Why is $\sup_{n \in \mathbb{N}}\Big|\frac{n-1}{n}z_n\Big|\le\sup_{n \in \mathbb{N}}\Big|\frac{n-1}{n}\Big|\sup_{n \in \mathbb{N}}\Big|z_n\Big|$

Question

$\sup_{n \in \mathbb{N}}\Big|\frac{n-1}{n}z_n\Big|\le\sup_{n \in \mathbb{N}}\Big|\frac{n-1}{n}\Big|\sup_{n \in \mathbb{N}}\Big|z_n\Big|$

when $z_n$ is a bounded sequence $\in \mathbb{C}$.

Answer 1

Let $a_n:=\frac{n+1}{n}$ and $c_n:=a_n|z_n|$. We want to estimate $\sup c_n$. Now, since $a_n\ge 0$,
$$ c_n\le (\sup_m a_m)|z_n|\le (\sup_m a_m)(\sup_k |z_k|), $$ so taking the $\sup$ of both sides yields $$ \sup_n c_n \le (\sup_m a_m)(\sup_k |z_k|),$$ as we wanted.