TITLE: Is there a nonnormal operator with spectrum strictly continuous?
QUESTION [8 upvotes]: Let $H$ be an infinite dimensional separable Hilbert space.  
Definition : An operator $A \in B(H)$ is normal if $AA^{*} = A^{*}A$.
Definition : The spectrum $\sigma(A)$ of $A \in B(H)$, is the set of all $\lambda \in \mathbb{C}$ such that $A - \lambda I$ is not bijective.
It decomposes as follows:
 - Point spectrum: $\sigma_{p}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \text{ not injective}   \}$
 - Continuous spectrum: $\sigma_{c}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \ \text{ injective with a dense nonclosed range}   \}$
 - Residual spectrum: $\sigma_{r}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \ \text{ injective with a nondense range}   \}$  
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.  
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 $[T^{*},T].e_{0} = e_{0}$.    


Is there a nonnormal operator with spectrum strictly continuous ?

Bonus questions : Can we exclude the compact operators ? How classify these operators ?

REPLY [4 votes]: Lets try the Volterra operator $(Vf)(t)=\int_0^tf(x)dx$ on $H=L^2[0,1].$ It is quasinilpotent (i.e. spectral radius is 0) so $\sigma(V)=\{0\}.$ Its range contains the set $\{f\in C^1[0,1]\mid f(0)=0\}$ which is dense. (To see this, note that $B=\{\sin\pi kx,\ k\in\mathbb N\}$ is an orthogonal base in $L^2[0,1].$) It means that $0$ belongs to continuous spectrum.