In control theory, the order of magnitude notation $\mathcal{O}(\varepsilon)$ is usually defined as:
- $f(\varepsilon)=\mathcal{O}(g(\varepsilon))$ if there exist positive constants $k$ and $c$ such that $|f(\varepsilon)|\leq k|g(\varepsilon)|$, $~\forall~ |\varepsilon|<c$.
I was wondering if the above definition is equivalent to
- $f(\varepsilon)=\mathcal{O}(g(\varepsilon))$ if $$\lim_{\varepsilon\to0}\sup \left| \frac{f(\varepsilon)}{g(\varepsilon}\right|<\infty.$$
If they are equivalent, can you give me some references that shows this?
It seems to me that the above definitions are equivalent. I checked several examples for which both definitions worked. For instance, for $f(\varepsilon)=\frac{\varepsilon^2}{1+\varepsilon}$ we obtain $f(\varepsilon)=\mathcal O(\varepsilon^2)$ as from the first definition $$\left|\frac{\varepsilon^2}{1+\varepsilon}\right| < \left|\frac{1}{1-a}\right|\varepsilon^2, \quad \forall~ |\varepsilon|<a<1$$ and from the second definition we have $$\lim_{\varepsilon\to0} \frac{\frac{\varepsilon^2}{1+\varepsilon}}{\varepsilon^2}=1<\infty.$$ We could similarly show that $g(\varepsilon)=\mathcal O(\varepsilon)$.
Yes, they're equivalent. I can't find a standard reference for this fact, but see Theorem 2.3 in the linked course notes
https://courses.marlboro.edu/pluginfile.php/34433/mod_page/content/1/week6.pdf
as well as the several related MSE questions