I'm going to be clear about definitions before I start so there's no ambiguity. Let D be a subset of the complex numbers and let $f: D \to \mathbb{R}^{+}$ be a positive real-valued map defined on D. We will write $f(x) = O(g(x))$ if $g: D \to \mathbb{R}^{+}$ and there exists a positive constant A such that:

$\displaystyle |f(x)| \leq Ag(x)$

for all x in D. If we have that $f(x) = O(g(x))$ and $g(x) = O(f(x))$, then we write $f(x) \asymp g(x)$. If D is unbounded (like the naturals or non-negative reals) then we will also write $f(x) \sim g(x)$ to mean:

$\displaystyle \lim_{|x| \to \infty} \frac{f(x)}{g(x)} = 1.$

The point of all this: I've occasionally used in proofs the intuition that $f(x) \sim g(x)$ implies $f(x) \asymp g(x)$, though the converse is definitely false. I've set about trying to convince myself with a proof, but I've only got as far as proving it for $D = \mathbb{N}$, and even putting D as the non-negative reals gets me close but not quite there. Any ideas?

fhas to be complex-valued andghas to be positive real-valued, especially if you're going to use them symmetrically. $\endgroup$ – Thierry Zell Jan 11 '11 at 14:34`$x \to \infty$`

. IfDis a subset of the reals, it seems fairly unambiguous; not so if it is a subset of the complexes (there are several different ways you can go to infinity in the complex plane, and they can give different limits). $\endgroup$ – Thierry Zell Jan 11 '11 at 14:49