# Is every locally bounded function asymptotically dominated by a real-analytic function?

A theorem of Poincaré states the following:

If $g:\mathbb{R} \to \mathbb{R}$ is continuously increasing, then there exists a real analytic function $f:\mathbb{R} \to \mathbb{R}$ such that $$\lim_{x \to +\infty} \frac{f(x)}{|g(x)|} = + \infty$$

Note that I was originally exposed to this theorem by Andrés E. Caicedo's answer here, which itself cites G.H Hardy's book Orders of Infinity.

Problem: Can the condition of "continuously increasing" be replaced with "locally bounded"?

The motivation here is simply to study how large the set of functions bounded by a real-analytic function can be. It isn't all functions, since functions with vertical asymptotes arbitrarily far into the positive real-line are of course problematic (e.g., see here). Locally bounded seems like a natural condition.

Below is what I've attempted. It's perhaps worth noting that in $\mathbb{R}$ "locally bounded" is equivalent to "bounded on compacts"

Suppose $M_n$ is an upper bound for $|g|$ on each $[n, n+1]$, $n \in \mathbb{N}_{\geq 0}$. Consider the set of points
(in $\mathbb{R}^{2}$) which can be written as $(n, 2^n M_n)$ for $n \in \mathbb{N}_{\geq 0}$. Then, as a consequence on this theorem, there exists a real analytic $f$ whose graph passes through all of those points. This implies that

$$\frac{f(n)}{|g(n)|} = \frac{2^n M_n}{|g(n)|} \geq 2^n \rightarrow + \infty$$

But the problem is that the variable is a natural number, not a real number, and we don't know whether or not the "real-analytic interpolation" is well-behaved enough to ensure that the limit exists when we replace $n$ with the continuous real-variable $x$.

Yes, it is possible. Let $g$ be our locally bounded function.
Let $M_n:= \sup_{x \in [0,n+1]} |g(x)|$.
Define $h(n):=M_{n}$ for $n \in \Bbb Z_{\geq 0}$, and define $h:[0,\infty) \to \Bbb R$ to be the linear interpolation of the points $(n,h(n))$.
Then $h$ is continuously increasing and $h(x) \geq |g(x)|$ for every $x$, so there exists an analytic function $f$ such that $\lim\frac{f}{h}=+\infty$ and hence $\lim \frac{f}{|g|}=+\infty$.