# Norm of the resolvent of bounded operator in Hilbert space

Consider the shift operator $$T (x) = (0, x_1,x_2, ... ),\ \parallel T \parallel =1$$ where $x\in l^2,\ x_i\in {\bf C}$ and $$\parallel x \parallel^2 = \sum_{j=1}^\infty x_j\overline{x_j}$$

Then we can calculate the resolvent of $T-\lambda I\ (\lambda \neq 0)$ :

If $\lambda =0$, then $$T^{-1}(y)=(y_2,y_3, ... ),\ \parallel T^{-1} \parallel =1$$

If $\lambda \neq 0$, then $$T_\lambda^{-1} y = (f_1(y_1),f_2(y_1,y_2), ... , f_n(y_1, ... , y_n),...),\ f_1(y_1)=-\frac{1}{\lambda} y_1,\ f_2=-\frac{1}{\lambda}(y_2-f_1(y_1)),\ ...,\ f_n=- \frac{1}{\lambda} (y_n-f_{n-1})$$

Here $T_\lambda^{-1}$ is bounded ?

• Normally the notation $T^{-1}$ would be reserved for a two-sided inverse of $T$-notice that your "$T^{-1}$" does have $T^{-1}T=1,$ but not so $TT^{-1}$. Commented Nov 19, 2013 at 2:49
• You're right so that I correct the terminology. Commented Nov 19, 2013 at 2:52

You can easily find point spectrum of the left shift operator it is open unit ball of $\mathbb{C}$. Since norm of left shift operator is $1$, then its spectrum contained in the closed unit ball of $\mathbb{C}$. Thus spectrum of the left shift operator is closed, contains open unit ball and conained inthe closed unit ball. Therefore the desired spectrum equals to the closed unit ball of $\mathbb{C}$. Recall that the right shift operator is adjoint of left shift and spectra of operator and its adjoint always coincide. Therefore spectrum of the right shift opertor is closed unit ball of $\mathbb{C}$. In other words $T_\lambda^{-1}$ exists and bounded iff $|\lambda|> 1$.
• Thank you for your explanation and interesting note. But I have question in Example 4. For irrational $\alpha$, $\alpha$-Rotation has the set of eigenvalues which is dense in unit circle. In further it is compact operator. As far as I know, only possible accumulation point of compact opertor is $0$. What is wrong ? Commented Nov 21, 2013 at 15:29