(Question edited according to Denis Serre comment).

While studying the action of dilating map of the circle on probability measures, I ran across the following operator:
$$\mathcal{K}^* : L^2_0(\mu)\to L^2_0(\mu)\quad u\mapsto \varphi' u\circ \varphi $$
where $\varphi$ is a $C^2$ dilating map of $S^1$, of degree $d$ say, $\mu$ is the unique absolutely continuous measure fixed by $\varphi$ (it has $C^1$ density) and $L^2_0$ means the subset of $L^2$ of functions having zero mean *with respect to Lebesgue measure*.

In the model case $\varphi(x)=dx \mod 1$, Fourier analysis shows easily that the spectrum is the closed disc of radius $d$. In fact every $\lambda$ such that $|\lambda| < d$ is an eigenvalue of the adjoint operator $\mathcal{K}$. In the general case it seems easier to work with $\mathcal{K}^*$ since the expression of $\mathcal{K}$ involves $\mu$.

My question is the following: what can be said in general on the spectrum of $\mathcal{K}^*$? references are welcome if they exist (note that this question led me far away from my usual domain of mathematics, so that this might be a dumb or very easy question -- or might be tough).

eigenvalue, because if $u\in L^2_0(S^1)$ and ${\mathcal K}^*u=\lambda u$, then $u=0$. – Denis Serre Oct 21 '10 at 14:05with respect to the Lebesgue measure. This is why I ask about $\mathcal{K}^*$: the expression of $\mathcal{K}$ involves $\mu$. – Benoît Kloeckner Oct 21 '10 at 16:06