- #36
nuuskur
Science Advisor
- 858
- 914
Let [itex]K\subseteq \mathbb C[/itex] be non-empty compact. Suppose [itex]\{\gamma_n \in K \mid n\in\mathbb N\} \subseteq K[/itex] is dense in [itex]K\ \ [/itex] ([itex]\mathbb C[/itex] is separable). Let [itex]D : e_n\mapsto \gamma_ne_n[/itex] be a diagonal operator (in the infinite case [itex]H\cong \mathbb C^\omega[/itex], otherwise [itex]H\cong \mathbb C^N[/itex]).
For any [itex]n\in\mathbb N[/itex] we have [itex]\gamma_n\in\sigma (D)[/itex], because [itex]D-\gamma_n\mathrm{id}[/itex] is not injective. If we look at the kernel, we have
[tex]
\sum \lambda _k(D-z_n\mathrm{id})(e_k) = \sum _{k\neq n} (\lambda _k\gamma_k - \gamma _n)e_k + (\lambda_n -1)\gamma _n e_n = 0.
[/tex]
We can clearly see [itex]\lambda _k \equiv 0[/itex] is not forced.
Since [itex]\sigma (D)[/itex] is closed, we also have [itex]K\subseteq \sigma (D)[/itex]. On the other hand, suppose [itex]\lambda\notin K[/itex]. Then
[tex]F \left ( \sum \lambda _ke_k\right ) := \sum\frac{\lambda _k}{\gamma _k-\lambda}e_k[/tex]
is the inverse of [itex]D-\lambda \mathrm{id}[/itex], thus [itex]\lambda\notin \sigma (D)[/itex]. Indeed, we can check for basis elements. We have
[tex]
(F \circ (D-\lambda \mathrm{id}))(e_k) = F((\gamma _k-\lambda)e_k) = \frac{\gamma _k-\lambda}{\gamma _k-\lambda}e_k = e_k
[/tex]
and
[tex]
((D-\lambda \mathrm{id}) \circ F)(e_k) = (D-\lambda \mathrm{id}) \left ( \frac{1}{\gamma _k-\lambda}e_k \right ) = e_k.
[/tex]
Altogether [itex]\sigma (D) = K[/itex].
For any [itex]n\in\mathbb N[/itex] we have [itex]\gamma_n\in\sigma (D)[/itex], because [itex]D-\gamma_n\mathrm{id}[/itex] is not injective. If we look at the kernel, we have
[tex]
\sum \lambda _k(D-z_n\mathrm{id})(e_k) = \sum _{k\neq n} (\lambda _k\gamma_k - \gamma _n)e_k + (\lambda_n -1)\gamma _n e_n = 0.
[/tex]
We can clearly see [itex]\lambda _k \equiv 0[/itex] is not forced.
Since [itex]\sigma (D)[/itex] is closed, we also have [itex]K\subseteq \sigma (D)[/itex]. On the other hand, suppose [itex]\lambda\notin K[/itex]. Then
[tex]F \left ( \sum \lambda _ke_k\right ) := \sum\frac{\lambda _k}{\gamma _k-\lambda}e_k[/tex]
is the inverse of [itex]D-\lambda \mathrm{id}[/itex], thus [itex]\lambda\notin \sigma (D)[/itex]. Indeed, we can check for basis elements. We have
[tex]
(F \circ (D-\lambda \mathrm{id}))(e_k) = F((\gamma _k-\lambda)e_k) = \frac{\gamma _k-\lambda}{\gamma _k-\lambda}e_k = e_k
[/tex]
and
[tex]
((D-\lambda \mathrm{id}) \circ F)(e_k) = (D-\lambda \mathrm{id}) \left ( \frac{1}{\gamma _k-\lambda}e_k \right ) = e_k.
[/tex]
Altogether [itex]\sigma (D) = K[/itex].
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