is the variance of a test function of a markov chain always increasing?

Question

Edits: Changed function to eigenfunction. I should have stated the problem with more explicit conditions. Anyways I realized the original formulation is not true, even when one starts at a single state: take the function on the $\mathbb{Z}$ with $f(1) =1$, $f(-1) = -1$, $f(0)= 0$, and $f(x) = \text{sgn}(x) \epsilon$ where $\epsilon$ is very small, and the simple random walk on $\mathbb{Z}$, then it will have the highest variance at $t=1$. One can easily adapt this example to the simple random walk on the $n$-cycle.

Given a function $f: \Omega \to \mathbb{R}$, where $\Omega$ is the state space of an ergodic finite state Markov chain, and let the chain start at a single state $x \in \Omega$. Assume $f$ is an eigenfunction of the chain. Is it true that $\mathbb{E}_t (f- \mathbb{E}_t f)^2$ is nondecreasing in $t$? Here $\mathbb{E}_t f$ denotes $\mathbb{E} P_t f$ where $P_t$ is the Markov transition kernel from time $0$ to time $t$. Note it's important to start with the point mass distribution at a single state, since otherwise one could choose an initial distribution that has an $f$-variance larger than the stationary (as pointed out by one of the commenters below).

Wilson's method gives a way to bound the variance of eigenfunctions in $t$ in a way that's reminiscent of the Martingale difference method, but since the variance at time $\infty$ is usually easy to calculate, if we know the above monotonicity result, we could bound the variance at finite time easily.

Answer 1

I can give you the answer for a certain class of Markov Chains, i.e. for reversible, irreducible and aperiodic Markov Chains on a finite state space.

Then the question is related to the spectral gap of the transition matrix $P:\Omega\times\Omega\to [0,1]$ of your chain. The transition matrix entry $P(x,y)$ is the probability of going from state $x$ to $y$. Thus one has $\sum_{y\in \Omega} P(x,y)=1$ If your chain is irreducible ($\forall x,y\in \Omega \exists n\in \mathbb{N}: P^n(x,y)>0$) and aperiodic (means that $\forall x\in \Omega$ the period defined by the greatest common divisor of the set $T(x)=\{n\in \mathbb{N}: P^n(x,x)>0\}$ is $1$) then one has the following Theorem:

If $\Omega$ is finite and the chain defined by the transition matrix $P$ is irreducible and aperiodic than there exists a unique stationary distribution $\pi$ such that $\pi P =\pi$.

A chain satisfying a detailed balance relation $\pi(x) P(x,y) =\pi(y) P(y,x)$ is reversible.

Under these conditions the eigenvalues of $P$ are bounded in modulus by $1$ and the largest is $1$, the corresponding left eigenvector is $\pi$. Then the spectral gap is defined by $\gamma=1-\max_{\lambda\ne 1} |\lambda| $. Note that $0<\gamma<1$. Then it yields the following estimate for the variance of a test function $f:\Omega\to\mathbb{R}$ with respect to the stationary measure $\pi$

${\mathrm{Var}}_\pi(P^n f) \leq (1-\gamma)^{2t} {\mathrm{Var}}_\pi(f)$