Does there exist a continuous function $\: f : \mathbf{R} \to \mathbf{R} \:$ such that for
all real analytic functions $\: g : \mathbf{R} \to \mathbf{R} \:$, for all real numbers $x$,
there exists a real number $y$ such that $\: x < y \:$ and $\: g(y) < f(y) \:$?

This question has been asked several times over on mathoverflow, and I believe also on this site. Someone who is better than I at searching should be able to find the past answers. – Pete L. Clark May 06 '12 at 22:15

Has the version with "real analytic" replaced with "real entire" also been asked? – May 06 '12 at 22:16

2If by "real entire" you mean "entire, and real on $\mathbb R$", then the interpolation proof provides such a function. – Robert Israel May 06 '12 at 22:23

Well, I would tend to define "real entire" as "expressible (on $\mathbf{R}$) as a globally convergent power series with real coefficients", but I do know that that is equivalent to what you gave. – May 06 '12 at 22:27

fixed ${}{}{}\:$ – May 08 '12 at 03:46
3 Answers
No. Only if you require $g$ or its coefficients to be computable. Suppose there is such an $f$, then we could just pick the points $(n,(1+\sup\{ f(z))n1<z<n+1\}))$, for $n=1,2,3\ldots$ and interpolate.
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3How is it proven that real analytic interpolation is possible? (I'm about to edit to impose continuity on $f$.) – May 06 '12 at 22:08

4It's usually done as a corollary of MittagLeffler's theorem and the Weierstrass factorization theorem. But see also http://www.jstor.org/stable/2370666 – Robert Israel May 06 '12 at 22:21
Recently, I was reading Hardy's Orders of Infinity (available here or here):
Godfrey Harold Hardy. Orders of infinity. The Infinitärcalcül of Paul du BoisReymond. Reprint of the 1910 edition. Cambridge Tracts in Mathematics and Mathematical Physics, No. 12. Hafner Publishing Co., New York, 1971. MR0349922 (50 #2415).
The book discusses this result, so I figured it may be worth adding some comments:
Theorem (Poincaré). For any continuous increasing $\phi:\mathbb R\to\mathbb R$ we can always find a real analytic function $f:\mathbb R\to\mathbb R$ such that $\displaystyle \lim_{x\to\infty}\frac{f(x)}{\phi}=+\infty$.
This was published in the American Journal of mathematics, vol. 14, p. 214. Hardy presents a proof due to Borel, in Leçons sur les séries à termes positifs, p.27:
We may replace $\phi$ with an increasing function $\Phi$ that is always positive, is pointwise larger than $\phi$, and tends to infinity, and proceed to define $f$ and show that $f/\Phi\to\infty$. Take an increasing sequence of numbers $a_n\to\infty$, and another sequence $b_n$ with $$ a_1<b_2<a_2<b_3<a_3<\dots, $$ and define $$ f(x)=\sum_{n\ge 1}\left(\frac x{b_n}\right)^{\nu_n}, $$ where the positive integers $\nu_n$ are strictly increasing, and satisfy $\displaystyle \left(\frac{a_n}{b_n}\right)^{\nu_n}>\Phi^2(a_n)$. Then $f$ is entire and satisfies the required property.
In detail: The series converges because, given any positive $x$, the $n$th root of the $n$th term is at most $x/b_n\to 0$. If $x\in[a_n,a_{n+1})$, then $f(x)>(a_n/b_n)^{\nu_n}$, so $$ f(x)>\Phi^2(a_{n+1})>\Phi^2(x). $$ It follows that $f/\Phi^2\ge 1$ for $x\ge a_1$, and since $\Phi(x)\to\infty$, then also $f/\Phi\to\infty$, as wanted.
Hardy mentions this while discussing a result of du BoisReymond: Given functions $f,g\to\infty$, positive, and increasing, write $f\succ g$ iff $f/g\to\infty$.
Theorem (du BoisReymond). Given any "ascending scale" $(f_n)_{n\in\mathbb N}$, that is, a sequence of functions $f_n:\mathbb R\to\mathbb R$, all positive and increasing to infinity, and such that $f_1\prec f_2\prec f_3\dots$, there is a function $f$ that increases faster than any function in the scale, that is, such that $f\succ f_n$ for all $n$.
This result was generalized by several authors, beginning with Hadamard, and eventually led to Hausdorff work on what we now call Hausdorff gaps.
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1Do you know if the "continuous increasing" condition for the theorem of Poincare can be replaced with "bounded on compact intervals"? – MathematicsStudent1122 Sep 04 '17 at 01:58


1It can, since every locally bounded function is bounded by a continuous and increasing function. See https://math.stackexchange.com/questions/2416117/iseverylocallyboundedfunctionasymptoticallydominatedbyarealanalyticfu/2416126 – Bananach Sep 06 '17 at 09:47

@Bananach Yes, of course, thank you. I actually noticed that (and should have noticed it right away when I left that comment), although the function you get is only eventually increasing, which is all that matters here. Globally increasing is in general impossible. – Andrés E. Caicedo Sep 06 '17 at 11:28
Just take $f(x) = \tan(x)$ (defining $f(x) = 0$, say, when $x$ is an integer multiple of $\pi/2$. But this has nothing to do with "growing too fast".
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