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][1]) 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][2] 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][3]** 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][4]).
- Let $p$ be a non-constant polynomial (see [here][5]).   
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 ?


  [1]: http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.mmj/1028999975
  [2]: http://en.wikipedia.org/wiki/Decomposition_of_spectrum_%28functional_analysis%29
  [3]: http://en.wikipedia.org/wiki/Volterra_operator
  [4]: https://math.stackexchange.com/questions/446466/is-there-a-nonnormal-operator-with-spectrum-strictly-continuous/
  [5]: https://math.stackexchange.com/questions/447278/does-the-nontrivial-commutants-of-the-volterra-operator-admit-a-strictly-continu/