How to find operator norm of shift operator Shift operator is $S:\ell^2(\mathbb{Z}) \to \ell^2(\mathbb{Z})$ defined by
$$
(Sa)_n = a_{n-1}
$$
for $a = (a_n)_{n = 1}^\infty \in \ell^2(\mathbb{Z})$.
Let $I$ be the identity operator, then How to find the norm of $I-S-S^2$ ?
What I know is following:

*

*$\|I-S-S^2\| \leq \|I\| + \|S\| + \|S\|^2 = 3$

*$\|I-S-S^2\| \geq \sqrt{11/3}$ because of $(\cdots, 0, 1,1,-1, 0,\cdots)$
 A: $S$ is normal (it's a unitary), so $I-S-S^2$ is also normal. In particular,
$$
\|I-S-S^2\|=\max\big\{|\lambda|:\ \lambda\in\sigma(I-S-S^2)\big\}. 
$$
It is well known that $\sigma(S)=\mathbb T$. By the Spectral Mapping Theorem,
$$
\sigma(I-S-S^2)=\big\{1-\lambda-\lambda^2:\ \lambda\in\mathbb T\big\}. 
$$
Thus
$$
\|I-S-S^2\|=\max\big\{|1-\lambda-\lambda^2|:\ \lambda\in\mathbb T\big\}. 
$$
We may rewrite this as
$$
\|I-S-S^2\|=\max\big\{|1-e^{it}-e^{2it}|:\ t\in[-\pi,\pi]\big\}. 
$$
As Ryszard Szwarc noted, $1-e^{it}-e^{2it}=-e^{it}\,(1+e^{it}-e^{-it})$, so $$|1-e^{it}-e^{2it}|=|1+2i\sin t|=(1+4\sin^2t)^{1/2}.$$ Thus
$$
\|I-S-S^2\|=\sqrt5.
$$

An alternative way to the operator theory part is to recognize that $S$ is multiplication by $\overline z$ on $C(\mathbb T)$. Then one is looking at norm of the multiplication operator by $1-e^{-it}-e^{-2it}$, and the norm of a multiplication operator is the infinity norm of the function. Which leads to the same estimate $\max\{|1-e^{it}-e^{-2it}|:\ t\}$.
