The Smirnov class is the class of quotients by outer bounded functions

Quotients by outer bounded functions lie in $N^{+}$. Assume $f=g\mathrm{/}h$ with $g,h\in H^{\mathrm{\infty }}$, $h$ outer and $h\equiv ̸0$. By [L2], $f$ is holomorphic and in $N(\mathbb{D})$, and $\log \mathrm{\mid }g\mathrm{\mid }\le P[\log \mathrm{\mid }{g}^{\ast }\mathrm{\mid }]$, while outer-ness of $h$ gives $\log \mathrm{\mid }h\mathrm{\mid }=P[\log \mathrm{\mid }{h}^{\ast }\mathrm{\mid }]$; subtracting the two displays gives, using linearity of the Poisson integral, $\log \mathrm{\mid }f\mathrm{\mid }=\log \mathrm{\mid }g\mathrm{\mid }-\log \mathrm{\mid }h\mathrm{\mid }\le P[\log \mathrm{\mid }{g}^{\ast }\mathrm{\mid }-\log \mathrm{\mid }{h}^{\ast }\mathrm{\mid }]=P[\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }],$ with the identification $f^{\ast }=g^{\ast }\mathrm{/}{h}^{\ast }$ a.e. and $\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }=\log \mathrm{\mid }{g}^{\ast }\mathrm{\mid }-\log \mathrm{\mid }{h}^{\ast }\mathrm{\mid }\in L^1$ by [L2]. Hence $f\in N^{+}(\mathbb{D})$ by [L1].

A Smirnov function is such a quotient. Assume $f\in N^{+}(\mathbb{D})$, so by [L1] $\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }\in L^1$ and $\log \mathrm{\mid }f\mathrm{\mid }\le P[\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }]$. Put $\phi :={\log }^{+}\mathrm{\mid }{f}^{\ast }\mathrm{\mid }+1\ge 1$, an $L^1$ function, and $h:=[e^{-\phi }]$. By [L3], $h$ is outer with $0<h(0)=e^{-\int \phi }\le e^{-1}<1$ and $\mathrm{\mid }h\mathrm{\mid }=e^{-P[\phi ]}\le 1$, so $h\in H^{\mathrm{\infty }}$ and $h$ is outer. Let $g:=fh$, holomorphic on $\mathbb{D}$. Then $\log \mathrm{\mid }g\mathrm{\mid }=\log \mathrm{\mid }f\mathrm{\mid }-P[\phi ]\le P[\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }]-P[\phi ]=P[\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }-{\log }^{+}\mathrm{\mid }{f}^{\ast }\mathrm{\mid }-1]=P[-{\log }^{-}\text{\hspace{0.17em}⁣}\mathrm{\mid }{f}^{\ast }\mathrm{\mid }-1]\le -1,$ because $-{\log }^{-}\mathrm{\mid }{f}^{\ast }\mathrm{\mid }-1\le -1$ pointwise, and the Poisson integral of a function $\le -1$ is $\le -1$. Hence $\mathrm{\mid }g\mathrm{\mid }\le e^{-1}<1$ and $g\in H^{\mathrm{\infty }}(\mathbb{D})$, while $f=g\mathrm{/}h$ with $h$ outer, $h(0)>0$, $\mathrm{\mid }h\mathrm{\mid }\le 1$.

Non-uniqueness and normalization. If $f=g\mathrm{/}h$ with $g,h\in H^{\mathrm{\infty }}$ and $h$ outer, and $\phi \in H^{\mathrm{\infty }}$ is outer, both products $g\phi ,h\phi$ are bounded holomorphic by [L5]. The denominator is outer by [L3] and is zero-free. Therefore $(g\phi )\mathrm{/}(h\phi )=f$ is another representation with an outer denominator. Taking, for example, the constant outer multiplier $2$ gives a distinct pair, so the representation is not unique. Step 2.1 already constructs a denominator with $h(0)>0$ and $\mathrm{\mid }h\mathrm{\mid }\le 1$. A merely zero-free bounded multiplier preserves the quotient but need not preserve an outer denominator.

Assembly. Step 1.1 proves the backward implication, steps 2.1 and 3.1 the forward implication with the stated normalization, and step 3.1 records the non-uniqueness. This proves the equivalence of membership in $N^{+}(\mathbb{D})$ with the quotient representation.