Consider the bounded linear operator $M\_a$ defined by $M\_au(x)=\frac{1}{\sqrt{a}}u\left(\frac{x}{a}\right)$, for $a>1$. On $L^2(\mathbb{R})$, it is easy to see that this is a unitary operator and that (either directly or by an application of Stone's theorem to the continuous one-parameter group) it has spectrum the entire unit circle:$$\sigma(M\_a)=\sigma(M\_a^\*)=\{|z|=1\}.$$ Furthermore, this spectrum is all continuous spectrum, as it is not difficult to show that there are no eigenvalues.


My question concerns what happens when you consider this operator on a weighted space, such as $L^2(\mathbb{R},e^{-x^2}dx)$ for example. Here the operator $M\_a$ is still bounded but is no longer normal. We may compute the spectrum by computing the norm $$\|M\_a\|=\sqrt{\|M\_a^\*M\_a\|}=\sqrt{\sup\_{x\in\mathbb{R}}e^{-(a^2-1)x^2}}=1$$ and remarking that $\{|z|<1\}$ are eigenvalues corresponding to eigenvectors of the form $x^s$.


That leaves the boundary $\{|z|=1\}$. Is this continuous or residual spectrum? And what are the $\sigma\_r$ and $\sigma\_c$ of $M\_a^\*$?


(Apologies if this is obvious, I am new to spectral theory.)



Spectrum decomposition of the scaling operator on weighted spaces
It is continuous spectrum, since the adjoint has no eigenvalues. In fact $M\_a^\*v(y)=\sqrt {a} v(ay) e^{(1-a^2)y^2}$ (duality with respct to the measure $e^{-y^2} dy$) and assuming $M\_a^\* v=\lambda v$ we get the functional equation
$$\sqrt av(ay)e^{(1-a^2)y^2}=\lambda v(y) \quad {\rm or}
\quad v(ay)e^{-a^2y^2}=\mu v(y)e^{-y^2}
$$
with $\nu=\lambda/\sqrt a$. This equation gives $v(y)e^{-y^2}=|y|^s w(y)$ where $a^s=\mu$ and $w$ satisfies $w(ay)=w(y)$. Then $w(y)=h(\log y)$ for $y>0$ with $h$ periodic of period $\log a$ and then $v(y)=|y|^s e^{y^2}w(y)$ is never in $L^2(e^{-y^2}dy)$.


