Inequality $|f'(0)|\leq 2|f(0)| \log{ \frac{1}{f(0)}}\leq 1 − |f(0)|^2.$ How to prove the following:
If $f ∈ H(D)$ and $0 < |f(z)| < 1$ for all $z ∈ D$, then
$$|f'(0)|\leq 2|f(0)| \log{ \frac{1}{|f(0)|}}\leq 1 − |f(0)|^2.$$
It seems to me that I should use Harnack's inequality on harmonic function log|u|, but I didn't succeed in that. Therefore, any help is welcome. Thanks in advance.
 A: Let's rotate $f$ s.t $f(z)=a_0+a_1z+...$ has $a_0 >0$ as that doesn't change anything in what we have to prove. Let $g = \log f$ where we use the branch s.t $b_0=g(0)=\log a_0$ negative real number ($0<a_0=f(0) <1$).
$-\Re g =-\log |f| >0$ so the coefficients of $g(z)=b_0+b_1z+..$ satisfy the classical Herglotz-Caratheodory inequalities $|b_k| \le 2|b_0|$, hence in particular $|b_1| \le -2\log a_0$
(easy direct proof of those coefficients bounds without Herglotz representation or subordination theory from which the inequalities are trivial is to notice that if $h=\frac{g}{b_0}, B=\frac{h-1}{h+1}$ is a Schwarz function - geometric function theory sense not distribution sense - so $B(0)=0, |B|<1$, hence $|B'(0)| \le 1$ and that translates to $|b_1| \le 2 |b_0|$ and then the convex trick of replacing $g$ with its $n$th symmetric $nk(z)=g(z)+g(\omega z)+...g(\omega^{n-1}z), \omega^n=1$ primitive root of unity of order $n$ preserves $b_0$ and the class of functions with positive (or negative) real part and with $z^n \to z$ transforms $b_n$ into $b_1$ so we can get the required inequality fom the $b_1$ case)
Coming back to our problem, by differentiating we get $g'f=f'$ so $b_1a_0=a_1$ hence $|f'(0)|=|a_1| \le -2a_0\log a_0$ which is the required first inequality.
Second inequality holds in general for any $|a_0| <1$ and follows from the fact that $r(x)=1-x^2+2x\log x$ is decreasing on $(0,1)$ as $r'' >0$ which implies $r' <0$ as $r'(1)=0$ which implies $r$ decreasing and $r(1)=0$
