Let $f$ be a function of bounded type (Nevanlinna class) in the unit disk $\mathbb{D}$. Suppose:
- $f$ is analytic and zero-free in $\mathbb{D}$;
- Both $f$ and $1/f$ are of bounded type in $\mathbb{D}$;
- $f$ has non-tangential boundary values with $|f(e^{i\theta})| = 1$ for almost every $\theta$.
Question: Under conditions (1)–(3), is it true that $f$ has no inner factor in its canonical factorization (so $f$ is outer, and hence $f \equiv e^{i\alpha}$ for some real $\alpha$)?
I would particularly appreciate a reference in Garnett's Bounded Analytic Functions (Chapter II) or Koosis's The Logarithmic Integral (Chapter VI) that states this exact combination of hypotheses.
Why this formulation? In the disk, the canonical factorization is $f = B \cdot S \cdot O$ (Blaschke product, singular inner function, outer function).
- Condition (1) removes the Blaschke factor ($B \equiv 1$).
- If both $f$ and $1/f$ are of bounded type, then $f$ is in the Smirnov class $N^+$. Combined with $|f| = 1$ a.e. on the boundary, this should force the outer part to have boundary modulus $1$ a.e., hence be constant.
I'm looking for a clean theorem/lemma that bundles these implications.
Optional context: A vertical strip $\Sigma_\delta = {|\Re s - \tfrac12| \le \delta}$ maps conformally to $\mathbb{D}$ via $$ \zeta = \phi(s) = \frac{ e^{\frac{\pi}{\delta}\left(s-\tfrac12\right)} - 1 } { e^{\frac{\pi}{\delta}\left(s-\tfrac12\right)} + 1 }. $$
so a function $F$ of bounded type in $\Sigma_\delta$ with $|F(\tfrac12+it)| = 1$ a.e. yields $f(\zeta) := F(\phi^{-1}(\zeta))$ satisfying the above conditions. I only need the disk statement.
Pointers to a precise statement (or a counterexample if one hypothesis is unnecessary) would be very helpful.
