 Solution

By the definition of the subdifferential, $p\in \partial J(u)$ meansRearrangement bounds $⟨v,p⟩-J(v)$ by $⟨u,p⟩-J(u)$, and equality is attained at $v=u$. Taking the supremum in the definition of the convex conjugate gives $J^{\ast }(p)=⟨u,p⟩-J(u)$. Conversely this equality bounds every member of that supremum and rearranges to the subgradient inequality. ThereforeApply the same argument to $J^{\ast }$ and use the Fenchel-Moreau theorem $J^{\ast \ast }=J$. With the canonical Hilbert identification of the bidual, the equality is also equivalent to $u\in \partial J^{\ast }(p)$. This proves subgradient inversion under convex conjugacy:The conditions include finiteness at the points in question; expressions involving $+\infty$ are not subgradients merely by formal subtraction.