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.