 Solution

As in part (c), every admissible transport map has nonnegative displacement $D(x)=T(x)-x$ with ${\int }_0^{1}D(x)dx=1$. The square root is a strictly concave function, so the reversed Jensen inequality givesThe map $T^{†}(x)=x+1$ is admissible and attains one. ThereforeThus “worst” means largest cost among admissible transport maps. Equality in the Jensen inequality for a strictly concave function forces $D(x)$ to be constant almost everywhere, so this maximizer is unique up to a $\mu$-null set. For comparison, the admissible reflection $T(x)=2-x$ has smaller costThe upper bound also holds for every transport plan.