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The biconjugate of $f$ is the convex conjugate of its convex conjugate:It is the supremum of all affine minorants. For a proper function with an affine minorant, the Fenchel-Moreau theorem identifies it with the closed convex envelope; a proper lower semicontinuous convex function equals its biconjugate.

For a proper extended-real function with an affine minorant, biconjugate gives its greatest lower-semicontinuous convex minorant. Geometric separation of its epigraph turns containing closed half-spaces into affine lower bounds. Vertical half-spaces are recovered by adding multiples of a domain-separating inequality to one fixed affine minorant.

The concave biconjugate is an upper envelope of $F$, and equals $F$ for a proper closed concave function under the usual duality hypotheses. It equals $-(-F{)}^{\ast \ast }$ in terms of the ordinary biconjugate. If $e=D{F}_{\ast }(\tau )$, differentiability and concavity of $F_{\ast }$ imply $F_{\ast }(\tau )+F_{\ast \ast }(e)=\tau :e$. For a finite differentiable dual the equality follows because $\tau$ minimizes $s:e-F_{\ast }(s)$.

The convex conjugate records a function using slopes rather than input positions. Let $f:{\mathbb{R}}^n\to \mathbb{R}\cup \{+\infty \}$ be a proper extended-real function: it is finite somewhere and never takes $-\infty$. Its convex conjugate iswhere $p\cdot x$ is the dot product and $\mathrm{dom}f=\{x:f(x)<+\infty \}$ is the effective domain. An infinite value of $f$ excludes that input from the supremum. The value $f^{\ast }(p)$ can be $+\infty$, but cannot be $-\infty$ because an input with finite $f(x)$ supplies a finite lower bound.

The convex conjugate is automatically a convex function, even when $f$ is not convex: each fixed input gives an affine function $p\mapsto p\cdot x-f(x)$, and their supremum is convex. It is also lower semicontinuous. It need not be proper: for $f(x)=-x^2$ on $\mathbb{R}$, $px-f(x)=px+x^2$ is unbounded above for every $p$. The definition and the automatic convexity are discussed in Boyd and Vandenberghe, Convex Optimization (2004), §3.3.1.

The best lower bound with a prescribed slope. Whenever $f^{\ast }(p)$ is finite, the affine functionis an affine minorant of $f$. Indeed, an affine function $p\cdot x+b$ lies below $f$ everywhere exactly when $b\le {\inf }_x(f(x)-p\cdot x)=-f^{\ast }(p)$. Thus $-f^{\ast }(p)$ is the largest possible intercept at slope $p$. The convex conjugate trades positions for slopes: $x$ specifies where to inspect $f$, whereas $p$ specifies which global affine minorant to use. Taking the supremum over all these affine minorants produces the biconjugate; the Fenchel-Moreau theorem says this recovers a proper convex function when it is lower semicontinuous.

A quadratic, computed by completing the square. For the convex quadratic function $f(x)=x^2/2$,Consequently $f^{\ast }(p)=p^2/2$, and the unique maximizing input is $x=p$. The affine minorant ${\ell }_p(x)=px-p^2/2$ is the tangent line at $(p,p^2/2)$: its slope is $p$ and its intercept is $-f^{\ast }(p)=-p^2/2$. For example, $p=2$ gives $f^{\ast }(2)=2$ and the tangent line $2x-2$, touching at $(2,2)$. This is the same curve described either by its point values or by the intercepts of its tangent lines.

A corner turns into a constraint on slopes. For the absolute value $f(x)=|x|$, if $|p|\le 1$ then $px-|x|\le 0$, with value zero at $x=0$. If $p>1$, taking $x\to +\infty$ makes $(p-1)x\to +\infty$; if $p<-1$, taking $x\to -\infty$ does the same. Therefore its convex conjugate is zero on $[-1,1]$ and $+\infty$ outside: it is the indicator functional of a constraint set for this interval. For $|p|<1$ the unique maximizing input is zero; for $p=1$ every $x\ge 0$ maximizes, and for $p=-1$ every $x\le 0$ maximizes. The corner's many supporting slopes are exactly the subdifferential $\partial f(0)=[-1,1]$.

A supremum need not be attained. For a differentiable convex function finite on all of ${\mathbb{R}}^n$, an input maximizes $p\cdot x-f(x)$ exactly when $p=\nabla f(x)$. If $f$ is strictly convex, there is at most one such input; existence is a separate issue. For the exponential function $f(x)=e^x$, the derivative is positive. At $p>0$, solving $p=e^x$ gives $x=\log p$ and $f^{\ast }(p)=p\log p-p$. At $p=0$, the supremum of $-e^x$ is zero as $x\to -\infty$, with no maximizing input. At $p<0$, $px-e^x\to +\infty$ as $x\to -\infty$. Thus a strictly convex differentiable function can have both unattained finite conjugate values and infinite conjugate values.

The Legendre-Fenchel transform and biconjugate areThe Fenchel-Moreau theorem says that every proper convex function which is lower semicontinuous equals its biconjugate. More generally, if a proper extended-real $f$ has an affine minorant, thenwhere the right side is the greatest lower semicontinuous convex minorant. This is biconjugation as closed convexification. The affine-minorant hypothesis ensures that the closed convexification is proper; an unqualified statement including arbitrary improper functions would need separate conventions.

Here are the essential proof steps. The Fenchel–Young inequality gives $f^{\ast \ast }\le f$. Every term in the supremum defining $f^{\ast \ast }$ is an affine minorant of $f$, and conversely any affine minorant $⟨p,x⟩+a\le f(x)$ has $a\le -f^{\ast }(p)$. Thus $f^{\ast \ast }$ is exactly the supremum of all affine minorants, hence is convex and lower semicontinuous.

Put $g=\mathrm{clconv}f$. Its epigraph is the closed convex hull of the epigraph of $f$. Applying the half-space representation of a closed convex set in ${\mathbb{R}}^{n+1}$ recovers that epigraph from its containing half-spaces. A containing half-space writtenhas $q\le 0$, since epigraphs extend upwards. If $q<0$, it is precisely the epigraph inequality of an affine minorant, $t\ge (⟨v,x⟩-a)/(-q)$.

Vertical half-spaces with $q=0$ must also be accounted for. Choose one affine minorant ${\ell }_0$ of $g$, which exists because $g$ is proper and closed: strictly separate $(x_0,g(x_0)-1)$ from its epigraph for a finite domain point $x_0$; the separating coefficient of $t$ cannot be zero, since that would not distinguish points with the same $x_0$. If a vertical containing inequality is $⟨v,x⟩\le a$, thenis still an affine minorant on the domain of $g$. At a point violating the vertical inequality, these minorants tend to infinity as $L\to \infty$. Thus vertical domain restrictions are also recovered by the supremum of affine minorants.

Consequently $g$ equals that supremum. Every affine minorant of $g$ is below $f$, while the epigraph of any affine minorant of $f$ contains its closed convex epigraph hull. Therefore $f$ and $g$ have the same affine minorants, completing $g=f^{\ast \ast }$. The theorem in the previous solution supplies the geometric separation step behind biconjugation.

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.

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.