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



> 
> $$f(z)=-\dfrac{\log(1-z)}{z}$$
> 
> 
> 


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 $\mathbb{D}$.


Thank.


1. Duren Peter L., Univalent functions (Grundlehren der mathematischen Wissenschaften 259), Springer-Verlag Berlin and Heidelberg GmbH & Co. K (1983), [p.42](https://books.google.com/books?id=8mVIaZmsy0gC&pg=PA42).



Prove $f(z)=-\frac{\log(1-z)}{z}$ is convex in $\mathbb{D}$
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


