 Solution

Let $\Phi ={\phi }_{\epsilon }$ and let ${\Phi }^{\ast }$ denote its convex conjugate. Its Fenchel–Young gapis nonnegative by the Fenchel–Young inequality, and the hypothesis says $\int Hd{\pi }_{\epsilon }\le \epsilon$.

First justify the integrability needed for the certificate. Finite second moments and the Cauchy-Schwarz inequality imply, for every transport plan $\pi$,Since $\Phi \in L^1(\mu )$ and $H\in L^1({\pi }_{\epsilon })$, the identity ${\Phi }^{\ast }(y)=H(x,y)-\Phi (x)+x\cdot y$ proves ${\Phi }^{\ast }\in L^1(\nu )$ by the marginal distribution property. In particular, no subtraction of infinite integrals is being used.

SetThese are integrable Kantorovich potentials. The Fenchel–Young inequality gives $u(x)+v(y)\le \frac{1}{2}|x-y{|}^2$. More precisely, for every transport plan $\pi$,Taking the infimum in the first inequality therefore provesThe factor $1/2$ in the quadratic cost is essential for this exact gap identity.