I assume that $(X_k)_{k≥0}$ is a uniformly ergodic Markov chain on $X$ with transition density $q$ and stationary distribution $π$. Then for all $n ≥ 1$, let the $n$-step transition density $q^n$ be defined as
$q^n(x_0, x_{n}) = \int q^{n-1}(x_0, x_{n-1}) q(x_{n-1}, x_n) dx_{n-1}$
I then let $\phi$ be a bounded measurable function on $X$ and I then wish to show that the series
$\hat\phi(x) = \sum_{n=0}^\infty (\int \phi(x_n) q^n(x, x_n) dx_n - \pi(\phi))$
with, by convention, $q^0(x,x_n) = \delta_x(dx_0)$, is convergent for all $x \in X$. I would also like to show that $\hat\phi$ is a bounded solution of the Poisson equation associated to $\phi$, i.e. for all $x \in X$,
$\hat\phi(x) - \int q(x, z) \hat\phi(z) dz = \phi(x) - \pi(\phi)$.
I am not really sure how to tackle these problems, so any help would really be appreciated. I thought about using a theorem that states; "Assume that $(X_k)_{k≥0}$ is a uniformly ergodic Markov chain and that $\phi$ is bounded. Then there exist a bounded solution $\hat\phi$ of the Poisson equation associated to $\phi$."
Thank you!
To start, I think you need some fact that uniform ergodicity means some kind of exponential convergence on transition probabilities to the stationary distribution. Using this case you can show convergence of the series. I'm pretty sure some theorem from Meyn and Tweedie would come handy here.
Once convergence is shown, you should be able to insert the definition of $\hat \phi$ into the Poisson equation, and check whether you get what you want.