Compute $\|(iI + \frac{1}{2}U)^{-1}\|$

Question

Let $U \in L(\ell^2(\mathbb{Z}))$ be the bilateral shift defined by $U\delta_k=\delta_{k+1}$ for $k\in \mathbb{N}$. Let $I$ be the identity operator of $\ell^2(\mathbb{Z})$ which maps $x \mapsto x$. How do we compute $\|(iI + \frac{1}{2}U)^{-1}\|$?

Answer 1

The spectrum of the bilateral shift is $\mathbb T$. You have the $U$ is normal. Multiplying by a scalar, adding a scalar, and taking the inverse, still give you a normal operator. Those three operations apply naturally to the spectrum. So $$ \|(iI+\tfrac12 U)^{-1}\|=\max\left\{\frac1{|i+\lambda/2|}:\ \lambda\in\mathbb T\right\} =\frac1{\min\{|i+\lambda/2|:\ \lambda\in \mathbb T\}}=\frac1{\min\{|i-\lambda/2|:\ \lambda\in \mathbb T\}}. $$ The last minimum is achieved at $\lambda=i$, since it can be seen as the distance from $i$ to the circle of radius $1/2$ centered at the origin. So the minimum is $|i-i/2|=1/2$ and the norm is $2$.