21.11 Convex functions and convex bodies9A65EFDDFA3F

Proof. Assume first that the epigraph is convex. Given $a>F(n)$ and $b>F(r)$, choose finite $a'<a$ and $b'<b$ with $F(n)\leq a'$ and $F(r)\leq b'$. Convexity of the epigraph gives $F(tn+(1-t)r)\leq ta'+(1-t)b'<ta+(1-t)b$.

Conversely, take $(n,a),(r,b)$ in the epigraph. For every $\varepsilon>0$ the assumed inequality applied to $a+\varepsilon,b+\varepsilon$ gives $F(tn+(1-t)r)<ta+(1-t)b+\varepsilon$. Let $\varepsilon\downarrow0$. The resulting weak inequality says that the convex combination of these two points belongs to the epigraph. Endpoints cause no difficulty. This proof also covers values $-\infty$; points with value $+\infty$ contribute no points to the epigraph.

Proof. For every $x\in N$ and $a\in\mathbb R$, the inequality $a\geq\sup_iF_i(x)$ holds exactly when $a\geq F_i(x)$ for every $i$. Consequently \[\operatorname{epi}(\sup_iF_i)=\bigcap_i\operatorname{epi}F_i.\] An intersection of convex sets is convex, so the supremum is convex. This also covers extended values and an empty epigraph.

Proof. Let $H$ denote the infimum displayed in the statement, with the infimum of the empty set equal to $+\infty$. It is at most $F_i$ for every $i$: put weight $1$ on that index and weight $0$ on the others, choosing arbitrary points in their nonempty effective domains.

To prove convexity, take admissible representations of $x$ and $y$, multiply their weights by $t$ and $1-t$, and combine them. For each index $i$ with combined weight $\gamma_i>0$, replace its two points by their weighted barycenter. Convexity of $F_i$ makes its new contribution no larger than the sum of the old contributions. If $\gamma_i=0$, choose any point of its domain. This produces a representation of $tx+(1-t)y$. Taking approximations to the infima proves the strict-height convexity criterion proved above, including when an infimum is $-\infty$.

Finally, every convex function $G\leq F_i$ for all $i$ is at most the value of every admissible combination, by convexity applied to finite epigraph heights above the $F_i(x_i)$ and then taking their limits. Thus $G\leq H$. Therefore $H$ is the largest convex minorant of $\inf_iF_i$, which is precisely $F_1\land\cdots\land F_m$.

Proof. For a closed proper $F$, finite-dimensional Fenchel duality gives \[F(x)=F^{**}(x)=\sup_p\{\langle p,x\rangle-F^*(p)\}.\] For every $p$ with $F^*(p)<+\infty$, the expression in braces is an affine minorant of $F$, by the definition of conjugacy. Conversely, every affine minorant $\langle p,x\rangle+c$ satisfies $c\leq-F^*(p)$. Thus the supremum is exactly that of all affine minorants. If $F\equiv+\infty$, arbitrary constant minorants already have supremum $+\infty$. If $F\equiv-\infty$, there are no finite affine minorants and their supremum is $-\infty$.

Proof. Apply the closure theorem for proper convex functions. It proves that the lower semicontinuous regularization of $F$ is closed and proper and agrees with $F$ on the relative interior of its effective domain. Since $F$ is proper, this regularization is exactly the closure defined here.

Outside the closure of $\operatorname{Dom}F$, both functions are $+\infty$: a neighborhood misses the effective domain, so the lower semicontinuous regularization is $+\infty$ there. Every remaining point outside the relative boundary belongs to the relative interior of the domain. Therefore disagreement is possible only on that relative boundary, as asserted.

Proof. For proper closed functions this is the uniqueness assertion in the convex closure theorem. More explicitly, the closure of the epigraph over the relative interior of the effective domain is the whole epigraph: for a point of the epigraph, take convex combinations with a fixed epigraph point whose base is in that relative interior, and let the coefficient of the fixed point tend to zero. Thus identical values on the common relative interior give identical epigraphs and hence identical functions.

The improper cases in the definition of closedness are also determined: empty effective domain means the function identically $+\infty$; if one function is identically $-\infty$, the common relative interior is $N$ and the value agreement makes the other identical to it. These cases complete the proof.

Proof. On every open convex ball relatively compact in $C$, dense convergence for finite convex functions applies to the restriction of the given dense set and supplies a finite convex local limit with locally uniform convergence. Local limits agree on overlaps, because they are pointwise limits of the same sequence. They therefore define a finite function $f$ on $C$.

Every compact subset of $C$ is covered by finitely many smaller such balls. Taking the maximum of the finitely many uniform error bounds proves uniform convergence on that compact set. Passing to the limit in each convexity inequality proves convexity wherever the corresponding segment lies in $C$; in the usual convention that the domain of a convex function is convex, this is convexity on $C$. In particular the limiting values at the original dense set determine $f$ uniquely.

Proof. For proper $F$, all assertions are the finite-dimensional Fenchel theorem, whose separation proof is given earlier: $F^*$ is closed and proper, $(\operatorname{cl}F)^*=F^*$, and $F^{**}=\operatorname{cl}F$.

For completeness consider improper convex functions. If $F\equiv+\infty$, then $F^*\equiv-\infty$ and $F^{**}\equiv+\infty$. Otherwise an improper $F$ takes $-\infty$ somewhere, so every $F^*(p)$ is $+\infty$, its biconjugate is identically $-\infty$, and this is exactly the closure prescribed in the definition. Its conjugate is again identically $+\infty$. Neither conjugate is proper in these improper cases. This proves the equivalence of properness and all the identities also at the extended-valued extremes.

Proof. Put $H=\operatorname{cl}\operatorname{CE}F$. Since $H\leq F$, order reversal gives $H^*\geq F^*$. Fix $p$ with $F^*(p)$ finite. The affine function $\ell(x)=\langle p,x\rangle-F^*(p)$ lies below $F$, hence below its convex envelope and below its lower semicontinuous regularization, because $\ell$ itself is continuous. Thus $\langle p,x\rangle-H(x)\leq F^*(p)$ for all $x$, proving $H^*(p)\leq F^*(p)$.

If $F^*(p)=+\infty$, the reverse inequality already gives equality. If $F^*(p)=-\infty$, then $F$ is identically $+\infty$; its closed convex envelope is also identically $+\infty$, and the equality is immediate. This treats every possible value of the conjugate.

Proof. Subgradients at points outside the effective domain are empty by definition. On the relative interior, existence and the directional formula follow from local convex regularity applied in the affine span $A$ of the domain. To view a subgradient as a functional on $N$, extend its restriction to the direction space of $A$ by a linear functional. Any functional vanishing on that direction space may be added without changing the subgradient inequality. Thus for a direction outside that space the supremum in the directional formula is $+\infty$, just as the one-sided directional derivative is; for tangent directions it is the relative formula.

At an interior point of the domain, local Lipschitz continuity bounds all subgradients: if $|F(x+v)-F(x)|\leq L|v|$ for small $v$, the subgradient inequality for $v$ and $-v$ gives $|\langle p,v\rangle|\leq L|v|$, hence $|p|\leq L$. Conversely, suppose the subgradient set is nonempty at a point $x$ of the domain which is not interior. A supporting hyperplane to the closure of the convex domain at $x$ supplies a nonzero $q$ with $\langle q,y-x\rangle\leq0$ for every $y$ in the domain. If $p$ is a subgradient, then so is $p+tq$ for every $t\geq0$, so the set is unbounded. Supporting hyperplanes also exist when the domain lies in a proper affine subspace, by taking an annihilating functional. This proves the final equivalence.

Proof. If $p\in\partial F(x)$, the subgradient inequality gives $F^*(p)=\langle p,x\rangle-F(x)<+\infty$, proving the first inclusion. Now assume $F$ is closed and let $p$ lie in the relative interior of $\operatorname{Dom}F^*$. The conjugate is closed and proper by Fenchel duality, so it has a subgradient $x$ at $p$. The same duality theorem gives $p\in\partial F(x)$. Thus the relative interior is contained in the subgradient image. Taking closures in the two inclusions and using that a nonempty convex set has the same closure as its relative interior proves the equality of closures.

Proof. For every decomposition $p=\sum_i p_i$, the defining inequalities for the conjugates give $(\sum_iF_i)^*(p)\leq\sum_iF_i^*(p_i)$. It remains to prove the reverse inequality when the left side is finite. The case $r=1$ is immediate.

For $r\geq2$, put \[V(z_1,\ldots,z_{r-1})=\inf_x\left\{\sum_{i<r}F_i(x+z_i)+F_r(x)-\langle p,x\rangle\right\}.\] This is convex, being an infimum over the fibers of a convex epigraph. Its value at zero is $-(\sum_iF_i)^*(p)$, which is finite. Choose $x_0$ in the common relative interior in the hypothesis. The product of the effective domains has $(x_0,\ldots,x_0)$ in its relative interior. The linear map $(x_1,\ldots,x_r)\mapsto(x_1-x_r,\ldots,x_{r-1}-x_r)$ maps a relative neighborhood of this point onto a relative neighborhood of zero in its image affine span: a surjective linear map between finite-dimensional spaces has a continuous linear right inverse. Consequently zero lies in the relative interior of $\operatorname{Dom}V$.

The function $V$ cannot take $-\infty$ anywhere. If it did, combining that point with a point of its effective domain on the opposite side of zero would force $V(0)=-\infty$. It is therefore proper, and existence of supporting slopes gives $p_1,\ldots,p_{r-1}$ with $V(z)\geq V(0)+\sum_{i<r}\langle p_i,z_i\rangle$. Set $p_r=p-\sum_{i<r}p_i$. Substituting arbitrary $x_i=x+z_i$ in this inequality yields \[\sum_i\bigl(F_i(x_i)-\langle p_i,x_i\rangle\bigr)\geq V(0).\] Taking independent infima proves $\sum_iF_i^*(p_i)\leq-V(0)=(\sum_iF_i)^*(p)$. Together with the first inequality this is equality. If the conjugate on the left is $+\infty$, the first inequality forces every decomposition cost to be $+\infty$, so the formula still holds.

Proof. Apply the proved decreasing smooth approximation with fixed growth body. It gives smooth convex $F_j$ with bounded difference from the support function of $P$, decreasing pointwise, in fact locally uniformly, to $F$. The bounded-difference condition is precisely membership in $\mathcal E^\infty(N,P)$, and $\mathcal P(N,P)$ is the same upper-growth class in both formulations. That construction preserves the common growth body $P$, including when $P$ has empty interior; therefore the empty-interior case requires no full-dimensional approximation of $P$.

Proof. The measure normalization here is $n!$ times the Lebesgue measure of the subgradient image. The density formula for locally $C^{1,1}$ convex functions proves that this measure is $n!\det D^2F\,\mathrm dx$. Its hypotheses hold because $C^{1,1}$ implies that the gradient is locally Lipschitz, and the lattice coordinates identify $\mathrm dx$ with the normalized $\mathrm d\operatorname{vol}$. The Hessian is its almost-everywhere derivative, equivalently the density of the distributional Hessian. Thus the formula includes nonsmooth $C^{1,1}$ functions and has the required factor $n!$.

Qualification and proof. The topology on $\mathcal P(N_{\mathbb R},P)$ is pointwise convergence. Since all these functions are finite and convex, local uniform convergence of convex functions shows that $F_j\to F$ uniformly on compact subsets. Alexandrov-measure continuity now gives convergence against every compactly supported continuous test function, that is, weak convergence as Radon measures on this noncompact space.

Convergence against all bounded continuous functions would require an additional no-loss-of-mass condition and is not automatic here. For example, on $\mathbb R$ the functions $F_j(x)=\log(1+\exp(x-j))$ have common growth body $[0,1]$ and converge locally uniformly to $0$, while their real Monge--Ampère measures have total mass $1$ and move to $+\infty$. Their vague limit is zero. Thus the weak topology in this assertion must be read as the compactly supported test-function topology; if total masses also converge, a compact-cutoff argument upgrades it to the bounded-continuous topology.

Proof. The construction of mixed Alexandrov measures gives a positive, symmetric, additive and positively homogeneous measure in the $n$ arguments, with diagonal equal to the Alexandrov measure defined here. Repeatedly expand the diagonal measure of $\sum_i\lambda_iF_i$ using this additivity and homogeneity. Each ordered choice $(i_1,\ldots,i_n)$ contributes $\lambda_{i_1}\cdots\lambda_{i_n}\operatorname{MA}_{\mathbb R}(F_{i_1},\ldots,F_{i_n})$, which is the asserted polynomial formula.

The earlier proof first verifies positivity for smooth convex functions using mixed determinants, and then uses convolution and local uniform continuity of the diagonal measure to pass to finite convex functions. In particular positivity is proved rather than inferred from a signed polarization formula. Uniqueness follows by recovering each coefficient from the diagonal polynomial, for instance by polarization. The normalization remains $n!$ on the diagonal.