Question

The statement "If f = o(g) then f = O(g)" can be abbreviated o⇒O. Prove this.

Answer 1

Question 1. Prove the statement "If $f = o(g)$ then $f = O(g)$".

Solution. Recall that $f(x) = o(g(x))$ as $x \to x_0$ iff for any $\varepsilon > 0$ there is $\delta > 0$, such that $|f(x)| \leq \varepsilon |g(x)|$ for all $x$, such that $|x - x_0| < \delta$. Also recall that $f(x) = O(g(x))$ iff there are $M > 0$ and $\delta > 0$, such that $|f(x)| \leq M |g(x)|$ for all $x$ with $|x - x_0| < \delta$. Now if $f(x) = o(g(x))$ as $x \to x_0$, then one can fix some $\varepsilon > 0$, find the corresponding $\delta > 0$ and set $M = \varepsilon$. Then $|f(x)| \leq M |g(x)|$ for all $x$, such that $|x - x_0| < \delta$. By definition, this means that $f(x) = O(g(x))$.