Do Minkowski sums have anything like calculus?

Question

Is there anything resembling differential calculus over the space of (nicely behaved) regions in $\mathbb{R}^d$, where addition is interpreted in terms of Minkowski sums? For example, it is known that Minkowski sums act linearly on the perimeter of two-dimensional convex bodies. Is there any sense in which one could define something like a (constant) gradient for the perimeter function? I would assume that if anything, said gradient would just be a circle centered at the origin.

Answer 1

A convenient way to think about it is to represent a convex body in terms of its support function (restricted to the unit sphere). Minkowski addition corresponds to the addition of support functions. The perimeter is the integral of the support function: $$L(h) = \int_{\mathbb{S}^1} h\, dx = \langle h, 1 \rangle_{L^2}.$$ So, the gradient of the perimeter functional is the constant function $1$, that is the unit disk centered at the origin, as you have assumed.

Answer 2

There is a theory for convex bodies. If $A, B \subset \mathbb{R}^n$ are convex bodies, whose interiors contain the origin, you can use set addition to define $A+tB$, for any $t \ge 0$, and $$ V(A,B) = \lim_{t\rightarrow 0+} \frac{A+tB}{t}. $$ This is known as the first mixed volume. $h_A$ and $h_B$ are the support functions, then $$ V(A,B) = \int_{S^{n-1}} h_B\,dS_A, $$ where $S^{n-1}$ is the unit sphere and $dS_A$ is known as the surface area measure of $A$. In this sense, the gradient of the volume function is this measure. If $B$ is the unit ball, then $V(A,B)$ is the surface area (perimeter if $n = 2$). If you differentiate again, you get higher order curvature measures. It turns out that $V(A+tB)$ is a polynomial of degree $n$, called the Steiner polynomial, and its coefficients are called mixed volumes.