Brunn–Minkowski inequality for compact convex sets

If one of $A,B$ has zero volume, say ${\lambda }_d(A)=0$, choose $a\in A$. By [F11], $A+B$ is compact and measurable. Then $a+B\subseteq A+B$, so [F6] gives ${\lambda }_d(A+B)\ge {\lambda }_d(B)$; the stated inequality follows after taking $d$th roots. If both volumes vanish the right side is zero. Thus assume both volumes are positive.

Put $a={\lambda }_d(A{)}^{1\mathrm{/}d}$, $b={\lambda }_d(B{)}^{1\mathrm{/}d}$, $A^{\mathrm{\prime }}=a^{-1}A$, $B^{\mathrm{\prime }}=b^{-1}B$, and $\lambda =b\mathrm{/}(a+b)$. By [F5], ${\lambda }_d(A^{\mathrm{\prime }})={\lambda }_d(B^{\mathrm{\prime }})=1$; [F11] makes these dilates and their convex combinations compact. It suffices to show that every convex combination $(1-\lambda )A^{\mathrm{\prime }}+\lambda B^{\mathrm{\prime }}$ has volume at least one, because $A+B=(a+b)((1-\lambda )A^{\mathrm{\prime }}+\lambda B^{\mathrm{\prime }})\mathrm{.}$ The cases $\lambda =0,1$ are immediate, so take $0<\lambda <1$ and rename $A^{\mathrm{\prime }},B^{\mathrm{\prime }}$ as $A,B$.

We prove the normalized claim by induction on $d$. In dimension one, $A$ and $B$ are intervals of length one, so their convex combination is an interval of length $(1-\lambda )+\lambda =1$. Now let $d\ge 2$ and assume the normalized claim in dimension $d-1$. The normalization calculation in step 2.1 is dimension-independent, so this induction hypothesis also gives the full weighted inequality in dimension $d-1$ for positive-volume compact convex sets. By [F10], positive volume implies each body has nonempty interior and contains an interior ball. By [F11], their coordinate projections are compact intervals and their slices are compact and measurable; [F2] makes the slice-volume functions measurable. Project each convex body onto the first coordinate; its projection is a nondegenerate compact interval. For an interior coordinate $x$, let $A_x\subseteq {\mathbb{R}}^{d-1}$ be its slice and $f(x)={\lambda }_{d-1}(A_x)$. Define $B_y$ and $g(y)$ likewise. These slice volumes are positive and finite in the open projection intervals. Write the projections as $[a_{-},a_{+}]$ and $[b_{-},b_{+}]$. An interior ball in $A$ has center with first coordinate $x_0$. If $a_{-}<x<x_0$, choose $p\in A$ with first coordinate $p_1<x$ and combine $p$ with the ball centered at $x_0$; the convex combination with weight $\lambda =(x-p_1)\mathrm{/}(x_0-p_1)\in (0,1)$ on the ball lies in the slice at $x$ and contains a $(d-1)$-ball of radius $\lambda \rho >0$. If $x_0<x<a_{+}$, choose $p_1>x$ and use weight $(p_1-x)\mathrm{/}(p_1-x_0)$ on the interior ball. At $x=x_0$, the original ball supplies a positive-radius slice. The same argument applies to $B$.

Convexity gives $A_{(1-t)x_0+t{x}_1}\supseteq (1-t)A_{x_0}+t{A}_{x_1}$ for interior $x_0,x_1$. The compact slices in [F11] let the induction hypothesis apply to these sums. It therefore makes $f^{1\mathrm{/}(d-1)}$ concave on the open projection interval; likewise $g^{1\mathrm{/}(d-1)}$. To see continuity of $\psi =f^{1\mathrm{/}(d-1)}$, fix an interior point $x_0$ and $a<x_0<b$ in the interval. For $x\downarrow x_0$, concavity gives the chord lower bound $\psi (x)\ge \frac{b-x}{b-x_0}\psi (x_0)+\frac{x-x_0}{b-x_0}\psi (b)\mathrm{.}$ Writing $x_0=\eta a+(1-\eta )x$ with $\eta =(x-x_0)\mathrm{/}(x-a)$ gives the upper bound $\psi (x)\le \frac{\psi (x_0)-\eta \psi (a)}{1-\eta }\mathrm{.}$ Both bounds tend to $\psi (x_0)$; exchanging left and right proves continuity from the other side. Thus $f,g$ are continuous and strictly positive on their open intervals. By [F2], $\int f=\int g=1$.

Define the cumulative functions $F_A(x)={\int }_{a_{-}}^{x}f(t)dt$ and $F_B(y)={\int }_{b_{-}}^{y}g(t)dt$ on their projection intervals. Since $A,B$ lie in bounded boxes, all their slices lie in one fixed bounded box in ${\mathbb{R}}^{d-1}$; [F6, F9] therefore bound $f,g$ uniformly. Their cumulative integrals consequently extend continuously to the projection endpoints with values zero and one, using $\int f=\int g=1$ from step 4.1. Positivity of $f,g$ on the open intervals makes $F_A,F_B$ strictly increasing there, so they map those intervals onto $(0,1)$. On every compact interior subinterval [F7] gives derivatives $F_A^{\mathrm{\prime }}=f$ and $F_B^{\mathrm{\prime }}=g$. The inverse theorem then gives that $\phi =F_B^{-1}\circ F_A$ is $C^1$ on the open interval and ${\phi }^{\mathrm{\prime }}(x)=f(x)\mathrm{/}g(\phi (x))\mathrm{.}$ Let $z(x)=(1-\lambda )x+\lambda \phi (x)$; then $z^{\mathrm{\prime }}(x)=(1-\lambda )+\lambda f(x)\mathrm{/}g(\phi (x))>0$.

Let $C=(1-\lambda )A+\lambda B$, which is compact by [F11], and let $c_z$ be the $(d-1)$-dimensional volume of its compact slice at coordinate $z$. The slice-volume function is measurable by [F2] and [F11]. The slice of $C$ at $z(x)$ contains $(1-\lambda )A_x+\lambda B_{\phi (x)}$. The induction hypothesis yields $c_{z(x)}\ge ((1-\lambda )f(x{)}^{1\mathrm{/}(d-1)}+\lambda g(\phi (x){)}^{1\mathrm{/}(d-1)}{)}^{d-1}\mathrm{.}$ Set $r=g(\phi (x))\mathrm{/}f(x)>0$. Multiplying this bound by $z^{\mathrm{\prime }}(x)$ and using [F3] twice gives $c_{z(x)}{z}^{\mathrm{\prime }}(x)\ge f(x)((1-\lambda )+\lambda r^{1\mathrm{/}(d-1)}{)}^{d-1}((1-\lambda )+\lambda \mathrm{/}r)\ge f(x)\mathrm{.}$ The last inequality follows by applying [F3] to both positive brackets.

Restrict $x$ to an open interval $J$ strictly inside $(a_{-},a_{+})$. The map $z$ is $C^1$ with positive derivative by step 5.1, and [F7] makes it a $C^1$ diffeomorphism onto its open image $z(J)$. Thus [F4] and step 6.1 give ${\lambda }_d(C)=\int c_{z}dz\ge {\int }_{z(J)}{c}_{z}dz={\int }_J{c}_{z(x)}{z}^{\mathrm{\prime }}(x)dx\ge {\int }_{J}f(x)dx\mathrm{.}$ Here $c_z$ is measurable and the equality is the one-dimensional change of variables. For all sufficiently large integers $m$, take $J_m=(a_{-}+1\mathrm{/}m,a_{+}-1\mathrm{/}m)$; these intervals increase to $(a_{-},a_{+})$. By [F8] and $\int f=1$, ${\lambda }_d(C)\ge 1$. This proves the normalized claim in dimension $d$.

The induction proves the normalized claim in every finite dimension. Undoing the normalization in step 2.1 gives ${\lambda }_d(A+B{)}^{1\mathrm{/}d}\ge a+b={\lambda }_d(A{)}^{1\mathrm{/}d}+{\lambda }_d(B{)}^{1\mathrm{/}d},$ including the zero-volume cases from step 1.1.