Let $H$ be an infinite dimensional separable Hilbert space.


**Definition**: The **commutant** $\mathcal{S}'$ of a subset $\mathcal{S} \subset B(H)$ is $ \{A \in B(H) : AB=BA \ , \ \forall B \in \mathcal{S} \} $.


**Definitions** : An operator $A \in B(H)$ is :


* **Irreducible** ([Halmos 1968](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-15/issue-2/Irreducible-operators/10.1307/mmj/1028999975.full)) if its commutant $\{ A\}'$ does not contain projections other than $0$ and $I$ ($A \ne A\_{1} \oplus A\_{2}$, $A$ generates $B(H)$ as von Neumann algebra : $\{A,A^{\*}\}''=B(H)$).
* **Nonnormal** if $\{ A\}'$ does not contain $A^{\*}$ (i.e. $AA^{\*} \ne A^{\*}A$).
* **Noncompact commuting** if $\{ A\}'$ does not contain a compact operator.


**Definition** : The **spectrum** $\sigma(A)$ of $A \in B(H)$ is $\{\lambda \in \mathbb{C} : A - \lambda I \text{ not bijective} \}$.  

It [decomposes](https://en.wikipedia.org/wiki/Decomposition_of_spectrum_(functional_analysis)) as follows:


* *Point spectrum*: $\sigma\_{p}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \text{ not injective} \}$
* *Continuous spectrum*: $\sigma\_{c}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \ \text{injective, dense nonclosed range} \}$
* *Residual spectrum*: $\sigma\_{r}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \ \text{ injective, nondense range} \}$


The spectrum of $A$ is **strictly continuous** if $\sigma(A) = \sigma\_{c}(A)$.


**Examples**:


* Let $S$ be the **bilateral shift** defined on $H = l^{2}(\mathbb{Z})$ by $S.e\_{n} = e\_{n+1} $.  

Its spectrum is *strictly continuous* : $\sigma(S) = \sigma\_{c}(S) = \mathbb{S}^{1}$.  

It's also a unitary operator ($SS^{\*} = S^{\*}S = I$), so a fortiori a *normal* operator.  

It is *noncompact commuting* and *reducible*.
* Let $T$ be the **unilateral shift** defined on $H = l^{2}(\mathbb{N})$ by $T.e\_{n} = e\_{n+1} $.  

Its spectrum is not *strictly continuous* because $0 \in \sigma\_{r}(T)$.  

It's a *nonnormal* operator because $[T^{\*},T].e\_{0} = e\_{0}$.  

It is *noncompact commuting* and *irreducible*.
* Let $V$ the **[Volterra operator](https://en.wikipedia.org/wiki/Volterra_operator)** defined on $H= L^{2}[0,1]$ by $(V.f)(t)=\int\_0^tf(x)dx$.  

Its spectrum is *strictly continuous* $\sigma(V) = \sigma\_{c}(V) = \{ 0\}$.  

It is *compact*, *irreducible* and *nonnormal* (see [here](https://math.stackexchange.com/questions/446466/is-there-a-nonnormal-operator-with-spectrum-strictly-continuous/)).
* Let $p$ be a non-constant polynomial (see [here](https://math.stackexchange.com/questions/447278/does-the-nontrivial-commutants-of-the-volterra-operator-admit-a-strictly-continu/)).  

Then $p(V)$ is *nonnormal*, *compact commuting* and *irreducible*.  

Its spectrum is *strictly continuous* $\sigma(p(V)) = \sigma\_{c}(p(V)) = \{ p(0)\}$.  

It's *compact commuting*, *nonnormal* and *irreducible*.
* Let $S \oplus V$ defined on $l^{2}(\mathbb{Z}) \oplus L^{2}[0,1]$.  

It is *reducible*, *compact commuting*, *nonnormal* and with spectrum *strictly continuous*.


*If you find a mistake, thank you let me know in comment.*



> 
> **The main question**: Is there an irreducible, noncompact commuting and nonnormal operator, with spectrum strictly continuous ?
> 
> 
> 


**Bonus questions** : How classify these operators ?



Is there an irreducible, noncompact commuting, nonnormal operator, with spectrum strictly continuous?
Essentially I think weighted shifts should be a sufficiently rich class of operators. Consider, for instance, the following example.


Take the doubly infinite sequence
$$ w\_k=\left\{\begin{array}{ll} 2 & \text{if } k<0 \\ 1 & \text{if } k\geq 0 \end{array}\right. $$
and let $W$ be the weighted shift on $\ell^2(\mathbb{Z})$ defined by
$$ We\_j=w\_je\_{j+1} \qquad \forall j\in\mathbb{Z}\qquad (\text{here }(e\_j)\text{ is the usual basis}).$$
Then


1. W has strictly continuous spectrum. This is an almost identical argument the usual bilateral shift.
2. W is not normal. The adjoint is given by $W^\ast e\_j=w\_{j-1}e\_{j-1}$ and so $$W^\ast We\_j=w\_j^2e\_{j},\quad \text{but}\quad WW^\ast e\_j = w\_{j-1}^2e\_j.$$
3. $W$ is irreducible. This follows from Corollary 2 of
R. Gellar, *Operators commuting with a weighted shift*, Proc. Amer. Math. Soc. **23** (1969), 538-545.
4. ~~I do not seem to be able to show~~ $W$ is non-compact-commuting. See Bill's comment below.


