# Real-valued bounded analytic functions on the unit disc

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 $$|f|$$.

• 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 $f=1$ – Ninad Munshi Sep 13 '19 at 11:44
• 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! – CS1994 Sep 13 '19 at 11:50
• @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, $\frac{\partial f}{\partial \bar{z}} \neq 0$ – Ninad Munshi Sep 13 '19 at 11:58
• 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 – Olivier Roche Sep 13 '19 at 12:03

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?
• MMP stands for Maximum Modulus Principle. According to this theorem if $|f(z)| \leq M$ for $|z|=R$ then $|f(z)| \leq M$ for $|z|\leq R$. – Kavi Rama Murthy Sep 13 '19 at 12:14
• Ah okay, so this immediately just gives that since $|f(z)| \leq1$ on $|z|=1$, then $|f(z)| \leq 1$ also for $|z|\leq 1$? – CS1994 Sep 13 '19 at 12:16
• 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 – Kavi Rama Murthy Sep 13 '19 at 12:20
• 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 – CS1994 Sep 13 '19 at 12:27