# Little-O Implies Big-O/General Result

## Theorem

Let $X$ be a topological space.

Let $V$ be a normed vector space over $\R$ or $\C$ with norm $\norm {\,\cdot\,}$

Let $f, g: X \to V$ be mappings.

Let $x_0 \in X$.

Let $f = \map o g$ as $x \to x_0$, where $o$ denotes little-O notation.

Then $f = \map O g$ as $x \to x_0$, where $O$ denotes big-O notation.

## Proof

From the definition of little-O notation:

there exists a neighborhood $U$ of $x_0$ such that $\norm {\map f x} \le \norm {\map g x}$ for all $x \in U$.

By definition of big-O notation, $f = \map O g$ as $x \to x_0$.

$\blacksquare$