I'm struggling to understand a proof in Katznelson (An Introduction to Harmonic Analysis, III - Lemma 1.3)
Basically, given $f \in L^1(T)$, take $F(z) = e^{-f - i \tilde{f}}$. Since this is holomorphic, it's harmonic and I follow why it's bounded, $|F| \leq 1$.
My first question is why take $f > 0 $, isn't $f \geq 0$ also fine since you still get $|F| \leq 1$ in that case?
It's also clear that $F$ has radial limit of modulus $e^{-f} > 0 $ almost everywhere.
My confusion mainly comes from how that implies that $\tilde{f}$ has a finite radial limit, namely using this fact and what has been shown about $F$ and $f$
$$e^{-i \tilde{f}} = F(z)e^{f(z)} $$
I feel like this is obvious in some way using continuity of $\tilde{f}$ that I am not seeing - any help would be appreciated.