Upper bound on difference of expected values between two close distributions

Question

Let $p,q$ be two distributions over the same finite set $X$ and the total variation distance $TV(p,q)=\epsilon$. If for some function $f:X\to\mathbb{R}$, $0\leq \sum_{x\in X} f(x) p(x)\leq 1$ and $0\leq \sum_{x\in X} f(x)q(x)\leq 1$ hold. What is the best upper bound of $$\Big|\sum_x f(x)(p(x)-q(x))\Big|$$ as a function of $\epsilon$? I can only get a trivial upper bound $2$ by triangle inequality, but the constraint that the total variance distance is $\epsilon$ is not used. Is it possible to get a better bound?

Answer 1

You can bound it by $$ \left| \sum_x f(x)(p(x) - q(x)) \right| \leq \max_x |f(x)| \cdot \sum_x |p(x) - q(x)| = \epsilon \cdot \max_x |f(x)| $$