Bound for Analytic Function and Derivatives

Lemma
Let $f$ be a complex function.

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

Let $r$ be a real number in $\R_{>0}$.

Let $\Gamma$ be a circle in $\C$ with center at $z_0$ and radius $r$.

Let $f$ be analytic on $\Gamma$ and its interior.

Let $t \in \C$ be such that $\left\lvert{t - z_0}\right\rvert < r$.

Then a real number $M$ exists such that, for every $n \in \N$:
 * $\displaystyle \left\lvert{f^{\left({n}\right)} \left({t}\right)}\right\rvert \le \frac {M r \, n!} {\left({r - \left\lvert{t - z_0}\right\rvert}\right)^\left({n + 1}\right)}$

Proof
We have:
 * $f$ is analytic on $\Gamma$ and its interior
 * $t$ is in the interior of $\Gamma$

Therefore:
 * $\displaystyle f^{\left({n}\right)} \left({t}\right) = \frac {n!} {2 \pi i} \int_\Gamma \frac {f \left({z}\right)} {\left({z − t}\right)^{\left({n + 1}\right)}} \rd z$ by Cauchy's Integral Formula for Derivatives

where $\Gamma$ is traversed counterclockwise.

We have that $f$ is bounded on $\Gamma$ by the lemma.

Therefore, there is a positive real number $M$ that satisfies:
 * $M \ge \left\lvert{f \left({z}\right)}\right\rvert$ for every $z$ on $\Gamma$

We have $\left\lvert{t - z_0}\right\rvert < r$.

Therefore:
 * $0 < r - \left\lvert{t - z_0}\right\rvert$

We observe that $r - \left\lvert{t - z_0}\right\rvert$ is the minimum distance between $t$ and $\Gamma$.

Therefore:
 * $\left({r - \left\lvert{t - z_0}\right\rvert}\right) \le \left\lvert{z − t}\right\rvert$ for every $z$ on $\Gamma$

We get:

Lemma (Analytic Function Bounded on Circle)
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 greater than zero.

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

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

Proof
Let:
 * $f_{\operatorname{Re}} \left({z}\right) = \operatorname{Re} \left({f \left({z}\right)}\right)$
 * $f_{\operatorname{Im}} \left({z}\right) = \operatorname{Im} \left({f \left({z}\right)}\right)$

Let $\left[{a \,.\,.\, b}\right]$, $a < b$, be a real interval.

Let $p$ be a continuous complex-valued function defined such that:
 * $\Gamma = \left\{{p \left({u}\right): u \in \left[{a \,.\,.\, b}\right]}\right\}$

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

Also, Real and Imaginary Part Projections are Continuous.

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

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

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

Therefore, $f_{\operatorname{Re}} \left({p \left({u}\right)}\right)$ and $f_{\operatorname{Im}} \left({p \left({u}\right)}\right)$ are continuous real functions by Composite of Continuous Mappings is Continuous.

$f_{\operatorname{Re}} \left({p \left({u}\right)}\right)$ and $f_{\operatorname{Im}} \left({p \left({u}\right)}\right)$ are bounded on $\left[{a \,.\,.\, b}\right]$ by Continuous Real Function is Bounded.

Therefore, $f \left({p \left({u}\right)}\right)$ is [Definition:Bounded Complex-Valued Function|bounded]] on $\left[{a \,.\,.\, b}\right]$ as $f \left({p \left({u}\right)}\right) = f_{\operatorname{Re}} \left({p \left({u}\right)}\right) + i f_{\operatorname{Im}} \left({p \left({u}\right)}\right)$ where $i = \sqrt{-1}$.

Accordingly, $f$ is bounded on $\Gamma$ as $\Gamma = \left\{{p \left({u}\right): u \in \left[{a \,.\,.\, b}\right]}\right\}$.