total variation distance and coupling

Question

How can we show that $||\gamma M- \beta M||_{TV}\leq||\gamma -\beta||_{TV}$ (total variation distance) for a transition matrix of a markov chain $M$ ?

$\gamma, \beta$ are two distributions on the state space $S$

Answer 1

\begin{align} \|\gamma M- \beta M\|_{\text{TV}}&=\frac{1}{2}\sum_{s\in S}\left|(\gamma M)(s)-(\beta M)(s)\right|\\ &=\frac{1}{2}\sum_{s\in S}\left|\sum_{s'\in S}(\gamma(s')-\beta(s')) M(s',s)\right|\\ &\le\frac{1}{2}\sum_{s'\in S}\left|\gamma(s')-\beta(s')\right|\sum_{s\in S}M(s',s) \\ &=\|\gamma - \beta\|_{\text{TV}}. \end{align}