Uniformly bounded derivatives implies globally analytic

Statement

Global statement

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

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

Then, $f$ is a globally analytic function: the Taylor series of $f$ about any point in $\mathbb{R}$ 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 $\mathbb{R}$ such that, for any fixed $a,b\in \mathbb{R}$, there is a constant $C$ (possibly dependent on $a,b$) such that for all nonnegative integers $n$, we have:

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

A point $x_0\in \mathbb{R}$ and a point $x\in \mathbb{R}$.

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 $\underset{n\to \mathrm{\infty }}{lim}{P}_n(f;x_0)(x)=f(x)$ 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: $\underset{n\to \mathrm{\infty }}{lim}{R}_n(f;x_0)(x)=0$

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:

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

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:

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

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 $n\to \mathrm{\infty }$, we get:

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

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.