PomeDefinition 21.7.1.B45BB086BF76 The Hausdorff distance between convex bodies $K,L\subset M_{\mathbb R}$ is
\[ d_{\mathrm H}(K,L)=\max \left\{ \sup _{p\in K}\operatorname {dist}(p,L), \sup _{q\in L}\operatorname {dist}(q,K)\right\} . \]
21.7 Lattice points and volume2479DAC58B22
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By theorem 21.9.5 below, the dimension of a space of sections on a projective toric manifold is the number of lattice points in a dilated polytope. We show that $k^{-n}$ times the number of lattice points in $kK$ tends to the volume of the convex body $K$, and that this remains true if the lattice points are prescribed only up to a sublinear distance from $kK$. The reason is that the boundary of a convex body has volume zero.
In this section we fix a basis of $M$. Its coordinates identify $M_{\mathbb R}$ with $\mathbb R^ n$ and $\operatorname {vol}$ with Lebesgue measure; we use the Euclidean norm of these coordinates on $M_{\mathbb R}$ and the dual norm on $N_{\mathbb R}$. For a convex body $K\subset M_{\mathbb R}$, $k>0$ and $\varepsilon >0$ put $kK=\{ kp:p\in K\} $ and
\[ K^\varepsilon =\{ p:\operatorname {dist}(p,K)\leq \varepsilon \} , \qquad K_\varepsilon =\{ p:\overline{B(p,\varepsilon )}\subset K\} , \]
and $K^0=K$. Since the distance to the compact set $K$ is attained, the outer parallel body $K^\varepsilon $ equals $K+\overline{B(0,\varepsilon )}$, the image of the compact convex set $K\times \overline{B(0,\varepsilon )}$ under addition; so it is a convex body. The inner set $K_\varepsilon =\bigcap _{|y|\leq \varepsilon }(K-y)$ is compact and convex but may be empty; it is empty whenever $K$ has empty interior and $n>0$, because a closed ball of positive radius has interior points.
Proposition 21.7.2.EDD78BC61BD7 If $n>0$, then for convex bodies $K,L\subset M_{\mathbb R}$
\[ d_{\mathrm H}(K,L)=\max _{|x|=1}|h_ K(x)-h_ L(x)|. \]
If convex bodies $K_ j$ converge to a convex body $K$ in this distance, then $\operatorname {vol}(K_ j)\to \operatorname {vol}(K)$. In particular,
\[ \lim _{\varepsilon \downarrow 0}\operatorname {vol}(K^\varepsilon ) =\operatorname {vol}(K)= \lim _{\varepsilon \downarrow 0}\operatorname {vol}(K_\varepsilon ). \]
In dimension zero the only convex body is the point, which has volume one.
Proof. Let $n>0$. If $d_{\mathrm H}(K,L)\leq \varepsilon $, then every $p\in K$ has a point $q\in L$ with $|p-q|\leq \varepsilon $, and for $|x|=1$ we get $\langle p,x\rangle \leq \langle q,x\rangle +\varepsilon \leq h_ L(x)+\varepsilon $; hence $h_ K\leq h_ L+\varepsilon $ on the unit sphere, and by symmetry $|h_ K-h_ L|\leq \varepsilon $ there. Conversely, suppose that $|h_ K-h_ L|\leq \varepsilon $ on the unit sphere. The support function of the convex body $L+\overline{B(0,\varepsilon )}$ is $h_ L(x)+\varepsilon |x|$. If some $p\in K$ were outside $L+\overline{B(0,\varepsilon )}$, lemma 21.2.3(1) would give a unit vector $x$ with $\langle p,x\rangle >h_ L(x)+\varepsilon \geq h_ K(x)$, which is impossible. Hence $K\subset L+\overline{B(0,\varepsilon )}$, and by symmetry $d_{\mathrm H}(K,L)\leq \varepsilon $. This proves the formula. It shows that $d_{\mathrm H}$ satisfies the triangle inequality, and by lemma 21.2.3(1) a convex body is determined by its support function; thus $d_{\mathrm H}$ is a metric.
Let $\varepsilon _ j=d_{\mathrm H}(K_ j,K)\to 0$. For large $j$ all $K_ j$ lie in the bounded set $K+\overline{B(0,1)}$. A point $p$ outside $K$ has positive distance from $K$, so $p\notin K_ j$ once $\varepsilon _ j<\operatorname {dist}(p,K)$. If $p$ is an interior point of $K$, choose $r>0$ with $\overline{B(p,r)}\subset K$. Then, for $|x|=1$,
\[ h_{K_ j}(x)\geq h_ K(x)-\varepsilon _ j \geq \langle p,x\rangle +r-\varepsilon _ j. \]
If $\varepsilon _ j<r$ and $p\notin K_ j$, strict separation would give a unit vector $x$ with $\langle p,x\rangle >h_{K_ j}(x)$, contradicting this inequality; hence $p\in K_ j$ for large $j$. Thus $\mathbf1_{K_ j}\to \mathbf1_ K$ outside the boundary of $K$. If $K$ has nonempty interior, its boundary is a null set by lemma 21.2.3(4); if $K$ has empty interior, it lies in a proper affine subspace and is itself a null set, and $\mathbf1_{K_ j}\to 0=\mathbf1_ K$ almost everywhere by the outside-point argument. In either case dominated convergence, with the dominating function $\mathbf1_{K+\overline{B(0,1)}}$, gives $\operatorname {vol}(K_ j)\to \operatorname {vol}(K)$.
The outer parallel bodies $K^\varepsilon $ have finite volume and decrease to the compact set $K$ as $\varepsilon \downarrow 0$, so continuity of measure from above gives the first limit. The inner sets $K_\varepsilon $ increase, as $\varepsilon $ decreases, to the interior of $K$, which differs from $K$ by a null set, as we saw in the previous paragraph; monotone convergence gives the second limit. In dimension zero both $K^\varepsilon $ and $K_\varepsilon $ equal the single point for every $\varepsilon $. $\square$
Theorem 21.7.3.831AEEB7084F Let $K\subset M_{\mathbb R}$ be a convex body, let $a_ k\in M_{\mathbb R}$ be arbitrary, and let $r_ k\geq 0$ satisfy $r_ k/k\to 0$. Then for every continuous function $f\colon M_{\mathbb R}\to \mathbb R$,
\[ \lim _{k\to \infty }\frac1{k^ n} \sum _{m\in (kK)^{r_ k}\cap (M+a_ k)}f(m/k) =\int _ K f(p)\, \mathrm dp. \]
In particular $k^{-n}\# (kK\cap M)\to \operatorname {vol}(K)$. Here the exponent $n$ and the measure are those of the ambient space, also when $K$ has empty interior. The convergence is uniform with respect to the translations $a_ k$.
Proof. For $n=0$ every set in question is the single point, and both sides equal its value under $f$. Suppose that $n>0$, and let $Q=[0,1)^ n$ be the half-open unit cube in the chosen coordinates. The cells
\[ \frac mk+\frac1k Q,\qquad m\in M+a_ k, \]
partition $M_{\mathbb R}$; each has volume $k^{-n}$, and its points lie within distance $c/k$ of its anchor $m/k$, where $c=\sqrt n$. Let $U_ k$ be the union of the cells whose anchors $m/k$ occur in the sum, that is, with $m/k\in K^{r_ k/k}$. Then
\[ K_{c/k}\subset U_ k\subset K^{(r_ k+c)/k}. \]
For the first inclusion, if $p\in K_{c/k}$, the anchor of the cell containing $p$ lies within $c/k$ of $p$, hence in $K$, so the cell is selected. For the second, every point of a selected cell lies within $c/k$ of its anchor, which lies within $r_ k/k$ of $K$. Since $K_{c/k}\subset K\subset K^{(r_ k+c)/k}$ and $(r_ k+c)/k\to 0$, proposition 21.7.2 gives $\operatorname {vol}(U_ k\mathbin \triangle K)\to 0$, and the volumes $\operatorname {vol}(U_ k)$ stay bounded.
For large $k$ the selected cells lie in the fixed compact set $K^1$. Let $\omega _ f$ be the modulus of continuity of $f$ on $K^1$. Since $k^{-n}f(m/k)$ is the integral of the constant $f(m/k)$ over the cell of $m$, we obtain
\[ \left|k^{-n}\sum _ m f(m/k)-\int _{U_ k}f(p)\, \mathrm dp\right| \leq \omega _ f(c/k)\operatorname {vol}(U_ k)\longrightarrow 0. \]
Since $f$ is bounded on $K^1$, the symmetric-difference estimate gives $\int _{U_ k}f\, \mathrm dp\to \int _ Kf\, \mathrm dp$. Taking $f=1$, $r_ k=0$ and $a_ k=0$ proves the counting formula. All the estimates depend only on the size of the cells, and not on their position, so they hold uniformly in the translations $a_ k$. $\square$
Corollary 21.7.4.CF4226A5034F Let $K\subset M_{\mathbb R}$ be a convex body, and suppose that subsets $S_ k\subset M$ satisfy
\[ kK\cap M\subset S_ k\subset (kK)^{r_ k}\cap M, \qquad r_ k\geq 0,\quad r_ k/k\to 0. \]
Then $k^{-n}\# S_ k\to \operatorname {vol}(K)$. In particular, a boundary error of bounded size does not affect the leading term of the count, even when $K$ has zero volume.
Proof. By theorem 21.7.3 with $f=1$ and $a_ k=0$, both $k^{-n}\# (kK\cap M)$ and $k^{-n}\# ((kK)^{r_ k}\cap M)$ tend to $\operatorname {vol}(K)$, and $\# S_ k$ lies between these two counts. $\square$
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