21.11.1 The notion of convex functionsFB8022E8BCA8

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.