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For a functional $E$ on a real Banach space, its convex conjugate is defined on the dual space byHere $⟨p,u⟩$ is the dual pairing. For complex spaces, use its real part, equivalently regard the space as real for convex analysis. The convex conjugate is a convex function, being the supremum of affine functions of $p$, even when $E$ itself is not convex.

For $p\in {\mathcal{U}}^{\ast }$, the definition of the convex conjugate giveswhere the substitution $w=u-v$ is a bijection of the independent pairs. ThereforeThis infimal convolution identity does not need convexity of $E,F$ or attainment of the inner infimum. The assumptions ensure that both functions have nonempty effective domains; their conjugates never take $-\infty$, so the separated sum is well defined, allowing $+\infty$.

Sincefor $x>0$, the second-derivative criterion shows that $f(x)=x^a/a$ is strictly convex. Put $b=a/(a-1)$, so $1/a+1/b=1$. For dual variable $p>0$, the stationary equation in the Legendre transform isIt is the unique global maximizer of $px-f(x)$, and henceApplying the same calculation with the conjugate exponent $b$ givesUnder the Legendre-Fenchel transform on all real dual variables, the same supremum additionally gives $f^{\ast }(p)=0$ for $p\le 0$.

Now take $a=r$, $b=s$, and apply the Fenchel–Young inequality $f(x)+f^{\ast }(y)\ge xy$. This yields Young's inequality for productsfor positive $x,y$ and conjugate exponents $r,s$.

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.

Use the sign conventionfor the Lagrangian function in constrained optimization. The Lagrangian dual problem iswhere $f^{\ast }$ is the convex conjugate. For this convex problem with affine equality constraints, the stationarity and feasibility parts of the Karush-Kuhn-Tucker conditions areThey say exactly that the displayed operator satisfiesThus its zeros are precisely the primal-dual optimal points, subject to the usual attainment assumptions.

For $u=(x,z)$ and $\tilde{u}=(y,s)$, the Euclidean inner product givesThe last two terms cancel by the defining property of the matrix transpose, and the first is nonnegative by part a. Hence $F$ is a monotone operator.

For a proper lower-semicontinuous convex function $f$, its proximal operator isThe squared norm is strongly convex, so the minimizer is unique. The subdifferential sum rule gives the necessary and sufficient conditionMore generally,The subgradient inversion rule for the convex conjugate says $u\in \partial f(x)$ exactly when $x\in \partial f^{\ast }(u)$. Hencewhich is precisely the proximal optimality conditionSince $x=y-tu$, this proves the generalized Moreau decomposition

The function $\varphi (x)={\max }_{v\in C}⟨x,v⟩$ is the support function ${\sigma }_C$. For a nonempty compact convex set,so its convex conjugate is the indicator function ${\iota }_C$. Applying the Moreau decomposition,Multiplication of an indicator function by a positive scalar does not change it, and its proximal operator is the Euclidean projection onto a convex set. Therefore

The Reuss elastic approximation assumes uniform stress. The local strain is the derivative of the convex conjugate $W^{\ast }(S,x)$; its mean is $D⟨W^{\ast }⟩(S)$. Under invertible differentiable duality, this mean determines $S=D{W}_R(E)$. For positive linear elasticity, $C_R=⟨C^{-1}{⟩}^{-1}$. Uniform stress is generally not accompanied by a compatible local displacement field, so this approximation is distinguished from an exact heterogeneous solution.

The support function of a nonempty set $C$ is ${\sigma }_C(x)={\sup }_{v\in C}⟨x,v⟩$. It is convex and positively homogeneous, and for a closed convex set its convex conjugate is the indicator function of $C$.