Boundedness of $f(x)/f'(x)$ for a $C^{\infty}$ function with all derivatives positive.

Question

A problem from a past university-level Brazilian Maths Olympiad (OBM) talks about a $C^\infty$ function $f: \mathbb{R}^+ \to \mathbb{R}^+$, where we have $f^{(k)}(x) > 0$ for all $k \in \mathbb{N}$, and $f(m)$ is an integer for all natural $m$. Then, it asks us to prove that $f(m)\geq 2^{m-1}$, but I'm not asking for a full proof in this question.

Rather, the question is about a statement which I believe to be true, but which I'm struggling to prove. In solving it, I showed that the condition that $f(x)/f'(x)$ is bounded on $[1, \infty)$, or equivalently in this case, that $\lim_{x \to \infty} {f'(x)}/{f(x)}>0$, would lead to a proof of what is asked. I'm not stating my proof since it does not directly concern my question, but if you are curious or find it necessary, feel free to ask.

Now, this would imply that $f$ grows faster or as fast as an exponential, and the rest is easier now. But is the said boundedness condition true for $f$, after all? And how could I show it?

Edit: a friend pointed out this question is an 'XY problem', and this is somewhat true. But I was mainly curious if $C^{\infty}$ and positive derivatives, together with integral points, implies the said boundedness, and the original problem was stated for additional context.

Answer 1

This is not true if you assume only the condition that the derivatives are positive (it seems plausible but unlikely to me that adding the integer values condition would change the answer).

You can construct a counterexample using a power series $$f(x)=\sum_{k=1}^\infty \frac{x^{n_k}}{n_k!}$$ for a sequence $(n_k)$ that grows fast enough. Let's suppose we have already chosen $n_1,\dots,n_M$ and points $x_1,\dots,x_M$ such that the polynomial $$f_M(x)=\sum_{k=1}^M \frac{x^{n_k}}{n_k!}$$ satisfies $f_M(x_i)/f_M'(x_i)>i$ for $i=1,\dots, M$. We want to choose $n_{M+1}$ and a point $x_{M+1}$ such that if we define $f_{M+1}(x)=f_M(x)+x^{n_{M+1}}/n_{M+1}!$ then $f_{M+1}(x_i)/f_{M+1}'(x_i)>i$ for $i=1,\dots, M+1$. To do this, note first that we can pick $n_{M+1}$ sufficiently large such that the term $x^{n_{M+1}}/n_{M+1}!$ will have negligible effect on the values of $f_{M+1}(x_i)$ and $f_{M+1}'(x_i)$ for $i\leq M$, so that their ratio will still be greater than $i$. Since $f_{M+1}$ is a polynomial of degree $n_{M+1}$ and $f'_{M+1}$ is a polynomial of degree $n_{M+1}-1$, we can then choose $x_{M+1}$ sufficiently large so that $f_{M+1}(x_{M+1})/f_{M+1}'(x_{M+1})>M+1$.

In the limit, then we have $f_M\to f$ and $f_M'\to f'$ since the power series is convergent, and so $f$ will satisfy $f(x_i)/f'(x_i)\geq i$ for all $i$. In particular, $f(x)/f'(x)$ will be unbounded as $x\to\infty$. However, all the derivatives of $f$ will be positive on $\mathbb{R}_+$ since all the coefficients of the power series are nonnegative.

Answer 2

As @Eric Wolfsey showed in his answer, it's not hard to build a power series that shows that the the first conditions (excluding the integral points) are not enough for it to be true. Here I complement his reasoning by showing an even simpler example where it fails.

Let's try to build a power series that grows slower than any function on the form $e^{Cx}$ (that is, $\lim_{x\to\infty}f(x)/e^{Cx} = 0$), but still suffices the condition for all derivatives to be positive. Then, it makes intuitive sense that, since the derivative of such exponential function is a multiple of itself, functions that grow slower than $e^{Cx}$ may have derivatives that grow slower than a multiple of themselves. In that case, since our functions are monotonically increasing, $f(x)/f'(x)$ should not be bounded. Let's see if this holds for the following function:

$$f(x)=\sum_{n=0}^{\infty}\frac{x^{n}}{\left(2n\right)!}$$

It is not obvious, but with some help from Mathematica, we actually have $f(x)=\cosh(\sqrt{x}) - 1$. So $f'(x) = \sinh(\sqrt{x})/(2\sqrt{x})$. Then, $$\lim_{x\to\infty}\frac{\cosh(\sqrt{x}) - 1}{e^{x}}=0$$

We also have: $$\lim_{x\to\infty}\frac{2\sqrt{x}(\cosh(\sqrt{x}) - 1)}{\sinh(\sqrt{x})} = +\infty$$ So $f(x)/f'(x)$ is not bounded.

However, since I assume that what the problem asks me to prove is true, it would still be interesting to find out how the integral points condition affects this result, and I would very much appreciate an answer.