Difference between revisions of "Uniformly bounded derivatives implies globally analytic"

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===Global statement===
 
===Global statement===
  
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Suppose <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a,b \in \R</math>, there is a constant <math>C</math> (possibly dependent on <math>a,b</math>) such that for all nonnegative integers <math>n</math>, we have:
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Suppose <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a<b \in \R</math>, there is a constant <math>C</math> (possibly dependent on <math>a,b</math>) such that for all nonnegative integers <math>n</math>, we have:
  
 
<math>|f^{(n)}(t)| \le C \ \forall t \in [a,b]</math>
 
<math>|f^{(n)}(t)| \le C \ \forall t \in [a,b]</math>

Latest revision as of 20:27, 12 July 2012

Statement

Global statement

Suppose f is an infinitely differentiable function on \R such that, for any fixed a<b \in \R, there is a constant C (possibly dependent on a,b) such that for all nonnegative integers n, we have:

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

Then, f is a globally analytic function: the Taylor series of f about any point in \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 \R such that, for any fixed a,b \in \R, there is a constant C (possibly dependent on a,b) such that for all nonnegative integers n, we have:

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

A point x_0 \in \R and a point x \in \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 \ne 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)| \le \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)| \le C \ \forall \ n

Plugging this in, we get that:

|R_n(f;x_0)(x)| \le 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)| \le 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.

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