$\sup_{k\geq0}|x_{n+k}-x_n|\leq\sup_{k,l\geq0}|x_{n+k}-x_{n+l}|\leq2\sup_{k\geq0}|x_{n+k}-x_n|$

Question

How to prove that $$\sup_{k\geq0}|x_{n+k}-x_n|\leq\sup_{k,l\geq0}|x_{n+k}-x_{n+l}|\leq2\sup_{k\geq0}|x_{n+k}-x_n|$$ for every real sequence $\{x_n\}$?

Answer 1

Let $n\in \Bbb N$. Since $\{|x_{n+k} - x_n| : k \geq 0\} \subset \{|x_{n+k} - x_{n+l}| : k,l \geq 0\}$ we have $$\sup_{k\geq0}|x_{n+k}-x_n|= \sup \{|x_{n+k} - x_n| : k \geq 0\}\\ \leq \sup \{|x_{n+k} - x_{n+l}| : k,l \geq 0\} = \sup_{k,l\geq0}|x_{n+k}-x_{n+l}|$$ Further, for every $k,l \geq 0$ we have $|x_{n+k} - x_{n+l}|\leq |x_{n+k} - x_n| + |x_{n+l} - x_n|$, thus $$\sup_{k,l\geq 0} |x_{n+k} - x_{n+l}| \leq \sup_{k,l\geq 0} |x_{n+k} - x_n| + |x_{n+l} - x_n| \leq \sup_{k,l\geq 0} |x_{n+k} - x_n| + \sup_{k,l\geq 0} |x_{n+l} - x_n|\\ = \sup_{k\geq 0} |x_{n+k} - x_n| + \sup_{l\geq 0} |x_{n+l} - x_n|= \sup_{k\geq 0} |x_{n+k} - x_n| + \sup_{k\geq 0} |x_{n+k} - x_n|\\ = 2 \sup_{k\geq 0} |x_{n+k} - x_n|$$