Bloch space

Remarks

With this norm, $\mathcal{B}$ is a Banach space. Example: let $f(z) = \log(1 - z)$ with the principal branch, so $f'(z) = -1/(1 - z)$. Since $\abs{1 - z} \ge 1 - \abs{z}$, one has $(1 - \abs{z}^2)\abs{f'(z)} \le (1 - \abs{z})(1 + \abs{z})/(1 - \abs{z}) < 2$, so $f \in \mathcal{B}$, although $f$ is unbounded. Non-example: $g(z) = 1/(1 - z)$ has $g'(z) = 1/(1 - z)^2$, and for real $z = r \in (0, 1)$ one gets $(1 - r^2)/(1 - r)^2 = (1 + r)/(1 - r) \to \infty$, so $g \notin \mathcal{B}$. Every bounded holomorphic function is in $\mathcal{B}$: if $\abs{f} \le 1$, the Schwarz–Pick theorem gives $(1 - \abs{z}^2)\abs{f'(z)} \le 1$. The closure of the polynomials in $\mathcal{B}$ is the little Bloch space $\mathcal{B}_0$, the set of $f$ with $(1 - \abs{z}^2)\abs{f'(z)} \to 0$ as $\abs{z} \to 1$.