Uniformly bounded derivatives implies globally analytic

Statement

Global statement

Suppose $f$ 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:

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

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

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:

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

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 $x \neq x_{0}$ .

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 $n^{t h}$ 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:

$| R_{n} (f; x_{0}) (x) | \leq ({max}_{t \in J} | f^{(n + 1)} (t) |) \frac{| x - x_{0} |^{n + 1}}{(n + 1)!}$

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:

${max}_{t \in J} | f^{(n + 1)} (t) | \leq C \forall n$

Plugging this in, we get that:

$| R_{n} (f; x_{0}) (x) | \leq C \frac{| x - x_{0} |^{n + 1}}{(n + 1)!}$

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

$lim_{n \to \infty} | R_{n} (f; x_{0}) (x) | \leq C lim_{n \to \infty} \frac{| x - x_{0} |^{n + 1}}{(n + 1)!}$

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.