The notion of analyticity I am working with is that a complex function $f:\Omega\subset\mathbb{C}\to\mathbb{C}$ is analytic in $\Omega$ if for any $z_0 \in \Omega$ $f(z) = \sum_{n=0}^\infty a_n(z – z_0)^n, a_n \in \mathbb{C}$ for $z$ in some neighbourhood of $z_0$ where the said powerseries is convergent.
My question is is that how do we know that an analytic function is bounded in this neighbourhood of $z_0$? Let $B(z_0, r)$ be the aforementioned neighbourhood of $z_0$. Then, we know immediately from the definition of analyticity and convergence of infinite series that $\forall z \in B(z_0, r):\exists M_z \in \mathbb{R}_{\geq 0}:|f(z)|\leq M_z < \infty$, as the series representation of $f$ would not converge otherwise. But if were to take supremum of $|f(z)|$ over all $z$ in $B(z_0, r)$, what would stop the local bound of $f$ from growing without a bound, i.e. that $\sup_{z\in B(z_0, r)}|f(z)| = \infty$?
Best Answer
We do not know that.
For example, $\frac{1}{1-z} = \sum_0^\infty z^n$ is not bounded on the disk of convergence $|z|<1$.
You can say that if you shrink the domain slightly to a compact subset of the disk of convergence (in my example, say, $|z|\leq 1 - \epsilon$) then $f$ will be bounded in modulus on this restricted domain simply because it is a continuous function on a compact set.