Growth of a function with every derivative everywhere increasing If a function $f: \mathbb{R} \to \mathbb{R}^{+}$, is infinitely differentiable everywhere, and is such that every derivative is monotonically strictly increasing everywhere, then what is the slowest-growing function that $f$ could be? What if the derivatives are just monotonically nondecreasing everywhere?
Feel free to add in additional properties (which you would specify) if this question is not well-formulated enough.
This question just popped into my head. I have suspicions and will work on it too.
 A: We will assume $f:\mathbb R\to \mathbb R^+$ to be the slowest-growing function, every derivative of which is monotonically strictly increasing everywhere.
For each $n\in\mathbb N_0$ we have that $f^{(n)}(x)$ is positive, because $f^{(0)}(x)=f(x)$ is positive by definition and if $n>0$, $f^{(n-1)}(x)$ is monotonically strictly increasing, which implies, that ${f^{(n-1)}}'(x)=f^{(n)}(x)$ is positive. Additionally, it follows that $f^{(n)}(x)$ is continuous for every $n \in \mathbb N_0$. A short proof:
$$\lim_{x \to x_0} {f^{(n)}(x)} = \lim_{x \to x_0} {f^{(n)}(x) - f^{(n)}(x_0) + f^{(n)}(x_0)} = \lim_{x \to x_0} {\frac {f^{(n)}(x) - f^{(n)}(x_0)} {x - x_0} (x - x_0) + f^{(n)}(a)} = \lim_{x \to x_0} {\frac {f^{(n)}(x) - f^{(n)}(x_0)} {x - x_0}} \cdot \lim_{x \to x_0} {(x - x_0)} + \lim_{x \to x_0} {f^{(n)}(x_0)} = f^{(n + 1)}(x_0) \cdot 0 + f^{(n)}(x_0)$$
$$\therefore \lim_{x \to x_0} {f^{(n)}(x)} = f^{(n)}(x_0)$$
For any $t > 0$ the function $f(x)$ is obviously the antiderivative of $f'(x)$ on the interval $x \in [0; t]$. By the fundamental theorem of calculus $f(t) = f(0) + \int_0^t{f'(u) du}$. We also know that $f'(u) > f'(0)$ for $u > 0$, because $f'(x)$ is monotonically increasing. Using the inequality $f(0) > 0$, we get
$$f(x) = f(0) + \int_0^x{f'(u) du} > f(0) + \int_0^x{f'(0) du} = f(0) + f'(0) (x - 0) > f'(0) x$$
$$\lim_{x \to +\infty} {f(x)} \ge \lim_{x \to +\infty} {f'(0) x} = +\infty$$
$$\lim_{x \to +\infty} {f(x)} = +\infty$$
Every derivative of $f(x)$ has to grow at least as fast because otherwise $f(x)$ would not be the slowest growing. $$\lim_{x \to +\infty} {f^{(n)}(x)} \ge \lim_{x \to +\infty} {f(x)} = +\infty$$
$$\lim_{x \to +\infty} {f^{(n)}(x)} = +\infty$$
Now consider $\phi(x):=f(\frac x 2)$, it is positive and monotonically strictly increasing. Its derivatives are also increasing as ${\phi^{(n)}}'(x) = \frac 1 {2^n}f^{(n+1)}(\frac x 2) > 0$.
Now consider the following limit.
$$\lim_{x \to +\infty} {\frac {\phi(x)} {f(x)}} = \lim_{x \to +\infty} {\frac {f(\frac x 2)} {f(x)}}$$
Both the numerator and the denominator grow unboundedly, therefore, we can apply l'Hôpital's rule here.
$$\lim_{x \to +\infty} {\frac {f(\frac x 2)} {f(x)}} = \lim_{x \to +\infty} {\frac 1 2 \frac {f'(\frac x 2)} {f'(x)}}$$
Applying l'Hôpital's rule $n$ times results in
$$\lim_{x \to +\infty} {\frac {\phi(x)} {f(x)}} = \lim_{x \to +\infty} {\frac 1 {2^n} \frac {f^{(n)}(\frac x 2)} {f^{(n)}(x)}}$$
For each $n\in\mathbb N$ we know that $f^{(n)}(x)$ is a positive monotonically strictly increasing function, therefore $0 < f^{(n)}(\frac x 2) < f^{(n)} (x)$ for positive $x$.
$$\therefore 0 < \frac {f^{(n)}(\frac x 2)} {f^{(n)} (x)} < 1 \Rightarrow 0 < \frac 1 {2^n} \frac {f^{(n)}(\frac x 2)} {f^{(n)}(x)} < \frac 1 {2^n}$$
$$\lim_{x \to +\infty} 0 \le \lim_{x \to +\infty} {\frac {\phi(x)} {f(x)}} \le \lim_{x \to +\infty} \frac 1 {2^n}$$
$$\lim_{n \to \infty} \lim_{x \to +\infty} 0 \le \lim_{n \to \infty} \lim_{x \to +\infty} {\frac {\phi(x)} {f(x)}} \le \lim_{n \to \infty} \lim_{x \to +\infty} \frac 1 {2^n}$$
$0$ is a constant, therefore, $\lim_{n \to \infty} \lim_{x \to +\infty} 0 = 0$.
$\lim_{x \to +\infty} {\frac {\phi(x)} {f(x)}}$ does not depend on $n$, therefore, $\lim_{n \to \infty} \lim_{x \to +\infty} {\frac {\phi(x)} {f(x)}} = \lim_{x \to +\infty} {\frac {\phi(x)} {f(x)}}$.
$\frac 1 {2^n}$ does not depend on $x$, therefore, $\lim_{n\to\infty} \lim_{x \to +\infty} \frac 1 {2^n} = \lim_{n \to \infty} \frac 1 {2^n} = 0$.
$$\therefore \lim_{x \to +\infty} {\frac {\phi(x)} {f(x)}} = 0$$
Therefore, $\phi(x)$ is a positive function on real numbers with monotonically strictly increasing derivatives. Also, as we have concluded $\phi(x) \in o(f(x))$. Thus there cannot be the slowest growing function with strictly increasing derivatives.$\blacksquare$
A: The OP property means that $f^{(k)}(x) \ge 0, k \ge 2$ (or strict inequality if needed); this means $f''$ is absolutely monotonic on $\mathbb R$ and by Bernstein's theorem it is (entire) and of the form $f(x)=\int_0^{\infty}e^{xt}d\mu(t)$ where $\mu$ is a positive measure on $[0,\infty)$ for which the integral is absolutely convergent for all $x \in \mathbb C$ (which is actually equivalent to absolute convergence for all $x >R$).
There is a trivial case when $f''=a \ge 0$ (in the $\ge$ case only of course) corresponding to $f(x)$ a quadratic which is positive on $\mathbb R$, but otherwise, the support of the measure cannot be concentrated at zero only so there is $\mu[\alpha, \beta] >0, \alpha >0$ and trivially $f''(x) > \mu [\alpha, \beta]e^{\alpha x}, x >0$ hence $f''$ is an entire function of at least exponenttial growth and so is $f$; by taking $f(x)=e^{\epsilon x}, \epsilon >0$ we obviously do not have a "lowest growth" as such
Note that the non-trivial part is absolute monotonicity to the left (ie on $(-\infty, a)$ as for $(a, \infty), a$ finite anything of the form $\sum {a_n(x-a)^n}$ with $a_n > 0$ radius of convergence infinite will do, so Bernstein theorem is stated that way usually - ie absolute monotonicity on the negative axis implies the integral representation and the continuation to the positive axis follows by analytic continuation - sometimes the results are stated also in terms of complete monotonicity - derivatives alternate in sign - on the positive axis or to the right more generally, and the results are equivalent with $x \to -x$)
