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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.