Suppose that $f$ is infinitely differentiable on $[a,b]$ and suppose that for any $a ≤ x ≤ b$ the Taylor series of $f$ has positive radius of convergence at $x$. Use the Baire Category Theorem to show that $f$ must be analytic on a subinterval of $[a,b ]$.

  • $\begingroup$ Hint. For each $x$ pick $n(x)$ such that Taylor series converges in $(x-\dfrac1{n(x)},x+\dfrac1{n(x)})$. Let $A_n=\{x\in(a,b):n(x)=n\}$. Note that $(a,b)$ is the union of the $A_n$ so the closure of at least one $A_n$ must contain an interval $(c,d)$. $\endgroup$ – Mirko Jul 28 '15 at 23:02
  • $\begingroup$ @Mirko: Don't you need to use $[a,b]$ in order to have a complete/compact (sub) space to apply Baire? $\endgroup$ – Gary. Jul 28 '15 at 23:12
  • $\begingroup$ Thank you Mirko. Your hint is helpful. How do we show Taylor series converges to $f$ in the interval $(c,d)$? We have to show f is analytic! $\endgroup$ – Mathsira Jul 28 '15 at 23:12
  • $\begingroup$ @Mathsira:: I think when the T-series of $f$ converges at x, it converges to $f(x)$ $\endgroup$ – Gary. Jul 28 '15 at 23:23
  • $\begingroup$ @Gary: Then we have an interval which f is analytic in before applying Baire Category according to Mirko hint ! $\endgroup$ – Mathsira Jul 28 '15 at 23:35

This is taken from the argument given in the link in my comment above. See the link for details.


For $k$ a positive integer, define $$ E_k=\bigl\{ \,y\in [a,b]\,\mid \, \sup_n |f^{(n)} (y)/ n!|^{1/n}< k\,\bigr\}. $$

For a given $y$, the quantity $\sup_n |f^{(n)} (y)/ n!|^{1/n}$ is finite, since the Taylor series about $y$ has positive radius of convergence. Use Baire to show some $E_k$ is not nowhere dense.

From the subinterval obtained via the appeal to Baire, argue that the Lagrange form of the Taylor remainder at a $y$ in this subinterval tends to $0$ for every $x$ sufficiently close to $y$.


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