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Let $(E,\mathscr E)$ be a measure space and $P:E \times\mathscr E\to [0,1]$ be a stochastic kernel - i.e. $$ P(x,A)\in [0,1] $$ for any $x\in E$ and $A\in \mathscr E$. On a set $b\mathscr E$ of bounded measurable functions with a norm $$\|f\| = \sup\limits_{x\in E}|f(x)|$$ define the action of the kernel as a linear operator $$ Pf(x) = \int\limits_E f(y)P(x,dy). $$ Let $\tilde P$ be another probability kernel and consider two norms $$ \|\tilde P - P\|' = \sup\limits_{A\in \mathscr E}\sup\limits_{x\in E}|\tilde P(x,A) - P(x,A)| $$ $$ \|\tilde P - P\|'' = \sup\limits_{f\in b\mathscr E\setminus\{0\}}\frac{\|(\tilde P - P)f\|}{\|f\|}. $$

It is easy to show that $\|\tilde P - P\|'\leq \|\tilde P - P\|''$ since an indicator function $1_A\in b\mathscr E$ for all measurable sets $A$. I wonder if the reverse inequality is true as well.

My idea was to consider a simple function $f(x) = \sum\limits_{i=1}^n f_i1_{E_i}(x)$ where $E_i$ is a partition of $E$. But if I am not missing anything, even for simple function the reverse inequality is not true.

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Indeed, the reverse inequality is not true. Take $x_0\in X$, and put $P_1(x,A):=\delta_x(A)$, and $P_2(x,A):=\frac 12(\delta_x(A)+\delta_{x_0}(A))$. We have \begin{align*} \lVert P_1-P_2\rVert&=\sup_{A\in\mathcal E}\sup_{x\in E}|P_1(x,A)-P_2(x,A)|\\ &=\sup_{A\in\mathcal E}\sup_{x\in E}|\delta_x(A)-\frac 12\delta_x(A)-\frac 12\delta_{x_0}(A)|\\ &=\frac 12\sup_{A\in\mathcal E, A\neq E}\sup_{x\in E}|\delta_x(A)-\delta_{x_0}(A)|\\ &=\frac 12 \end{align*} (indeed, since $|\delta_x(A)-\delta_{x_0}(A)|\leq 1$ this supremum is $\leq 1$, and it is reached for a set $A$ which contains $x_0$). Now if we assume $\{x_0\}$ measurable, consider the map $f$ defined by $f(x)=\begin{cases} 1&\mbox{ if }x\neq x_0\\\ -1&\mbox{ if }x=x_0. \end{cases}$ $f$ is measurable, bounded, and its norm is $1$. We have $$(P_1-P_2)(f)(x)=f(x)-\frac 12f(x)-\frac 12f(x_0)=1/2f(x)-1/2f(x_0)=1$$ so $\lVert P_1-P_2\rVert''\geq\frac{\lVert (P_1-P_2)(f)\rVert}{\lVert f\rVert}=1>\lVert P_1-P_2\rVert'$.

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thank you - I was confused about the definition of the total variation norm for a finite signed measure and the total variation distance between two probability measure which does not take factor $2$ into account –  Ilya Mar 30 '12 at 11:16

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