Closed numerical ranges

Let $x=(x_n{)}_n$ be a bounded complex sequence and let $M_{x} = \sup_n |x_n|$\mathrm{.}Byusinganormaloidoperatorrelatedtothesequence$ x=(x_{n}){n} $,weprovethat$$ \sup{\lambda \in $\mathbb{C}$, |\lambda| \leq M_x} \sup_n |x_n+\lambda| = 2M_x.$$