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Consider a pair of holomorphic functions $f,g \in \mathcal{O}(\Delta)$ on the complex unit disk $\Delta = \{|z| < 1\}$ that both satisfy $f(0) = g(0) = 0$ and $f'(0) = g'(0) = 1$. Does the domain $$ f(\Delta) \cdot g(\Delta) := \{f(z)g(w) \mid z, w \in \Delta \} \subset \mathbb{C} $$ contain the unit disk $\Delta$, with equality if and only if $f(\Delta) = g(\Delta) = \Delta$?

In this setting, it may be useful to recall Bloch's theorem, which states that $f(\Delta)$ and $g(\Delta)$ contain disks of a fixed (absolute) radius; however those disks need not of course be centered at the origin.

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    $\begingroup$ What is $\mathcal{O}(\Delta)$? $\endgroup$ – fedja Mar 22 '18 at 21:42
  • $\begingroup$ @fedja Just the ring of holomorphic functions on the complex unit disk $\Delta$. So this notation may as well be dropped out of the question. $\endgroup$ – Vesselin Dimitrov Mar 22 '18 at 21:47
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    $\begingroup$ Yes, I misread the question. I deleted the ans. $\endgroup$ – Alexandre Eremenko Mar 22 '18 at 23:05
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    $\begingroup$ I believe I can prove this (the reduction and the area argument that I indicated in the comment to Conrad's post, only one has to choose the area element in a more natural and intelligent way: $\frac{dm_2(z)}{|z|^2}$ slightly regularized near $0$ and $\infty$ to make it finite; also when integrating, switch to polar coordinates, use Cauchy-Schwarz, and integrate by parts). Too late to type the full solution though.and the next few days are as hectic as any so I hope somebody will figure it out with these hints in case I don't come back soon.. $\endgroup$ – fedja Mar 23 '18 at 2:51
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Let's start with the simple reduction. Notice that $f(\Delta)$ and $g(\Delta)$ are connected open sets containing small disks near the origin, so if one of them is unbounded, $f(\Delta)g(\Delta)=\mathbb C$.

Let $a\in\Delta\setminus\{0\}$ (we certainly have $0=f(0)g(0)$, so the origin is never problematic) be not in $f(\Delta)g(\Delta)$. Consider $f(\Delta)\ni 0$ and $(a/g)(\Delta)\ni\infty$. Those are connected but not necessarily simply connected disjoint open sets. Let $\Omega$ be the union of $f(\Delta)$ and all bounded connected components of $\mathbb C\setminus f(\Delta)$ (or, which is the same, the complement of the connected component of $\mathbb C\setminus f(\Delta)$ containing $\infty$). If $(a/g)(\Delta)$ intersects $\Omega$, it must intersect $f(\Delta)$ as well (you cannot reach a bounded connected component of a complement of an open set by a path (actually even an open sausage, if you want) from $\infty$ without crossing the set itself. Now replace $f$ with the conformal mapping $\varphi$ from $\Delta$ to $\Omega$ with $\varphi(0)=f(0)$, $\Phi=\varphi'(0)>0$. Then, by the Schwarz lemma, $\Phi\ge f'(0)=1$, so if we consider $\widetilde f=\Phi^{-1}\varphi$, we will have $\Phi^{-1}a\notin \widetilde f(\Delta)g(\Delta)$ and $\widetilde f$ is now univalent. Similarly we can make $g$ univalent.

Now comes the main

Lemma: Let $f$ be a (bounded and, if you want, analytic up to the boundary) univalent function such that $f(0)=0, f'(0)=1$. Let $A$ be the area on $\mathbb C\setminus\{0\}$ given by $dA(z)=|z|^{-2}dm_2(z)$, which is invariant under $z\mapsto az (a\ne 0)$ and $z\mapsto z^{-1}$. Then $$ A(f(\Delta\setminus r\Delta))\ge 2\pi\log\frac{1}{r}+o(1)\text{ as }r\to 0^+\,. $$ Proof: Let $S=\Delta\setminus r\Delta$. We have (since $f$ is univalent) $$ A(f(S))\times 2\pi\log\frac 1r=\left[\int_S\frac{|f'|^2}{|f|^2}\,dm_2\right]\left[\int_S\frac{1}{|z|^2}\,dm_2\right] \\ \ge \left[\int_S\frac{|f'|}{|f|}\,\frac{dm_2(z)}{|z|}\right]^2=I^2\,. $$ Note that $$ I=\int_{[0,2\pi]}d\theta\int_r^1d\rho \frac{|f'(\rho e^{i\theta})|}{|f(\rho e^{i\theta})|}\ge \int_{[0,2\pi]}d\theta\int_r^1 d\rho \frac d{d\rho}(\log|f(\rho e^{i\theta})|) \\ = \int_{[0,2\pi]}d\theta\log|f(e^{i\theta})|-\int_{[0,2\pi]}d\theta\log|f(re^{i\theta})|=I_1-I_2\,. $$ However, under our assumptions we have $I_1=0$ while $I_2=2\pi\log r+O(r)$ as $r\to 0^+$ whence the lemma.

Now life gets easy. Notice that for sufficiently small $r>0$ both $f(S)$ and $(a/g)(S)$ lie in the annulus $\{w\in\mathbb C\,:\,(1-o(1))r\le|w|\le (1+o(1))|a|r^{-1}$. (we use both the boundedness and the univalence properties here). But the invariant area of this annulus is only $4\pi\log\frac 1r+\log|a|+o(1)$, so the images must overlap somewhere, thus finishing the story.

Cute question, by the way :-)

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    $\begingroup$ @VesselinDimitrov You are most cordially welcome, Vess! $\endgroup$ – fedja Mar 24 '18 at 23:30
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Note that if the image of f contains a circle (not disk) of radius r and g attains a w with absolute value 1/r (or has a radial limit of absolute value 1/r), by continuity it must attain values with absolute values between 0 and 1/r and then we can fill the unit disk with products of values of f and g.

Now by circular symmetrization it can be proved that the supremum of such r is at least 1/4 in your case (see Hayman, Multivalent Functions, second edition, Thm 4.13 p. 131), so any counterexample must be with functions less than 4 in absolute value (and that can be tightened as there are inequalities relating the supremum of f and the radius r above - again from Hayman's book, chapter 4) so I think your assertion has a strong chance of being true.

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    $\begingroup$ This approach is a bit too crude to deal with the case of small perturbations of the disk. However, it would suffice to show that the integral $\int_{\Delta}\frac{|f'|^2}{(1+|f|^2)^2}dm_2$ is minimized by $f(z)=z$. I suspect that this extremal problem has been investigated by someone somewhere though I have no idea how to handle it. $\endgroup$ – fedja Mar 23 '18 at 1:12
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    $\begingroup$ Also, it will suffice to consider the case of univalent functions but, as the title shows, Vesselin knows this trivial reduction himself. $\endgroup$ – fedja Mar 23 '18 at 1:15

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