# Uniformly bounded derivatives implies globally analytic

## Statement

### Global statement

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:

Then, is a globally analytic function: the Taylor series of about any point in converges to . In particular, the Taylor series of about 0 converges to .

## Facts used

## Examples

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 .

## Proof

**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**:

where .

**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.