Let $l_2(a)$ be a hilbert space defined with following inner product:
$\langle x_n,y_n\rangle = \sum a^k x_k y_k$. (It's a weighted sequence space with the weights $\omega_i = a^i$).
It's elements are the sequences for which the norm is defined and finite. ( $\langle x_n,x_n \rangle = c ).$
The left and right shift operators are defined as usual: $S_r(({x_0 , x_1, x_2, … x_n , …})) = (0, x_0, …)$
$S_l(({x_0 , x_1, x_2, … x_n , …})) = (x_1, x_2, …)$
Now, I was asked to find $\sigma(S_r), \sigma(S_l)$. My attempt was this:
I found the operator norm of both operators. $\lvert S_r\rvert = a$, $\lvert S_l\rvert = 1/a$.
Then, I figured that $S_l$ has eigenvalues for every $\lambda$ satisfying $\lvert \lambda \rvert < 1/a$. Because the specturm is closed and bounded by the operator norm, i figured that $\sigma(S_l)$ is the closed ball with radius $1/a$.
Additionaly, i found that $S_r ^ * = aS_l$, and therefore $Im(S_r – \lambda I)^\bot = Ker(aS_l – \bar{\lambda} I) = Ker(a(S_l – (\bar{\lambda}/a) I)) = Ker(S_l – (\bar{\lambda} / a) I) \neq 0 \Longleftrightarrow \bar{\lambda} / a < 1/a$.
So i deduced that if $\lvert\lambda\rvert < 1$ so $\lambda \in \sigma(S_r)$. which doesn't make sense to me if $a < 1$ and then the spectrum has to be bounded by $\lvert S_r \rvert = a$.
Thanks alot.
Best Answer
There is a natural map $U : \ell^2_{a}\rightarrow \ell^2$ given by $$ U(x_0,x_1,x_2,\cdots) = (a^{0/2}x_0,a^{1/2}x_1,a^{2/2}x_2,a^{3/2}x_3,\cdots) $$ $U$ is a unitary map between the two Hilbert spaces. If $S_a$ is the shift operator on $\ell^2_a$, then $S_a$ may be pulled back to $\ell^2$ by looking at $T : \ell^2 \rightarrow \ell^2$ defined by $$ T = US_aU^{-1}, $$ which is described by \begin{align} (x_0,x_1,x_2,\cdots) & \mapsto (x_0,a^{-1/2}x_1,a^{-2/2}x_2,\cdots) \\ & \mapsto (0,x_0,a^{-1/2}x_1,a^{-2/2}x_2,\cdots) \\ & \mapsto (0,a^{1/2}x_0,a^{2/2}a^{-1/2}x_1,a^{3/2}a^{-2/2}x_2,\cdots) \\ & = \sqrt{a}(0,x_0,x_1,x_2,\cdots). \end{align} That is, $T=\sqrt{a}S$, where $S$ is the ordinary shift operator on $\ell^2$.