Let $K$ be a Markov kernel from $\mathcal{X}$ to $\mathcal{Y}$, i.e., $K(\cdot|x)$ is a probability measure on $\mathcal{Y}$ for all $x\in \mathcal{X}$.
Let $\mu$ and $\nu$ be two probability measures on $\mathcal{X}$. Moreover, let $K\mu$ and $K\nu$ be the probability measures on $\mathcal Y$ induced by $K$ with input distributions $\mu$ and $\nu$, respectively. That is, $K\mu(A) = \int\_{\mathcal X} \mu(dx) K(A|x)$ for any measurable set $A\subset \mathcal Y$.


It is known that
$$\sup\_{\mu, \nu}\frac{\|K\mu-K\nu\|}{\|\mu-\nu\|} = \sup\_{x, x'\in \mathcal{X}} \|K(\cdot|x)- K(\cdot|x')\|,$$
where $\|\cdot\|$ is the $\ell\_1$ distance. This result is largely attributed to Dobrushin.
I'm trying to prove a similar result as follows.
Let $\kappa>1$ be given. I wish to show that
$$\sup\_{\mu, \nu}\frac{\|K\mu-\kappa K\nu\|}{\|\mu-\kappa \nu\|} \leq \sup\_{x, x'\in \mathcal{X}} \|K(\cdot|x)- K(\cdot|x')\|.$$


Any thoughts?
I suspect that this problem (or some versions of it) has already appeared in some literature.



An inequality relating $\ell_1$ distance of input and output of a Markov krnel
Let me assume that $X$ and $Y$ are topological spaces (satisfying the usual conditions), and denote by $M(X), M(Y)$ the respective Banach spaces of finite (signed) Borel measures endowed with the total variation norm. I will also need the closed codimension 1 subspaces $M\_0(X),M\_0(Y)$ that consist of the measures of total weight 0 (i.e., the ones for which the masses of the positive and the negative parts in the Hahn decomposition are equal). We are given a Markov operator $K$ from $X$ to $Y$ (as it has already been pointed out, the standard probabilistic notation for the action of a Markov operator on measures is $\mu\mapsto \mu K$).


First, about the norm of the restriction of $K$ to the subspace $M\_0(X)$. Since for the "elementary measure" $\mu=\delta\_x-\delta\_{x'}$ associated with any $x\neq x'\in X$ one has
$$
\frac{\| \mu K \|}{\|\mu \|} = \frac12 \|(\delta\_x - \delta\_{x'}) K \| \;,
$$
and any measure from $M\_0(X)$ can be decomposed into an integral of elementary ones, the norm of $K$ as an operator from $M\_0(X)$ to $M\_0(Y)\subset M(Y)$ is the supremum of the RHS's of the above formula over all $x\neq x'\in X$.


I don't see what is here to be attributed to Dobrushin - or to anyone else. By the way, when stating your claim you mistakenly assumed that $\|\delta\_x - \delta\_{x'}\|=1$ - this is how some probabilists (who apparently don't believe in numbers greater than 1) define the total variation, but you explicitly stated that your total variation is the $\ell^1$ one.


Now, when passing to the measures $\mu-\kappa\nu$ you leave the subspace $M\_0(X)$, so that (by elementary functional analysis) you are asking about the norm of the operator $K$ on the whole space $M(X)$ which is 1 for **any** Markov operator. For an explicit couterexample to your claim consider the situation when the $K$-image of $M(X)$ is one-dimensional, i.e., **all** measures $\delta\_x K$ are the same. In this case your RHS is 0 (as the whole subspace $M\_0(X)$ is in the kernel of $K$), whereas the LHS is 1.


