A homogeneous Banach space $B$ on group $\mathbb T=\mathbb R/2\pi\mathbb Z$ is a linear subspace $B$ of $L^1(\mathbb T)$ having a norm $\|\ \|_B\ge\|\ \|_{L^1}$ under which it is a Banach space, and having the following properties:
- If $f\in B$ and $\tau\in\mathbb T$, then $f_\tau\in B$ and $\|f_\tau\|_B=\|f\|_B$ (where $f_\tau=f(t-\tau)$).
- For all $f\in B\quad $, $\tau, \tau_0\in\mathbb T,\quad $$\lim_{\tau\to\tau_0}\|f_\tau-f_{\tau_0}\|=0.$
We consider a homogeneous Banach space $B$ on $\mathbb T$ and assume $e^{int}\in B$ for all $n$. Let $\mu\in B^\ast$ and define $$ \hat\mu(n)=\overline{\langle e^{int},\mu \rangle},\quad n\in\mathbb Z.$$ Let $\sigma_n(\mu)\in B^\ast$ be $$ \sigma_n(\mu)=\sum_{-n}^n\left(1-\frac{|j|}{n+1}\right)\hat\mu(j)e^{ijt}. $$ Then it is claimed that the linear operator $\Sigma_n$ on $B^\ast$ $$ \Sigma_n:\mu\mapsto\sigma_n(\mu) $$ has norm $$ \|\Sigma_n\|^{B^\ast}=1. $$
It is evident that $\Sigma_n$ is bounded, because: \begin{align} \|\Sigma_n\|^{B^\ast}&=\sup_{\|\mu\|_{B^\ast}=1}\left\| \sum_{-n}^n\left(1-\frac{|j|}{n+1}\right)\hat\mu(j)e^{ijt} \right\|_{B^\ast}\\ &=\sup_{\|\mu\|_{B^\ast}=1}\sup_{\|f\|_B=1}\left|\int_{\mathbb T}f\sum_{-n}^n\left(1-\frac{|j|}{n+1}\right)\langle e^{ijt},\mu \rangle e^{-ijt}dt\right|\\ &\le \sup_{\|f\|_B=1}\int_{\mathbb T}\left|f\right|\sum_{-n}^n\left(1-\frac{|j|}{n+1}\right)\|e^{ijt}\|_B dt\\ &\le \sup_{\|f\|_B=1}\|f\|_{L^1}\sum_{-n}^n\left(1-\frac{|j|}{n+1}\right)\|e^{ijt}\|_B \\ &\le \sup_{\|f\|_B=1}\|f\|_{B}\sum_{-n}^n\left(1-\frac{|j|}{n+1}\right)\|e^{ijt}\|_B \\ &= \sum_{-n}^n\left(1-\frac{|j|}{n+1}\right)\|e^{ijt}\|_B\\ &<\infty \end{align} But how can I prove that the norm of $\Sigma_n$ is $1$?
This is from page 37 of Yitzhak Katznelson's book AN INTRODUCTION TO HARMONIC ANALYSIS