21.4 Slope bodies and bounded growth8CD47210DEE8

Proof. If $F\leq h_ P+C$, conjugation reverses the inequality and gives $F^*\geq h_ P^*-C=\chi _ P-C$ by theorem 21.3.2, so $\operatorname {dom}F^*\subset P$. Every subgradient of $F$ lies in $\operatorname {dom}F^*$ by the same theorem, so the second condition implies the third. If all slopes of $F$ lie in $P$ and $p\in \partial F(x)$, the subgradient inequality at $x$, evaluated at $0$, gives

Thus $F\leq h_ P+F(0)$, which proves the first condition and, since $h_ P(0)=0$, the identity $\max _ x(F(x)-h_ P(x))=F(0)$. If $p_ x\in \partial F(x)$ and $p_ y\in \partial F(y)$, the subgradient inequalities give $\langle p_ y,y-x\rangle \geq F(y)-F(x)\geq \langle p_ x,y-x\rangle $, so $|F(y)-F(x)|\leq \max _{p\in P}|p|\, |y-x|$.

Conjugation reverses inequalities, and $F^{**}=F$ because the finite convex function $F$ is continuous. Hence $h_ P-C\leq F\leq h_ P+C$ holds if and only if $\chi _ P-C\leq F^*\leq \chi _ P+C$, that is, if and only if $F^*=+\infty $ outside $P$ and $|F^*|\leq C$ on $P$. In that case $K_ F=P$, so $F\in \mathcal E(N_{\mathbb R},P)$ (by the convention if $P$ has empty interior), and $\int _ PF^*\, \mathrm dp\leq C\operatorname {vol}(P)<\infty $; thus $\mathcal E^\infty \subset \mathcal E^1$. The inclusions $\mathcal E^1\subset \mathcal E\subset \mathcal P$ hold by definition. If $P$ has empty interior and $n>0$, then $P$ lies in a proper affine subspace, because its relative interior is nonempty and open in its affine span (lemma 21.2.3(4)); hence $\operatorname {vol}(P)=0$, every integral over $P$ vanishes, and $\mathcal E^1=\mathcal E=\mathcal P$.

Now let $F,G$ be finite convex functions with $K_ F$ compact. By proposition 21.3.3(4), $(F+G)^*(p)$ is finite exactly when $p=q+r$ with $q\in \operatorname {dom}F^*$ and $r\in \operatorname {dom}G^*$, so

It remains to pass to closures. Since addition $M_{\mathbb R}\times M_{\mathbb R}\to M_{\mathbb R}$ is continuous, $K_ F+K_ G$ lies in the closure of $\operatorname {dom}F^*+\operatorname {dom}G^*$. Conversely, if $q_ k+r_ k\to s$ with $q_ k\in \operatorname {dom}F^*$ and $r_ k\in \operatorname {dom}G^*$, then a subsequence of $(q_ k)$ converges to some $q\in K_ F$, because $K_ F$ is compact, and along it $r_ k\to s-q\in K_ G$; hence $s\in K_ F+K_ G$. This proves $K_{F+G}=K_ F+K_ G$. Finally, if $K_ F=P$ and $K_ G=P'$, then $F\preceq h_ P$ and $G\preceq h_{P'}$ by the first part, and $h_ P+h_{P'}=h_{P+P'}$; hence $F+G\in \mathcal P(N_{\mathbb R},P+P')$ and $K_{F+G}=P+P'$, so that $F+G\in \mathcal E(N_{\mathbb R},P+P')$. $\square$

Proof. If $n=0$, then $A*\rho _\varepsilon =A$ and there is nothing to prove; let $n\geq 1$. Recall that $\rho $ is smooth, nonnegative, even, supported in the unit ball, and has integral one. Since $A$ is continuous, being locally Lipschitz by proposition 21.2.4, the functions $A*\rho _\varepsilon $ are smooth by section 2.5, and they converge to $A$ locally uniformly by lemma 2.5.1(1). They are convex, because $A*\rho _\varepsilon (x)=\int A(x-y)\rho _\varepsilon (y)\, \mathrm dy$ and the convexity inequalities of the functions $x\mapsto A(x-y)$ can be integrated against $\rho _\varepsilon (y)\, \mathrm dy\geq 0$. For fixed $x$, consider the function

which is finite because $A$ is continuous and $\rho $ has compact support. The substitution $y\mapsto \varepsilon y$ shows that $g(\varepsilon )=A*\rho _\varepsilon (x)$ for $\varepsilon >0$, and $g(0)=A(x)$. For each $y$, the function $\varepsilon \mapsto A(x-\varepsilon y)$ is convex, so $g$ is convex; and $g$ is even, because $\rho $ is even. A convex even function on $\mathbb R$ is nondecreasing on $[0,+\infty )$: if $0\leq a<b$, then $a=\lambda b+(1-\lambda )(-b)$ with $\lambda =(a+b)/(2b)\in (0,1]$, so that $g(a)\leq \lambda g(b)+(1-\lambda )g(-b)=g(b)$. Hence $A*\rho _\varepsilon (x)$ is nondecreasing in $\varepsilon >0$ and bounded below by $g(0)=A(x)$. $\square$

Proof. Let $A_ j=\max \{ F,h_ P-j\} $. These are convex functions, decreasing in $j$, with $A_ j\sim h_ P$, and $A_ j=F$ on every compact set on which $h_ P-F\leq j$; hence $A_ j\to F$ locally uniformly. With the mollifiers $\rho _\varepsilon $ of lemma 21.4.3, choose $\varepsilon _ j\downarrow 0$ and set $F_ j=A_ j*\rho _{\varepsilon _ j}$. By that lemma, these functions are smooth and convex, and the monotonicity in $\varepsilon $ together with $A_{j+1}\leq A_ j$ gives $F_{j+1}=A_{j+1}*\rho _{\varepsilon _{j+1}}\leq A_ j*\rho _{\varepsilon _{j+1}}\leq A_ j*\rho _{\varepsilon _ j}=F_ j$, and $F_ j\geq A_ j\geq F$. By proposition 21.4.2, $F$ and $h_ P$ are Lipschitz with constant $\max _{p\in P}|p|$, hence so is $A_ j$, and therefore $|F_ j-A_ j|\leq C\varepsilon _ j$ with $C=\max _{p\in P}|p|\int |y|\rho (y)\, \mathrm dy$. Consequently $F_ j\sim A_ j\sim h_ P$, so $F_ j\in \mathcal E^\infty (N_{\mathbb R},P)$, and $F_ j\to F$ locally uniformly.

If $F_ j\downarrow F$ pointwise, then

and the conjugates $F_ j^*$ increase, because conjugation reverses inequalities. Finally, if a sequence of finite convex functions on $N_{\mathbb R}$ converges pointwise to a finite function, it converges locally uniformly by proposition 21.2.6, applied with $\Omega $ and the dense subset both equal to $N_{\mathbb R}$, and conversely local uniform convergence implies pointwise convergence. $\square$

Proof. If $\alpha =\sum _ i\lambda _ i\beta _ i$ is a convex combination, then $\langle \alpha ,x\rangle \leq \max _ i\langle \beta _ i,x\rangle \leq \log \sum _ i\mathrm e^{\langle \beta _ i,x\rangle }$, so the function is nonnegative. Conversely, suppose that $\alpha $ is not in the convex hull. The hull is compact, being the image of the standard simplex under the linear map $(\lambda _ i)\mapsto \sum _ i\lambda _ i\beta _ i$, and convex, so lemma 21.2.3(1), applied in $M_{\mathbb R}$, gives $v\in N_{\mathbb R}$ with $\langle \alpha ,v\rangle >\max _ i\langle \beta _ i,v\rangle $. Since the logarithmic sum differs from $\max _ i\langle \beta _ i,x\rangle $ by a number in $[0,\log m]$, the value of the function at $x=tv$ is at most $t(\max _ i\langle \beta _ i,v\rangle -\langle \alpha ,v\rangle )+\log m$, which tends to $-\infty $ as $t\to +\infty $. For the last assertion, put $x_ j=\log |z_ j|^2$. Then $|z^\gamma |^2=\mathrm e^{\langle \gamma ,x\rangle }$ for every $\gamma \in \mathbb Z^ n$, including exponents with negative entries, so the displayed quotient equals the exponential of minus the function above; and $z\mapsto x$ maps $(\mathbb C^*)^ n$ onto $\mathbb R^ n$. $\square$