Uniformly bounded derivatives implies globally analytic: Difference between revisions

Revision as of 14:51, 7 July 2012

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 $f$ .

Revision as of 14:51, 7 July 2012 (view source) Vipul (talk \| contribs) (Created page with "==Statement== 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...")		Revision as of 14:51, 7 July 2012 (view source) Vipul (talk \| contribs) No edit summary Newer edit →
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	<math>\|f^{(n)}(x)\| \le C \ \forall x \in [a,b]</math>		<math>\|f^{(n)}(x)\| \le C \ \forall x \in [a,b]</math>

	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 0.		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>.