For any $y\in E$, $|x-y|\leq|x-z|+|z-y|$. Taking the infimum over $y$ gives $d_E(x)\leq|x-z|+d_E(z)$; reverse $x,z$ for the Lipschitz estimate. Taking the infimum over the closure gives the same value by continuity of $y\mapsto|x-y|$. A point has zero distance precisely when points of $E$ approach it, proving the zero-set identity. Inclusion enlarges the set over which the infimum is taken, so $E\subset F$ implies $d_E\geq d_F$. Conversely, if that inequality holds, every $x\in E$ has $d_F(x)=0$, hence lies in closed $F$. The infimum over a union is the smaller of the two infima. Finally an open ball and its closure have the same closure, so their distance functions coincide.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

Write $\delta=d_H(K,L)$. Choose $y\in L$ nearest to $x$ and $z\in K$ nearest to $y$. Then $d_K(x)\leq|x-y|+|y-z|\leq d_L(x)+\delta$. Reversing the sets proves the global upper bound. On $K$, $|d_K-d_L|=d_L$; on $L$, it equals $d_K$. Maximizing there recovers the defining $\delta$, proving both equalities. The uniform norm triangle inequality gives the metric triangle inequality; symmetry and nonnegativity are immediate, and zero distance implies equality of the closed zero sets. Finally $L+\varepsilon B=\{d_L\leq\varepsilon\}$ by attainment of nearest points, so the two inclusions are precisely the two directed distance bounds. In the maximum on $K\cup L$ we mean a pointwise supremum, not a Lebesgue essential supremum on a possibly null set.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

A linear functional takes its maximum on the ball in direction $u$ at $c+ru/|u|$, giving $c\cdot u+r|u|$ (also valid for $u=0$). On a segment its maximum is $\max(a\cdot u,b\cdot u)$; on a convex hull it is $\max_i v_i\cdot u$, because a convex combination cannot exceed that maximum. Finally maximize $(x+y)\cdot u$ independently over $K,L$. The same convex-combination argument gives the union formula. For the half-space representation, every point of $K$ satisfies every displayed inequality. If $z\notin K$, a separating hyperplane supplies a unit $u$ with $z\cdot u>h_K(u)$, excluding $z$ from the intersection. Inclusion plainly implies support inequalities; conversely, those inequalities place every point of $K$ in every supporting half-space of $L$, hence in $L$. In particular, equality of support functions determines the convex set uniquely.

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The condition $d_H(K,L)\leq\varepsilon$ is equivalent to $K\subset L+\varepsilon B$ and $L\subset K+\varepsilon B$. By separation, inclusion of convex sets is equivalent to the corresponding support-function inequalities. The support of $L+\varepsilon B$ is $h_L+\varepsilon$ on the unit sphere. Thus the two inclusions are exactly $|h_K-h_L|\leq\varepsilon$. Width is $h(u)+h(-u)$, so the triangle inequality gives the stated bound. The equality with the uniform distance-function difference holds even without convexity, by the distance-function exercise. For the unit circle $S^1$ and closed unit disk $B\subset\mathbb R^2$, the support functions are both identically one on $S^1$, but $d_H(S^1,B)=1$, attained by the distance of the origin to the circle. Thus support functions encode the convex hull and cannot recover holes in a general compact set.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

For $y\in K$ and unit $u$, $x\cdot u-h_K(u)\leq(x-y)\cdot u\leq|x-y|$, so every expression is at most $d_K(x)$. If $x\notin K$, let $p$ be a nearest point. Convexity and minimality give $(x-p)\cdot(y-p)\leq0$ for every $y\in K$: differentiate $|x-p-t(y-p)|^2$ at $t=0+$. Thus $u=(x-p)/|x-p|$ supports $K$ at $p$, and $x\cdot u-h_K(u)=|x-p|$, attaining the bound. If $x\in K$ all terms are nonpositive, proving the first formula. For an interior point, $B(x,r)\subset K$ is equivalent to $x\cdot u+r\leq h_K(u)$ for every unit $u$. Thus the largest contained centered ball has radius $\min_u(h_K(u)-x\cdot u)$. This is exactly $\operatorname{dist}(x,\partial K)$: every smaller centered ball stays inside, while a larger one passes beyond a nearest boundary point along the ray from $x$. Negating gives the second formula; boundary points have value zero.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

Hausdorff convergence and the distance-function bound give the first condition, while nearest points in $K_n$ give the second. Conversely, if $\max_{x\in K_n}d_K(x)$ stays above some $\eta>0$ along a subsequence, choose maximizing points and extract a convergent subsequence in $Q$. Its limit lies in $K$ by condition 1, a contradiction. If the reverse directed distance stays positive, choose $y_j\in K$ witnessing it and pass to $y_j\to y\in K$. Condition 2 gives $z_n\in K_n$ with $z_n\to y$, so $d_{K_{n_j}}(y_j)\leq|y_j-y|+|y-z_{n_j}|\to0$, another contradiction.

For inclusion, approximate $x\in A$ by $x_n\in A_n\subset B_n$ and apply condition 1 for $B_n\to B$. For convexity, approximate $x,y\in K$ by $x_n,y_n\in K_n$; each $(1-t)x_n+ty_n$ lies in convex $K_n$, so its limit lies in $K$.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

The union inequality follows immediately from the two neighborhood inclusions for each pair. For the hulls, compactness follows from Carathéodory: every point is a convex combination of at most $d+1$ points, so the hull is the continuous image of the compact product $K^{d+1}$ times the coefficient simplex. If $\delta=d_H(K,L)$, replace each $x_i\in K$ in a convex combination by $y_i\in L$ with $|x_i-y_i|\leq\delta$. The resulting combination moves by at most $\delta$. Reversing the roles proves the estimate.

For the counterexample in the plane, take $K_n=K=[-1,1]\times\{0\}$ and $L_n=\{(t,t/n):-1\leq t\leq1\}$. Then $d_H(L_n,K)\leq1/n$, but $K_n\cap L_n=\{0\}$ for every $n$, whereas $K\cap K=K$. The Hausdorff distance between these intersections is one. Thus convexity alone does not make intersection continuous.

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For the decreasing case, compactness of $K_1$ and the finite-intersection property give a nonempty compact intersection $K$. Any limit of $x_j\in K_{n_j}$ lies in every closed $K_m$, since eventually $n_j\geq m$; and every $x\in K$ belongs to all $K_n$. The sequential criterion applies.

For the increasing case the closure $K$ of the union is compact under confinement, and any cluster point stays in $K$. Given $x\in K$ and $\varepsilon>0$, choose $y\in K_m$ with $|x-y|<\varepsilon$. Then $d_{K_n}(x)<\varepsilon$ for every $n\geq m$, so nearest points supply the other sequential condition. Without confinement, $K_n=[-n,n]$ has no compact Hausdorff limit.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

The restrictions $d_{K_n}|_Q$ are uniformly bounded by $\operatorname{diam}Q$ and 1-Lipschitz. Arzelà–Ascoli gives a uniformly convergent subsequence with limit $f\geq0$. Put $K=\{x\in Q:f(x)=0\}$. Choose $y_n\in K_n$ and pass to a convergent subsequence; uniform convergence and Lipschitz continuity show its limit belongs to $K$, so $K$ is nonempty and compact. The 1-Lipschitz property gives $f(x)\leq d_K(x)$ for $x\in Q$.

Fix $x\in Q$ and choose nearest points $z_n\in K_n$. Along a subsequence $z_n\to z\in Q$. Uniform convergence gives $f(z)=0$, while $f(x)=\lim|x-z_n|=|x-z|\geq d_K(x)$. Therefore $f=d_K$ on $Q$. Finally the Hausdorff distance equals the supremum of the distance-function difference already on $K_n\cup K\subset Q$, so uniform convergence on $Q$ proves the claimed Hausdorff convergence. Convexity of the sets is not needed.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

For $x\in Q\setminus\Omega$, both $d_{Q\setminus\Omega}(x)$ and $q_\Omega(x)$ are zero. If $x\in\Omega$, the nearest point of its full complement lies on $\partial\Omega\subset Q\setminus\Omega$, so the two distances again coincide. The distance-function characterization of $d_H$ may take its supremum on $Q$, because both compact complements lie there. Outside $Q$ both $q$ functions vanish, giving the displayed equality and independence of the containing ball. The positive set of $q_\Omega$ is precisely $\Omega$, so this is a metric on bounded open sets, including the empty set.

Since $q_\varnothing=0$, the distance to the empty set is $\sup q_\Omega$, the largest possible clearance from the boundary, equivalently the supremal contained-ball radius. For $U$, $q_U(x)=(1-|x|)_+$. Inside the punctured ball the clearance is $\min(|x|,1-|x|)$, with value zero at $0$. At $0$ the two clearances differ by one, and nowhere can they differ by more than one. Hence $d_H^c(U,U\setminus\{0\})=1$, even though the compact closures have Hausdorff distance zero.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

The assertion is automatic if $F$ is empty. Otherwise, by compactness, $a=\min_F q_\Omega>0$. Uniform convergence of the clearance functions gives $q_{\Omega_n}\geq a/2$ on $F$ for large $n$, hence $F\subset\Omega_n$. In particular every fixed $x\in\Omega$ eventually lies in $\Omega_n$. Thus $1_{\Omega\setminus\Omega_n}\to0$ pointwise and is dominated by the integrable indicator of the containing ball. Dominated convergence gives the first measure assertion. Since $|\Omega_n|\geq|\Omega|-|\Omega\setminus\Omega_n|$, the lower-semicontinuity assertion follows. Apply the first argument to the compact support of the function for the final statement. Nothing here controls the extra measure in $\Omega_n\setminus\Omega$.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

Every point of $[0,1]$ is within $1/(2n)$ of $G_n$, with equality at the grid-interval midpoints; the reverse directed distance is zero. This proves the first formula. With containing ball (interval) $Q=[0,1]$, the complements of $\Omega_n$ and $\varnothing$ are $G_n$ and $[0,1]$, giving the second formula. Each grid has measure zero but its compact Hausdorff limit has measure one. Conversely $|\Omega_n|=1$ while its complementary Hausdorff limit is empty. Removing finitely many points changes neither the closure of the open interval nor its indicator almost everywhere. The same sequence therefore records different information under compact-closure convergence, convergence of indicators, and complementary Hausdorff convergence.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

A compact subset of either candidate limit is contained in $U$, hence in every $\Omega_n$. Both candidates have closure $\overline U$. A compact $G$ outside $\overline U$ has a positive distance from it, so for large $n$ it is outside $\overline{\Omega_n}$. Thus both distinct open sets are limits under this definition. The complement of the closure in the second condition is essential; replacing it by the complement of the open set would give a different convergence notion.

For the measure claim, a point of $\Omega$ eventually belongs to every $\Omega_n$, and a point outside $\overline\Omega$ eventually belongs to none. The indicators therefore converge pointwise away from $\partial\Omega$. That boundary has measure zero. On any bounded region, dominated convergence applies to $|1_{\Omega_n}-1_\Omega|^p\leq1$, proving local $L^p$ convergence. If all sets lie in one bounded region, the convergence is global in $L^p$.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

The function $|1_E-1_F|$ is the indicator of $E\triangle F$, so every positive finite power has the same integral. The triangle inequality follows from $E\triangle G\subset(E\triangle F)\cup(F\triangle G)$; zero distance means equality modulo null sets. The essential supremum is one precisely when the symmetric difference has positive measure.

Under weak-* convergence, testing against $1$ and $1_E$, both integrable on $Q$, gives $|E_n|\to|E|$ and $|E_n\cap E|\to|E|$. Thus $$|E_n\triangle E|=|E_n|+|E|-2|E_n\cap E|\to0.$$ The first formula gives every finite $L^p$ convergence. Select a subsequence with $|E_{n_j}\triangle E|\leq2^{-j}$. The sum of these indicator differences is integrable, hence finite almost everywhere, so outside a null set membership differs only finitely often. This proves subsequential almost-everywhere convergence without making a claim about the full sequence.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

For continuous $\varphi$, the integral of $1_{E_n}-1/2$ over each cell is zero. Subtracting the left-endpoint value of $\varphi$ on each cell bounds the total integral against $\varphi$ by its modulus of continuity at $1/n$, which tends to zero. The functions $1_{E_n}-1/2$ are uniformly bounded, so approximation in $L^1$ extends convergence to every $L^1$ test function. This is weak-* convergence to the constant density $1/2$.

However $|1_{E_n}-1/2|=1/2$ almost everywhere, so its $L^p$ norm is always $1/2$. Any strong limit would also be the weak limit $1/2$, which is impossible. Each component interval of $E_n$ has length $1/(2n)$, so the largest clearance is $1/(4n)$. Hence $d_H^c(E_n,\varnothing)=1/(4n)\to0$. The fractional weak limit of indicators and the empty domain limit describe different phenomena.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

The support-distance identity gives $h_K-\delta_n\leq h_{K_n}\leq h_K+\delta_n$. Since $h_K\geq r$, this implies $$(1-\delta_n/r)h_K\leq h_{K_n}\leq(1+\delta_n/r)h_K,$$ which is the claimed pair of inclusions. Both $K_n$ and $K$ lie between the same inner and outer dilates, so $$|K_n\triangle K|\leq\big[(1+\delta_n/r)^d-(1-\delta_n/r)^d\big]|K|\to0.$$ Volume convergence follows as well. For a nonempty compact $F\subset\operatorname{int}K$ (the empty case is automatic), there is a uniform support margin $\eta=\min_{x\in F,u\in\mathbb S^{d-1}}(h_K(u)-x\cdot u)>0$. When $\delta_n<\eta$, every point of $F$ has positive support margin in $K_n$, proving the interior inclusion.

If instead the Hausdorff limit is lower-dimensional, the volume still tends to zero: all $K_n\subset K+\delta_nB$, whose volume tends to $|K|=0$: for every fixed $\varepsilon>0$ the inclusion $K_n\subset K+\varepsilon B$ holds eventually, and $|K+\varepsilon B|\downarrow|K|$ as $\varepsilon\downarrow0$. This uses continuity from above of finite Lebesgue measure and does not require $\delta_n$ to decrease monotonically. The multiplicative sandwich itself requires the interior ball and is not asserted in that case.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

Put $\delta=d_H(K,L)=\max_u|h_K(u)-h_L(u)|$. Since $s_K(x)=\max_u(x\cdot u-h_K(u))$, the elementary bound for the difference of two maxima gives $|s_K(x)-s_L(x)|\leq\delta$. At $x\in\partial K$, $s_K(x)=0$ and $|s_L(x)|=\operatorname{dist}(x,\partial L)$; hence that distance is at most $\delta$. Reverse the roles to obtain $d_H(\partial K,\partial L)\leq\delta$.

Every compact convex body is the convex hull of its boundary (an interior point lies on a chord with boundary endpoints). Nonexpansiveness of convex hull gives $d_H(K,L)\leq d_H(\partial K,\partial L)$, proving equality. For a finite grid in the real line, $\partial G_n=G_n$, whereas $\partial[0,1]=\{0,1\}$. Along even $n$, the distance between these boundaries is $1/2$, although $d_H(G_n,[0,1])=1/(2n)\to0$.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

Write $\delta=d_H(K,L)$. Given $x,y\in K$, choose $x',y'\in L$ within $\delta$ of them. Then $|x-y|\leq|x'-y'|+2\delta$. Take suprema and reverse the roles for the diameter bound. If $L\subset B(c,R)$, then $K\subset L+\delta B\subset B(c,R+\delta)$; taking infima over enclosing balls and reversing the roles proves the radius bound.

For convex bodies, a ball $B(c,r)$ contained in $K$ gives $h_K(u)\geq c\cdot u+r$. The support-distance estimate implies $h_L(u)\geq c\cdot u+r-\delta$, hence $B(c,r-\delta)\subset L$ when $r\geq\delta$. If $r<\delta$ the required inequality $r(L)\geq r-\delta$ is automatic. Optimize over inscribed balls and reverse roles. Concentric balls whose radii differ by $\delta$ attain all three constants.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

Inclusion gives the inequality. Take unit disks with centers $(-a,0),(a,0)$, where $0<a<1$. Both support functions equal $1$ in direction $(0,1)$. In their lens, adding the inequalities $(x\pm a)^2+y^2\leq1$ gives $y^2\leq1-a^2-x^2$. The uppermost point is $(0,\sqrt{1-a^2})$, so the intersection support is strictly below $1$.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

The support function of $-K$ at $u$ is $h_K(-u)$. Hence the support of $K+(-K)$ on the sphere is precisely the width function. A convex body is determined by its support function, and $h_{wB}=w$. More explicitly, if $K_i$ has width $w_i$ and $\lambda_i\geq0$, then $h_{\sum_i\lambda_iK_i}(u)+h_{\sum_i\lambda_iK_i}(-u)=\sum_i\lambda_iw_i$. Whenever this sum is positive, the resulting body has that constant width; the all-zero combination degenerates to a point.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

For distinct $x,y\in K$, choose $u=(x-y)/|x-y|$; then $|x-y|\leq h_K(u)+h_K(-u)=w$. Choose contact points $x$ and $y$ for normals $u,-u$. Their projection difference is $w$, while their distance is at most $w$, so equality forces $x-y=wu$. Fixing $y$ shows that there is only one possible $x$, and conversely. The opposite contact chord has length $w$, so the diameter bound is attained. Thus all support faces are singletons and $K$ is strictly convex. Strict convexity rules out boundary segments; it does not rule out corners, where one contact point has several supporting normals.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

Translate the symmetry center to zero. Then $K=-K$, so $h_K(u)=h_K(-u)$ and the width condition gives $h_K=w/2$. This is the support function of $(w/2)B$.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

For a unit vector $u\in E$, $(\operatorname{proj}_E x)\cdot u=x\cdot u$, so the projected support function is the restriction of $h_K$ to $E$. The opposite-direction sum remains $w$. Barbier’s theorem applied in each plane gives $\pi w$. Equal shadow perimeters do not assert equal shadow areas.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

For any center $c$, write $H_c(u)=h_K(u)-c\cdot u$. The smallest enclosing radius about $c$ is $R_c=\max_u H_c(u)=\max_{x\in K}|x-c|$. Since $H_c(u)+H_c(-u)=w$, its minimum is $w-R_c$. If this minimum is positive, it is the largest inscribed radius about $c$, by the support criterion for inclusion. A body of positive constant width has interior: otherwise an orthogonal direction to its affine hull has width zero. An interior center therefore has $R_c<w$. Therefore the minimum of $R_c$ is strictly less than $w$. At a minimizing center $c_*$, the positive quantity $w-R_{c_*}$ is a contained-ball radius. For any other inscribed ball $B(c,r)$, the support inequalities give $r\leq\min_u H_c(u)=w-R_c\leq w-R_{c_*}$. Thus this ball is a largest inball, the same center gives a smallest enclosing ball, and $r(K)+R(K)=w$. Existence of a minimum follows because $R_c\to\infty$ as $|c|\to\infty$.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

The diameter bound gives the forward inclusion. If $z\notin K$, separate it by a support plane: $z\cdot u>h_K(u)$. Let $y$ be the contact point for $-u$. Then $(z-y)\cdot u>w$, so $z\notin B(y,w)$. For the spindle statement, every ball in the displayed representation contains the chosen two points; the intersection over all balls containing them is a subset of this particular intersection.

Checked draft. ChatGPT 6 Astra; mathematical and exposition review, 2026-10-15; human mathematical review pending

For any contact point $X$ with outer normal $n(\theta)$, the support inequality gives $h(\theta+s)\geq X\cdot n(\theta+s)$, with equality at $s=0$. Differentiating this scalar inequality at its minimum yields $X\cdot t(\theta)=h^{\prime}(\theta)$; also $X\cdot n(\theta)=h(\theta)$. Thus the contact point is unique and $$X(\theta)=h(\theta)n(\theta)+h^{\prime}(\theta)t(\theta).$$ Every boundary point lies on some supporting line, so this parametrization covers the boundary. Differentiate with $n^{\prime}=t$ and $t^{\prime}=-n$ to obtain $X^{\prime}=(h+h^{\prime\prime})t=\rho t$. Since $\rho>0$, increasing $\theta$ traverses the boundary counterclockwise once, and $ds=\rho\,d\theta$. Periodicity gives $$P=\int_0^{2\pi}\rho\,d\theta=\int_0^{2\pi}h\,d\theta.$$ Green’s formula and $\det(n,t)=1$ give $$A=\frac12\int_0^{2\pi}\det(X,X^{\prime})\,d\theta=\frac12\int_0^{2\pi}h(h+h^{\prime\prime})\,d\theta.$$ Periodic integration by parts yields $A=\frac12\int_0^{2\pi}(h^2-(h^{\prime})^2)\,d\theta$. This derivation does not assume in advance that the unknown contact-point map is differentiable.