F. Riesz factorization of a Hardy-space function

The Blaschke condition. By the $H^p$ clause of [L2] applied to $f\in H^p$, the liminf hypothesis holds and ${\sum }_n(1-\mathrm{\mid }{a}_n\mathrm{\mid })<+\mathrm{\infty }$; hence $B$ is a well-defined Blaschke product with $\mathrm{\mid }B\mathrm{\mid }\le 1$, each $B_N$ extends to the closed disc with $\mathrm{\mid }{B}_N\mathrm{\mid }=1$ on $\mathbb{T}$, and each $g_N=f\mathrm{/}{B}_N$ is holomorphic on $\mathbb{D}$ by [L3].

Bounding the $N$-th quotient. Fix $0<r<R<1$, $0<\epsilon <1$ and $N$. By [L1] applied to the holomorphic function $g_N$, ${\int }_{\mathbb{T}}\mathrm{\mid }{g}_N(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta )\le {\int }_{\mathbb{T}}\mathrm{\mid }{g}_N(R\zeta ){\mathrm{\mid }}^{p}dm(\zeta )={\int }_{\mathbb{T}}\frac{\mathrm{\mid }f(R\zeta ){\mathrm{\mid }}^p}{\mathrm{\mid }{B}_N(R\zeta ){\mathrm{\mid }}^p}dm(\zeta )\mathrm{.}$ Since $B_N$ is continuous on $\bar{\mathbb{D}}$ with $\mathrm{\mid }{B}_N\mathrm{\mid }=1$ on $\mathbb{T}$ by [L2], there is $\rho <1$ with $\mathrm{\mid }{B}_N(R\zeta )\mathrm{\mid }\ge 1-\epsilon$ for all $\zeta \in \mathbb{T}$ and all $R\in (\rho ,1)$; for such $R$, using [L1] again, ${\int }_{\mathbb{T}}\mathrm{\mid }{g}_N(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta )\le (1-\epsilon {)}^{-p}{\int }_{\mathbb{T}}\mathrm{\mid }f(R\zeta ){\mathrm{\mid }}^{p}dm(\zeta )\le (1-\epsilon {)}^{-p}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p\mathrm{.}$

The quotient. The function $g=f\mathrm{/}B$ is holomorphic off the zeros of $B$; at each $a$ occurring $m$ times, $g$ is bounded near $a$ and hence extends holomorphically by [L3], and $g(a)\mathrm{\ne }0$ because the order of the zero of $f$ at $a$ equals the multiplicity $m$ with which $a$ is listed. Thus $g$ is holomorphic and zero-free on $\mathbb{D}$, and $\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\mid }g\mathrm{\mid }$ because $\mathrm{\mid }B\mathrm{\mid }\le 1$.

Passing to the limits in the correct order. Fix $N$ and $r<1$. In step 1.2 choose $R$ close enough to $1$ for this $N$ and $\epsilon$, then let $\epsilon \downarrow 0$. This gives $\int \mathrm{\mid }{g}_N(r\zeta ){\mathrm{\mid }}^{p}dm\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p$, independently of $N$. The identities $g_N=b_{a_{N+1}}{g}_{N+1}$ and $\mathrm{\mid }{b}_a\mathrm{\mid }\le 1$ show that $\mathrm{\mid }{g}_N{\mathrm{\mid }}^p$ increases with $N$; off the zeros of $B$, $g_N=f\mathrm{/}{B}_N\to f\mathrm{/}B=g$. A fixed circle contains only finitely many zeros, a null set, so monotone convergence [L5] gives $\int \mathrm{\mid }g(r\zeta ){\mathrm{\mid }}^{p}dm\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p$. Taking the supremum over $r$ yields $\mathrm{\parallel }g{\mathrm{\parallel }}_{H^p}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$. For finite or empty zero lists, the products stabilize, and the same argument gives $B=1$, $g=f$ when the list is empty.

Equality and assembly. Since $\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\mid }g\mathrm{\mid }$ pointwise by step 2.1, the radial means satisfy ${\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm\le {\int }_{\mathbb{T}}\mathrm{\mid }g(r\zeta ){\mathrm{\mid }}^{p}dm$ for every $r$, so $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{H^p}$ by [L1]; combined with step 3.1 this gives $\mathrm{\parallel }g{\mathrm{\parallel }}_{H^p}=\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$ with $g\in H^p(\mathbb{D})$. Steps 1.1, 2.1 and 3.1 together prove all the asserted clauses.