\mathrm{\infty }}(\mathbb{T},m;\mathbb{C})$ with $u=P[f]$, the norm identity $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^{\mathrm{\infty }}}=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and the weak-star convergence of the radial functions against $L^1$; no norm convergence is claimed at $p=\mathrm{\infty }$, in accordance with the fact that [L1] asserts norm convergence only for finite exponents. The Axiom of Choice was used exactly through step 1.2: ${\mathrm{A}\mathrm{C}}_{\omega }$ for reflexivity of $L^p$ and the ultrafilter lemma, DC and HB for the weak-subsequence criterion of [L4], while the case $p=\mathrm{\infty }$ additionally rests on the $h^1$ representation theorem [L3], itself licensed by AC. This proves all the assertions of the Statement.