Let $f: \overline{\mathbb{D}} \to \mathbb{R}^+$ be a real (positive) valued function on the closed unit disc that is bounded and analytic on $\mathbb{D}$ (open unit disc) and $$\lim_{|z| \to 1}f(z) = 1.$$ Can we conclude that $f(z) \leq 1$ in the whole disc?
I'm aware of a Blaschke product type argument that says that a function that is of unit modulus on the boundary, and bounded and analytic in the interior of the disc must have finitely many zeroes and hence must be a finite Blaschke product - however, can we ensure that a Blaschke product is always real valued? And if so, what does this say about the bound on $f$ in the interior of the disc?
EDIT: An alternative formulation to the above, would be to consider the function $|f|$ where $f$ is an analytic, bounded function on the interior of the disc and continuous and bounded on the boundary with $$\lim_{|z| \to 1} |f(z)| =1.$$ Similar to the above, the question would now concern the bound on $|f|$.
The answer to the edited question is YES.
Let $\epsilon >0$. Then there exists $\delta >0$ such that $|f(z)| \leq 1+\epsilon$ for $|z| \geq 1-\delta$. By MMP applied to the disk of radius $1-\delta$ we get $|f(z)| \leq 1+\epsilon$ whenever $|z| \leq 1-\delta$. Can you finish the proof now?