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Regarding the following theorem: If $f$ is infinitely differentiable on an interval $I$, and $f^{(n)}(x)\ge0$ for all $n\in\mathbb N$ and $x\in I$, then $f$ is analytic on $I$.

This theorem is proven for non negative derivatives. My question: what if some or all of the derivatives of the function are negatives ? Does that imply that the function is not analytical, and why ?

David C. Ullrich
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user460753
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    What question??? – Robert Israel Aug 09 '19 at 23:01
  • Let $I = [0,1)$ and $f(x) = \sum_{n=0}^\infty c_n x^n$ converging for $|x| < 1$ then $\forall n, x \in I, f^{(n)}(x) \ge 0$. With $g(x) = -f(x)$ then $\forall n, x \in I, g^{(n)}(x) \le 0$. From there what you need is to prove the converse : if $f$ is smooth and $\forall n, x \in I, f^{(n)}(x) \ge 0$ then its Taylor series at $x=0$ converges for $|x| < 1$ and $f$ is analytic on $I$. – reuns Aug 10 '19 at 05:02
  • There are definitely infinitely differentiable functions on intervals that are not analytic. So some hypothesis on the derivatives are needed. But it's good enough if all but finitely many derivatives are nonnegative, since then some derivative of $f$ is analytic, which implies that $f$ itself is analytic. – Greg Martin Aug 10 '19 at 05:53

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