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

I understand, or at least think I understand, the nature of a function that is "little o": If $f$ is a function between Banach spaces E and F, then it is "little-o" if

$$|x|\rightarrow 0 \implies \frac{|f(x)|}{|x|} \rightarrow 0$$

Thus the evaluation of $f$ at $x$ approaches $0$ faster than $x$ itself. I have read other posts on here, such as this one that give a different definition. Also, textbook authors don't seem to be in agreement either. For instance, in their advanced calculus text, Loomis and Sternberg declare a function to be "little o" if it satisfies essentially the definition I just gave but also add the condition that $f(0) = 0$. On the other hand, Marsden et. al. in "Manifolds, Tensor Analysis and Applications" define a "little o" function as any continuous function $f:E\rightarrow F$ such that $$ \lim_{x\rightarrow 0}\frac{f(x^k)}{|x|^k} = 0 $$

Is there any hope of reconciling these definitions? They seem to be saying approximately the same thing, but not quite.

share|cite|improve this question
I've never head of a function being simply "little o", period. It is always "little o of something". Namely, $f$ is $o(g(x))$ iff $\frac{|f(x)|}{|g(x)|}\to 0$ for the limit in question. – Henning Makholm Oct 23 '11 at 4:48
@HenningMakholm See, for instance Loomis and Sternberg Advanced Calculus p137; this is how they define "little o" – ItsNotObvious Oct 23 '11 at 5:24
Well, pics or it didn't happen! And if it did happen, I'd be rather wary of trusting that test for anything of importance. – Henning Makholm Oct 23 '11 at 5:33
The book's available at – Gerry Myerson Oct 23 '11 at 6:48
What Loomis and Sternberg actually do is they define a set of functions, which set they denote by a symbol that looks to me like some sort of Gothic little-oh, by saying $f$ is in that family if $f(0)=0$ and $\|f(x)\|/\|x\|\to0$ as $x\to0$. So they never say "$f$ is little-oh;" rather, "$f$ is in little-oh." – Gerry Myerson Oct 23 '11 at 21:53
up vote 4 down vote accepted

This is more of a long comment to the comment of Henning Makholm. The objective of the $o$ and $O$ notations is to compare growth (asymptotic behavior) of 2 functions $f$ and $g$. The function $g$ doesn't have to be $x$ or $x^k$. Defining $f(x) = \underset{x\to a}{o}(g(x))$ as saying $|f(x)|/|g(x)| \underset{x\to a}{\to} 0$ is also bad because it leads to writing nonsense when $g$ has zeros.

A good definition would be $f(x) = \underset{x\to a}{o}(g(x))$ if there exists a nonnegative function $\varepsilon$ and a neighborhood $U$ of $a$ such that $\varepsilon(x) \underset{x\to a}{\to} 0$ and for every $x$ in $U\setminus\{a\}$, $|f(x)| = \varepsilon(x)|g(x)|$.

Similarly $f(x) = \underset{x\to a}{O}(g(x))$ if there exists a nonnegative bounded function $\varepsilon$ and a neighborhood $U$ of $a$ such that for every $x$ in $U\setminus\{a\}$, $|f(x)| = \varepsilon(x)|g(x)|$.

share|cite|improve this answer
That, and if you want a correct definition, add the condition that $\varepsilon$ is defined on a neighborhood of $a$ minus $\{a\}$ (or that $|f|=\varepsilon\cdot|g|$ on a neighborhood of $a$ minus $\{a\}$). – Did Oct 23 '11 at 17:27
Thanks for pointing that out. – AFK Oct 23 '11 at 17:34
Your revised version was still not quite correct and I took on me to modify it. If you do not like the result, please go back to the previous version. Compare with this. – Did Oct 23 '11 at 17:42
Your edit is fine. I didn't mention explicitly the neighborhood not to be redundant but it is indeed better to make the quantification explicit. – AFK Oct 23 '11 at 20:51

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.