Bound for Analytic Function and Derivatives/Analytic Function Bounded on Circle

Lemma

Let $f$ be a complex function.

Let $z_0$ be a point in $\C$.

Let $\Gamma$ be a circle in $\C$ with center at $z_0$ and radius in $\R_{>0}$.

Let $f$ be analytic on $\Gamma$.

Then $f$ is bounded on $\Gamma$.

Proof

Let:

$\map {f_{\Re} } z = \map \Re {\map f z}$
$\map {f_{\Im} } z = \map \Im {\map f z}$

Let $\closedint a b$, $a < b$, be a closed real interval.

Let $p$ be a continuous complex-valued function defined such that:

$\Gamma = \set {\map p u: u \in \closedint a b}$

$f$ is continuous on $\Gamma$ as $f$ is analytic on $\Gamma$ by the definition of analytic.

Therefore, $f_{\Re}$ and $f_{\Im}$ are continuous by the corollary to Composite of Continuous Mappings is Continuous.

Observe that $f_{\Re}$ and $f_{\Im}$ are real-valued functions that are continuous.

Also, $p$ is a continuous function defined on a set of real numbers.

Therefore $\map {f_{\Re} } {\map p u}$ and $\map {f_{\Im} } {\map p u}$ are continuous real functions by the corollary to Composite of Continuous Mappings is Continuous.

$\map {f_{\Re} } {\map p u}$ and $\map {f_{\Im} } {\map p u}$ are bounded on $\closedint a b$ by Continuous Real Function is Bounded.

Therefore $\map f {\map p u}$ is bounded on $\closedint a b$ as $\map f {\map p u} = \map {f_{\Re} } {\map p u} + i \map {f_{\Im} } {\map p u}$ where $i = \sqrt {-1}$.

Accordingly, $f$ is bounded on $\Gamma$ as $\Gamma = \set {\map p u: u \in \closedint a b}$.

$\blacksquare$