$\newcommand{\vol}{\operatorname{vol}}$Let $K$ be a convex body in $\Bbb R^n$ (a convex body is a convex, compact subset of $\Bbb R^n$ with nonempty interior). The mean width of $K$ is defined as $b(K)=\dfrac{2}{\vol_{n-1}(\partial B_2^n)} \int_{S^{n-1}} h_K(u) \,du$, where $h_K(u)=\sup_{x\in K}\langle x,u\rangle$ is the support function of $K$ and $du$ is shorthand for $dH^{n-1}(u)$. Is there a closed-form expression for the mean width of a disjoint union of convex bodies? Or perhaps a closed form expression for the support function of a disjoint union?

For example, for convex bodies $K$ and $L$ the volume (and surface area) satisfy a valuation property $\vol_n(K\cup L)=\vol_n(K)+\vol_n(L)-\vol_n(K\cap L)$ (and similarly for surface area), even when $K\cup L$ is not convex. Is this true for the mean width as well, i.e. is it true that $b(K\cup L)+b(K\cap L)=b(K)+b(L)$? If not, does one of the directions in the inequality always hold? Or, by what formula or inequality are the support function $h_{K\sqcup L}$ of a disjoint union and $h_K,h_L$ related?

My attempt at a solution: By Groemer's extension theorem, the mean width can be extended uniquely to a valuation on the class of polyconvex sets, so it satisfies the valuation property $b(K\cup L)+b(K\cap L)=b(K)+b(L)$. (seems fishy)

Thank you in advance for your help.


1 Answer 1


these are actually two questions.

  1. If you define the "mean width" for non-convex bodies by the same integral over $h_K(x)=\sup_{v\in K} \langle x,v\rangle$, the "mean width" of the union of two convex bodies will in general depend on the concrete shape and the relative position of the bodies. Consider a simple example:

    Let $K$ be a unit disk centered at $(-d/2,0)$ and $L$ be a unit disk centered at $(d/2,0)$, $d>2$ Then $b(K)=b(L)=2$ and $b(K\cap L)=b(\emptyset)=0$. However, if an element of $S^1$ is represented by $u\in [0,2\pi]$:

    \begin{align*} b(K\cup L)&=\frac{1}{\pi}\int_0^{2\pi} (1+ d/2 |\cos u|) du \\ &= 2 + \frac{2d}{\pi} \int_0^{\pi/2} \cos u du = 2+ \frac{2d}{\pi} \; . \end{align*}

    Hence there can't be a formula expressing this integral just in terms of $b(K)$, $b(L)$ and $b(K\cap L)$.

  2. The second question is, whether $b$ can be extended from the class of convex bodies to the class of arbitrary unions of convex bodies, such that additivity holds. The answer is yes. The mean width is a valuation for convex bodies, as the support function mapping is a valuation, s. e.g. Rolf Schneider: Convex Bodies: The Brunn–Minkowski Theory, p. 330. See also p. 331 for an explicit additive extension of the support function and thus the mean width. (Groemer's extension theorem might have been used instead to prove the existence of such an extension.)

    The extension of the support function looks like $$h_K(x) = \sum_{\lambda \in {\mathbb R}} (\chi(H_{x,\lambda}\cap K) - \lim_{\mu\searrow \lambda} \chi(H_{x,\mu}\cap K)) \; ,$$ and the integral definition of the mean width with this extension of the support function yields the extension of the mean width.

  • $\begingroup$ Thank you for the informative answer. I did not see this in time to give you the points bounty. I wonder if there is a way I can still give you the points? $\endgroup$
    – user3816
    Oct 17, 2016 at 18:20
  • 1
    $\begingroup$ Never mind the points, we do like mathematics, don't we? $\endgroup$ Oct 18, 2016 at 17:40
  • $\begingroup$ haha yes we do. thank you again. $\endgroup$
    – user3816
    Oct 18, 2016 at 21:20
  • $\begingroup$ I am trying to obtain the integrand (support function) you computed for $b(K\cup L)$. For a fixed angle $u$, I looked at the intersection of the line $y=(\tan u)x$ with the boundary of $L$ (the outer intersection point) to get an $x$ value, and thus a $y$ value from the equation of the circle $\partial L$. One can then use the pythagorean theorem to get $h_{K\cup L}(u)$? $\endgroup$
    – user3816
    Oct 18, 2016 at 21:50
  • $\begingroup$ nevermind, i figured it out. ill post the solution later for others who may come across this question. thanks again, stefan! $\endgroup$
    – user3816
    Oct 19, 2016 at 1:21

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