Locally analytic not implies globally analytic - Revision history

Vipul: Created page with "==Statement== It is possible to have a function $f$ defined on all of $\R$ such that $f$ is a locally analytic function everywhere on
2012-07-08T14:07:35Z

Created page with "==Statement== It is possible to have a function <math>f</math> defined on all of <math>\R</math> such that <math>f</math> is a locally analytic function everywhere on <ma..."

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==Statement==

It is possible to have a function <math>f</math> defined on all of <math>\R</math> such that <math>f</math> is a [[locally analytic function]] everywhere on <math>\R</math> but <math>f</math> is ''not'' a [[globally analytic function]].

==Proof==

Consider the function:

<math>f(x) := \frac{1}{x^2 + 1}</math>

This is locally analytic everywhere, but not globally analytic. In fact, the radius of convergence of its power series at any point <math>x_0</math> is <math>\sqrt{x_0^2 + 1}</math>.