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While recently studying spectral theory from the book of Davies (Spectral Theory and Differential Operators) I came along the notion of cyclicity for general self-adjoint (unbounded) operators. If is defined in the following way:

The operator $T$ on the Hilbert space $\mathcal{H}$ is called cyclic iff there exists $v \in \mathcal{H}$, such that $$\mathcal{H} = \mathrm{clos}\left(\mathrm{span}\{ (T-z)^{-1}v \quad\vert z\in \mathbb{C}\setminus\mathbb{R}\}\right)$$

It is then shown, that for finite dimensional operatos this coincides with the classical definition of cyclicity. Furthermore, Davies notes that a self-adjoint matrix is cyclic if and only if it has distinct eigenvalues. This may be easily seen via the companion matrix.

Now, my question: Is there a analogous result for general operators, such as: "A self-adjoint operator is cyclic iff all eigenvalues are of multiplicity one and the continuous/singular spectrum is of the form..."?

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  • $\begingroup$ Is this definition related to the one that for the $C^*$ algebra $\mathcal A$ generated by $T$ and $\Bbb 1$ one has $\mathcal A \cdot v$ is dense in $\mathcal H$? $\endgroup$ – s.harp Jul 3 '17 at 16:38
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Let $\mathcal{H}=L^2(\mathbb{R},\mu)$ where $\mu$ is a finite Lebesgue measure on $\mathbb{R}$. The operator $T : \mathcal{D}(T) \subseteq \mathcal{H}\rightarrow\mathcal{H}$ defined by $(Tf)(x) = xf(x)$ on the domain $\mathcal{D}(T)$ consisting of all $f\in L^2$ for which $xf \in L^2$ is a densely-defined selfadjoint linear operator with a cyclic vector $v(x)\equiv 1$. To see why this is true, suppose $f \in \mathcal{H}$ satisfies $$ ((T-\lambda I)^{-1}v,f)=0,\;\;\; \lambda\in\mathbb{C}\setminus\mathbb{R}. $$ Then $$ \int_{\mathbb{R}}\frac{1}{x-\lambda}f(x)d\mu(x) = 0,\;\;\lambda\notin\mathbb{R}, $$ and $$ 0=\int_{\mathbb{R}}\frac{1}{x-\lambda}-\frac{1}{x-\overline{\lambda}}f(x)d\mu(x) = \int_{\mathbb{R}}\frac{\Im\lambda}{|x-\lambda|^2}f(x)d\mu(x). $$ This is the Poisson integral of a measure $d\nu=fd\mu$ for which $$ \int_{\mathbb{R}}\frac{1}{1+x^2}d|\nu| < \infty. $$ From this it follows that $d\nu=0$, or $f=0$ a.e.. $[d\mu]$. Hence, $f=0$, which proves that $v\equiv 1$ is cyclic under $T$, using your stated definition of cyclic.

The finite-dimensional cases are special cases of such an operator $T$ as given above, where $\mu$ is a discrete measure with a finite number of atoms. The case you are suggesting is also a special case of this $T$. However, the operator $$ T : L^2[0,1]\rightarrow L^2[0,1],\;\; (Tf)(x)=xf(x), $$ is an example that does not fit the pattern you are suggesting because it has no eigenvalues--its spectrum is continuous and equal to $[0,1]$.

The example given here is essentially the model for the general case. The non-cyclic case can essentially be reduced to multiple orthogonal copies of model operators.

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