I am reading the following paper: http://arxiv.org/pdf/1602.03849v2.pdf
I am struggling with one part. The general setup is this:
Let $x\in X=\mathbb{T}^d$, where $\mathbb{T}^d$ is the d dimensional torus. Let $\rho$ be a Borel prob. measure on SL$_d(\mathbb{Z})$.
For $f$ Borel on $X$, $f\geq 0$, and $x\in X$, define the Markov operator $Pf(x)=\int_{SL_d(\mathbb{Z})}f(gx)d\rho(g).$
Under some assumptions on $\rho$, we have that $P$ has a spectral gap in $L^2(X,\nu)$ where $\nu$ is the Lebesgue measure on $X$. Moreover, it is easy to see that $\int_X Pfd\nu = \int_X fd\nu$.
What I am struggling to show is the following:
For any $f\in L^2(X,\nu)$, there exists $g\in L^2(X,\nu)$ such that $f=g-Pg+\int fd\nu$.
So far I have taken then obvious choice for $g=\sum\limits_{n\in\mathbb{N}} P^nf$ (assuming $\int fd\nu=0$, and using the spectral gap to expand $g=(I-P)^{-1}f$). I am however struggling to show that this is in $L^2$. Could anyone please help me? Best wishes, Ian.
Does not the statement that "$P$ has a spectral gap in $L^2(X,\nu)$" mean that there is a constant $\gamma\in(0,1)$ such that if $f\in L^2(X,\nu)$ with $\int f\,d\nu=0$ then $\|Pf\|_2\le\gamma \|f\|_2$? If this is so then $ \|\sum_{n=0}^\infty P^nf\|_2\le\|f\|_2\sum_{n=0}^\infty\gamma^n=\|f\|_2/(1-\gamma)$ and your suggested $g$ is indeed an element of $L^2(X,\nu)$.