Poisson-Jensen inequality for Hardy functions

The Jensen inequality. Assume first $f(z)\mathrm{\ne }0$ and take $R<1$ sufficiently close to $1$ that $f(Rz)\mathrm{\ne }0$, which follows from continuity and $f(z)\mathrm{\ne }0$. By [L2] the function $F(w)=f_R({\phi }_z(w))$ is holomorphic on a neighbourhood of the closed unit disc with $F(0)=f(Rz)\mathrm{\ne }0$; applying [L3] and dropping the nonnegative zero terms gives $\log \mathrm{\mid }f(Rz)\mathrm{\mid }\le \frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }{f}_R({\phi }_z(e^{i\theta }))\mathrm{\mid }d\theta$ (if $F$ has boundary zeros, use the limiting form of [L3]). By the change of variables [L4], the right side equals ${\int }_{\mathbb{T}}\log \mathrm{\mid }{f}_R(\zeta )\mathrm{\mid }P(z,\zeta )dm(\zeta )={\int }_{\mathbb{T}}P(z,\zeta )\log \mathrm{\mid }f(R\zeta )\mathrm{\mid }dm(\zeta )\mathrm{.}$

The case $f(z)=0$. If $f(z)=0$, then $\log \mathrm{\mid }f(z)\mathrm{\mid }=-\mathrm{\infty }\le P[\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }](z)$ by the convention on $\log 0$; the inequality holds trivially.

Positive logarithmic parts converge in L1. For finite p set $C=\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p$, so $\int \mathrm{\mid }{f}_R{\mathrm{\mid }}^{p}dm\le C$ and $\int \mathrm{\mid }{f}^{\ast }{\mathrm{\mid }}^{p}dm\le C$ by [L1]. Given $\epsilon >0$, choose $L>0$ large enough that $\log t\le \epsilon t^p$ for all $t>e^L$; indeed [L1] with $q=p\mathrm{/}2$ gives $\log t\mathrm{/}{t}^p\le (2\mathrm{/}p)t^{-p\mathrm{/}2}$ for $t\ge 1$, and it suffices to take L with $(2\mathrm{/}p)e^{-Lp\mathrm{/}2}\le \epsilon$. Writing $u_R={\log }^{+}\mathrm{\mid }{f}_R\mathrm{\mid }$, $u^{\ast }={\log }^{+}\mathrm{\mid }{f}^{\ast }\mathrm{\mid }$, the tails satisfy $\int (u_R-L{)}_{+}dm\le \epsilon C,\int (u^{\ast }-L{)}_{+}dm\le \epsilon C\mathrm{.}$ The truncated functions $\min (u_R,L)$ converge almost everywhere to $\min (u^{\ast },L)$ and are bounded by L, so [L5] gives L1 convergence by dominated convergence. Therefore ${\mathrm{lim\hspace{0.17em}sup}}_{R\uparrow 1}\mathrm{\parallel }{u}_R-u^{\ast }{\mathrm{\parallel }}_1\le 2\epsilon C$, and letting epsilon decrease to zero proves the claim. For p infinity, all positive logarithms are bounded by ${\log }^{+}\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, so dominated convergence applies directly. These limit statements hold along every sequence tending to one, hence for the stated radial limit.

Pass to the Jensen inequality. For f(z) nonzero, let R increase to one in step 1.1. Its left side tends to $\log \mathrm{\mid }f(z)\mathrm{\mid }$. The weight $P(z,\cdot )$ is bounded by [L5], so step 1.3 gives convergence of the weighted positive logarithmic integrals. Fatou's lemma gives $\int P(z,\zeta ){\log }^{-}\mathrm{\mid }{f}^{\ast }(\zeta )\mathrm{\mid }dm\le {\mathrm{lim\hspace{0.17em}inf}}_{R\uparrow 1}\int P(z,\zeta ){\log }^{-}\mathrm{\mid }f(R\zeta )\mathrm{\mid }dm\mathrm{.}$ Subtracting this inequality from the positive-part limit bounds the limsup of the Jensen right side by $P[\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }](z)$. This quantity is finite by [L1] and the bounded weight, so $\log \mathrm{\mid }f(z)\mathrm{\mid }\le P[\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }](z)$. Step 1.2 covers zeros of f. Only CC boundary existence and the logarithmic tail estimate were used; no AC Hardy representation is invoked.

Equality for outer functions. If $f$ is outer, then by [L6] the identity $\log \mathrm{\mid }f(z)\mathrm{\mid }=P[\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }](z)$ holds for every $z$, so equality holds in the Poisson-Jensen inequality; thus the inequality is an identity for all nonzero outer functions.