# Spectrum decomposition of the scaling operator on weighted spaces

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.)

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 $$L^2(e^{-y^2}dy)$$.