48
$\begingroup$

My question is the following: Let $f\in C^\infty(a,b)$, such that $f^{(n)}(x)\ne 0$, for every $n\in\mathbb N$, and every $x\in (a,b)$. Does that imply that $f$ is real analytic?

EDIT. According to a theorem of Serge N. Bernstein (Sur les fonctions absolument monotones, Acta Mathematica, 52 (1928) pp. 1–66) if $f\in C^\infty(a,b)$ and $f^{(n)}(x)\ge 0$, for all $x\in(a,b)$, then $f$ extends analytically in a ball centered at $(a,0)\in\mathbb C$ and radius $b-a$!

$\endgroup$
9
  • 3
    $\begingroup$ And if you add a suitable analytic function to your example? $\endgroup$ Dec 26, 2013 at 0:03
  • 3
    $\begingroup$ I think $e^{-1/x^2}$ has perfectly reasonable derivatives near 0. If you add $e^x$, you even get something where the derivatives are far from 0. $\endgroup$ Dec 26, 2013 at 1:15
  • 14
    $\begingroup$ I don't understand the votes to close. $\endgroup$ Dec 26, 2013 at 4:12
  • 5
    $\begingroup$ @AnthonyQuas: Nope, if we take $f(x) = e^{-1/x^2} + e^x$, then $f'$ has a zero near $x \approx -0.59118$. $\endgroup$ Dec 26, 2013 at 6:32
  • 9
    $\begingroup$ The pattern might not be obvious from Nate Eldredge's comment, but higher derivatives of $\exp(-1/x^2)$ have increasingly large oscillations increasingly close to $0$. Adding something with small derivatives doesn't change this property. $\endgroup$ Dec 26, 2013 at 9:18

2 Answers 2

55
$\begingroup$

If $f$ is $C^{\infty}$ every derivative is continuous, so the hypothesis on $f$ implies that each derivative $f^{(n)}$ has constant sign. Such functions were studied by S. Bernstein and called regularly monotonic. In particular he proved in 1926 that a regularly monotonic function is real analytic.

This 1971 AMM article by R.P. Boas provides a proof, more history, and further results along these lines. See also this 1975 PAMS article of J. McHugh.

$\endgroup$
11
  • 3
    $\begingroup$ A very readable proof of Bernstein's Theorem appears on p. 437 of Volume I (second edition) of Apostol's Calculus. It also seems to be one of the rare occasions one needs the integral form of the remainder formula in Taylor's Theorem. $\endgroup$ Dec 26, 2013 at 20:13
  • 1
    $\begingroup$ I have always preferred the integral formula for the remainder term, because it's an explicit formula. $\endgroup$
    – Deane Yang
    Dec 26, 2013 at 20:19
  • $\begingroup$ @Ted: Thanks. I was thinking of putting some material on analytic functions into my Honors Calculus notes, but a lot of the basic facts have rather unrewardingly technical proofs. However this is a beautiful theorem. I'll look to see what Apostol does. $\endgroup$ Dec 26, 2013 at 20:58
  • 1
    $\begingroup$ Deane and Ted: The integral form of the remainder is also used to prove the analyticity of Newton's function $(1+x)^{\alpha}$. Come to think of it, this may not be a coincidence... $\endgroup$ Dec 26, 2013 at 21:11
  • 1
    $\begingroup$ Don't get me wrong. I'm glad to see the need for it.:) @Pete, yes, I knew that it was needed for the generalized binomial theorem. Mike Spivak and I batted that one around when I convinced him to de-emphasize the integral form in the latest edition of his book. Waits for missiles to be hurled. $\endgroup$ Dec 26, 2013 at 22:36
22
$\begingroup$

Yes, any such function is analytic.

Assume contrary, let $f$ be such a function. Note that if it is analytic at two points then it has to be analytic everywhere between. So by taking restriction, we may assume that the function is not analytic in any subinterval.

We can assume that $0$ is a point in the interval.

Assume the Taylor series of $f$ at $0$ converges in the $\varepsilon$-neighborhood of $0$. Denote by $\bar f$ its sum. The monotonicity of $f^{(n)}$ gives a bound on the error $f(x)-\bar f(x)$ on one side from $0$; it follows that $\bar f(x)$ converges to $f(x)$ if $0<x<\varepsilon$ or $\varepsilon<x<0$, a contradiction.

It remains to consider the case when the Taylor series of $f$ at $0$ diverges in any neighborhood of $0$. In this case, for any $\varepsilon >0$, there is arbitrary large $n$ such that $|f^{(n)}(0)|>\tfrac{n!}{\varepsilon^n}$. Applying monotonicity of $f^{(n)}$ and integrating, we get that $|f^{(k)}(x)|>2^n$ for any $k\le n$ and some $-4{\cdot}\varepsilon<x<4{\cdot}\varepsilon$, a contradiction.

$\endgroup$
5
  • 4
    $\begingroup$ You said that Note that if it is analytic at two points then it has to be analytic everywhere between. Let $f(x)=\exp(\frac{1}{x^2-1})$, for $|x|<1$ and $f(x)=0$, for $|x|\ge 1$. The $f$ is analytic in $\mathbb R\smallsetminus\{-1,1\}$, and not at $x=\pm$. So, for example it is analytic around $x=0$ and around $x=2$, but not in the interval $(0,2)$. $\endgroup$
    – smyrlis
    Dec 26, 2013 at 8:47
  • 1
    $\begingroup$ @smyrlis, you have to use that the is yours; namely, all its derivatives are monotonic. $\endgroup$ Dec 26, 2013 at 16:06
  • $\begingroup$ I can not understand your comment. $\endgroup$
    – smyrlis
    Dec 26, 2013 at 16:13
  • $\begingroup$ @smyrlis, at the end points Taylor series converge to $f$. Since the coefficients in the Taylor series are monotonic, the Taylor series at any point between also converge... $\endgroup$ Dec 26, 2013 at 16:33
  • 9
    $\begingroup$ That's true - not so obvious though. $\endgroup$
    – smyrlis
    Dec 26, 2013 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.