Closed numerical ranges

Abstract

Let $x=(x_n{)}_n$ be a bounded complex sequence and let $M_x={\sup }_n|x_n|$. By using a normaloid operator related to the sequence $x=(x_n{)}_n$, we prove that $$\underset{\lambda \in \mathbb{C},|\lambda |\le M_x}{\sup }\underset{n}{\sup }|x_n+\lambda |=2M_x.$$