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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 $|f|$.

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    $\begingroup$ Edit: Reading the question again, the only analytic function with a purely real image would be a constant so yes, that inequality is trivially true because that means $f=1$ $\endgroup$ Sep 13, 2019 at 11:44
  • $\begingroup$ Thank you @NinadMunshi! I have actually added an edit to the question in response to your previous comment which adds a bit more of a substantial query. Though, your answer may well be enough for my problem! $\endgroup$
    – CS1994
    Sep 13, 2019 at 11:50
  • $\begingroup$ @OlivierRoche No that is not analytic. One definition of analytic is that $f\in \text{ker}\left(\frac{\partial}{\partial \bar{z}}\right)$ but for your function, $\frac{\partial f}{\partial \bar{z}} \neq 0$ $\endgroup$ Sep 13, 2019 at 11:58
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    $\begingroup$ I hoped the function $z \mapsto 2 - |z|$ would give a counterexample but this fails since it isn't continuously differentiable in $0$. I think the function $f : z \mapsto 2 - \cos (|z|\cdot \frac{\pi}{2})$ might do the trick, but I'm too lazy to check if $f$ is analytic. :D $\endgroup$ Sep 13, 2019 at 12:03

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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?

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  • $\begingroup$ I'm sorry, what is MMP? (My complex analysis is abysmal) $\endgroup$
    – CS1994
    Sep 13, 2019 at 12:13
  • $\begingroup$ MMP stands for Maximum Modulus Principle. According to this theorem if $|f(z)| \leq M$ for $|z|=R$ then $|f(z)| \leq M$ for $|z|\leq R$. $\endgroup$ Sep 13, 2019 at 12:14
  • $\begingroup$ Ah okay, so this immediately just gives that since $|f(z)| \leq1$ on $|z|=1$, then $|f(z)| \leq 1$ also for $|z|\leq 1$? $\endgroup$
    – CS1994
    Sep 13, 2019 at 12:16
  • $\begingroup$ Unfortunately you have to use a more complicated argument. This is because $f$ is only given to be analytic in the open set $(|z|<1)$ and the circle $|z|=1$ is not inside this open set , so MMP cannot be applied directly. @CS1994 $\endgroup$ Sep 13, 2019 at 12:20
  • $\begingroup$ Could the Blaschke product argument not work in this scenario? Here we have an analytic function $f$ in the open disc that has the limit as stated in the question. This implies that $f$ must have finite number of zeroes and be of the form of a finite Blaschke product (see Fatou for the well-known result). A property of a finite Blaschke product is that its modulus is at most 1 in the interior of the disc. The proof of Fatou's result looks similar to the logic you are using here. @kaviramamurthy $\endgroup$
    – CS1994
    Sep 13, 2019 at 12:27

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