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:

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

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:

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

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 $x\neq x_{0}$.

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 $n^{th}$ 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:

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

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:

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

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.