I'm trying to prove following inequality. For any probability distribution $P$ и ${Q}$ on $U$ and function $f(x): U \rightarrow [0; B]$ it is true that: $$ |E_{\tilde P}[f(x)] − E_{\tilde Q}[f(x)]| \leq \frac B2 ||P − Q||$$ where $||\cdot||$ – total variation distance. I can use discrete distributions for proof.
What I have for now: $$|\sum_{x\in U} f(x)(p_P(x) - p_Q(x))| \leq \frac B2||P - Q||$$ Denote $g(x) = \frac{f(x)}{B}$, then $g: U \rightarrow [0;1]$: $$|\sum_{x\in U} g(x)(p_P(x) - p_Q(x))| \leq \frac 12\sum_{x\in U: p_P(x)>p_Q(x)}[p_P(x) - p_Q(x)]$$ Now open the summation on the left with the consideration that difference of measures can be positive or negative: $$|\sum_{p_P(x)>p_Q(x)} g(x)(p_P(x) - p_Q(x)) - \sum_{p_Q(x)>p_P(x)} g(x)(p_Q(x) - p_P(x))| \leq $$
$$\leq \sum_{p_P(x)>p_Q(x)} g(x)(p_P(x) - p_Q(x)) \leq \sum_{p_P(x)>p_Q(x)} (p_P(x) - p_Q(x))$$
What I did wrong?