For analytic function $f:\mathbb{C}\to\mathbb{C}$ with


I want to prove that it's a convex map in the unit disk $\mathbb{D}$, i.e. that $f$ maps the unit disk conformally onto a convex domain. I know $${\bf Re}\left(1+z\dfrac{f''(z)}{f'(z)}\right)>0$$ for $z\in\mathbb{D}$, if and only if function be convex in $\mathbb{D}$ [1], but I couldn't use it in my problem. Is the any way to prove that $f$ is convex in $\mathbb{D}$.


  1. Duren Peter L., Univalent functions (Grundlehren der mathematischen Wissenschaften 259), Springer-Verlag Berlin and Heidelberg GmbH & Co. K (1983), p.42.
  • 3
    $\begingroup$ What is a convex function from $\mathbb{D}$ to $\mathbb{C}$? $\endgroup$
    – abx
    Jan 10 '18 at 8:25
  • $\begingroup$ @abx Thanks for your attention. I added the reference! $\endgroup$
    – Nosrati
    Jan 10 '18 at 9:37
  • 3
    $\begingroup$ I took the freedom and added the definition of convexity from Duren's book into the body of the question. $\endgroup$
    – Dirk
    Jan 10 '18 at 9:50
  • $\begingroup$ I don't know a high level argument, but calculating the image curve of the boundary of $\mathbb{D}$ should do the trick… $\endgroup$
    – Dirk
    Jan 10 '18 at 10:24
  • 2
    $\begingroup$ Numerical plotting makes it clear that this is true, with the minimum value of $g(z)=\textbf{Re}(1+zf''(z)/f'(z))$ being $g(-1)\approx 0.294$. Also $f(\mathbb{D})$ fills most of the strip where $x\geq\ln(2)$ and $|y|\leq\pi/2$, but with the left hand end made blunt. You probably knew that anyway, but you should really include it in the question. $\endgroup$ Jan 10 '18 at 10:28

The normalization of f, (f-1)/2 is the image of the fundamental convex function z/(1-z) under the operator H(g) = (2/z)* primitive g. It is a theorem of Libera (Goodman, Univalent Functions 1983, vol 2, page 156) that while H doesn't preserve the S class (schlicht), it does preserve the classes CV (convex S), ST (starlike S), CC (close to convex S), so the answer to your question follows. The proof is based on a lemma about p-valent starlike functions (here we need the 2-valent primitive of g, for g in one the S classes above) and a result giving a (necessary) and sufficient condition for a function to belong to P (functions on the unit disk with ReP > 0 and P(0)=1, so it is not obvious

  • $\begingroup$ oops, the normalization is 2*(f-1) as f is 1+z/2+... $\endgroup$
    – Conrad
    Jan 15 '18 at 23:45

Here is a proof which makes use of Maple, namely, its DirectSearch package created by Sergey Moiseev. The GlobalOptima command finds the global minimum of $${\Re}\left(1+z\dfrac{f''(z)}{f'(z)}\right)$$ in the unit disk with the absolute error less than or equal to $10^{-12}$:

f := -log(1-z)/z:
A := evalc(Re(eval(1+z*(diff(f, z, z))/(diff(f, z)), z = x+I*y))):
DirectSearch:-GlobalOptima(A, {x^2+y^2 <= 1}, tolerances = 10^(-12));

[ 0.294349724781047,[x=- 0.999999999999916,y= 0.0000004101682696],498]

B := evalc(Re(eval(1+z*(diff(f, z, z))/(diff(f, z)), z = x+I*sqrt(-x^2+1)))):
DirectSearch:-GlobalOptima(B, {x >= -1, x <= 1}, tolerances = 10^(-12));

[ 0.294349724781044,[x=- 0.999999999999999],126]

These outputs mean that the global minimum equals $0.2943497247810$. I am pretty sure that the results imply the affirmative answer to the question. However, colleagues are welcome to discuss it.

Addition. Just for illustration, here is a plot of A on the unit disk.

plot3d(A, y = -sqrt(-x^2+1) .. sqrt(-x^2+1), x = -1 .. 1, axes = frame);

enter image description here

In fact, the function is unbounded from above around $z=1$.

  • $\begingroup$ @MyGlasses: Why do you think so? What place in the above answer is doubtful to you? $\endgroup$
    – user64494
    Jan 11 '18 at 14:01
  • 1
    $\begingroup$ @MyGlasses: Let us consider your function $A(z):=\Re(\dots)$ on $S:=\{z:|z| \le 1 \} \setminus \{z:|z-1| < \frac 1 4 \}$. The one is harmonic on $S$. Therefore, its minimum is attained on the boundary $\partial S$. Because $A(z)=A(\overline{z})$, the point $z=-1$ is an extremum of $A(z)$ on the boundary $\partial S$. This extremum is the global minimum. The value of $A(-1)$ is calculated numerically. Hope these analytic arguments would be satisfactory to you. $\endgroup$
    – user64494
    Jan 11 '18 at 14:52


The original article of Libera1 is in Proceedings AMS, 16, 1965, p 755-758. freely available at the link above

The proof there is pretty much the one I read in the Goodman book

1R. J. Libera. Some classes of regular univalent functions, Proc. Amer. Math. Soc. 16 (1965), 755-758; doi: 10.1090/S0002-9939-1965-0178131-2


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