# 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:

- $f_{\Re} \paren z = \Re \paren {f \paren z}$
- $f_{\Im} \paren z = \Im \paren {f \paren z}$

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

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

- $\Gamma = \left\{{p \paren u : u \in \closedint {a} {b}}\right\}$

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

Also, Real and Imaginary Part Projections are Continuous.

Therefore, $f_{\Re}$ and $f_{\Im}$ are continuous by 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, $f_{\Re} \paren {p \paren u}$ and $f_{\Im} \paren {p \paren u}$ are continuous real functions by Composite of Continuous Mappings is Continuous.

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

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

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

$\blacksquare$