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For a logarithmically homogeneous barrier on a proper cone, the convex conjugate with reversed argument defines a barrier on the interior dual cone. The inverse gradient relation makes $y=-\mu \nabla F(s)$ equivalent to $s=-\mu \nabla F_{†}(y)$. This gives a joint primal-dual central path without assuming the barrier is identical on a self-dual cone.

The lower conjugate uses an infimum, unlike the supremum defining a convex conjugate. For any $F$, $F_{\ast }(\tau )=-(-F{)}^{\ast }(-\tau )$ and $F(e)\le \tau :e-F_{\ast }(\tau )$. The inequality is informative only when the lower conjugate is finite. If the infimum is attained uniquely and the dual is differentiable, $D{F}_{\ast }(\tau )$ is its minimizing argument. This sign convention is useful for upper comparison-energy bounds.

For $t>0$, the generalized Moreau decomposition is ${\mathrm{prox}}_{tf}(y)=y-t{\mathrm{prox}}_{t^{-1}{f}^{\ast }}(y/t)$, where $f^{\ast }$ is the convex conjugate.

In the Voigt elastic approximation, every local strain is $E$. The corresponding local stress is ${\sigma }_V(x)=D_{\epsilon }W(E,x)$. Differentiating the averaged strain energy density under the integral givesThe uniform strain need not generate an equilibrated local stress; this calculation concerns the stated approximation, not an assertion that it solves the heterogeneous boundary problem.

For the Reuss elastic approximation, use the convex conjugateUnder differentiable strictly convex duality, the local strain produced by uniform stress $S$ is $e_R(x)=D_S{W}^{\ast }(S,x)$. Set $F(S)=⟨W^{\ast }(S,x)⟩$. Averaging the local strains gives $E=DF(S)$. At the maximizing stress in $W_R(E)=F^{\ast }(E)$, the stationarity condition is precisely $DF(S)=E$. ConsequentlyThus the derivative of each approximate energy is its own mean stress, includingFor positive linear elasticity, $W=e:C(x)e/2$ and $W^{\ast }=S:C(x{)}^{-1}S/2$, giving the useful check $C_V=⟨C⟩$ and $C_R=⟨C^{-1}{⟩}^{-1}$. More general convex duality gives the same conjugacy relation with subgradients when differentiability or strict convexity is absent.

Let $C=\mathrm{conv}\{a,b\}$. Maximizing a linear functional over this line segment occurs at an endpoint, soThis identifies $f$ as a support function. Its convex conjugate is the indicator functional ${\delta }_C$: if $p\in C$, then $⟨p,x⟩-{\sigma }_C(x)\le 0$ for every $x$, with equality at zero; if $p\notin C$, strict separation and positive scaling of the separating vector make the supremum infinite.

For a convex function, equality in the Fenchel–Young inequality characterizes its subdifferential. HenceThis is support-function subgradients as exposed faces. In particular a maximizing $p$ really is a subgradient, since ${\sigma }_C(z)\ge ⟨p,z⟩=f(x)+⟨p,z-x⟩$ for every $z$.

Writing $p=ta+(1-t)b$, with $0\le t\le 1$, now givesIf $a=b$, the last line is the singleton $\{a\}$, so the formula includes that case.

Compute the convex conjugate of $f_z$ at a general pair $(v,y)$. With $w=z+u-x$, the independent variables become $(x,w)$ and $u=w+x-z$, soTherefore the Lagrange dual function isUsing strong duality from the preceding solution givesEquivalently the conjugate of an infimal convolution is $F^{\ast }=k^{\ast }+h^{\ast }$, and the continuous convex $F$ equals its biconjugate.

For practical subgradient computation, minimize the known convex dual objectiveEvery optimizer, and only an optimizer, belongs to $\partial F(z)$ by Fenchel–Young inequality. The full characterization isThis is infimal-convolution dual subgradients; the set is nonempty because $F$ is finite convex everywhere. If both conjugates are differentiable at the optimizer, solve $\nabla k^{\ast }(y)+\nabla h^{\ast }(y)=z$. For nonsmooth conjugates, use the displayed aggregate subdifferential or a convex minimization algorithm. Replacing it by $\partial k^{\ast }(y)+\partial h^{\ast }(y)$ requires the usual subdifferential sum rule qualification; it is not automatically justified solely by knowing the two conjugates.

LetThe proximal operator optimality condition saysBy subgradient inversion under convex conjugacy, which uses Fenchel–Young inequality and $f^{\ast \ast }=f$ from the Fenchel-Moreau theorem,Consequently $z-p=\tau v\in \tau \partial f^{\ast }(p)$, meaning $p=B_{\tau f^{\ast }}(z)$. Uniqueness of the backward subgradient step identifies the result:This is the scaled Moreau decomposition. It requires one backward step on $f$, with reciprocal parameter $1/\tau$ and scaled input $z/\tau$, followed by a scalar multiplication and subtraction. No separate proximal computation of the convex conjugate is needed.

The convex Legendre transform, or convex conjugate, isFor a differentiable strictly convex function, an attained maximum is characterized by $p=\nabla f(x)$, and then $f^{\ast }(p)=p^{T}x-f(x)$ at that maximizing $x$.

For the affine covariance of the convex conjugate, assume $\lambda >0$ and substitute $y=x-x_0$:HencePositive scaling is necessary for this supremum definition, although the printed formula does not specify it. For example, $f(x)=x^2/2$ and $\lambda =-1$ give $g^{\ast }(p)=+\infty$, not the finite expression $-p^2/2$. If the transform is instead defined as a stationary value using a locally invertible gradient, the same algebraic identity holds for any $\lambda \ne 0$: the critical-point relation is $p=\lambda \nabla f(y)$, and substitution of that stationary $y$ gives the displayed formula. That extension does not assert a supremum for negative scaling.

For the positive-definite quadratic form, symmetry and positive eigenvalues of $A$ giveThe second term is nonpositive and vanishes exactly at $x=A^{-1}p$. Therefore

For an extended-real function define its Fenchel conjugate by $f^{\ast }(p)={\sup }_x(⟨p,x⟩-f(x))$ and its biconjugate by $f^{\ast \ast }(x)={\sup }_p(⟨p,x⟩-f^{\ast }(p))$. The Fenchel-Moreau theorem states, in the standard proper-envelope setting,Here the right side is the largest lower semicontinuous convex function below $f$, equivalently the function whose epigraph is the closed convex hull of $\mathrm{epi}f$. It is enough to assume $f$ is proper and has an affine minorant, ensuring this envelope is proper. In particular, for a proper convex function that is lower semicontinuous, $\boxed{f^{\ast \ast }=f}$.

First, $⟨p,x⟩-f^{\ast }(p)\le f(x)$ by the definition of the convex conjugate. The biconjugate is a supremum of continuous affine functions, so it is convex, lower semicontinuous and no greater than $f$. Second, the best intercept for an affine minorant with slope $p$ is $-f^{\ast }(p)$: $⟨p,x⟩+a\le f(x)$ for all $x$ precisely when $a\le -f^{\ast }(p)$. Thus $f^{\ast \ast }$ is the supremum of all affine minorants.

To prove that no part of the closed convex envelope is missed, set $E=\mathrm{cl}\mathrm{conv}(\mathrm{epi}f)$. The half-space representation of a closed convex set from part (a), applied in ${\mathbb{R}}^{n+1}$, separates any $(x_0,t_0)\notin E$ from $E$ by an inequality $a\cdot x+bt\le d$. Because $E$ is upward closed, $b\le 0$. If $b<0$, division by $-b$ gives an affine minorant $\ell$ with $\ell (x_0)>t_0$.

A vertical separator has $b=0$. Let ${\ell }_0$ be an existing affine minorant. Combine $a\cdot x\le d$ with ${\ell }_0(x)\le t$ to obtainSince $a\cdot x_0-d>0$, a sufficiently small positive $\epsilon$ makes this affine function exceed $t_0$. Thus vertical half-spaces can be approximated by nonvertical epigraph supports. Every point below the envelope is excluded by an affine minorant, so the supremum of these minorants is exactly the envelope. This is the decisive use of part (a).

Properness and the minorant convention matter for unrestricted extended-real functions. For example, $f(x)=-x^2$ on $\mathbb{R}$ has no affine minorant; $f^{\ast }\equiv +\infty$ and $f^{\ast \ast }\equiv -\infty$. With the corresponding improper-envelope convention its closed convex envelope is also $-\infty$. The theorem should not silently describe such an envelope as proper. The identically $+\infty$ function is another degenerate case, handled separately by extended-real conventions.

Let $r(u)=\Vert u-g{\Vert }_2^2$ and fix the reference noise budget $\sigma$. Define the convex perturbation functionPositive $z$ increases the allowed squared noise level. The feasible epigraph set $r(u)\le \sigma +z$ is convex and closed, so this is a proper lower semicontinuous jointly convex perturbation. Writing the inequality indicator as a supremum over its multiplier givesThe Lagrange dual function is $d(\lambda )={\inf }_{u}L(u,\lambda )$ for $\lambda \ge 0$. In the signed convex conjugate convention of part (a), $y=-\lambda$; positive $y$ gives dual objective $-\infty$ because the perturbation can be made arbitrarily large.

The feasible ball is nonempty and compact, so lower semicontinuity of total variation gives a primal minimizer. For $\sigma >0$, $u=g$ satisfies $r(g)=0<\sigma$, providing the Slater condition. Total variation is finite everywhere in the given discretization and hence continuous. Strong duality and dual attainment give a finite optimal ${\lambda }^{\ast }\ge 0$. The pair satisfies the Karush-Kuhn-Tucker conditionsEquivalently,Thus the set of saddle points is nonempty. Compactness handles primal attainment, while strict feasibility supplies a finite multiplier; these are distinct steps. An inactive constraint can yield ${\lambda }^{\ast }=0$. At $\sigma =0$, feasibility still forces $u=g$, but this strict-feasibility argument does not apply.

Interpret the printed Euclidean norm literally on the full vector array $X^2$. It is a single global norm, not the sum of pixelwise gradient lengths in discrete isotropic total variation. Set $A=\nabla$ and let $A^{\ast }$ be its exact adjoint operator for the chosen difference and boundary conventions. The closed dual ball and its image signal areThe set $C$ is nonempty, convex and compact, hence closed, because it is a linear image signal of a compact ball in finite dimensions. Euclidean norm duality givesthe support function of $C$.

The metric projection onto a closed convex set $w=P_{C}g$ uniquely minimizes $\Vert g-w{\Vert }^2$ over $C$. Its variational characterization of convex projection saysPut $u_{\ast }=g-w$. This inequality says precisely that ${\sigma }_C(u_{\ast })=⟨u_{\ast },w⟩$. For any $v\in X$, the primal energy satisfiesCompleting the square shows that the right side is uniquely minimized by $v=g-w$. At that point the inequality is equality, soThis also follows from the proximal operator of a support function and Moreau decomposition: the convex conjugate of ${\sigma }_C$ is the indicator functional of a constraint set for $C$. The squared fidelity is strictly convex and coercive, so the primal minimizer exists and is unique.

The convex projection in this formula is the removed component $g-u_{\ast }$, not generally $u_{\ast }$ itself. A metric convex projection onto one fixed closed convex set is idempotent. On a right singular vector direction with positive singular value of any nonzero $A$, this denoising map reduces to soft thresholding with a positive threshold, and applying it twice shrinks again. It therefore cannot be such a convex projection for all data. This qualifies the printed convex projection wording while giving the required global gradient-norm projection residual.

Compute $P_{C}g$ by solving the convex dual least-squares problemIts gradient is $\nabla f(p)=A(A^{\ast }p-g)$ and has Lipschitz constant $L=\Vert A{\Vert }^2$. Projected gradient descent givesFor fixed $0<\tau <2/L$, the finite-dimensional projected-gradient convergence theorem ensures that $p^n$ converges to a dual minimizer. Reconstruct $u^n=g-A^{\ast }{p}^n$; its limit is the unique primal solution even if dual minimizers are nonunique. If $A=0$, the primal solution is simply $g$ and no iteration is needed. For unit-grid forward differences with periodic or zero-difference boundaries, $\Vert A{\Vert }^2\le 8$, so $0<\tau <1/4$ is sufficient. Grid-spacing factors or other boundary stencils change this bound.

The normalization here is global. If a pixelwise sum of gradient lengths had instead been intended, the dual feasible set would be a product of pixelwise balls and convex projection would normalize each block separately; it is a different regularizer and should not be silently substituted.

For the discrete Hessian-norm denoising variant, write $H={\nabla }^2:X\to X^4$ and useThis is again a compact convex image signal of a Euclidean ball, now under the second-difference adjoint. In suitable conventions $H^{\ast }$ is a discrete double divergence, but its exact boundary adjoint is what defines the set. It lies in $(\ker H{)}^{\perp }$, so components in $\ker H$ are preserved by denoising. Interior second derivatives annihilate affine image signals; whether all such image signals remain in the kernel depends on the boundary convention. With a pixelwise Hessian norm, the corresponding balls would instead be four-component blocks.

Normalize the constant relative risk aversion utility as $U(c)=c^{1-R}/(1-R)$. The printed specification of $U^{\prime }$ alone also permits an additive constant $K$; that would add $K/\rho$ to the value, so the stated homogeneity holds for the normalized value. SetMultiplying initial wealth, the historical maximum, dollar investments, and consumption by $a>0$ multiplies every term of the wealth equation by $a$, including the tax term. It maps admissible policies bijectively and multiplies the normalized reward by $a^{1-R}$. Therefore the value scales asThe historical maximum means ${\bar{w}}_t=\max ({\bar{w}}_0,{\sup }_{0\le s\le t}{w}_s)$ when initially $w<\bar{w}$.

Inside the state region $w<\bar{w}$, the maximum is locally constant. The Hamilton-Jacobi-Bellman equation isAt $w=\bar{w}$, the finite-variation contribution in the Itô formula is $(V_{\bar{w}}-\tau V_w)d\bar{w}$. The high-water mark tax boundary condition is consequentlyFor an increasing, strictly concave function of wealth, optimizing the two controls givesThe consumption convex conjugate isSubstitution of the homogeneity derivatives gives the reduced equation and boundary conditionThe controls in scaled variables are $c^{\ast }/\bar{w}=(v^{\prime }{)}^{-1/R}$ and ${\theta }^{\ast }/\bar{w}=-(\mu -r)v^{\prime }/({\sigma }^2{v}^{\prime \prime })$.

Use the wealth-variable Legendre dualSince $v^{\prime \prime }<0$, the inverse map satisfies $J^{\prime }=-x$ and $J^{\prime \prime }=-1/v^{\prime \prime }>0$. Inserting $v=J-z{J}^{\prime }$ into the reduced Hamilton-Jacobi-Bellman equation gives a linear Euler equationFor the Euler differential equation operator on the left, its action on $z^t$ is $Q(t)z^t$. Write $b=1-1/R$. Under ${\gamma }_M=-Q(b)>0$, the particular solution is $-\tilde{U}(z)/Q(b)$. For the nondegenerate case $\kappa \ne 0$, the two characteristic roots are $-\alpha <0<t_{+}$, with $t_{+}>1$ because $Q(0)=-\rho$ and $Q(1)=-r$. Thus the general solution also contains $A(z/z_{\ast }{)}^{-\alpha }+B{z}^{t_{+}}$.

Here $z_{\ast }=v^{\prime }(1)$ and the relevant dual domain is $z\ge z_{\ast }$. The inverse wealth ratio must have $x=-J^{\prime }(z)\to 0$ as $z\to \infty$. The particular solution and the negative-root term have derivatives tending to zero, whereas $B{z}^{t_{+}}$ has an unbounded derivative unless $B=0$. Hence the admissible dual solution has the claimed formThe quadratic-root formulation presupposes a nonzero market price of risk. If $\kappa =0$, the dual equation becomes first order; it is treated directly or by an appropriate nondegenerate limit rather than by assuming two quadratic roots.

The inverse map and the tax boundary provide two equations at $z_{\ast }$:Because ${\tilde{U}}^{\prime }(z)=-z^{-1/R}$, these becomeLet $D=\alpha R+R-1$. Eliminating $z_{\ast }^{1-1/R}/{\gamma }_M$ gives $-DA=\tau z_{\ast }$. Also $D=R(\alpha +b)>0$, since $Q(b)<0$ puts $b$ strictly between the two roots. The derivative boundary then gives $z_{\ast }^{-1/R}/{\gamma }_M=(D+\alpha \tau )/D$. Therefore the constants areThere is an important admissibility qualification in the printed conclusion. With $x(z)=-J^{\prime }(z)$,The second term is negative, but decays faster than the first because $\alpha +1>1/R$. Direct differentiation at the boundary givesA wealth-variable Legendre dual of a concave value must be a convex function. Thus the printed smooth tax-paying solution requires $\alpha \tau <1$, with the limiting case $\alpha \tau =1$ allowed as a degenerate boundary. The assumptions ${\gamma }_M>0$ and $0<\tau <1$ do not imply this: for example $R=2$, $r=0.03$, $\rho =0.1$, $\kappa =0.2$, and $\tau =0.5$ give $\alpha =(5+\sqrt{105})/4>2$ and ${\gamma }_M=0.07$, hence negative boundary dual curvature.

For higher tax the investor can avoid raising the historical maximum. The wealth-cap investment boundary replaces the tax-paying equality by $J^{\prime \prime }(z_{\ast })=0$, alongside $J^{\prime }(z_{\ast })=-1$. Solving these equations givesThese are the same constants with $\tau$ replaced by ${\tau }_{\mathrm{eff}}=1/\alpha$. More generally the admissible two-regime expression usesin the constants, while the claimed formulas themselves describe the tax-paying regime. In the cap regime the portfolio volatility vanishes at $x=1$ and the consumption rate there is $r+{\kappa }^2(\alpha +1)/(2R)>r$, so the wealth drift points inward and no new maximum is required. The remaining boundary inequality is $(V_{\bar{w}}-\tau V_w)/V_w=1/\alpha -\tau \le 0$, consistent with avoiding costly maximum increases. For $z>z_{\ast }$, dual curvature is positive because the negative term in $J^{\prime \prime }$ decays faster than the positive term. Thus this qualification repairs an actual missing parameter restriction rather than a TeX transcription error.

In congestion control, resource $j$ observes only its own aggregate load $y_j={\sum }_{r:j\in r}{x}_r$ and generates the nonnegative feedback price ${\mu }_j=p_j(y_j)$. User $r$ receives the sum $M_r={\sum }_{j\in r}{\mu }_j$ of prices along its route. Its desired increase $w_r$ competes with the congestion charge $x_r{M}_r$, while ${\kappa }_r$ sets the adjustment speed. This is local information: a resource needs the rates passing through it, and a user needs the feedback along its own route, rather than the complete network state.

There is a hypothesis to make explicit before claiming global convergence. If “increasing” means nondecreasing, the printed assumptions allow $p_j\equiv 0$, in which case ${\dot{x}}_r={\kappa }_r{w}_r>0$ and there is no equilibrium point. The intended theorem holds when every route contains at least one resource whose price is not identically zero, with nonempty routes and nonnegative, continuous, nondecreasing prices. Strictly increasing prices satisfy this condition automatically. We prove the theorem under this necessary qualification; without it the zero-price example disproves the literal claim.

Let $P_j(y)={\int }_0^y{p}_j(u)du$. Each $P_j$ is a nonnegative convex function. For positive rates define the primal congestion potentialThe weighted logarithmic utility is a strictly concave function, and subtracting the resource costs preserves strict concavity. Thus $U$ has at most one maximizer and at most one critical point on the positive orthant.

It also attains its maximum there. A nonzero nondecreasing price is bounded below by some positive constant at all sufficiently large loads; hence its primitive satisfies $P_j(y)\ge c_{j}y-K_j$ with $c_j>0$. Choose $c_j=0$ at zero-price resources. The route coverage assumption gives $c={\min }_r{\sum }_{j\in r}{c}_j>0$, soThis bounds all superlevel sets of $U$. Within a bounded set, approaching $x_r=0$ makes $U\to -\infty$. Consequently every nonempty superlevel set is a compact set contained in the positive orthant. A maximizing sequence stays in one such set, giving a unique interior maximizer $x^{\ast }$. Its critical-point equations areThe rate dynamics are the gradient flow with diagonal mobility ${\dot{x}}_r={\kappa }_r{x}_r{g}_r(x)$. Along any positive trajectory,with equality only at $x^{\ast }$. Thus $V(x)=U(x^{\ast })-U(x)$ is a strict Lyapunov function. The initial superlevel set traps the trajectory away from both infinity and the boundary, ensuring global continuation. Positive initial rates remain positive: variation of constants in ${\dot{x}}_r+{\kappa }_r{M}_r(t)x_r={\kappa }_r{w}_r$ makes this explicit. Starting with finite nonnegative rates also gives positive rates at every later positive time.

For completeness, mere continuity of $p_j$ suffices for convergence; one need not silently assume differentiable prices. On the trapping compact set, both the vector field and $D(x)={\sum }_r{\kappa }_r{x}_r{g}_r(x{)}^2$ are continuous and bounded. The trajectory satisfies a uniform Lipschitz condition in time, so $D(x(t))$ is uniformly continuous. Also ${\int }_0^{\infty }D(x(t))dt<\infty$, since $U$ increases to a finite limit. The decay of a nonnegative uniformly continuous integrable function applies: otherwise disjoint intervals of a fixed positive width would each contribute a fixed positive amount. Every accumulation point therefore has $D=0$ and is $x^{\ast }$. Compactness then proves the global convergenceThis is the dissipation argument behind the LaSalle invariance principle. If uniqueness of trajectories is desired under continuous prices, put $z_r=2\sqrt{x_r/{\kappa }_r}$. The transformed equation is $\dot{z}={\nabla }_{z}U(\kappa z^2/4)$. The transformed potential is concave: it is a sum of positive multiples of $\log z_r$, minus convex nondecreasing $P_j$ composed with nonnegative quadratic sums. Its gradient is monotone decreasing, so the squared distance between two solutions cannot increase. This proves uniqueness for positive initial data without a locally Lipschitz function hypothesis on $p_j$.

For the alternative price dynamics, restrict first to positive prices on resources used by at least one route, and discard unused resources. Define the dual congestion potentialIts domain has ${\mu }_j\ge 0$ and $M_r>0$ for every route. Strictly increasing $q_j$ make each $Q_j$ strictly convex, and $-\log M_r$ is convex. Thus $H$ is a strictly convex function. Each $q_j$ is eventually bounded below by a positive constant, so its integral grows at least linearly, dominating the negative logarithms as $\Vert \mu \Vert \to \infty$. If a route price tends to zero in a bounded set, $H\to \infty$. Therefore $H$ attains a unique minimum on its domain including the admissible boundary faces.

That minimum has every used-resource price positive. At a face ${\mu }_j=0$, all route prices are still positive and the inward derivative isHence the minimum cannot lie on that face. At the unique interior minimum,Since the price dynamics are ${\dot{\mu }}_j=-{\kappa }_j{\mu }_j{\partial }_{j}H$, their positive equilibrium points are exactly these critical points. There is one equilibrium in the positive price domain. An unused resource instead has its only nonnegative equilibrium at zero.

The positive-domain qualification matters here even with strictly increasing $q_j$. If zero prices are admitted, the multiplicative factor ${\mu }_j$ permits boundary equilibria of multiplicative price dynamics. Take one route using two resources, $w=1$, and $q_1(a)=q_2(a)=a$. Both $({\mu }_1,{\mu }_2)=(1,0)$ and $(0,1)$ are equilibria, with $x=1$, as is the positive equilibrium $(1/\sqrt{2},1/\sqrt{2})$, with $x=1/\sqrt{2}$. Thus the literal assertion of uniqueness on the whole nonnegative orthant is false. A boundary equilibrium need not satisfy load equals supplied capacity at its zero-price resources.

The relationship between the two intended equilibria is inverse resource response. The first system has ${\mu }_j=p_j(y_j)$; the positive equilibrium of the second has $y_j=q_j({\mu }_j)$. If $q_j=p_j^{-1}$ on the relevant ranges, these are the same resource equations, and both systems also impose $x_r{M}_r=w_r$. They therefore have the same rates and prices. For strictly increasing continuous $p_j$ with $p_j(0)=0$ and unbounded range, the ordinary inverses satisfy all the stated conditions on $q_j$. With arbitrary independently chosen $p_j$ and $q_j$, the equilibria need not agree. Under inverse response, $P_j$ and $Q_j$ are convex conjugates on the nonnegative half-line, and $H$ supplies the Lagrangian dual problem for maximizing the weighted logarithmic utility minus resource costs, up to constants independent of $\mu$.

Write $h=f\square g$ for the infimal convolution. It is proper by the permitted hypothesis. If $h(x_0)$ and $h(x_1)$ are finite, choose $y_0,y_1$ within $\epsilon$ of their respective infima. For $0<t<1$, put $x_t=(1-t)x_0+t{x}_1$ and $y_t=(1-t)y_0+t{y}_1$. Convexity of $f$ and $g$ givesLet $\epsilon \downarrow 0$. If either endpoint value is infinite, the desired inequality is automatic. Thus the infimal convolution is convex, without assuming the infimum is attained.

For its convex conjugate, replace a negative infimum by a supremum and then change variables $u=x-y$:The two suprema separate because $u$ and $y$ are independent. Properness of $f$ and $g$ makes each supremum strictly greater than $-\infty$, so this separation remains valid when one or both are $+\infty$. Therefore $(f\square g{)}^{\ast }=f^{\ast }+g^{\ast }$. This is the conjugate of an infimal convolution. The bounded-domain assumption is unnecessary for these two calculations; the assumed properness and lower semicontinuity of $h$ will be used when applying subgradient inversion under convex conjugacy.

For the convex conjugate, use a dual scalar $q$ to avoid confusing it with the fixed centre $z$. Differentiating $qx-\lambda (x-z{)}^2/2$ gives $q-\lambda (x-z)=0$, so the unique maximizing point is $x=z+q/\lambda$. Substitution yieldsThe positivity of $\lambda$ makes the maximized quadratic strictly concave and the convex conjugate finite for every real $q$.

For a real Banach space $X$ and a proper convex function $E:X\to \mathbb{R}\cup \{+\infty \}$, the convex conjugate is the extended-valued function on the continuous dual spaceHere proper means that $E$ never takes $-\infty$ and is finite somewhere. The pairing is the duality pairing, or the inner product after identification in a Hilbert space. The conjugate is convex and lower semicontinuous because it is a supremum of continuous affine functions. Its definition gives the Fenchel–Young inequality $E(x)+E^{\ast }(p)\ge ⟨x,p⟩$; the Fenchel-Moreau theorem recovers $E$ as $E^{\ast \ast }$ under the stated closed-convex hypotheses.

For the absolute value function, $px-|x|\le 0$ for all $x$ if $|p|\le 1$, with equality at $x=0$. If $|p|>1$, taking $x=t\mathrm{sign}(p)$ with $t\to \infty$ makes the expression diverge. Thus its convex conjugate isThis is the indicator functional of a constraint set for the dual unit interval, not a zero-one indicator function.

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.

Write $D=\nabla$ and use the real pixelwise inner products. Let $D^{\ast }$ be the adjoint operator, defined by $⟨Du,p⟩=⟨u,D^{\ast }p⟩$. For each nontrivial grid direction its one-dimensional formula iswith the analogous formula in $j$ for $D_y^{+\ast }{p}^2$. Their sum is $D^{\ast }p$; unused components $p_{N,j}^1$ and $p_{i,N}^2$ do not contribute. For $N=1$, $D=D^{\ast }=0$. If discrete divergence is defined, its sign convention is $\mathrm{div}=-D^{\ast }$, as in the adjoint of a discrete forward gradient.