# Uniformly bounded derivatives implies globally analytic

## Statement

### Global statement

Suppose $f$ is an infinitely differentiable function on $\R$ such that, for any fixed $a, there is a constant $C$ (possibly dependent on $a,b$) such that for all nonnegative integers $n$, we have:

$|f^{(n)}(t)| \le C \ \forall t \in [a,b]$

Then, $f$ is a globally analytic function: the Taylor series of $f$ about any point in $\R$ converges to $f$. In particular, the Taylor series of $f$ about 0 converges to $f$.

## Facts used

1. Max-estimate version of Lagrange formula

## Examples

The functions $\exp, \sin, \cos$ all fit this description.

If $f = \exp$, we know that each of the derivatives equals $\exp$, so $f^{(n)}(t) = f(t)$ for all $t \in [a,b]$. Since $\exp$ is continuous, it is bounded on the closed interval $[a,b]$, and the upper bound for $\exp$ thus serves as a uniform bound for all its derivatives. (In fact, since $f$ is increasing, we can explicitly take $C = \exp(b)$).

For $f = \sin$ or $f = \cos$, we know that all the derivatives are $\pm \sin$ or $\pm \cos$, so their magnitude is at most 1. Thus, we can take $C = 1$.

## Proof

Given: $f$ is an infinitely differentiable function on $\R$ such that, for any fixed $a,b \in \R$, there is a constant $C$ (possibly dependent on $a,b$) such that for all nonnegative integers $n$, we have:

$|f^{(n)}(t)| \le C \ \forall t \in [a,b]$

A point $x_0 \in \R$ and a point $x \in \R$.

To prove: The Taylor series of $f$ at $x_0$, evaluated at $x$, converges to $f(x)$.

Proof: Note that if $x_0 = x$, there is nothing to prove, so we consider the case $x \ne x_0$.

In order to show this, it suffices to show that $\lim_{n \to \infty} P_n(f;x_0)(x) = f(x)$ where $P_n(f;x_0)(x)$ denotes the $n^{th}$ Taylor polynomial of $f$ at $x_0$, evaluated at $x$.

This in turn is equivalent to showing that the remainder approaches zero:

'Want to show: $\lim_{n \to \infty} R_n(f;x_0)(x) = 0$

where $R_n(f;x_0)(x) = f(x) - P_n(f;x_0)(x)$.

Proof of what we want to show: By Fact (1), we have that:

$|R_n(f;x_0)(x)| \le \left(\max_{t \in J} |f^{(n+1)}(t)|\right) \frac{|x - x_0|^{n+1}}{(n + 1)!}$

where $J$ is the interval joining $x_0$ to $x$. Let $a = \min \{ x,x_0 \}$ and $b = \max \{ x, x_0 \}$. The interval $J$ is the interval $[a,b]$.

Now, from the given data, there exists $C$, dependent on $x$ and $x_0$ but not on $n$, such that:

$\max_{t \in J} |f^{(n+1)}(t)| \le C \ \forall \ n$

Plugging this in, we get that:

$|R_n(f;x_0)(x)| \le C \frac{|x - x_0|^{n+1}}{(n + 1)!}$

Now taking the limit as $n \to \infty$, we get:

$\lim_{n \to \infty} |R_n(f;x_0)(x)| \le C \lim_{n \to \infty} \frac{|x - x_0|^{n+1}}{(n + 1)!}$

Since exponentials grow faster than power functions, the expression under the limit goes to zero, so we are left with a right side of zero, hence the left side limit is zero, and we are done.