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I'm studying from a textbook and came across an exercise to prove the following, which it calls the Bernstein's 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 $I$.

I'm trying to find more information and/or a proof of it, though most of my search comes up with the Cantor-Bernstein or Schröder-Bernstein theorems, which don't seem to be the same as this. Is anyone more knowledgable of this proof?

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1 Answer 1

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By shifting and rescaling, assume without loss of generality that $[-1,1]\subseteq I$, and it suffices to show that $f$ is analytic at 0. By Taylor's theorem, we have $$f(x) = \sum_{i=0}^k \frac{f^{(i)}(0)}{i!}x^k+R_k(x)$$ for all $x\in I$, where the remainder $R_k$ has the integral form $$R_k(x) = \int_0^x \frac{f^{(k+1)}(t)}{k!}(x-t)^k\ dt$$ Since $f^{(k+1)}\geq 0$, it follows that $R_k(x)\geq0$ for all $x\in[0,1]$. Also, since $\sum_{i=0}^k \frac{f^{(i)}(0)}{i!}x^k \geq 0$ for all $x\in[0,1]$, we have $R_k(x)\leq f(x)$ for all $x\in[0,1]$. In particular, $R_k(1) \leq f(1)$ for all $k$. Now, using the fact that $\frac{x-t}{1-t} \leq x$ for $0\leq t\leq x\leq 1$, we have \begin{align*} R_k(x) &= \int_0^x \frac{f^{(k+1)}(t)}{k!}(1-t)^k \frac{(x-t)^k}{(1-t)^k}\ dt\\ &\leq \int_0^x \frac{f^{(k+1)}(t)}{k!}(1-t)^k x^k\ dt\\ &= x^k \int_0^x \frac{f^{(k+1)}(t)}{k!}(1-t)^k\ dt\\ &\leq x^k \int_0^1 \frac{f^{(k+1)}(t)}{k!}(1-t)^k\ dt \\ &= x^k R_k(1) \\ &\leq x^k f(1) \end{align*} Thus $\lim_{k\to\infty} R_k(x)=0$ for every $x\in[0,1)$. Similarly, $\lim_{k\to\infty} R_k(x)=0$ for $x\in(-1,0]$. Thus we obtain a power series expansion converging to $f(x)$ on the interval $(-1,1)$, as desired.

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