User:IssaRice/Little o notation

Definition

Definition (little o near a point). Let $f:\mathbf {R} \to \mathbf {R}$ and $g:\mathbf {R} \to \mathbf {R}$ be two functions, and let $a\in \mathbf {R}$ . We say that $f$ is little o of $g$ near $a$ iff for every $\epsilon >0$ there exists $\delta >0$ such that $|x-a|<\delta$ implies $|f(x)|<\epsilon |g(x)|$ . Some equivalent ways to say the same thing are:

Notation	Comments
$f$ is little o of $g$ near $a$
$f(x)\in o(g(x))$ as $x\to a$	In this notation, we think of $o(g(x))$ as a set.
$f(x)=o(g(x))$ as $x\to a$
$f\in o(g)$ near $a$
$f=o(g)$ near $a$

Definition (little o at infinity). Let $f:\mathbf {R} \to \mathbf {R}$ and $g:\mathbf {R} \to \mathbf {R}$ be two functions. We say that $f$ is little o of $g$ at infinity iff for every $\epsilon >0$ there exists $M$ such that for all $x$ , $x>M$ implies $|f(x)|<\epsilon |g(x)|$ .

Can we write just $f\in o(g)$ or $f=o(g)$ or $f(x)\in o(g(x))$ or $f(x)=o(g(x))$ ?

Expand to see solution:

In general we can't because for this notation to make sense, we also need to know where the argument

x

is going. In algorithms, we have

x\to \infty

, but in analysis (e.g. in some definitions of differentiability) we have

x\to 0

.

If we are being a little pedantic, what is wrong with saying " $f\in o(g)$ as $x\to a$ "?

Expand to see solution:

We are saying

x\to a

, but we haven't clarified what

x

is. Instead, we are relying on the reader to assume that

x

is an argument to

f

and

g

.

Properties

Proposition. Let $f:\mathbf {R} \to \mathbf {R}$ and $g:\mathbf {R} \to \mathbf {R}$ be two functions, and suppose $g(x)\neq 0$ for all $x\in \mathbf {R}$ . Then f is little o of g near a if and only if $\lim _{x\to a}{\frac {f(x)}{g(x)}}=0$ .