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(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).

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  • $\begingroup$ When $\phi(x)=dx$, it is correct that every $\lambda$ with $|\lambda|< d$ is in the spectrum, but it is not an eigenvalue, because if $u\in L^2_0(S^1)$ and ${\mathcal K}^*u=\lambda u$, then $u=0$. $\endgroup$
    – Denis Serre
    Commented Oct 21, 2010 at 14:05
  • $\begingroup$ @Denis: thanks for the correction, I mixed up things; in fact, these $\lambda$ are (in the model case) eigenvalues of the adjoint operator $\mathcal{K}$. By the way, in the general case, the space that comes up is in fact $L^2_0(\mu)$ where $\mu$ is the unique absolutely continuous measure fixed by $\varphi$ (which has $C^1$ density), and the $0$ means that $u$ has mean zero with respect to the Lebesgue measure. This is why I ask about $\mathcal{K}^*$: the expression of $\mathcal{K}$ involves $\mu$. $\endgroup$
    – Benoît Kloeckner
    Commented Oct 21, 2010 at 16:06
  • $\begingroup$ @Benoit: I'm just trying to understand why your map preserves $L^2_0$ - can you briefly explain? $\endgroup$
    – Łukasz Grabowski
    Commented Oct 21, 2010 at 18:18
  • $\begingroup$ @Lukasz Grabowski: $\varphi$ is a covering map, so you can divide the interval $[0,1)$ into $d$ intervals $I_i=[a_i,a_{i+A})$ such that the restriction $\varphi_i$ of $\varphi$ to $I_i$ is a homeomorphism to $[0,1)$. Then you can cut $\int \varphi' u\circ\varphi$ into $d$ integrals on the $I_i$, perform a change of variable using $\varphi_i$, and you get what you want. $\endgroup$
    – Benoît Kloeckner
    Commented Oct 22, 2010 at 12:14

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