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You wrote:

I can verify that $\Psi$ is continuously differentiable, $\Psi(t,\tau)>0$ for all $t,\tau\in\mathbb{R}$, and of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$.

[...] these properties should be sufficient to guarantee that a stationary distribution $\pi$ exists and is unique, and that $f_k \to \pi$ (in the T.V. sense) for any initial $f_0$.

Of course, this is not so. E.g., if $\Psi(t,s)=g(t-s)$, where $g$ is the standard normal pdf, then (considering, for instance the Fourier transform, one can easily see that) there is no stationary distribution. Also, then for any initial $f_0$ and each real $t$ we have $f_k(t)\to0$ as $k\to\infty$.

You have now added more conditions:

let's suppose for the moment that the Markov chain is irreducible, positive recurrent, aperiodic, and Harris. And of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$

saying then the following:

These properties should be sufficient to guarantee that a stationary distribution $\pi$ exists and is unique, and that $f_k \to \pi$ (in the T.V. sense) for any initial $f_0$. Moreover, all moments of $\pi$ are finite and the $m^\text{th}$ moment of $f_k$ converges to the $m^\text{th}$ moment of $\pi$ as $k\to\infty$.

However, the latter conclusion will still fail to hold in general -- because the the state space of the chain can be nonlinearly transformed in an arbitrary manner.

More specifically, suppose (say) that the support set of the stationary distribution $\pi$ of an (irreducible positive recurrent aperiodic Harris) Markov chain $(X_k)$ is not bounded from above, so that $$G(x):=\pi\big((x,\infty)\big)>0$$ for all real $x$. Let then $$Y_k:=f(X_k),$$ where $$f(x):=\int_0^x\frac{du}{G(u)}$$ for real $x$, with $\int_0^x:=-\int_x^0$ for real $x<0$. Then $(Y_k)$ is an (irreducible positive recurrent aperiodic Harris) Markov chain with stationary distribution $\pi_f:=\pi f^{-1}$, the pushforward of $\pi$ under the map $f$. Moreover, \begin{align} \int_{[0,\infty)}y\,\pi_f(dy)&=\int_{[0,\infty)}f(x)\,\pi(dx) \\ &=\int_{[0,\infty)}\pi(dx)\,\int_0^x\frac{du}{G(u)} \\ &=\int_0^\infty\frac{du}{G(u)}\,\int_{(u,\infty)} \pi(dx) \\ &=\int_{[0,\infty)}\frac{du}{G(u)}\,G(u)=\infty. \end{align} So, the first moment of $\pi_f$ cannot be finite.

Similarly one can deal with the case when the support set of the stationary distribution $\pi$ has a finite limit point.

You wrote:

I can verify that $\Psi$ is continuously differentiable, $\Psi(t,\tau)>0$ for all $t,\tau\in\mathbb{R}$, and of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$.

[...] these properties should be sufficient to guarantee that a stationary distribution $\pi$ exists and is unique, and that $f_k \to \pi$ (in the T.V. sense) for any initial $f_0$.

Of course, this is not so. E.g., if $\Psi(t,s)=g(t-s)$, where $g$ is the standard normal pdf, then (considering, for instance the Fourier transform, one can easily see that) there is no stationary distribution. Also, then for any initial $f_0$ and each real $t$ we have $f_k(t)\to0$ as $k\to\infty$.

You have now added more conditions:

let's suppose for the moment that the Markov chain is irreducible, positive recurrent, aperiodic, and Harris. And of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$

saying then the following:

These properties should be sufficient to guarantee that a stationary distribution $\pi$ exists and is unique, and that $f_k \to \pi$ (in the T.V. sense) for any initial $f_0$. Moreover, all moments of $\pi$ are finite and the $m^\text{th}$ moment of $f_k$ converges to the $m^\text{th}$ moment of $\pi$ as $k\to\infty$.

However, the latter conclusion will still fail to hold in general -- because the the state space of the chain can be nonlinearly transformed in an arbitrary manner.

More specifically, suppose (say) that the support set of the stationary distribution $\pi$ of an (irreducible positive recurrent aperiodic Harris) Markov chain $(X_k)$ is not bounded from above, so that $$G(x):=\pi\big((x,\infty)\big)>0$$ for all real $x$. Let then $$Y_k:=f(X_k),$$ where $$f(x):=\int_0^x\frac{du}{G(u)}$$ for real $x$, with $\int_0^x:=-\int_x^0$ for real $x<0$. Then $(Y_k)$ is an (irreducible positive recurrent aperiodic Harris) Markov chain with stationary distribution $\pi_f:=\pi f^{-1}$, the pushforward of $\pi$ under the map $f$. Moreover, \begin{align} \int_{[0,\infty)}y\,\pi_f(dy)&=\int_{[0,\infty)}f(x)\,\pi(dx) \\ &=\int_{[0,\infty)}\pi(dx)\,\int_0^x\frac{du}{G(u)} \\ &=\int_0^\infty\frac{du}{G(u)}\,\int_{(u,\infty)} \pi(dx) \\ &=\int_0^\infty\frac{du}{G(u)}\,G(u)=\infty. \end{align} So, the first moment of $\pi_f$ cannot be finite.

Similarly one can deal with the case when the support set of the stationary distribution $\pi$ has a finite limit point.

You wrote:

I can verify that $\Psi$ is continuously differentiable, $\Psi(t,\tau)>0$ for all $t,\tau\in\mathbb{R}$, and of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$.

[...] these properties should be sufficient to guarantee that a stationary distribution $\pi$ exists and is unique, and that $f_k \to \pi$ (in the T.V. sense) for any initial $f_0$.

Of course, this is not so. E.g., if $\Psi(t,s)=g(t-s)$, where $g$ is the standard normal pdf, then (considering, for instance the Fourier transform, one can easily see that) there is no stationary distribution. Also, then for any initial $f_0$ and each real $t$ we have $f_k(t)\to0$ as $k\to\infty$.

You wrote:

I can verify that $\Psi$ is continuously differentiable, $\Psi(t,\tau)>0$ for all $t,\tau\in\mathbb{R}$, and of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$.

[...] these properties should be sufficient to guarantee that a stationary distribution $\pi$ exists and is unique, and that $f_k \to \pi$ (in the T.V. sense) for any initial $f_0$.

Of course, this is not so. E.g., if $\Psi(t,s)=g(t-s)$, where $g$ is the standard normal pdf, then (considering, for instance the Fourier transform, one can easily see that) there is no stationary distribution. Also, then for any initial $f_0$ and each real $t$ we have $f_k(t)\to0$ as $k\to\infty$.

You have now added more conditions:

let's suppose for the moment that the Markov chain is irreducible, positive recurrent, aperiodic, and Harris. And of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$

saying then the following:

These properties should be sufficient to guarantee that a stationary distribution $\pi$ exists and is unique, and that $f_k \to \pi$ (in the T.V. sense) for any initial $f_0$. Moreover, all moments of $\pi$ are finite and the $m^\text{th}$ moment of $f_k$ converges to the $m^\text{th}$ moment of $\pi$ as $k\to\infty$.

However, the latter conclusion will still fail to hold in general -- because the the state space of the chain can be nonlinearly transformed in an arbitrary manner.

More specifically, suppose (say) that the support set of the stationary distribution $\pi$ of an (irreducible positive recurrent aperiodic Harris) Markov chain $(X_k)$ is not bounded from above, so that $$G(x):=\pi\big((x,\infty)\big)>0$$ for all real $x$. Let then $$Y_k:=f(X_k),$$ where $$f(x):=\int_0^x\frac{du}{G(u)}$$ for real $x$, with $\int_0^x:=-\int_x^0$ for real $x<0$. Then $(Y_k)$ is an (irreducible positive recurrent aperiodic Harris) Markov chain with stationary distribution $\pi_f:=\pi f^{-1}$, the pushforward of $\pi$ under the map $f$. Moreover, \begin{align} \int_{[0,\infty)}y\,\pi_f(dy)&=\int_{[0,\infty)}f(x)\,\pi(dx) \\ &=\int_{[0,\infty)}\pi(dx)\,\int_0^x\frac{du}{G(u)} \\ &=\int_0^\infty\frac{du}{G(u)}\,\int_{(u,\infty)} \pi(dx) \\ &=\int_{[0,\infty)}\frac{du}{G(u)}\,G(u)=\infty. \end{align} So, the first moment of $\pi_f$ cannot be finite.

Similarly one can deal with the case when the support set of the stationary distribution $\pi$ has a finite limit point.

You wrote:

I can verify that $\Psi$ is continuously differentiable, $\Psi(t,\tau)>0$ for all $t,\tau\in\mathbb{R}$, and of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$.

[...] these properties should be sufficient to guarantee that a stationary distribution $\pi$ exists and is unique, and that $f_k \to \pi$ (in the T.V. sense) for any initial $f_0$.

Of course, this is not so. E.g., if $\Psi(t,s)=g(t-s)$, where $g$ is the standard normal pdf, then (considering, for instance the Fourier transform, one can easily see that) there is no stationary distribution. Also, for any initial $f_0$ and each real $t$ we have $f_k(t)\to0$ as $k\to\infty$.

You wrote:

I can verify that $\Psi$ is continuously differentiable, $\Psi(t,\tau)>0$ for all $t,\tau\in\mathbb{R}$, and of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$.

[...] these properties should be sufficient to guarantee that a stationary distribution $\pi$ exists and is unique, and that $f_k \to \pi$ (in the T.V. sense) for any initial $f_0$.

Of course, this is not so. E.g., if $\Psi(t,s)=g(t-s)$, where $g$ is the standard normal pdf, then (considering, for instance the Fourier transform, one can easily see that) there is no stationary distribution. Also, then for any initial $f_0$ and each real $t$ we have $f_k(t)\to0$ as $k\to\infty$.