Uniformly bounded derivatives implies globally analytic: Difference between revisions

Revision as of 14:49, 8 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:

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

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

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 $f^{(n)} (x) = f (x)$ for all $x \in [a, b]$ . 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.

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 $C = 1$ .

@@ Line 8: / Line 8: @@
 Then, <math>f</math> is a [[globally analytic function]]: the [[Taylor series]] of <math>f</math> about any point in <math>\R</math> converges to <math>f</math>. In particular, the Taylor series of <math>f</math> about 0 converges to <math>f</math>.
+==Facts used==
+# [[uses::Max-estimate version of Lagrange formula]]
 ==Examples==
 The functions <math>\exp, \sin, \cos</math> all fit this description.
+If <math>f = \exp</math>, we know that each of the derivatives equals <math>\exp</math>, so <math>f^{(n)}(x) = f(x)</math> for all <math>x \in [a,b]</math>. Since <math>\exp</math> is continuous, it is bounded on the closed interval <math>[a,b]</math>, and the upper bound for <math>\exp</math> thus serves as a uniform bound for all its derivatives.
+For <math>f = \sin</math> or <math>f = \cos</math>, we know that all the derivatives are <math>\pm \sin</math> or <math>\pm \cos</math>, so their magnitude is at most 1. Thus, we can take <math>C = 1</math>.