# Unbounded Bloch function

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?

• $z \mapsto \mathrm{argtanh} z$ is such an example Jan 18 at 0:48

$$f(z) = \mathrm{argtanh}\,z$$ 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 $$g=\frac{z}{1-z}, c=-1/2$$)