If $\alpha_n\to 0$, $x_n=O(\alpha_n)$, and $y_n=O(\alpha_n)$, then $x_ny_n=o(\alpha_n)$.

Question

If $\alpha_n\to 0$, $x_n=O(\alpha_n)$, and $y_n=O(\alpha_n)$, then $x_ny_n=o(\alpha_n)$.

I know that $x_n=O(\alpha_n)$ means there exists a constant $C$ and integer $n_0$ such that $|x_n|\leq C|\alpha_n|$ for all $n\geq n_0$. Any solutions or hints are greatly appreciated.

Answer 1

By assumption there are $M_{1},M_{2} \geq 0$ such that $|x_{n}| \leq M_{1}|\alpha_{n}|$ and $|y_{n}| \leq M_{2}|\alpha_{n}|$ for large $n$, so $|x_{n}y_{n}| \leq M_{1}M_{2}|\alpha_{n}|^{2}$ for large $n$. Since $\alpha_{n} = o(1)$ by assumption, we are done.