Relationship between big O notation and exponential type

Question

Let $f: \mathbb{R} \to \mathbb{R}$, $C\in \mathbb{R}$. What, if any, is the difference between "$ f = O(e^{Cx}) $" and "$f$ is of exponential type $C$"? If they're different, is it possible to express the latter using big O notation?

Answer 1

In short, they are similar but not quite the same. In memory I have only seen "of exponential type" being used for complex functions $f: \mathbb{C} \to \mathbb{C}$, but presumably the analogous definition has uses for real functions. To say that $f: \mathbb{C} \to \mathbb{C}$ (usually assumed to be complex differentiable) is of exponential type $C$ (for some $C > 0$) means that for all $\epsilon > 0$, there exists $M_\epsilon > 0$ and $R_\epsilon > 0$ such that, for all $r > R_\epsilon$ $$ |f(re^{i\theta})| \leq M_\epsilon e^{(C+\epsilon)r} $$ or equivalently for all $z \in\mathbb{C}$ with $|z| > R_\epsilon$ $$ |f(z)| \leq M_\epsilon e^{(C+\epsilon)|z|}. $$

The same definition makes sense for $f: \mathbb{R}\to\mathbb{R}$ if we remove the $e^{i\theta}$ in the argument of $f$. That is, we may say that $f:\mathbb{R}\to\mathbb{R}$ is of exponential type $C$ if for all $\epsilon > 0$ there exist $M_\epsilon, R_\epsilon > 0$ such that for all $x \in \mathbb{R}$ with $|x| > R_\epsilon$ $$ |f(x)| \leq M_\epsilon e^{(C+\epsilon)|x|}. $$ So, by the definition of $O(g(x))$, this says that $f(x) \in O_\epsilon(e^{(C+\epsilon)|x|})$ as $x \to \infty$ and as $x \to -\infty$ for all $\epsilon > 0$. (The subscript on $O_\epsilon$ just indicates that the coefficient $M_\epsilon$ in the inequality depends on $\epsilon$.) So in general, no, $f$ being of exponential type $C$ isn't the same as $f$ being $O(e^{Cx})$ or even $O(e^{C|x|})$ as $x \to a$ for some $a$; instead, being of exponential type includes the asymptotic behavior of $f$ at both $\pm \infty$ while at the same time doesn't assume $f \in O(e^{C|x|})$ (only $O_\epsilon(e^{(C+\epsilon)|x|})$ for all $x$, which is a weaker condition).

Note, of course, it may be that certain authors consider a real function $f$ to be of exponential type if a different definition holds (e.g., if the defining inequality condition holds only as $x \to \infty$ and/or without being dependent on a parameter $\epsilon$), in which case it could of course be equivalent to $O(e^{Cx})$ at $\infty$.