Analytic Poisson integrals are exactly the measures with vanishing negative coefficients

Expansion of the kernel. Fix $z\in \mathbb{D}$ and $\zeta \in \mathbb{T}$, put $\xi :=\phi (\zeta )\in \mathbb{C}$, so that $\mathrm{\mid }\xi \mathrm{\mid }=1$ and ${\xi }^{-1}=\bar{\xi }$, and put $w:=z{\xi }^{-1}$, so $\mathrm{\mid }w\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }<1$. Dividing numerator and denominator of $(\xi +z)\mathrm{/}(\xi -z)$ by $\xi$ and using $P(z,\zeta )=\mathrm{Re}[(\xi +z)\mathrm{/}(\xi -z)]$ gives $P(z,\zeta )=\mathrm{Re}[(1+w)\mathrm{/}(1-w)]$; since $(1+w)\mathrm{/}(1-w)=1+2{\sum }_{n\ge 1}{w}^n$ for $\mathrm{\mid }w\mathrm{\mid }<1$, and $w^n=z^n{\xi }^{-n}=z^n{e}_{-n}(\zeta )$ while ${\bar{w}}^n={\bar{z}}^n{\xi }^n={\bar{z}}^n{e}_n(\zeta )$, taking real parts yields $P(z,\zeta )=1+{\sum }_{n\ge 1}(z^n{e}_{-n}(\zeta )+{\bar{z}}^n{e}_n(\zeta )),$ a series that converges absolutely and uniformly in $\zeta \in \mathbb{T}$ for fixed $z$, because $\mathrm{\mid }{z}^n{e}_{-n}(\zeta )\mathrm{\mid }+\mathrm{\mid }{\bar{z}}^n{e}_n(\zeta )\mathrm{\mid }=2\mathrm{\mid }z{\mathrm{\mid }}^n$ and ${\sum }_n\mathrm{\mid }z{\mathrm{\mid }}^n<\mathrm{\infty }$.

(ii) implies (i). Assume (ii) and put $g(z):={\sum }_{n\ge 0}\widehat{\mu }(n)z^n$. Since $\mathrm{\mid }\widehat{\mu }(n)\mathrm{\mid }\le \mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})$ for every $n$, the series converges for $\mathrm{\mid }z\mathrm{\mid }<1$; its sum $g$ is analytic on the open unit disc by [L6] and hence holomorphic there by A complex function is holomorphic if and only if it is analytic, and (ii) says $P[\mu ]=g$. Thus (i) holds.

The $H^1$ bound. For every $z\in \mathbb{D}$ and every radius $0<r<1$, [L1] and [L4] give $\mathrm{\mid }P[\mu ](z)\mathrm{\mid }\le {\int }_{\mathbb{T}}P(z,\zeta )d\mathrm{\mid }\mu \mathrm{\mid }(\zeta )$ and hence, integrating over the circle $z=r{\zeta }^{\mathrm{\prime }}$ against $m$, ${\int }_{\mathbb{T}}\mathrm{\mid }P[\mu ](r{\zeta }^{\mathrm{\prime }})\mathrm{\mid }dm({\zeta }^{\mathrm{\prime }})\le {\int }_{\mathbb{T}}({\int }_{\mathbb{T}}P(r{\zeta }^{\mathrm{\prime }},\eta )dm({\zeta }^{\mathrm{\prime }}))d\mathrm{\mid }\mu \mathrm{\mid }(\eta )=\mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T});$ the interchange is Tonelli's theorem applied to the nonnegative product-measurable integrand $({\zeta }^{\mathrm{\prime }},\eta )\mapsto P(r{\zeta }^{\mathrm{\prime }},\eta )$, and the inner integral equals the unit mass of the kernel by [L2] together with translation invariance of $m$, the substitution showing ${\int }_{\mathbb{T}}P(r{\zeta }^{\mathrm{\prime }},\eta )dm({\zeta }^{\mathrm{\prime }})=\frac{1}{2\pi }{\int }_0^{2\pi }{P}_r(\theta )d\theta =1$.

Termwise integration and the two-sided expansion. For fixed $z\in \mathbb{D}$ the partial sums of the series of step 1.1 converge uniformly on $\mathbb{T}$ to the continuous function $\zeta \mapsto P(z,\zeta )$, so integrating term by term against the finite complex measure $\mu$ gives $P[\mu ](z)=\mu (\mathbb{T})+{\sum }_{n\ge 1}\left(\right({\int }_{\mathbb{T}}{e}_{-n}d\mu )z^n+({\int }_{\mathbb{T}}{e}_{n}d\mu \left){\bar{z}}^n\right)={\sum }_{n\ge 0}\widehat{\mu }(n)z^n+{\sum }_{n<0}\widehat{\mu }(n){\bar{z}}^{-n},$ since $\mu (\mathbb{T})=\int e_{0}d\mu =\widehat{\mu }(0)$ and, for $n\ge 1$, $\int e_{n}d\mu =\widehat{\mu }(-n)$ by the substitution $m=-n$. This identity holds for every finite complex Borel measure $\mu$ and every $z\in \mathbb{D}$.

Fourier coefficients of the radial traces. Fix $0<r<1$ and $m\in \mathbb{Z}$, and regard a torus element ${\zeta }^{\mathrm{\prime }}$ also as the unit-circle point $\phi ({\zeta }^{\mathrm{\prime }})$, writing ${\zeta }^{\mathrm{\prime }n}:=e_n({\zeta }^{\mathrm{\prime }})$. Since ${\overline{\phi ({\zeta }^{\mathrm{\prime }})}}^{\mathrm{\mid }n\mathrm{\mid }}=\phi ({\zeta }^{\mathrm{\prime }}{)}^{-\mathrm{\mid }n\mathrm{\mid }}=e_{-\mathrm{\mid }n\mathrm{\mid }}({\zeta }^{\mathrm{\prime }})$, evaluating step 2.1 at $z=r\phi ({\zeta }^{\mathrm{\prime }})$ gives $P[\mu ](r{\zeta }^{\mathrm{\prime }})={\sum }_{n\ge 0}\widehat{\mu }(n)r^n{\zeta }^{\mathrm{\prime }n}+{\sum }_{n<0}\widehat{\mu }(n)r^{\mathrm{\mid }n\mathrm{\mid }}{\zeta }^{\mathrm{\prime }n}={\sum }_{n\in \mathbb{Z}}\widehat{\mu }(n)r^{\mathrm{\mid }n\mathrm{\mid }}{\zeta }^{\mathrm{\prime }n},$ a series that converges uniformly in ${\zeta }^{\mathrm{\prime }}\in \mathbb{T}$ because ${\sum }_n\mathrm{\mid }\widehat{\mu }(n)\mathrm{\mid }{r}^{\mathrm{\mid }n\mathrm{\mid }}\le \mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T}){\sum }_n{r}^{\mathrm{\mid }n\mathrm{\mid }}<\mathrm{\infty }$ by [L4]; the function ${\zeta }^{\mathrm{\prime }}\mapsto P[\mu ](r{\zeta }^{\mathrm{\prime }})$ is therefore continuous and its $m$-th Fourier coefficient is ${\int }_{\mathbb{T}}P[\mu ](r{\zeta }^{\mathrm{\prime }})e_{-m}({\zeta }^{\mathrm{\prime }})dm({\zeta }^{\mathrm{\prime }})={\sum }_{n\in \mathbb{Z}}\widehat{\mu }(n)r^{\mathrm{\mid }n\mathrm{\mid }}{\int }_{\mathbb{T}}{e}_{n-m}dm=\widehat{\mu }(m)r^{\mathrm{\mid }m\mathrm{\mid }},$ by termwise integration and orthonormality.

Equivalence of (ii) and (iii). If (iii) holds, the second sum in the expansion of step 2.1 vanishes termwise, so $P[\mu ](z)={\sum }_{n\ge 0}\widehat{\mu }(n)z^n$ for every $\mathrm{\mid }z\mathrm{\mid }<1$, which is (ii). Conversely assume (ii). For $m<0$ and $0<r<1$, the right side of (ii) is the uniformly convergent series ${\sum }_{k\ge 0}\widehat{\mu }(k)r^k{\zeta }^{\mathrm{\prime }k}$ on $\mathbb{T}$ (uniform convergence as in step 3.1, using $\mathrm{\mid }\widehat{\mu }(k)\mathrm{\mid }\le \mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})$), so integrating it against $e_{-m}$ and using orthonormality gives ${\int }_{\mathbb{T}}P[\mu ](r{\zeta }^{\mathrm{\prime }})e_{-m}({\zeta }^{\mathrm{\prime }})dm({\zeta }^{\mathrm{\prime }})=0$; by step 3.1 this equals $\widehat{\mu }(m)r^{\mathrm{\mid }m\mathrm{\mid }}$, hence $\widehat{\mu }(m)=0$ for every $m<0$, which is (iii).

(i) implies (iii), and the Taylor coefficients. Assume (i). By [L6] the holomorphic function $P[\mu ]$ equals its Taylor series $P[\mu ](z)={\sum }_{k\ge 0}{a}_k{z}^k$ throughout $\mathbb{D}$, and this series converges uniformly on the closed subdisc $\mathrm{\mid }z\mathrm{\mid }\le r$ for every fixed $0<r<1$. Fix such an $r$ and $m\in \mathbb{Z}$. Evaluating at $z=r\phi ({\zeta }^{\mathrm{\prime }})$ gives the uniformly convergent series $P[\mu ](r{\zeta }^{\mathrm{\prime }})={\sum }_{k\ge 0}{a}_k{r}^k{\zeta }^{\mathrm{\prime }k}$; integrating against $e_{-m}$ and using orthonormality, its $m$-th Fourier coefficient is $a_m{r}^m$ when $m\ge 0$ and $0$ when $m<0$. Comparing with the identity of step 3.1 gives $a_m{r}^m=\widehat{\mu }(m)r^m$ for $m\ge 0$, so $a_m=\widehat{\mu }(m)$ for every $m\ge 0$ (take any $r\in (0,1)$), and $0=\widehat{\mu }(m)r^{\mathrm{\mid }m\mathrm{\mid }}$ for $m<0$, so $\widehat{\mu }(m)=0$ for every $m<0$, which is (iii).

Assembly. Steps 4.1, 1.2 and 4.2 prove that (i), (ii) and (iii) are equivalent: (ii) and (iii) are equivalent by step 4.1, (ii) implies (i) by step 1.2, and (i) implies (iii) by step 4.2, while (iii) implies (ii) again by step 4.1. If the equivalent conditions hold, then by step 4.2 the Taylor coefficients of $P[\mu ]$ are $a_m=\widehat{\mu }(m)$ for $m\ge 0$, its holomorphy is (i), and step 1.3 gives ${\sup }_{0<r<1}{\int }_{\mathbb{T}}\mathrm{\mid }P[\mu ](r{\zeta }^{\mathrm{\prime }})\mathrm{\mid }dm({\zeta }^{\mathrm{\prime }})=:\mathrm{\parallel }P[\mu ]{\mathrm{\parallel }}_{H^1}\le \mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})<+\mathrm{\infty }$ by [L7], so $P[\mu ]\in H^1(\mathbb{D})$ with the asserted bound.