I'm studying the Bloch space $\mathcal{B}$; this is the set of all analytic fuctions $f$ defined on the open unit disc $\mathbb{D}$ such that $$\sup_{|z|<1}(1-|z|^2)|f'(z)|<\infty$$ My question is more curious: can you give an example of an unbounded Bloch function?

  • 1
    $\begingroup$ $z \mapsto \mathrm{argtanh} z$ is such an example $\endgroup$
    – Didier
    Jan 18 at 0:48

2 Answers 2


$f(z) = \mathrm{argtanh}\,z$ is analytic with $|f(z)| \to +\infty$ as $z\to 1$. Since $f'(z) = \frac{1}{1-z^2}$, it satisfies $(1-|z|^2)|f'(z)| \leqslant 1$.


Take $f(z)=\log (1-z)$ (principal branch so $f(0)=0$); then $f'(z)=-\frac{1}{1-z}$ and since $|-\frac{1}{1-z}| \le \frac{1}{1-|z|}$ one gets that $||f||_B\le 2$ (and actually by taking $z=r \to 1$ one gets $2$

Note that there is a theorem of Pommerenke which completely characterizes Bloch functions in terms of Schlicht functions ($g$ univalent, $g(0)=0, g'(0)=1$), namely $f$ is Bloch if and only if there is $g$ Schlicht and $c \in \mathbb C$ st $f=c\log g'+f(0)$ where we choose the branch for which $\log g'(0)=\log 1=0$; one can construct many unbounded Bloch functions this way (the one above is for $g=\frac{z}{1-z}, c=-1/2$)


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