Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could you help me to show: $g(x,\epsilon)+f(x,\epsilon)=O(|\phi(x,\epsilon)|+|\psi(x,\epsilon)|)$ but $g(x,\epsilon)+f(x,\epsilon)\neq O(\phi(x,\epsilon)+\psi(x,\epsilon))$ (both when $\epsilon\to0$), where $O$ stands for the big Oh notation?

share|cite|improve this question
I suppose the assumption is that $g(x,\epsilon)=O(\phi(x,\epsilon))$ and likewise for $f$ with $\psi$? – Raskolnikov Feb 21 '11 at 18:47
Try $f(x)=g(x)=\phi(x)=x$ and $\psi(x)=-x$. – Yuval Filmus Feb 21 '11 at 19:02

So, assuming $g(x,\epsilon)=O(\phi(x,\epsilon))$ and $f(x,\epsilon)=O(\psi(x,\epsilon))$ this means that $\exists \delta_1,\delta_2,M_1,M_2$ such that

$$\forall |\epsilon|<\delta_1 : |g(x,\epsilon)| < M_1 |\phi(x,\epsilon)|$$


$$\forall |\epsilon|<\delta_2 : |f(x,\epsilon)| < M_2 |\psi(x,\epsilon)| \; .$$

This implies that

$$\begin{eqnarray} \forall |\epsilon| < \min(\delta_1,\delta_2) : & |g(x,\epsilon) + f(x,\epsilon)| & \leq |g(x,\epsilon)| + |f(x,\epsilon)| \\ & & < \max(M_1,M_2)(|\phi(x,\epsilon)| + |\psi(x,\epsilon)|) \; , \end{eqnarray} $$

using the triangle inequality. Or in other words $g(x,\epsilon)+f(x,\epsilon)=O(|\phi(x,\epsilon)|+|\psi(x,\epsilon)|)$.

Now, using Yuval Filmus' suggestion, you can see that for the choice $f(x)=g(x)=\phi(x)=x$ and $\psi(x)=-x$, $2x=O(2|x|)$, which is correct. The other identity can not be correct however since $2x \neq O(0)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.