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.