I recently got my hands on a copy of *Regular Variation* by Bingham, Goldie, and Teugels ("BGT), and it's been an absolute revelation for my research. The thing is, my current work centers around complex-valued functions of a real-variable. BGT mention in their appendices that a theory of functions of regular variation can be developed for such functions, and point to Elliot (1979)'s *Probabilistic Number Theory* as one such example; I have that book, and there isn't anywhere near as much there as I was expecting.

My research involving asymptotics is not centered on functions of regular variation, but rather, utilizes their properties. As such, it would be most useful to know *exactly which* aspects of this subject (Karamata Theory) hold in the case of complex-valued functions of a real variable, and those which do not.


First, recall that for any $\alpha\in\mathbb{R}$, we write $R_{\alpha}$ to denote the set of all regularly varying functions of index $\alpha$; these are functions of the form $f\left(x\right)=x^{\alpha}\varphi\left(x\right)$, where $\varphi$ is slowly varying at $\infty$.

Of particular interest to me are the following results:

> **BGT - Definition**: If $f$ is well-defined, real-valued, and locally bounded on $\left[X,\infty\right)$, and tends to $\infty$ as
> $x\rightarrow\infty$, the **generalized inverse** of $f$ is given by:
$$f^{\leftarrow}\left(x\right)\overset{\textrm{def}}{=}\inf\left\{ y\in\left[X,\infty\right):f\left(y\right)>x\right\} $$
 This function is
> then defined on $\left[f\left(X\right),\infty\right)$ and is monotone
> increasing to $\infty$.

> **BGT - Theorem 1.5.1.12**: If $f\in R_{\alpha}$ with $\alpha>0$, there exists $g\in R_{1/\alpha}$ with:
> $$f\left(g\left(x\right)\right)\sim g\left(f\left(x\right)\right)\sim x\textrm{ as }x\rightarrow\infty$$ Moreover, $g$, which is called an
> **asymptotic inverse** of $f$ is determined uniquely, up to asymptotic
> equivalence, and one such inverse is $f^{\leftarrow}\left(x\right)$.
 
> **BGT - Theorem 1.5.13 (de Bruijn (1959)**: If $\varphi$ varies slowly at $\infty$, there is a slowly varying function $\varphi^{\#}$, unique
> up to asymptotic equivalence, so that:
> 
> 
> $$\varphi^{\#}\left(x\varphi\left(x\right)\right)\sim\frac{1}{\varphi\left(x\right)}$$
> $$\varphi\left(x\varphi^{\#}\left(x\right)\right)\sim\frac{1}{\varphi^{\#}\left(x\right)}$$
> 
> as $x\rightarrow\infty$. Moreover, $\varphi^{\#\#}\sim\varphi$.
> $\varphi^{\#}$ is then called the **de Bruijn conjugate** of $\varphi$
> and $\left(\varphi,\varphi^{\#}\right)$ is called a **conjugate
> pair**.

> **BGT - Proposition 1.5.14**: If $\left(\varphi,\varphi^{\#}\right)$ is a conjugate pair of slowly varying functions, and $A,B,\nu$ are
> positive real constants, then, the following are also conjugate pairs:
> 
> $$\left(\varphi\left(Ax\right), \varphi^{\#}\left(Bx\right)\right)$$
> $$\left(A\cdot\varphi\left(x\right),\frac{1}{A}\cdot\varphi^{\#}\left(x\right)\right)$$
> $$\left(\left(\varphi\left(x^{\nu}\right)\right)^{1/\nu},\left(\varphi^{\#}\left(x^{\nu}\right)\right)^{1/\nu}\right)$$

I can't see how the given definition of the generalized inverse could be modified to apply to complex-valued functions of a real variable. Nevertheless, BGT have an appendix in which a method of deriving the de Bruijn conjugate of a function is given, in which it is assumed that $$\varphi\left(x\right)=\exp\left(h\left(\ln x\right)\right)$$ where $h$ is a holomorphic function defined on a complex domain $D$ that contains all sufficiently large real numbers. This suggests to me that there is probably a means of extending the above results to apply to complex-valued functions of a real variable.

Finally, I am interested in generalizing the following approximation result, to the case where $g$ is complex-valued:

> **BGT - Theorem 2.3.11**: Let $g:\left(0,\infty\right)\rightarrow\left(0,\infty\right)$ be a
> function ($g$ need not be continuous, nor even measurable), and
> suppose that: $$\alpha=\limsup_{x\rightarrow\infty}\frac{\ln g\left(x\right)}{\ln x}<\infty$$ ($\alpha$ is called the upper order
> of $g$). Then, there is an $f\in R_{\alpha}$ so that:
> $$\limsup_{x\rightarrow\infty}\frac{f\left(x\right)}{g\left(x\right)}=1$$