Let A A and B B be measurable subsets of ℝ n \mathbb{R}^n . Then

Vol ( A + B ) 1 / n ≥ Vol ( A ) 1 / n + Vol ( B ) 1 / n Vol(A + B)^{1/n} \geq Vol(A)^{1/n} + Vol(B)^{1/n}

where A + B = { a + b : a ∈ A , b ∈ B } A + B = \{ a + b : a \in A, b \in B \} . This is the Brunn-Minkoswki inequality. It’s most often mentioned in the case where A A and B B are convex, but it’s true in the vast generality of measurable sets (except for ∅ \emptyset , where it obviously fails).

Here I want to explain just a few basic things about this inequality and its consequences. For instance, it easily implies the famous isoperimetric inequality: that among all sets with a given surface area, the one that maximizes the volume is the ball.

There’s a fantastic expository paper that covers everything I’m about to say and much, much more:

Richard Gardner, The Brunn-Minkowski inequality. Bulletin of the American Mathematical Society 39 (2002), 355-405.

Thanks to Mark Meckes for mentioning this paper to me years ago. Cognoscenti like him may prefer this earlier, longer version. Both versions contain, among other things, a nice short proof of the inequality (Section 4 of the published version and Section 6 of the longer version). But I won’t give the proof here, nor will I mention the conditions for equality to hold.

The term A + B A + B is known as a Minkowski sum. You can think of A + B A + B as A A smeared out in a pattern dictated by B B :

A + B = ⋃ b ∈ B ( A + b ) . A + B = \bigcup_{b \in B} (A + b).

For instance, if A A is an egg shape (green) and B B is a quadrilateral (blue) then A + B A + B looks like the entire shape here:

(When I wrote those words, I didn’t mean to invoke the mental image of smearing an egg around a table. But maybe it’s not such a bad image to have.) Or again, if A A is egg-like and B B is a line segment then A + B A + B looks like this:

Of course, there’s also a complementary viewpoint: A + B A + B is B B smeared out in a pattern dictated by A A :

A + B = ⋃ a ∈ A ( a + B ) . A + B = \bigcup_{a \in A} (a + B).

In any case, an important observation is that

A + B A + B can have large volume even if A A and B B both have volume zero.

For instance, this happens in the plane when A A is a horizontal line segment and B B is a vertical line segment. There’s obviously no hope of getting an equation for Vol ( A + B ) Vol(A + B) in terms of Vol ( A ) Vol(A) and Vol ( B ) Vol(B) . But this example suggests that we might be able to get an inequality, stating that Vol ( A + B ) Vol(A + B) is at least as big as some function of Vol ( A ) Vol(A) and Vol ( B ) Vol(B) .

The Brunn-Minkowski inequality does this, but it’s really about linearized volume, Vol 1 / n Vol^{1/n} , rather than volume itself. If length is measured in metres then so is Vol 1 / n Vol^{1/n} . There’s also a much cruder corollary of Brunn-Minkowski that concerns actual volume. Take the Brunn-Minkowski inequality

Vol ( A + B ) 1 / n ≥ Vol ( A ) 1 / n + Vol ( B ) 1 / n Vol(A + B)^{1/n} \geq Vol(A)^{1/n} + Vol(B)^{1/n}

and raise each side to the power of n n . The right-hand side can be expanded using the binomial theorem, and throwing away all except the first and last terms gives

Vol ( A + B ) ≥ Vol ( A ) + Vol ( B ) Vol(A + B) \geq Vol(A) + Vol(B)

as a corollary of Brunn-Minkowski.

This is a much weaker statement than the Brunn-Minkowski inequality, because in deriving it we threw away a huge number of positive terms. But it’s interesting in its own right, and has an easy direct proof. Here’s an argument I was told by Joe Fu. I’ll indent it for easy skippability.

Let A A and B B be measurable subsets of ℝ n \mathbb{R}^n , and assume they’re both bounded. (Once we’ve proved the bounded case, it’s not hard to deduced the general case, but I’ll leave that as a puzzle.) We’re going to prove that A + B A + B contains almost-disjoint translates of A A and of B B , where by “almost disjoint” I mean that their intersection has measure zero. The inequality will follow. The idea is that we shift A A as high as possible and B B as low as possible. Formally, define h : ℝ n → ℝ h: \mathbb{R}^n \to \mathbb{R} by as h ( x 1 , … , x n ) = x n h(x_1, \ldots, x_n) = x_n . (That’s h h for “height”, but any other nonzero linear functional would work just as well.) Choose b + ∈ B b^+ \in B maximizing h ( b + ) h(b^+) and a − ∈ A a^- \in A minimizing h ( a − ) h(a^-) . It’s enough to show that ( A + b + ) ∩ ( a − + B ) (A + b^+) \cap (a^- + B) is a subset of a hyperplane, since then it has measure zero. So, let x x be any point in this intersection. We can write x = a + b + x = a + b^+ for some a ∈ A a \in A , and so h ( x ) = h ( a + b + ) = h ( a ) + h ( b + ) ≥ h ( a − ) + h ( b + ) = h ( a − + b + ) . h(x) = h(a + b^+) = h(a) + h(b^+) \geq h(a^-) + h(b^+) = h(a^- + b^+). But we can also write x = a − + b x = a^- + b for some b ∈ B b \in B , and a similar argument gives h ( x ) ≤ h ( a − + b + ) h(x) \leq h(a^- + b^+) . So we’ve shown that every x ∈ ( A + b + ) ∩ ( a − + B ) x \in (A + b^+) \cap (a^- + B) belongs to the hyperplane h − 1 ( h ( a − + b + ) ) h^{-1}(h(a^- + b^+)) , and that completes the proof.

That was a very weak consequence of the Brunn-Minkowski inequality, but here’s a more powerful and interesting one. In a slogan:

Brunn-Minkowski for small balls is the isoperimetric inequality.

In other words, if you take any A ⊆ ℝ n A \subseteq \mathbb{R}^n , and take B B to be a ball of radius tending to zero, then the Brunn-Minkowski inequality becomes the isoperimetric inequality in ℝ n \mathbb{R}^n .

To explain this, I first need to explain what the isoperimetric inequality is. The word “isoperimetric” suggests the two-dimensional version: that among all subsets of ℝ 2 \mathbb{R}^2 with a prescribed perimeter, the one with the greatest area is the disc. But I mean the general n n -dimensional version, which I’ll now state.

Every reasonable subset A ⊆ ℝ n A \subseteq \mathbb{R}^n has a volume (its Lebesgue measure), which I’ll write now as V n ( A ) V_n(A) . It also has a surface area, which I’ll write as V n − 1 ( A ) V_{n - 1}(A) . The isoperimetric inequality states that for nice enough subsets A ⊆ ℝ n A \subseteq \mathbb{R}^n (and I’ll come back to what that means),

V n ( A ) ≤ V n ( r B n ) V_n(A) \leq V_n(r B^n)

where the radius r r is chosen so that

V n − 1 ( A ) = V n − 1 ( r B n ) . V_{n - 1}(A) = V_{n - 1}(r B^n).

We can easily solve for r r in this last equation, using the fact that V n − 1 V_{n - 1} is homogeneous of degree n − 1 n - 1 . After a little bit of elementary algebraic rearranging, we find an equivalent form of the isoperimetric inequality that doesn’t have an “ r r ” in it:

V n ( A ) 1 / n V n − 1 ( A ) 1 / ( n − 1 ) ≤ V n ( B n ) 1 / n V n − 1 ( B n ) 1 / ( n − 1 ) \frac{V_n(A)^{1/n}}{V_{n - 1}(A)^{1/(n -1)}} \leq \frac{V_n(B^n)^{1/n}}{V_{n - 1}(B^n)^{1/(n - 1)}}

for all A ⊆ ℝ n A \subseteq \mathbb{R}^n .

This can be understood as follows.

Loosely speaking, the isoperimetric inequality says that the ratio of a shape’s volume to its surface area is greatest for balls. But we shouldn’t take the word “ratio” literally: for if we simply divided volume by surface area then we’d get a quantity measured in metres, so we could make it bigger simply by scaling the set up! What we want this ratio to be is a dimensionless quantity.

Since volume is measured in metres n {}^n and surface area is measured in metres n − 1 {}^{n - 1} , we can get a dimensionless quantity by linearizing both things before we take the ratio. In other words, take the n n th root of the volume and the ( n − 1 ) (n - 1) th root of surface area, so that both are measured in metres. Then the ratio will come out to be dimensionless.

(If you’re feeling perverse, you could instead take the ( 17 / n ) (17/n) th power of the volume and the ( 17 / ( n − 1 ) ) (17/(n - 1)) th power of the surface area, then divide one by the other. This would be a dimensionless quantity. But since it’s just the 17th power of the ratio in the last paragraph, and taking the 17th power is an order-preserving operation, it wouldn’t change the meaning of the inequality.)

To get much further, we need to ask: what is surface area? The usual idea is that the volume of a thin shell around A A , of thickness ε ≈ 0 \varepsilon \approx 0 , should be approximately ε \varepsilon times the surface area. So, we define

V n − 1 ( A ) = lim ε → 0 + V n ( A ε ) − V n ( A ) ε , V_{n - 1}(A) = \lim_{\varepsilon \to 0+} \frac{V_n(A_\varepsilon) - V_n(A)}{\varepsilon},

where A ε A_\varepsilon means the set of all points within distance ε \varepsilon of A A . Since we’re talking about Minkowski sums, it’s useful to observe that

A ε = A + ε B n A_\varepsilon = A + \varepsilon B^n

where B n B^n means the n n -dimensional unit ball.

I confess that I don’t know exactly when the limit above exists; I guess there are measurable sets A A for which it doesn’t. And among those sets for which the limit does exist, I don’t know when it’s the right definition of surface area. But certainly everything works as it should when A A is convex and compact, since then V n ( A ε ) V_n(A_\varepsilon) is actually a polynomial in ε \varepsilon whose linear term is the surface area. (That’s part of Steiner’s theorem.) So let’s just assume that A A is convex and compact.

Now we’ll use this formula to calculate the surface area of the unit ball. Taking A = B n A = B^n ,

V n − 1 ( B n ) = lim ε → 0 V n ( ( 1 + ε ) B n ) − V n ( B n ) ε = V n ( B n ) ⋅ lim ε → 0 ( 1 + ε ) n − 1 ε = n V n ( B n ) , \begin{aligned} V_{n - 1}(B^n) & = \lim_{\varepsilon \to 0} \frac{V_n((1 + \varepsilon)B^n) - V_n(B^n)}{\varepsilon} \\ & = V_n(B^n) \cdot \lim_{\varepsilon \to 0} \frac{(1 + \varepsilon)^n - 1}{\varepsilon} \\ & = n V_n(B^n), \end{aligned}

using the fact that V n V_n is homogeneous of degree n n . So,

V n − 1 ( B n ) = n V n ( B n ) . V_{n - 1}(B^n) = n V_n(B^n).

For instance, in ℝ 3 \mathbb{R}^3 , the volume of the unit ball is ( 4 / 3 ) π (4/3) \pi and the surface area of the unit sphere is 3 ⋅ ( 4 / 3 ) π = 4 π 3 \cdot (4/3) \pi = 4\pi .

Back to Brunn-Minkowski! I promised to show you that the isoperimetric inequality is just the Brunn-Minkowski inequality in the case where one of the two sets involved is a small ball. We’ve put the isoperimetric inequality into a form where it involves ( 1 / n ) (1/n) th powers, which also appear in the Brunn-Minkowski inequality. So things are looking promising… Here goes.

Take A ⊆ ℝ n A \subseteq \mathbb{R}^n (definitely measurable, probably convex and compact). The Brunn-Minkowski inequality tells us that for all ε > 0 \varepsilon \gt 0 ,

V n ( A + ε B n ) 1 / n ≥ V n ( A ) 1 / n + V n ( ε B n ) 1 / n . V_n(A + \varepsilon B^n)^{1/n} \geq V_n(A)^{1/n} + V_n(\varepsilon B^n)^{1/n}.

Rearranging, this is equivalent to

V n ( A + ε B n ) − V n ( A ) ε ≥ ( V n ( A ) 1 / n + ε V n ( B n ) 1 / n ) n − V n ( A ) ε . \frac{V_n(A + \varepsilon B^n) - V_n(A)}{\varepsilon} \geq \frac{(V_n(A)^{1/n} + \varepsilon V_n(B^n)^{1/n})^n - V_n(A)}{\varepsilon}.

Now let ε → 0 \varepsilon \to 0 . The left-hand side is just the definition of V n − 1 ( A ) V_{n - 1}(A) . The right-hand side is recognizable as a derivative, give or take a constant factor. So the Brunn-Minkowski inequality for (vanishingly) small balls states that for nice enough A ⊆ ℝ n A \subseteq \mathbb{R}^n ,

V n − 1 ( A ) ≥ n V n ( A ) ( n − 1 ) / n V n ( B n ) 1 / n . V_{n - 1}(A) \geq n V_n(A)^{(n - 1)/n} V_n(B^n)^{1/n}.

This looks like a mess until we remember the formula for the surface area of the unit ball: V n − 1 ( B n ) = n V n ( B n ) V_{n - 1}(B^n) = n V_n(B^n) . Using this to get rid of the n n in the inequality, and doing one last bit of elementary rearrangement, we get

V n ( A ) 1 / n V n − 1 ( A ) 1 / ( n − 1 ) ≤ V n ( B n ) 1 / n V n − 1 ( B n ) 1 / ( n − 1 ) . \frac{V_n(A)^{1/n}}{V_{n - 1}(A)^{1/(n -1)}} \leq \frac{V_n(B^n)^{1/n}}{V_{n - 1}(B^n)^{1/(n - 1)}}.

And that’s exactly the isoperimetric inequality.