Robert Casella Independence Sampler Result In the second edition of the Robert & Casella book (Monte Carlo Statistical Methods), the authors have a result, Theorem 7.8, on the independent Metropolis-Hastings sampler:  Letting $f$ be the density of the target measure an $g$ the density of the proposal, if a minimization condition is satisfied $f(x)\leq M g(x)$ for all $x$ in the support of $f$, then the chain is uniformly ergodic and
$$
\| K^n(x,\cdot) - f\|_{TV} \leq 2 (1-1/M)^n
$$
The proof begins with an odd result (equation 7.9),
\begin{equation}
\|K(x,\cdot) -f\|_{TV} = 2 \sup_{A} |\int_{A} K(x,y) - f(y) dy|.
\end{equation}
Where is this factor of 2 coming from?  Ignoring the factor of 2 for a moment, the authors then obtain
\begin{equation}
\|K(x,\cdot) -f\|_{TV} \leq 2 (1- 1/M),
\end{equation}
which I agree with, except, again, for the factor of 2.  They then say that, by induction, one can obtain the result.
The best that I can obtain is
$$
(2\|K^n(x, \cdot) - f\|_{TV})\leq (2\|K(x, \cdot) - f\|_{TV})^n \leq (2(1-1/M))^n,
$$
which is based on a computation in Roberts & Rosenthal (2004) in Probability Surveys.
This calculation has been driving me nuts.
 A: You are correct that the factor $2$ does not make sense as written. It should be
\begin{equation}
\|K(x,\cdot) -f\|_{TV} \stackrel{\text{def}}{=} \sup_{A} \left|\int_{A} K(x,y) - f(y) \text{d}y\right|
= \dfrac{1}{2} \int_{A} |K(x,y) - f(y)| \text{d}y
\end{equation}
which only involves a $1/2$ for the L¹ distance (as recalled in the Wikipedia page about TV). In my current version of the book, there is actually no 2, so it is quite possible we have removed the 2 when realising it is a typo as written! Apologies about this typo.
For the proof of this inequality, the easiest approach is to use coupling: since 
$$K(x,y)\ge\frac{1}{M} f(y)$$ the kernel can be written as
$$K(x,y)=\frac{1}{M} f(y)+\left(1-\frac{1}{M}\right)\frac{MK(x,y)-f(y)}{M-1}$$ Therefore, at each transition, the next generation is from $f$ with probability $1/M$, thus is not from $f$ with probability at most $$\left(1-\frac{1}{M}\right)\qquad\qquad(1)$$This means that the total variation distance between $f$ and $K(x,\cdot)$ is at most (1). Similarly, by a geometric argument, the generation at stage $n$ is not from $f$ with probability at most $$\left(1-\frac{1}{M}\right)^n$$
