AN ESTIMATE ON CONVEX BODIES IN THE PLANE

TERENCE TAO

Abstract. We prove a lemma of Beck concerning convex bodies in the plane.

1. Introduction

Let B be a convex body in the plane C. By smoothing B a little bit we may assume that the boundary ∂B of B is given as a smooth curve {p(θ) : θ ∈ S^{1}}, where p is a smooth periodic function on S^{1} such that at p(θ) the outward normal of B is equal to θ. The perimeter of B is then given as

∫_{S^{1}} |p′(θ)| dθ.

Assume that B contains a disk. For any θ ∈ S^{1}, define l(θ) to be the chord of B perpendicular to θ and containing the point p(θ) − θ; since B contains a unit circle, this chord is well-defined. Let |l(θ)| denote the length of l(θ). The chord l(θ) divides B into two regions; let D(θ) denote the region which contains p(θ).

The point of this note is to prove the estimate

∫_{S^{1}} |p′(θ)| dθ ∼ ∫_{S^{1}} |l(θ)|^{2} dθ  (1)

uniformly in the convex body B; this was first shown in [1].

We first observe that both sides are ≳ 1. For the left-hand side this is just the isoperimetric inequality. For the left, all one needs to do is observe that there is an arc I ⊂ S^{1} of length 1/200 ≤ |I| ≤ 1/100 on which ∫_{I} |p′(θ)| dθ ≳ 1, and observe that one necessarily then has l(θ) ≳ 1 on this interval I.

The key geometric lemma is the following.

Lemma 1.1. Let θ_{0} ∈ S^{1} be such that 2^{j} ≤ |l(θ_{0})| ≤ 2^{j+1} for some integer j ≥ −10. Then there exists an open arc I ⊂ S^{1} of length |I| ∼ 2^{−j} containing θ_{0} such that |l(θ)| ≳ 2^{j} for all θ ∈ I, and such that ∫_{I} |p′(θ)| dθ ∼ 2^{j}.

Proof Without loss of generality we may take θ_{0} = i, p(θ_{0}) = i, and l(θ_{0}) to be an interval [−a, b] in the real line for some a, b > 0. Then a + b ∼ 2^{j}. Without loss of generality we may assume b ∼ 2^{j}.

We may reparameterize a portion of ∂B as

{x + if(x) : 0 ≤ x ≤ b}

where f is a smooth concave real function with f(0) = 1, f′(0) = 0, and f(b) = 0. From concavity we see that f′(b) ≤ −1/b, while −2/b ≤ f′(b/100). From the intermediate value theorem, we may thus find b/100 ≤ c ≤ b such that −2/b ≤ f′(c) ≤ −1/b.

1991 Mathematics Subject Classification. 42B15, 35L05.

Now set I to be (a small neighbourhood of) the arc joining θ_{0} = i with p^{−1}(c + if(c)). From our bound on f′(c) we see that |I| ∼ 2^{−j}. From our bound on c we see that ∫_{I} |p′(θ)| dθ ∼ c ∼ 2^{j}. Also for any θ ∈ I we see that D(θ) must contain a significant fraction (∼ 1) of the line segment from i to c + if(c), which implies that |l(θ)| ≳ 2^{j} as desired.

We first prove the ≲ side of (1). For any interval I of the type in Lemma 1.1, we have

∫_{I} |p′(θ)| dθ ≲ ∫_{I} |l(θ)|^{2} dθ.

The collection of all such intervals I covers the set |l(θ)| ≥ 2^{−10}. We now apply a standard covering lemma:

Lemma 1.2. For any collection {I} of open arcs in S^{1}, there is a sub-collection which covers the same set, and such that every point is covered at most twice.

Proof By compactness we may take the collection to be finite. If a point is covered by three arcs, then one of them is contained in the union of the other two, and we can remove it. We continue removing arcs in this manner until every point is only covered twice.

Using this lemma and then summing over I, we obtain

∫_{|l(θ)|≥2^{−10}} |p′(θ)| dθ ≲ ∫ |l(θ)|^{2} dθ.

It remains to control the portions where |l(θ)| < 2^{−10}. But for any such θ, the curve p(θ) must turn by at least \frac{9}{10}π inside D(θ). Thus one cannot have three disjoint chords l(θ) of this type. Also, for each such chord the perimeter of D(θ) is O(1). Combining these statement we see that

∫_{|l(θ)|≥2^{−10}} |p′(θ)| dθ ≲ 1,

and the claim follows.

Now we prove the ≳ side of (1). We estimate

∫_{S^{1}} |l(θ)|^{2} dθ ≲ 1 + ∑_{j≥10} 2^{2j}|{θ : |l(θ)| ∼ 2^{j}}|.

For each j, we partition S^{1} into dyadic arcs I ∈ I_{j} of length 2^{−j}. By Lemma 1.1, for each θ_{0} such that |l(θ_{0})| ∼ 2^{j}, there exists an arc I ∈ I_{j} such that dist(I, θ_{0}) ≲ 2^{j} and

∫_{I} |p′(θ)| dθ ≳ 2^{j}.

Thus we have

|{θ : |l(θ)| ∼ 2^{j}}| ≲ 2^{j}#{I ∈ I_{j} : ∫_{I} |p′(θ)| dθ ≳ 2^{j}} ≲ ∑_{k≥0} 2^{−j}#{I ∈ I_{j} : ∫_{I} |p′(θ)| dθ ∼ 2^{j+k}}.

Thus it suffices to show that

∑_{k≥0} ∑_{j≥10} 2^{j}#{I ∈ I_{j} : ∫_{I} |p′(θ)| dθ ∼ 2^{j+k}} ≲ ∫_{S^{1}} |p′(θ)| dθ.

The left-hand side is comparable to

∑_{k≥0} 2^{−k} ∑_{j≥10} ∑ ; I∈I_{j}:∫_{I} |p′(θ)| dθ∼2^{k}/|I| ∫_{I} |p′(θ)| dθ.

Fix k, and let j and I vary. Observe that any point θ can be contained in at most O(1) intervals I (basically because the dyadic intervals nest, and 2^{k}/|I| is a decreasing function of I). Thus we can bound the above by

≲ ∑_{k≥0} ∫_{S^{1}} |p′(θ)| dθ

and the claim follows.

References

[1] J. Beck, On a lattice point problem of L. Moser, Combinatorica 8 (1988), 21–47 and 159–176.

Department of Mathematics, UCLA, Los Angeles, CA 90024

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