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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.

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    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.2011-10-23
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    @HenningMakholm See, for instance Loomis and Sternberg Advanced Calculus p137; this is how they define "little o"2011-10-23
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    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.2011-10-23
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    The book's available at http://www.math.harvard.edu/~shlomo/docs/Advanced_Calculus.pdf2011-10-23
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    Can you quote the definition from Marsden verbatim? What is the $k$ in the definition? Are you missing a quantifier (existential/universal) for $k$ in the definition?2011-10-23
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    @SrivatsanNarayanan Marsden et. al. don't explicitly define $k$, but I assume that it's a positive integer. Their definition is exactly: $o(e^k)$ will denote a continuous function of $e$ defined in a neighborhood of the origin of a normed vector space $E$ satisfying $\lim_{e \rightarrow 0} \frac{o(e^k)}{|e|^k}$2011-10-23
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    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."2011-10-23
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    @Gerry So, are they defining a concept that is not intended to have any relation to what most people mean when they say "little o" and "big O"? If so, that's very perplexing. Are there any other references that you are aware of that define the sets that L&S are referring to?2011-10-23
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    The "script little o" class of functions they define is certainly not standard notation. It corresponds to the class of functions that is usually denoted "$o(x)$" (together with an usually implicit specification that one is concerned about the limit $x\to 0$).2011-10-24
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    On page 139, L&S define $f(x)=o(x)$ to mean $f$ is in the family $o$. So it amounts to the same thing as what most people mean, only so far as I have looked in the book they don't extend the notation to $f(x)=o(g(x))$. They have a very specific application in mind, trying to define differentials in terms of infinitesimals with rigor (and without non-standard analysis), and it's all they need for that purpose. I note that this was the assigned text for a course I took in 1968-69.2011-10-24

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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 $U\setminus\{a\}$, $|f(x)| = \varepsilon(x)|g(x)|$.

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    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 $\{a\}$).2011-10-23
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    Thanks for pointing that out.2011-10-23
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    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](http://math.stackexchange.com/questions/72412/big-and-small-o-notation-help/72421#72421).2011-10-23
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    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.2011-10-23