Convergence of Taylor Series of Function Analytic on Disk

Theorem
Let $F$ be a complex function.

Let $x_0$ be a point in $\R$.

Let $R$ be an extended real number greater than zero.

Let $F$ be analytic at every point $z \in \C$ satisfying $\left\lvert{z - x_0}\right\rvert < R$.

Let $f = F {\restriction_{\R}}$ be a real function.

Then:
 * the Taylor series of $f$ about $x_0$converges to $f$ at every point $x \in \R$ satisfying $\left\lvert{x - x_0}\right\rvert < R$

Corollary: Taylor Series reaches closest Singularity
Let the singularities of a function be the points at which the function is not analytic.

Let $F$ be analytic everywhere except at a finite number of singularities.

Let $R \in \R_{>0}$ be the distance from $x_0$ to the closest singularity of $F$.

Then:
 * the Taylor series of $f$ about $x_0$ converges to $f$ at every point $x \in \R$ satisfying $\left\lvert{x - x_0}\right\rvert < R$

Corollary: Taylor Series of Analytic Function has infinite Radius of Convergence
Let $F$ be analytic everywhere.

Then:
 * the Taylor series of $f$ about $x_0$ converges to $f$ at every point in $\R$

Proof
Let $r$ be a real number satisfying:
 * $0 < r < R$

Let $x$ be a real number satisfying:
 * $\left\lvert{x - x_0}\right\rvert < r$

$f$ has a Taylor series expansion about $x_0$ with radius of convergence greater than zero as $f$ is analytic at $x_0$.

The Taylor's formula with remainder for $f$ about $x_0$ is:
 * $f \left({x}\right) = \displaystyle \sum_{i \mathop = 0}^n \frac {\left({x - x_0}\right)^i} {i!} f^{\left({i}\right)} \left({x_0}\right) + R_n \left({x}\right)$

where
 * $R_n \left({x}\right) = \dfrac 1 {n!} \displaystyle \int_{x_0}^x \left({x - t}\right)^n f^{\left({n \mathop + 1}\right)} \left({t}\right) \mathrm d t$

Our first aim is to prove:
 * $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$

For the case $x = x_0$, the interval of integration in the expression for $R_n \left({x}\right)$ has zero length.

Therefore, $R_n \left({x}\right) = 0$.

Accordingly, $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ is true for this case.

Now we consider the case $x \ne x_0$.

We have:
 * $0 < r - \left\lvert{x - x_0}\right\rvert$ as $\left\lvert{x - x_0}\right\rvert < r$

Observe that:
 * $\left\lvert{x - x_0}\right\rvert \ge \left\lvert{t - x_0}\right\rvert$

Therefore:
 * $r - \left\lvert{x - x_0}\right\rvert \le r - \left\lvert{t - x_0}\right\rvert$
 * $0 < r - \left\lvert{x - x_0}\right\rvert \le r - \left\lvert{t - x_0}\right\rvert$
 * $0 < \left\lvert{r - \left\lvert{x - x_0}\right\rvert}\right\rvert\le \left\lvert{r - \left\lvert{t - x_0}\right\rvert}\right\rvert$

We have:

Let $y \in \R$ be equal to $x_0 + r$ if $x > x_0$ and $x_0 - r$ if $x < x_0$.

Note that $y > x$ if $x > x_0$ and $y < x$ if $x < x_0$.

The general situation is:
 * $x_0 \le t \le x < y$ if $x > x_0$
 * $y < x \le t \le x_0$ if $x < x_0$

Let us study $\left\lvert{x - t}\right\rvert$ in the expression above for the bound for $\left\lvert{R_n \left({x}\right)}\right\rvert$:

Also, we have:

We combine these two results to get:

We use this result in the expression for the bound for $\left\lvert{R_n \left({x}\right)}\right\rvert$:

We have:
 * $\displaystyle \frac r {\left\lvert{x - x_0}\right\rvert} > 1$ as $\left\lvert{x - x_0}\right\rvert < r$ and $x \ne x_0$

Therefore:
 * $\displaystyle \lim_{n \mathop \to \infty} \frac n {\left({\frac r {\left\lvert{x - x_0}\right\rvert} }\right)^n} = 0$ by Lemma 3

Letting $n$ approach $\infty$ in the expression for the bound for $\left\lvert{R_n \left({x}\right)}\right\rvert$, we get:

So:

Accordingly, $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ is true for the case $x \ne x_0$.

Thus, $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ holds for every $x$ satisfying $\left\lvert{x - x_0}\right\rvert < r$ where $r < R$.

Since we can choose $r$ as close to $R$ as we like, we conclude that $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ holds for every $x$ that satisfies $\left\lvert{x - x_0}\right\rvert < R$.

Therefore, the Taylor series expansion of $f \left({x}\right)$ about $x_0$ converges to $f \left({x}\right)$ for every $x$ that satisfies $\left\lvert{x - x_0}\right\rvert < R$.

Lemma 1 (Bound for Analytic Function and Derivatives)
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 positive real number $M$ exists such that:
 * $\displaystyle \left\lvert{f^{\left({n}\right)} \left({t}\right)}\right\rvert \le \frac {M r \, n!} {\left({r - \left\lvert{t - z_0}\right\rvert}\right)^\left({n + 1}\right)}$

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

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

where $\Gamma$ is traversed counterclockwise.

We have that $f$ is bounded on $\Gamma$ by Lemma 2.

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 2 (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 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 $\Gamma = \left\{{p \left({u}\right): u \in \left[{a \,.\,.\, b}\right]}\right\}$.

Lemma 3
Let $y > 1$.

Then:
 * $\displaystyle \lim_{n \mathop \to \infty} \frac n {y^n} = 0$