This is a preliminary version. Just a short proof that the stationary distributin is unique (under the right assumptions). I’ll use the same symbols for distributions the their underlying pdfs and which of both is meant can be inferred from context.
Uniqueness Proof
Suppose we have two stationary distributions $\pi_0, \pi_1$ of some Markov chain $K$, i.e. for $i=0,1$ we have
\[\pi_i(y) = \int_X \pi_i(x) \ K(x,y) \ dx.\]Now consider their ratio $\frac{\pi_0}{\pi_1}$ and assume it’s minimized at $x’$ (the minimum exists if the underlying space is compact for instance), i.e.
\[x' := \mathop{\arg \min}_{x} \frac{\pi_0(x)}{\pi_1(x)}.\]With this in hand we clearly have $\pi_0(x’) = \frac{\pi_0(x’)}{\pi_1(x’)} \pi_1(x’)$ (basically multiplying by a “nutritious” one.). Computing the difference and applying the characteristic property given in the first equation above yields
\[0 \ = \ \pi_0(x') - \tfrac{\pi_0(x')}{\pi_1(x')} \pi_1(x') \ = \ \int \underbrace{ K(x,x')}_{>0} \cdot \underbrace{ \Big[ \pi_0(x) - \tfrac{\pi_0(x')}{\pi_1(x')} \pi_1(x) \Big]}_{\geq 0 \ \text{(since $x'$ is min)}} \ dx.\]By definition of $x’$ as the minimizer of the quotient we conclude that second integrand must be non-negative. Assuming the transition kernel isn’t vanishing it implies further that the second integrand must in fact be zero, and thus
\[\pi_0(x) = \tfrac{\pi_0(x')}{\pi_1(x')} \pi_1(x).\]Therefore both candidates merely differ by a scaling factor. However since we deal with normalized probability densities here we can deduce that the scaling factor is $1$ which concludes the uniqueness proof.