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:

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

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 $f^{(n)}(x)=f(x)$ for all $x\in [a,b]$. 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.

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 $C=1$.