19.7.1 Convex bodiesA0D1AD337F41

Proof. Fix $R>0$ and let $\mathcal K(R)$ be the nonempty compact convex subsets of the closed ball $\overline B(0,R)$. For $K\in\mathcal K(R)$ its support function $h_K$ on the unit sphere is bounded by $R$ in absolute value and is $R$-Lipschitz. A diagonal selection on a countable dense subset of the sphere, followed by the common Lipschitz bound, gives a uniformly convergent subsequence $h_{K_j}\to h$ from every sequence.

Let $K=\{x:\langle u,x\rangle\leq h(u)\text{ for all unit }u\}$. This is closed, convex and contained in $\overline B(0,R)$. It is nonempty: select $x_j\in K_j$ and pass to a convergent subsequence; the limit satisfies all the inequalities. For every fixed $u$, select $x_j\in K_j$ maximizing $\langle u,x\rangle$ and pass to a subsequence. Its limit lies in $K$ and attains $h(u)$, so $h=h_K$.

For compact convex sets, $d_{\mathrm{Haus}}(K,L)=\sup_{|u|=1}|h_K(u)-h_L(u)|$. One inequality follows by approximating each point of either body within its Hausdorff distance. For the other, an upper bound $\delta$ on the support-function difference gives $K\subset L+\delta\overline B$ and the reverse inclusion by separation of a point from a closed convex set. Hence $K_j\to K$ in Hausdorff distance. We have proved compactness of $\mathcal K(R)$. Every sufficiently small closed Hausdorff ball about a given body lies in some $\mathcal K(R)$ and is closed there, so is compact. In dimension zero the space has just one member.

Proof. Let $K_j\to K$ in Hausdorff distance. All bodies lie in a fixed ball. Every point outside $K$ eventually lies outside $K_j$, because its distance from $K$ is positive. If $K$ has interior, every point $x$ of its interior eventually lies in $K_j$: choose $\rho>0$ with $x+\rho\overline B\subset K$; uniform convergence of support functions then gives $\langle u,x\rangle<h_{K_j}(u)$ for every unit $u$, which by separation implies $x\in K_j$.

The boundary of a convex body has Lebesgue measure zero, as proved in the relative-boundary lemma. Thus the indicator functions converge almost everywhere, and dominated convergence on the fixed ball proves volume convergence. If $K$ has empty interior, it lies in a proper affine subspace and has measure zero; the outside-point argument alone again gives indicator convergence almost everywhere. In dimension zero the normalized volume is always $1$.

Proof. Hausdorff convergence immediately supplies points $x_i\in K_i$ tending to each $x\in K$, and forces every convergent selection from a subsequence to land in $K$.

Conversely, the two conditions first imply that the $K_i$ are uniformly bounded. Otherwise choose a subsequence containing points $y_i$ tending to infinity, and, by the first condition, points $x_i\in K_i$ tending to a fixed $x\in K$. Choose $R$ with $K\subset B(x,R)$. Convexity gives $z_i\in[x_i,y_i]\subset K_i$ with $|z_i-x|=R$ for large $i$. A subsequence of these sphere points converges outside $K$, contradicting the second condition.

If the Hausdorff distance did not tend to zero, one of two failures would occur along a subsequence. Either there are $y_i\in K_i$ at a fixed positive distance from $K$; uniform boundedness gives a convergent subsequence, contradicting the second condition. Or there are $x_i\in K$ at a fixed positive distance from $K_i$; pass to $x_i\to x\in K$ and use the first condition to find $y_i\in K_i$ with $y_i\to x$, again a contradiction. This proves the converse and explains why convexity is needed in the boundedness step.

Proof. For compact sets $K,L,K',L'$, \[d_{\mathrm{Haus}}(K+L,K'+L')\leq d_{\mathrm{Haus}}(K,K')+d_{\mathrm{Haus}}(L,L').\] Indeed, given $x+y\in K+L$, select $x'\in K'$ and $y'\in L'$ at their respective distances from $x,y$; compactness ensures minimizers exist. The triangle inequality gives the stated bound on the distance to $K'+L'$. Interchange the pairs for the reverse bound. A Minkowski sum of two nonempty compact convex sets is nonempty, compact and convex. Hence this estimate proves continuity of the asserted map.

Proof. We prove the stated multiplicative form by induction on $n$. If either body has zero volume, its right-hand side is zero and positivity suffices. In dimension zero all volumes are $1$. In dimension one the bodies are intervals: their Minkowski combination has length $(1-t)a+tb\geq a^{1-t}b^t$ by the weighted arithmetic--geometric mean inequality.

Assume the result in dimension $n-1$ and let $A=K'$, $B=K$, $C=(1-t)A+tB$ have positive-volume input bodies. Slice them by the first coordinate and put $f(x)=\operatorname{vol}_{n-1}(A_x)$, $g(y)=\operatorname{vol}_{n-1}(B_y)$ and $h(s)=\operatorname{vol}_{n-1}(C_s)$. Slice inclusion and the induction hypothesis give \[h((1-t)x+ty)\geq f(x)^{1-t}g(y)^t.\] On the interiors of their projection intervals, $f$ and $g$ are positive and continuous. Continuity follows from the proved Hausdorff continuity of slices and volume continuity in dimension $n-1$; endpoint values do not affect integrals.

Normalize $f,g$ to have integral $1$, and divide $h$ by $(\operatorname{vol}A)^{1-t}(\operatorname{vol}B)^t$. Their inequality has the same form. Let $T$ be the increasing map matching the cumulative integrals of $f$ and $g$. On the two open projection intervals it is continuously differentiable and satisfies $T'(x)=f(x)/g(T(x))$. The map $S(x)=(1-t)x+tT(x)$ is increasing, so change of variables and the slice inequality give \[\int h\geq\int h(S(x))S'(x)\,\mathrm dx\geq\int f(x)^{1-t}g(T(x))^t\left((1-t)+t\frac{f(x)}{g(T(x))}\right)\,\mathrm dx\geq\int f=1.\] The last inequality is the weighted arithmetic--geometric mean inequality applied to $1$ and $f(x)/g(T(x))$. The changes of variables can first be made on compact subintervals and then passed to the full intervals by monotone convergence. Undoing normalization and using Fubini proves $\operatorname{vol}C\geq(\operatorname{vol}A)^{1-t}(\operatorname{vol}B)^t$, as required.