Let $f: [0,1] \rightarrow \mathbb{R}_{\geq 0}$ be non-increasing with $\int_{0}^{1}f(x)\,dx = 1$ and $\int_{0}^{1}xf(x)\,dx := \mu \in [0,1].$
Is there a nice upper bound for $\int_{0}^{1}|f(x) - 1|\,dx$ in terms of $\mu$? If so, is there an $f$ for which the bound is tight? Thinking of $f$ as a probability density on $[0,1]$ with mean $\mu$, this would give an upper bound on (twice) the total variation distance between the $f$ distribution and the uniform distribution.