# Asymptotic behaviour of smooth functions with $f^{(n)}(x) > 0$.

Let $f:[0, \infty) \to \mathbb{R}$ be infinitely differentiable with $f^{(n)}(x) > 0$ for all $n \geq 0$ and $x \geq 0$.

What can we say about the asymptotic behaviour of such a function? Can we find a good asymptotic lower bound?

For example, is there an $f$ such that $f \in o(e^{\epsilon x})$ as $x \to \infty$ for all $\epsilon > 0$? Is $f$ necessarily logarithmically convex when $x$ is sufficiently large?

A trivial observation is that $f$ asymptotically dominates any polynomial.

Remark: I'm having difficulty constructing explicit functions which satisfy the conditions, except the "obvious" ones like $ce^{bx}$ ($c,b>0$) or $\exp \exp (x)$.

• I'm not sure if it's easy, but I'd suspect that $f(x) = 1 + \sum_{n = 1}^\infty \frac{1}{n^{n^n}} x^n$ is $o(e^{\epsilon x})$ for every $\epsilon > 0$. – Marcus M Jul 10 '17 at 18:06

Let $(a_n)_{n\in\Bbb N_0}$ be any sequence of positive reals. If $f(x):=\sum_{n=0}^\infty a_nx^n$ converges for all $x\in[0,\infty)$, then $f$ has the desired property. This is equivalent to $\limsup\sqrt[n]{a_n}=0$. In particular, $a_n$ is allowed converge to $0$ arbitrarily fast. As the coefficients of the Taylor expansion of $e^{\epsilon x}$ are $\sim \frac{\epsilon^n}{n!}$, letting $a_n=\frac1{n!^2}$ is one of many possible choices to achieve $f(x)=o(e^{\epsilon x})$ for all $\epsilon>0$.

Logarithmic convexity: The second derivative of $\ln f(x)$ is $\frac{f''(x)f(x)-f'(x)^2}{f(x)^2}$ and we are interested in the inequality $f''(x)f(x)\stackrel ?>f'(x)^2$. The coefficient of $x^n$ in $f''f-f'^2$ is \begin{align}\sum_{i+j=n+2} j(j-i-1)a_{i}a_{j} &=\sum_{i+j=n+2, i\le j} \bigl(j(j-i-1)+i(i-j-1)\bigr)a_{i}a_{j}\\ &=\sum_{i+j=n+2, i\le j}\bigl((i-j)^2-i-j\bigr)a_ia_j \end{align} Note that $-2ia_i^2$ occurs in this if $n=2i-2$ is even. Letting $a_n$ be small with few exceptions allows us to create a function where $f''f-f'^2$ will assume negative values for arbitrarily large $x$. Think of something $a_{n}=\frac1k$ if $n=(k!)!$ and $a_n=\frac1{n^{n^{n^n}}}$ otherwise, or something along that line - it should be straightforward to show the claim with such an explicit example.

For any sequence $a_i > 0$ that decreases sufficiently quickly,

$$f(x) = \sum_{i=0}^\infty a_i x^i$$

satisfies the conditions. For example, with $a_i = 1/(i!)^2$ you get the modified Bessel function $$f(x) = I_0(2 \sqrt{x})$$

which is asymptotic as $x \to \infty$ to $$\frac{1}{2\sqrt{\pi} x^{1/4}} \exp\left(2 \sqrt{x}\right)$$

and in particular is $o(\exp(\epsilon x))$ for all $\epsilon > 0$.