# Does the nontrivial commutants of the Volterra operator admit a strictly continuous spectrum?

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

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: $\sigma_{r}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \ \text{ injective with a nondense range} \}$

Definition : Let $H= L^{2}[0,1]$ and $V \in B(H)$ the Volterra operator defined as follows : $$(V.f)(t)=\int_0^tf(x)dx$$

Some properties of the Volterra operator $V$ :

• $V$ is compact, its spectrum $\sigma(V)=\{0\}$, its norm $\Vert V \Vert = 2/\pi$.

• $V$ is nonnormal ($VV^{*} \ne V^{*}V$) with spectrum strictly continuous ($\sigma(V) = \sigma_{c}(V)$), see here.

• The closed invariant subspaces of $V$ are $L^{2}[a,1]$, with $a \in [0,1]$ see Barria 1981.

• The commutant $\{ V \}'$ of $V$ is the strongly closed algebra generated by $V$, see Erdos 1982.

Remark : $\forall \lambda \in \mathbb{C}$, $V_{\lambda} := V+\lambda I \in \{ V \}'$ and $\sigma(V_{\lambda}) = \sigma_{c}(V_{\lambda}) = \{ \lambda \}$

Definition : Let $S=\{\lambda_{1}, ... ,\lambda_{r} \}$ be a finite subset of $\mathbb{C}$ and let $V_{S} \in \{ V \}'$ defined as follows: $$V_{S} := (V+\lambda_{1} I).(V+\lambda_{2} I)...(V+\lambda_{r} I)$$

Preliminary questions :

• What is the spectrum of $V_{S}$ ( $S$ or $\{ \prod \lambda_{i} \}$ or anything else) ?
• Is it true that $\sigma(V_{S}) = \sigma_{c}(V_{S})$ ?

Main question :

Does a nontrivial commutant operator of the Volterra operator admits a strictly continuous spectrum (i.e. $\mathbb{C} I \not\ni A \in \{ V \}' \Rightarrow \sigma(A) = \sigma_{c}(A)$) ?

Let me answer the preliminary questions.

1) For commuting bounded operators $A,B$ we have $r(A+B)\leq r(A)+r(B)$ and $r(AB)\leq r(A)r(B),$ where $r$ is the spectral radius. Hence $r(p(V))=0$ for every polynomial $p$ such that $p(0)=0.$ It implies that $\sigma(V_S)=\prod\lambda_i.$

2) It is true.

For any polynomial $p\not\equiv 0,$ the operator $p(V)$ has zero kernel. Indeed, let $p(x)=x^n+a_1 x^{n-1}+\dots+a_n$ and assume that $p(V)\varphi=0.$ Then $Vq(V)\varphi=-a_n\varphi,$ where $p(x)=xq(x)+a_n.$ Since $r(Vq(V))=0,$ we have $a_n=0.$ Since the kernal of $V$ is zero, $q(V)\varphi=0.$ Polynomial $q$ has degree $n-1$ so we can use induction.

Hence $p(V)$ has no eigenvalues for $p\not\equiv const$.

The adjoint to $V$ is $(V^*f)(y) = \int_y^1 \! f(t) \, dt.$ Hence $p(V^*)$ has zero kernel for $p\not\equiv 0$ as well. The formula $Ran\, p(V)^\perp=\ker(\overline p(V^*))$ implies that the range of $p(V)$ is dense for every $p\not\equiv 0,$ that is, residual spectrum of $p(V),$ is empty for all $p.$

As for the main question, I believe that norm-limits of $p_n(V)\in\{V\}'$ satisfy the property. But the strong limits of $p_n(V)$ probably not.

• Because of the "probably not", I still can't accept this answer. @YuriiSavchuk, can you improve this point ? – Sebastien Palcoux Aug 6 '13 at 10:14