# Is there a nonnormal operator with spectrum strictly continuous?

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 ?

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.
• The set $\{f \in C^1[0,1]: f(0) = 0\}$ is still dense in $L^2[0,1]$, even the continuous compactly supported functions are dense. – user38355 Jul 18 '13 at 14:49
• Ok, you are right ! Here is a more conventional approach: let $1$, $s_{n} = sin(2\pi nx)$ and $c_{n} = cos(2\pi nx)$ (with $n \in \mathbb{N}^{*}$) be an orthonormal basis of $H = L^{2}[0,1]$. Then $V.c_{n} = s_{n}/2\pi n$, $V.s_{n} = (1-c_{n})/2\pi n$ and $V.1 = x= 1/2 + \sum s_{n}/2\pi n$. Now $2V.(1-c_{1}-...-c_{r}) \to 1$. So $1 \in \overline{Im(V)}$, and finally $V$ is dense range. – Sebastien Palcoux Jul 18 '13 at 22:21