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?



Does f(x)~g(x) imply $f(x) \asymp g(x)$?
The result is false if $D$ is not closed: take a boundary point $a\not\in D$ and let $f(x)=|x-a|$ and $g(x)=1+f(x)$.


If $D$ is closed, it is true. There is some $R>0$ so that $\tfrac12 f(x) < g(x) < 2f(x)$ whenever $|x| > R$. On the other hand, on the compact set $\{x\in D\colon |x|\le R\}$ there are bounds $0 < m \le f(x) \le M$ and $0 < m \le g(x) \le M$. The result $f(x)\asymp g(x)$ follows immediately.


