# A non-exponentially bounded analytic function?

A function $f:\mathbb R\to\mathbb R$ is said to be exponentially bounded if there is an $n$ such that for sufficiently large $x\in\mathbb R$, $\exp(\exp(\cdots \exp(x)))>f(x)$ (where the $\exp$ is repeated $n$ times).

You know what analytic means. Is there a "classical" (or easy to describe) non-exponentially bounded function?

I ask because this is related to an open question in model theory, namely whether the real field, expanded with a non-exponentially-bounded function, can be o-minimal. Most of what we know about o-minimal expansions are related to analytic functions, so I'm interested in what we could be looking for.

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$\frac{1}{\cos^2 x+\cos^2 ax}$ can spike as high as you want if $a\in\mathbb R$ is chosen appropriately. – fedja Sep 14 '13 at 3:12
But for any specific $a$, the result is exponentially bounded, I think? – Richard Rast Sep 14 '13 at 13:02
No. It depends on the Liouvillian character of $a$ how far up each spike goes and that is uncontrollable. – fedja Sep 14 '13 at 13:04
Interesting. This doesn't fit the original model-theoretic goal (because it has periodic-like behavior), but I'm still curious. Do you have a reference for this? It's not obvious at all to me that this function has interesting behavior, and I haven't seen it before. – Richard Rast Sep 14 '13 at 13:07
You mean "Can I give a formal construction of $a$ so that the growth is faster than ...?". Yes, but later, when I come back home. If it doesn't fit your goal, it is now your turn to refine the question, not mine to refine the answer :-) – fedja Sep 14 '13 at 13:10

## 2 Answers

Sorry for the delay: the evening was busier than I thought it would.

Note that if $a$ is irrational, then the denominator never vanishes. On the other hand, if $a=\frac pq$ in simplest terms and $p,q$ are odd, then $x=\frac\pi 2q$ is a zero. Now let $F(x)$ be any given continuous function. We shall construct a sequence of nested intervals $I_k$ for $a$ and a sequence of numbers $q_k$ so that the spikes of our function $f_a$ are above $F$ at $\frac\pi 2q_k$ whenever $a\in I_k$. Since we also need to escape a rational value, we'll fix some enumeration $r_k$ of rationals and ensure that $r_k\notin I_k$.

Put $q_1=1$ and take $I_1$ to be a small interval $[1+\delta_1,1+2\delta_1]$ with very small $\delta>0$ so that $r_1\notin I_1$. Then, if $\delta_1$ is small enough, $f_a(\frac\pi 2q_k)>F(\frac\pi 2q_k)$.

Now choose any rational fraction $p_2/q_2$ with odd $p_2,q_2$ and $q_2\ge 2$ contained in the interior of $I_1$ and put $I_2=[\frac{p_2}{q_2}+\delta_2,\frac{p_2}{q_2}+2\delta_2]$. Again, we can choose $\delta_2>0$ so that $I_2\subset I_1$, $r_2\notin I_2$, and $f_a(\frac\pi 2q_2)>F(\frac\pi 2q_2)$. Now do $I_3$ using $q_3\ge 3$, $I_4$, etc. in the same way.

The nested interval lemma then yields $a$ such that $f_a>F$ on a sequence tending to infinity (i.e., beating any prescribed growth control), though this bad sequence itself is hard to discern explicitly.

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Very interesting. Now, is it completely obvious that this function is analytic? It looks like it should have poles (possibly complex), which I guess means it could be analytic everywhere but never having an everywhere-convergent power series? Sorry if I'm missing something obvious, it is a bit late. – Richard Rast Sep 15 '13 at 3:01
@RichardRast Interesting, it is a bit shocking to me to realize that there is indeed a difference between real analytic functions and analytic complex functions defined in a domain containing $\mathbb R$. – Matemáticos Chibchas Sep 15 '13 at 3:13
It is real analytic (analytic in a neighborhood of $\mathbb R$), not entire (analytic in $\mathbb C$). The complex poles exist and come very close to the real axis. However, $\frac 1{1+e^x}$ is not entire either and every entire function that is given by an elementary formula is, indeed, bounded by a tower of exponents, so to create a fast growing entire function, you need some tools other than combining standard objects from high school. Power series will do it in no time, but, again, not with coefficients given by explicit formulae. – fedja Sep 15 '13 at 12:47
That's a strong statement at the end there. Can it be made precise? – Richard Rast Sep 16 '13 at 3:08
@Richard Rast Probably yes, though it should be done with extreme caution (you can plug my example in, divided by $n!$, say, as the coefficients of Taylor series and show that there exists an $a$ such that the function is entire and grows fast, so the meaning of "explicit" should be made way more restrictive than just "an elementary function in the usual sense of Liouville" if you want to make sense of it). Let's just say that if we use $F(n)$ where $F$ is a function from Hardy domain, then it seems to be true as stated (though I haven't checked the details). – fedja Sep 16 '13 at 3:19

Well, if you really insist on the domain over all real numbers- just take $f(x)=e^{-x}.$

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Well, fine. That's not really what I meant. Let me fix the question. – Richard Rast Sep 14 '13 at 2:07