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Statement of the problem:

For the conformal mapping $f:\mathbb{D}\to\mathbb{D}$ , we suppose that the domain $f(\mathbb{D})$ is convex. Prove that for $\mathbb{D}_r=\{z\in \mathbb{C}:|z|<r\}$ the domain $f(\mathbb{D}_r)$ is convex

My approach: I considered the function $$g(z)=f^{-1}(tf(z)+(1-t)f(0)), t\in[0,1]$$ This function is holomorphic , maps the unit disc to itself, and $g(0)=0$. So we can apply Schwarz's lemma. After using it, we have that $$|f^{-1}(tf(z)+(1-t)f(0))|\leq|z|$$ which means that $$g(\mathbb{D}_r)\subset \mathbb{D}_r,\Longrightarrow f^{-1}(tf(z)+(1-t)f(0))\in \mathbb{D}_r \Longrightarrow tf(z)+(1-t)f(0) \in f(\mathbb{D}_r)$$ so all of these line segments belong to our domain, but in order to show that the domain is convex I need to prove that this happens for every $f(z),f(w)$ and not only for $f(z),f(0).$ At this stage I considered the function $$\phi^{-1}\circ g\circ \phi:\mathbb{D}\to \mathbb{D}, \phi(z)=\frac{z-w}{1-\bar{w}z},z,w \in \mathbb{D}$$ and i applied Schwarz's lemma again and I tried to prove it for all $z,w \in \mathbb{D}$ but got stuck.

I would be grateful if you give me just the smallest possible hint and not a whole solution.

Thanks in advance

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1 Answer 1

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This is Study's theorem. You can find a proof, e.g., in Duren's book Univalent functions. If we may assume that $f$ is defined and $C^1$ on the closed disc $\bar D$ the proof is quite easy: Look at the images $\gamma_r$ of concentric circles $\partial D_r$ $\,(0<r\leq1)$, given by $$\gamma_r:\quad t\mapsto w(t):=f\bigl(r e^{it}\bigr)\qquad(0\leq t\leq2\pi)\ .\tag{1}$$ Such a curve $\gamma_r$ is convex iff its tangent argument $$t\mapsto\theta(t):={\rm arg}\bigl(w'(t)\bigr)={\rm Im}\bigl(\log w'(t)\bigr)$$ is monotonically increasing. Therefore the convexity condition amounts to $$\theta'(t)={\rm Im}\left({w''(t)\over w'(t)}\right)\geq0\ .\tag{2}$$ If $f(D)$, hence $\gamma_1$, is convex then $(2)$ translates via $(1)$ into a condition involving $f'$, $f''$ on $\partial D$. The maximum principle then guarantees that this condition holds througout $D$, and this in turn implies that all $\gamma_r$ are convex.

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  • $\begingroup$ I searched and found it was Study's theorem. I fully understand your answer ! Thanks for that. I found a later paper though were it states that Radό proved the theorem using Schwarz's lemma. Do you have any idea on this ? $\endgroup$ Commented Nov 12, 2018 at 10:58
  • $\begingroup$ Found it. I was close , yet so far... Thanks for your insight $\endgroup$ Commented Nov 12, 2018 at 11:45

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