$\sup_{m \geq n} \, \operatorname{ess sup} |u_n - u_m| = \operatorname{ess sup} \, \sup_{m \geq n} |u_n - u_m|$

78 Views Asked by At

Let $u_n, u \in L^{\infty}(E)$ and let $u_n \rightarrow u$ almost everywhere. Is it true that $$ \DeclareMathOperator{\esssup}{ess sup} \sup_{m \geq n} \, \esssup_E |u_n - u_m| = \esssup_E \, \sup_{m \geq n} |u_n - u_m| \qquad ?$$

1

There are 1 best solutions below

0
On

No. Take $u_n(t)=\begin{cases}0&\text{, if }t\neq n\\1&\text{, if }t=n\end{cases}$.