# Conformal mapping maps the unit disc in a convex domain.

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 $$\mathbb{D}_r=\{z\in \mathbb{C}:|z| the domain $$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.

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 $$\partial D_r$$ $$\,(0, given by $$\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.