SANTALO’S INEQUALITY

TERENCE TAO

1. Introduction

One way to study a symmetric convex body^{1} A in R^{d} is by introducing its polar body A^{∘}, defined^{2} by

A^{∘} := {ξ ∈ R^{d} : |⟨x, ξ⟩| < 1 for all x ∈ A}.

Equivalently, A^{∘} is the largest subset of R^{d} such that ⟨A, A^{∘}⟩ ⊆ I, where I ⊂ R is the open interval I := {x ∈ R : |x| < 1} and ⟨A, B⟩ := {⟨x, ξ⟩ : x ∈ A, ξ ∈ B}. It is easy to verify that A^{∘} is another symmetric convex body. We then define the Mahler volume M(A) of A to be the quantity M(A) := mes(A)mes(A^{∘}).

We record some basic properties of the polar body and the Mahler volume:

Proposition 1.1. Let A, Ã be symmetric convex bodies in R^{d}, and let A′ be a symmetric convex body in R^{d′}, and let 1 ≤ p, q ≤ ∞ be such that 1/p + 1/q = 1. We define the p-sum set

A +_{p} Ã := ⋃_{0<θ<1}((1 − θ)^{1/p} · A + θ^{1/p} · Ã)

the p-intersection set

A ∩_{p} Ã := ⋃_{0<θ<1}((1 − θ)^{1/p} · A) ∩ (θ^{1/p} · Ã)

and the p-product set

A ×_{p} A′ := ⋃_{0<θ<1}(1 − θ)^{1/p} · A ×_{p} θ^{1/p} · Ã

(note that these correspond to the usual sum set, intersection, and Cartesian product when p = ∞, and that all three operations are associative). We also recall the Gamma function

Γ(x) := ∫_{0}^{∞} e^{s}x^{s−1}ds for x > 0.

• (Duality) We have (A^{∘})^{∘} = A, and thus M(A^{∘}) = M(A).

1A convex body is an open bounded convex set; it is symmetric if −A = A.

2A more invariant approach would be to let A be a subset of an arbitrary finite-dimensional vector space V, in which case the polar body A^{∘} would be a subset of the dual vector space V^{∗}, and one would not need the Euclidean inner product. However we will not need this formulation in our discussion here.

• (Affine invariance) If T : R^{d} → R^{d} is any invertible linear transformation, then (T A)^{∘} = (T^{∗})^{−1}(A^{∘}), and hence M(T A) = M(A).

• (Monotonicity) One has A ⊆ Ã if and only if Ã^{∘} ⊇ A^{∘}.

• (Sum and intersection) We have (A +_{p} Ã)^{∘} = A^{∘} ∩_{q} Ã^{∘} and (A ∩_{p} Ã)^{∘} = A^{∘} +_{q} Ã^{∘}.

• (Product) We have (A ×_{p} A′)^{∘} = A^{∘} ×_{q} (A′)^{∘} and

Γ((d + d′)/p + 1)mes(A ×_{p} A′) = Γ(d/p + 1)mes(A)Γ(d′/p + 1)mes(A′). (1)

In particular we have

M(A ×_{p} A′) = M(A ×_{q} A′) = \frac{Γ(d/p + 1)Γ(d′/p + 1)Γ(d/q + 1)Γ(d′/q + 1)}{Γ((d + d′)/p + 1)Γ((d + d′)/q + 1}M(A)M(A′). (2)

Proof We prove only the estimate (1), leaving the rest to Exercise 1. Define the norm ∥x∥_{A} := inf{t > 0 : x ∈ t · A}, and observe from Fubini’s theorem that

∫_{R^{d}} e^{−∥x∥_{A}^{p}} dx = ∫_{R^{d}} ∫_{t>0:x∈t·A} pt^{p−1}e^{−t^{p}} dtdx

= ∫_{0}^{∞} ∫_{t·A} pt^{p−1}e^{−t^{p}} dtdx

= mes(A) ∫_{0}^{∞} pt^{p+d}e^{−t^{p}} \frac{dt}{t}

= mes(A) ∫_{0}^{∞} s^{d/p+1}e^{−s} fracdss

= mes(A)Γ(d/p + 1).

Similarly for A′ and A × pA′. Since ∥(x, x′)∥_{A×_{p}A′}^{p} = ∥x∥_{A}^{p} + ∥x∥_{A′}^{p}, we have

∫_{R^{d}×R^{d′}} e^{−∥(x,x′)∥_{A×pA′}^{p}} dxdx′ = (∫_{R^{d}} e^{−∥x∥_{A}^{p}} dx)(∫_{R^{d′}} e^{−∥x′∥_{A′}^{p}} dx′)

and the claim follows.

We recall that the Gamma function is such that Γ(n+1) = n! for all integers n ≥ 0, that Γ(x + 1) = xΓ(x) for all x > 0, and that Γ(1/2) = \sqrt{π}, that log Γ is convex, and we have the Stirling approximation Γ(x) = (e + o(1))^{−x}x^{x} for large x (Exercise 2).

One can iterate (2) using the associativity of ×_{p} to conclude that

M(A_{1} ×_{p} … ×_{p} A_{n}) = \frac{∏_{j=1}^{n} Γ(d_{j}/p + 1)Γ(d_{j}/q + 1)M(A_{j})}{Γ(d/p + 1)Γ(d/q + 1)}

where each A_{j} is a symmetric convex body in R^{d_{j}} and d = d_{1} + … + d_{j}. Also note that for the interval I we have I = I^{∘} and thus M(I) = 4. Thus if one considers the d-dimensional ball B^{d} := I ×_{2} … ×_{2} I and the d-dimensional cube I^{d} := I ×_{∞} … ×_{∞} I or the unit octahedron O_{d} := I ×_{1} … ×_{1} I, then we have

M(B^{d}) = \frac{Γ(3/2)^{2d}4^{d}}{Γ(d/2 + 1)^{2}} = (2πe + o(1))^{d}d^{−d}

and

M(I^{d}) = M(O^{d}) = \frac{Γ(1)^{d}Γ(2)^{d}4^{d}}{Γ(d + 1)Γ(1)} = (4e + o(1))^{d}d^{−d} = (\frac{4}{2π} + o(1))^{d}M(B^{d}).

It is conjectured by Mahler that these are the two extreme cases, thus

M(I^{d}) ≤ M(A) ≤ M(B^{d})

for arbitrary symmetric convex bodies A.

Define the circumradius R(A) > 0 and the inradius r(A) > 0 of A to be the best quantities such that

r(A) · B^{d} ⊆ A ⊆ R(A) · B^{d}  (3)

and define the eccentricity e(A) ≥ 1 of A to be the quantity e(A) := R(A)/r(A). From Proposition 1.1 we easily see that

e(A)^{−d}M(B^{d}) ≤ M(A) ≤ e(A)^{d}M(B^{d})  (4)

while from John’s theorem [3] we see that we can ensure e(A) ≤ \sqrt{d} after an invertible linear transformation. This gives the somewhat weak estimate d^{−d/2}M(B^{d}) ≤ M(A) ≤ d^{d/2}M(B^{d}). Naive application of Fourier-analytic methods give even worse bounds (see Exercise 3), which seems to indicate some limitations of Fourier methods in obtaining bounds that are only polynomial in the doubling constant.

A convenient tool to study Mahler’s conjecture is Steiner symmetrization, which is in fact useful for many other problems in convex geometry also.

Definition 1.2 (Steiner symmetrization). If A is a set in R^{d}, we define the Steiner symmetral S_{0}(A) of A with respect to 0 to be the set

S_{0}(A) := \frac{1}{2} · (A − A) = \{\frac{x − y}{2} : x, y ∈ A\}.

More generally, if π : R^{d} → R^{d} is an orthogonal projection to a subspace of R^{d}, we define the Steiner symmetral S_{π}(A) of A with respect to π to be the set

S_{π}(A) := ⋃_{z∈π(A)} z + S_{0}(A ∩ π^{−1}(z)) = \{\frac{x − y}{2} + z : x, y ∈ A; z = π(x) = π(y)\}.

The facts about Steiner symmetrization which are relevant to Mahler’s conjecture are contained in the following lemma.

Lemma 1.3. [6] Let A, B be a symmetric convex bodies in R^{d} and let π : R^{d} → R^{d} be an orthogonal projection. Then S_{π}(A) is also an symmetric convex body. The operation A ↦ S_{π} is monotone (if A ⊆ B, then S_{π}(A) ⊆ S_{π}(B)), homogeneous (S_{π}(λ·A) = λ·S_{π}(A) for all λ > 0) and preserves the unit ball B^{d}. Furthermore, if π^{⊥}(x) := x−π(x) is the complementary projection to π, then we have the inclusions

⟨S_{π}(A), S_{π^{⊥}}(B)⟩ ⊆ S_{0}(⟨A, B⟩)  (5)

and

S_{π^{⊥}}(A^{∘}) ⊆ S_{π}(A)^{∘}.  (6)

Finally, we have the inequalities mes(S_{π}(A)) ≥ mes(A), M(S_{π}(A)) ≥ M(A), and

r(A) ≤ r(S_{π}(A)) ≤ R(S_{π}(A)) ≤ R(A).

In particular e(S_{π}(A)) ≤ e(A).

Proof The claims in the first two sentences are rather easy and are left to Exercise 4. To prove (5), we simply observe that if x, y ∈ A and z = π(x) = π(y), and ξ, η ∈ B and ζ = π^{⊥}(ξ) = π^{⊥}(η), then

⟨\frac{x − y}{2} + z, \frac{ξ − η}{2} + ζ⟩ = \frac{1}{2}(⟨x, ξ⟩ − ⟨y, ζ⟩)

(this is best seen by decomposing all vectors into components in the range of π and components in the range of π^{⊥}). Since S_{0}(I) = I and A^{∘} is the largest set such that ⟨A, A^{∘}⟩ ⊆ I, the inclusion (6) then follows immediately from (5). From the Brunn-Minkowski inequality we have mes(S_{0}(A)) ≥ mes(A); the corresponding claim for S_{π} then follows from Definition 1.2 Cavalieri’s principle (or Fubini’s theorem). From (6) this implies that mes(S_{π}(A)^{∘}) ≥ mes(A^{∘}), and hence M(S_{π}(A)) ≥ M(A). Finally, the claims about the inradius and circumradius follow from applying the monotonicity of Steiner symmetrization to the inclusions (3).

We can now obtain the upper bound in Mahler’s conjecture:

Theorem 1.4 (Santaló’s inequality). [8] For any symmetric convex body A ⊂ R^{d} we have M(A) ≤ M(B^{d}).

Proof [6] From Lemma 1.3 we see that Steiner symmetrization increases Mahler volume and decreases eccentricity. From this fact and (4), we see that to prove Santaló’s inequality it suffices to find a sequence of Steiner symmetrizations A_{0} = A, A_{1} = S_{π_{1}}(A_{0}), A_{2} = S_{π_{2}}(A_{1}), … whose eccentricity converges to 1. But this can be done in a number of ways, see for instance Exercise 7. This concludes the proof.

The lower bound to Mahler’s conjecture is still open, and considered very difficult. One reason for this is that in addition to the obvious minimizers M(I^{d}) and M(O^{d}), there also exist other convex bodies such as O^{d′} ×_{∞}O^{d−d′} or I^{d′} ×_{1}I^{d−d′} of the same Mahler volume, and so the variational structure of M at level M(I^{d}) = M(O^{d}) is extremely complex; it is unclear whether there is a procedure analogous to Steiner symmetrization to convert a general convex body into one of these minimizing bodies. The best known result is due to Bourgain and Milman [2], who established that M(A) ≥ C^{−d}M(B^{d}). The proof of this result requires some deep results from functional analysis and will not be given here. We will however give a slightly weaker result of Kuperberg, which has a simpler proof and improves upon the lower bound in (4). We first need a preliminary lemma, which is roughly speaking the analogue of Lemma 1.3 for Kuperberg’s argument:

Lemma 1.5. [5] Let A be a symmetric convex body in R^{d}. Then M(A) ≥ 2^{−d}M(A∩_{2} A^{∘}).

Proof We introduce the symmetric convex body K ⊂ R^{d} × R^{d} defined by K := T (A ×_{2} A^{∘}), where T : R^{d} × R^{d} → R^{d} × R^{d} is the shear map T (x, y) := (x, x + y). On one hand, since det(T) = 1, we have from Lemma 1.1 that

mes(K) = mes(A ×_{2} A^{∘}) = \frac{Γ(d/2 + 1)mes(A)Γ(d/2 + 1)mes(A^{∘})}{Γ(d + 1)} = M(A)/\binom{d}{d/2}

where \binom{n}{k} := Γ(n + 1)/Γ(k + 1)Γ(n − k + 1) is the generalized binomial coefficient. On the other hand, observe that K ∩π^{−1}(0) = (A∩_{2} A^{∘})×{0} and π(K) = A+_{2} A^{∘}, where π is the projection map π(x, y) := y. Since K is convex, for any 0 ≤ θ ≤ 1 and x ∈ θ·π(K), we see that K ∩π^{−1}(x) contains a translate of (1−θ)·(K ∩π^{−1}(0)). Thus by Cavalieri’s principle or Fubini’s theorem, K has at least as large a volume as π(K) ×_{1} (K ∩ π^{−1}(0)). Thus

mes(K) ≥ mes((A ∩_{2} A^{∘}) ×_{1} (A +_{2} A^{∘}))

= \frac{Γ(d + 1)Γ(d + 1)}{Γ(2d + 1)}mes(A ∩_{2} A^{∘})mes(A +_{2} A^{∘})

= \frac{Γ(d + 1)^{2}}{Γ(2d + 1)}mes(A ∩_{2} A^{∘})mes((A ∩_{2} A^{∘})^{∘})

= M(A ∩_{2} A^{∘})/\binom{2d}{d}.

Comparing this with our formula for mes(K), we see that we will be done as soon as we show that

\binom{2d}{d} ≤ 2^{d}\binom{d}{d/2}.

When d/2 is an integer this follows immediately from an inspection of Pascal’s triangle. When d/2 is a half-integer, this follows from the observation that \binom{d}{d/2} ≥ \binom{d}{⌊d/2⌋} (which is a consequence of the log-convexity of Γ) and another inspection of Pascal’s triangle.

We then induct using this lemma to obtain

Theorem 1.6. [5] For any symmetric convex body A ⊂ R^{d} we have M(A) ≥ (2 log_{2} e(A))^{−d}M(B^{d}). In particular, from John’s theorem and affine invariance we have M(A) ≥ (log_{2} d)^{−d}M(B^{d}).

Proof We shall induct on the eccentricity e(A). When e(A) ≤ 2 the claim follows already from (4), so suppose that 2^{2^{m}} < e(A) ≤ 2^{2^{m+1}} and the claim has already been proven for e(A) ≤ 2^{2^{m}}. Without loss of generality we may normalize r(A) = e(A)^{−1/2} and R(A) = e(A)^{1/2}, thus

e(A)^{−1/2} · B^{d} ⊆ A ⊆ e(A)^{1/2} · B^{d}

which on taking polars implies

e(A)^{−1/2} · B^{d} ⊆ A^{∘} ⊆ e(A)^{1/2} · B^{d}.

In particular, if we let Ã := A ∩_{2} A^{∘}, then Ã contains \frac{1}{\sqrt{2}}e(A)^{−1/2} · B^{d} and thus has inradius at least \frac{1}{\sqrt{2}}e(A)^{−1/2}. Also, observe that if x ∈ ((1 − θ)^{1/2} · A) ∩ (θ^{1/2} · A^{∘}) for some 0 < θ < 1 then |x|^{2} = ⟨x, x⟩ ≤ (1 − θ)^{1/2}θ^{1/2} ≤ \frac{1}{2} and hence A ∩_{2} A^{∘} has a circumradius at most \frac{1}{\sqrt{2}}. This shows that e(Ã) ≤ e(A)^{1/2}, and hence by induction

hypothesis M(Ã) ≥ (log_{2} e(A))^{−d}M(B^{d}). The claim then follows from Lemma 1.5.

(1) Prove the remaining claims in Proposition 1.1.

(2) Prove the stated claims about the Gamma function Γ(x). (To compute Γ(1/2), one way is to start by considering the integral ∫_{R^{2}} e^{−π|x|^{2}} dx and compute it in both Cartesian and polar co-ordinates, and use a quadratic change of variables). Conclude in particular Stirling’s formula

N! = N^{N+1/2}e^{−N+O(1)}  (7)

for any integer N. The constant term O(1) can be refined to \frac{1}{2} log 2π + O(1/N), and in fact has a full asymptotic series development (see e.g. [7]), but the comparatively crude estimate (7) will suffice for our applications.

(3) This exercise assumes knowledge about the Euclidean Fourier transform f̂(ξ) := ∫_{R^{d}} e^{−2πix·ξ}f(x) dx. Observe that if A is a convex body then there exists an absolute constant c > 0 such that |ˆ1_{A}(ξ)| > c for all ξ ∈ c · A^{∘}. Using this and Plancherel’s theorem, establish the (very poor) bound M(A) ≤ C^{d}. This exercise hints at the connection between Santalo’s inequality and the Fourier transform. One can also use the Fourier transform to obtain some lower bounds on M(A) by considering iterated convolutions of ˆ1_{A}, but the bounds obtained this way are extremely poor (double-exponential in d).

(4) Prove the remaining claims in Lemma 1.3.

(5) Show that if π : R^{d} → R^{d} is an orthogonal projection to a hyperplane in R^{d} then mes(S_{π}(A)) = mes(A) for all convex bodies A.

(6) Show that if A is a symmetric convex body then R(A^{∘}) = 1/r(A), r(A^{∘}) = 1/R(A), and e(A^{∘}) = e(A).

(7) Let A be a symmetric convex body with inradius 1 and circumradius R > 1. Show that A is contained in a symmetric convex body of the form S := (V + B^{d}) ∩ (R · B^{d}) for some hyperplane V. Show that there exists a finite sequence of Steiner symmetrization operations which, when applied consecutively to S, produces a body of circumradius at most R − c(R) for some c(R) > 0 which only goes to zero at the limit R → 1. Using this fact (and rescalings of it), show that given any symmetric convex body A one can find an infinite sequence of Steiner symmetrizations which when applied to this body, give a sequence of bodies whose eccentricity converges to 1. (In fact it is possible to get the eccentricity less than O(1) after no more than 3d Steiner symmetrizations; see [4]).

(8) [1] Let B be a symmetric convex body in R^{d} with inradius r(B), and let V be a hyperplane in R^{d}. Show that mes_{d−1}(B ∩ V) ≤ \frac{d}{r(B)}mes_{d}(B). More generally, if V is a n-dimensional linear subspace of R^{d}, show that mes_{n}(B ∩ V) ≤ \frac{d!}{n!r(B)^{n−d}} mes_{d}(B).

(9) [1] Let B be a symmetric convex body in R^{d}, and let V be an n-dimensional linear subspace in R^{d}. Let π : R^{d} → V be the orthogonal projection of R^{d} to V, and let V^{⊥} be the orthogonal complement of V in R^{d}. Show that

mes_{d}(B) ≤ \binom{d}{n}mes_{n}(π(B))mes_{d−n}(B ∩ V^{⊥}).

(Hint: the right-hand side is also the volume of π(B) +_{1} (B ∩ V^{⊥})).

References

[1] Y. Bilu, The (α + 2β)-inequality on a torus, J. London Math. Soc. (2) 57 (1998), no. 3, 513–528.

[2] J. Bourgain, V. Milman, New volume ratio properties for convex symmetric bodies in R^{n}, Invent. Math. 88 (1987), 319–340.

[3] F. John, Extremum problems with inequalities as subsidiary conditions, Studies and Essays presented to R. Courant on his 60th birthday, Jan 8. 1948, Interscience Publishers Inc., New York, NY 1948, 187–204.

[4] B. Klartag, V. Milman, Isomorphic Steiner symmetrization, Invent. Math. 153 (2003), 463– 485.

[5] G. Kuperberg, A low-technology estimate in convex geometry, Internat. Math. Res. Notices 9 (1992), 181–183.

[6] M. Meyer, A. Pajor, On Santaló’s inequality, Lecture Notes in Mathematics 1376, Springer-Verlag, Berlin, 1989, 261–263.

[7] H. Robbins, A remark on Stirling’s formula, Amer. Math. Monthly 62 (1955) 26–29.

[8] L. Santaló, Un invariante afin para los cuerpos convexos del espacio de n dimensiones, Portugalie Math. 8 (1949), 155–161.

Department of Mathematics, UCLA, Los Angeles CA 90095-1555

E-mail address: tao@math.ucla.edu