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.

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    $\begingroup$ I suspect this implies $f$ is at least exponential - but if you restrict the domain to just $\mathbb{R}^+$ you can basically get as slow as you like. $\endgroup$ – Dark Malthorp May 9 '20 at 22:43
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    $\begingroup$ e.g. $f(x) = \sum_{n=0}^\infty \frac{x^n}{(a n)!}$ grows with a rate of $\exp(x^{1/a})$ and has all its derivatives positive on $(0,\infty)$. $\endgroup$ – Dark Malthorp May 9 '20 at 22:44
  • $\begingroup$ Those were my thoughts too. $\endgroup$ – user173897 May 10 '20 at 23:13
  • $\begingroup$ What do you mean by slowest-growing? Do you mean f is big-O of g for any g growing as specified in the question? Or is it a contest? $\endgroup$ – Fullfungo May 13 '20 at 4:00
  • $\begingroup$ @Fullfungo I think what makes this question interesting is the existence or non-existence of functions with sub-exponential growth and also having $f^{(n)}(x)$ increasing for all $n$. $\endgroup$ – Dark Malthorp May 14 '20 at 15:51

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.$\blacksquare$

  • $\begingroup$ Perhaps if you use Taylor's formula with Lagrange remainder, you can relax the condition of $f$ being analytic to being smooth. Indeed, $$ f(x) = f(0) + f'(0)x + \frac{1}{2}f''(\xi _x )x^2 \ge f'(0)x $$ for any $x$. Also, at the very end, you need to say more since $\lim_{x\to \infty} 1/2^n$ is not zero. $\endgroup$ – Gary May 13 '20 at 13:26
  • $\begingroup$ @Gary Taylor's formula works for points close to the chosen one, and the error term is bounded only if the $n+1$-st is bounded, which is not the case but is not a problem either. However, the function is equal to your expression only in a close region of $0$, at least I know this formulation. Do you have any sources about its behaviour on large $x$? $\endgroup$ – Fullfungo May 15 '20 at 2:27
  • $\begingroup$ You can apply the Taylor formula with Lagrange remainder for each $x>0$ for the interval $\left[ 0,x \right]$ and you obtain that remainder with a suitable $0<\xi_x<x$. From the numerical point of view, this formula is not very useful indeed, but nevertheless is true. Now, for what we want, we need only that the remainder is positive (the size does not matter). Fortunately that is true because $f''$ is positive for positive argument. $\endgroup$ – Gary May 15 '20 at 6:31
  • $\begingroup$ I have just seen that you used a different method which leads to the same conclusion and is even simpler. $\endgroup$ – Gary May 15 '20 at 6:52

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 $x \to -x$)


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