Uniformly bounded derivatives implies globally analytic
Suppose is an infinitely differentiable function on such that, for any fixed , there is a constant (possibly dependent on ) such that for all nonnegative integers , we have:
The functions all fit this description.
If , we know that each of the derivatives equals , so for all . Since is continuous, it is bounded on the closed interval , and the upper bound for thus serves as a uniform bound for all its derivatives. (In fact, since is increasing, we can explicitly take ).
For or , we know that all the derivatives are or , so their magnitude is at most 1. Thus, we can take .
Given: is an infinitely differentiable function on such that, for any fixed , there is a constant (possibly dependent on ) such that for all nonnegative integers , we have:
A point and a point .
To prove: The Taylor series of at , evaluated at , converges to .
Proof: Note that if , there is nothing to prove, so we consider the case .
In order to show this, it suffices to show that where denotes the Taylor polynomial of at , evaluated at .
This in turn is equivalent to showing that the remainder approaches zero:
'Want to show:
Proof of what we want to show: By Fact (1), we have that:
where is the interval joining to . Let and . The interval is the interval .
Now, from the given data, there exists , dependent on and but not on , such that:
Plugging this in, we get that:
Now taking the limit as , we get:
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.