A maximum principle for the Smirnov class: $N^{+}\cap L^p=H^p$

If $f\equiv 0$, its boundary function and every asserted norm are zero, so all conclusions hold. Assume henceforth $f\equiv ̸0$. The case $p<\mathrm{\infty }$: $H^p$ membership and the bound $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}\le \mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_p$. Assume $p<\mathrm{\infty }$ and $f^{\ast }\in L^p$. By [L1], $\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }\in L^1$ and $\log \mathrm{\mid }f(z)\mathrm{\mid }\le P[\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }](z)$, so [L2] (applied with $u=\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }$, for which $e^{pu}=\mathrm{\mid }{f}^{\ast }{\mathrm{\mid }}^p\in L^1$) gives $\mathrm{\mid }f(z){\mathrm{\mid }}^p\le P[\mathrm{\mid }{f}^{\ast }{\mathrm{\mid }}^p](z)$ for every $z\in \mathbb{D}$. Integrating over the circle $\mathrm{\mid }z\mathrm{\mid }=r$ and using Tonelli's theorem and the unit mass of the kernel [L3] gives ${\int }_{\mathbb{T}}\mathrm{\mid }f(r\zeta ){\mathrm{\mid }}^{p}dm(\zeta )\le {\int }_{\mathbb{T}}\mathrm{\mid }{f}^{\ast }{\mathrm{\mid }}^{p}dm$ for every $r<1$; hence $f$ is holomorphic with bounded radial means, that is $f\in H^p(\mathbb{D})$, and $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}\le \mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_p$.

The reverse inequality by Fatou. By [L1], $f(r\zeta )\to f^{\ast }(\zeta )$ $m$-almost everywhere; Fatou's lemma [L4] applied to the nonnegative functions $\mathrm{\mid }{f}_r{\mathrm{\mid }}^p$ gives $\mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_p^p=\int \mathrm{\mid }{f}^{\ast }{\mathrm{\mid }}^{p}dm\le {\mathrm{lim\hspace{0.17em}inf}}_r\int \mathrm{\mid }{f}_r{\mathrm{\mid }}^{p}dm\le {\sup }_r\int \mathrm{\mid }{f}_r{\mathrm{\mid }}^{p}dm=\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p$, using the definition of the $H^p$ norm as a supremum [L3]. Together with step 1.1 this gives $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}=\mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_p$.

The case $p=\mathrm{\infty }$. If $f^{\ast }\in L^{\mathrm{\infty }}$, then $\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }\le \log \mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty }}$ a.e., so $P[\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }]\le \log \mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty }}$ and the defining inequality of [L1] gives $\mathrm{\mid }f(z)\mathrm{\mid }\le \mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $z$, that is $f\in H^{\mathrm{\infty }}$ with $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty }}$. Conversely, $f^{\ast }$ is an a.e. limit of the radial functions $f_r$, so $\mathrm{\mid }{f}^{\ast }\mathrm{\mid }\le {\sup }_{z\in \mathbb{D}}\mathrm{\mid }f(z)\mathrm{\mid }=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ a.e. and $\mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$; hence $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty }}$.

Assembly and sharpness for $N$. Steps 1.1, 2.1 and 2.2 prove $N^{+}\cap L^p=H^p$ with equality of norms for every $0<p\le \mathrm{\infty }$: if $f\in N^{+}$ and $f^{\ast }\in L^p$ then $f\in H^p$ with the stated norm identity, and the reverse inclusion $H^p\subseteq N^{+}$ is in the definition of $N^{+}$. The sharpness assertion is [L5]: the reciprocal of a nonconstant singular inner function lies in $N$ with unimodular boundary values, so $N\cap L^{\mathrm{\infty }}\mathrm{\ne }{H}^{\mathrm{\infty }}$, showing that the maximum principle fails for $N$.