Big-O notation - Knowlpedia

Let $f,g$ be real or complex functions on $(0,\varepsilon_0)$ , with $g(\varepsilon)>0$ . The statement $f(\varepsilon)=O(g(\varepsilon))$ as $\varepsilon\downarrow0$ means that there are $C<\infty$ and $\varepsilon_1>0$ such that

|f(\varepsilon)|\le Cg(\varepsilon)\qquad(0<\varepsilon<\varepsilon_1).

The constants do not depend on $\varepsilon$ . Other limiting regimes use the corresponding eventual neighborhood. For a normed-space valued function, replace absolute value by its norm.

Reading an estimate

The notation asserts a bound, not an equality with a particular function. Thus $\varepsilon^2=O(\varepsilon)$ , but the latter statement hides the sharper order. In $A\lesssim B$ , the implicit constant must have the same declared independence as in a Big-O estimate.

Derivatives require separate control

A bound on values does not automatically bound derivatives. For example, $f(\varepsilon)=\varepsilon^2\sin(\varepsilon^{-3})$ is $O(\varepsilon^2)$ , whereas its derivative contains $-3\varepsilon^{-2}\cos(\varepsilon^{-3})$ . Parameter-uniform and derivative estimates must be stated separately.

References

NIST Digital Library of Mathematical Functions, §2.1: Definitions and elementary properties.