Property of the little-o notation in Calculus Apostol

Question

The definition of little $o$ in Calculus Apostol book is stated as:

From this definition, we can say that, $as\: x \rightarrow 0$ we have

$o(x^2)=x^3 \quad and\quad o(x)=x^3.$

which means that $o(x^2)=x^3=o(x)$ as $x\rightarrow 0$. We could even proceed to show that $o(x^2)=o(x^3)=o(x^4)=...=o(x^n)$ by letting them all equal to $x^m$ where $m>n$.

But then, for example, if we let $o(x^2)=f(x)$ as $x\rightarrow a$, we cannot proceed to show that $f(x)=o(x^3)$ since we would have show that

$$\lim_{x\to 0} f(x)/x^3=0$$

and we know that

$$\lim_{x\to 0} f(x)/x^2=0$$.

Which is not going so well. So is something wrong here?

Is the $=$ sign in the definition $f(x)=o(g(x))$ really the same with the $=$ we usually use?

Answer 1

From this definition, we can say that, $as\: x \rightarrow 0$ we have

$o(x^2)=x^3 \quad and\quad o(x)=x^3.$

NO! One should be careful that the expression $f(x)=o(g(x))$ in the definition can NOT be written as $o(g(x))=f(x)$.

Whenever, one sees the expression $f(x)=o(g(x))$ (as $x\to a$), it means nothing but $$ \lim_{x\to a}\frac{f(x)}{g(x)}=0\tag{*} $$

There are some subtleties when the notation $o(g(x))$ appears in different places of an expression.

(1) $f(x)=o(g(x))$ as in the definition means the function $f(x)$ on the left has the property $(*)$. The symbol $=$ here is not the usual "equal" sign.

(2) when $o(g(x))$ appears in an expression like $1+o(x^2)$, it simply means any instance of functions that has property $(*)$.

(3) A combination of (1) and (2). For example, the meanings of different $o(x^2)$ in an "equality" like expression $$ x^3+o(x^2)=o(x^2),\quad \textrm{as } x\to 0 $$ are different. On the left, $o(x^2)$ is understood as in (2) and thus $x^3+o(x^2)$ is a function; one the right, it means that the function on the left has the property $(*)$ with $g(x)=x^2$ and $a=0$.

(4) $o(x^3)=o(x^2)$ (as $x\to 0$) is a correct statement but $o(x^2)=o(x^3)$ is not:

if $f(x)$ is such that $\lim_{x\to 0}\frac{f(x)}{x^3}=0$, then $\lim_{x\to 0}\frac{f(x)}{x^2}=0$; but not the other way around. See also the discussion in Wikipedia here: https://en.wikipedia.org/wiki/Big_O_notation#Matters_of_notation

Answer 2

The use of the equal sign in $f(x)=o(g(x))$ is a sometimes-convenient convention. That equal sign doesn’t indicate that this is a “real” identity any more than the dot product in the mnemonic formula $\nabla\cdot f$ for $\operatorname{div}f$ is a “real” dot product. Treating it as such can get you into trouble, as the many times someone has asked about the difference between $f\cdot\nabla$ and $\nabla\cdot f$ on this site will attest.

In particular, this “equality” is not transitive. As was mentioned in a comment, $o(g(x))$ is properly the set of all functions that satisfy the criteria in the quoted definition. Loosely speaking they are the functions that vanish “faster” than $g(x)$ as $x\to a$. The expression $f(x)=o(g(x))$ really means $f(x) \in o(g(x))$. If you have $x^3\in o(x^2)$ and $x^3\in o(x)$, you can conclude that $o(x)\cap o(x^2)\ne\emptyset$, but not that these two sets are identical. Indeed, $x^2 \in o(x)$, but obviously $x^2\notin o(x^2)$. For brevity we often say “$f(x)$ is little-oh of $g(x)$” instead of “$f(x)$ is a member of little-oh of $g(x)$,” which could be how writing $=$ instead of $\in$ came into use.

When you see the term $o(g(x))$ as part of some expression, it’s a shorthand for “some unknown function $\varphi$ such that $\varphi(x)\in o(g(x))$.” Each instance of this term potentially stands for a different function. So, in particular, the expression $o(g(x))+o(g(x))$ in your comment to the question really means $$\varphi(x)+\psi(x), \text{ where } \varphi(x),\psi(x)\in o(g(x)),$$ with the identities of $\varphi$ and $\psi$ otherwise left unspecified. With this understanding, the addition rule $o(g(x))+o(g(x))=o(g(x))$ really means that for any two functions $\varphi(x), \psi(x)\in o(g(x))$, $\varphi(x)+\psi(x)\in o(g(x))$ (i.e., the sum of two functions that vanish faster than $g(x)$ also vanishes faster than $g(x)$). The proof of the expanded statement is trivial.