Prove $f(z)=o(\phi (z))$ implies $f(z)=O(\phi (z))$ as $ z \to 0$.
My proof is:
The assumed statement is $$\lim_{z \to 0} \frac{ f(z)}{\phi (z)}=0$$
That statement implies the function $\frac{ f(z)}{\phi (z)}$ is bounded by a constant $K>0$. Therefore, $\left|\frac{ f(z)}{\phi (z)}\right| \leq K$ which implies $|f(z)|<K|\phi (z)|$.
Is this true correct?