January 13, 2011

THE BEHAVIOR OF ITERATIONS OF THE INTERSECTION BODY OPERATOR IN A SMALL NEIGHBORHOOD OF THE

UNIT BALL

ALEXANDER FISH, FEDOR NAZAROV, DMITRY RYABOGIN, AND ARTEM ZVAVITCH

Abstract. The intersection body of a ball is again a ball. So, the unit ball B_{d} ⊂ ℝ^{d} is a fixed point of the intersection body operator acting on the space of all star-shaped origin symmetric bodies endowed with the Banach-Mazur distance. We show that this fixed point is a local attractor, i.e., that the iterations of the intersection body operator applied to any star-shaped origin symmetric body sufficiently close to B_{d} in Banach-Mazur distance converge to B_{d} in Banach-Mazur distance. In particular, it follows that the intersection body operator has no other fixed or periodic points in a small neighborhood of B_{d}.

1. Introduction

The notion of an intersection body of a star body was introduced by E. Lutwak [Lu1]: K is called the intersection body of L if the radial function of K in every direction is equal to the (d−1)-dimensional volume of the central hyperplane section of L perpendicular to this direction:

(1.1)  ρ_{K}(ξ) = vol_{d−1}(L ∩ ξ^{⊥}), ∀ξ ∈ S^{d−1},

where ρ_{K}(ξ) = sup{a : aξ ∈ K} is the radial function of the body K and ξ^{⊥} = {x ∈ ℝ^{d} : (x, ξ) = 0} is the central hyperplane perpendicular to the vector ξ. Using the formula for the volume in polar coordinates in ξ^{⊥}, we derive the following analytic definition of an intersection body of a star body: K is the intersection body of L if

ρ_{K}(ξ) = \frac{1}{d − 1}Rρ_{L}^{d−1}(ξ) := \frac{1}{d − 1} ∫_{S^{d−1}∩ξ^{⊥}} ρ_{L}^{d−1}(θ)dθ.

Here R stands for the spherical Radon transform. We refer the reader to books [Ga], [K] for more information on the definition and properties of intersection bodies of star bodies and their applications.

Let us denote by IL the intersection body of a body L. Let 𝕊_{d} be the set of all star-shaped origin symmetric bodies in ℝ^{d} endowed with the Banach-Mazur distance

d_{BM}(K, L) = inf{b/a : ∃ T ∈ GL(d) such that aK ⊆ T L ⊆ bK}.

1991 Mathematics Subject Classification. Primary: 44A12, 52A15, 52A21.

Key words and phrases. Convex body, Intersection body, Spherical Harmonics, Radon Transform.

Supported in part by U.S. National Science Foundation grants DMS-0652684, DMS-0800243, DMS-0808908.

We note that I(T L) = | det T |(T^{∗})^{−1}(IL), for all T ∈ GL(d) (see Theorem 8.1.6 in [Ga]), hence the action of I on 𝕊_{d} is well defined, and d_{BM}(I(T K), I(T L)) = d_{BM}(IK, IL).

The action of I on 𝕊_{2} is quite simple; since IL is just L rotated by π/2 and stretched 2 times, we have IL = L in 𝕊_{2}, so every point of 𝕊_{2} is a fixed point of I.

Let B_{d} be the unit Euclidean ball. We have

ρ_{I(B_{d})}(ξ) = vol_{d−1}(B_{d} ∩ ξ^{⊥}) = vol_{d−1}(B_{d−1}).

Thus, B_{d} is a fixed point of I in 𝕊_{d}. Question: Do there exist any other fixed or periodic points of I in 𝕊_{d}, d ≥ 3?

In this paper we show that there are no such points in a small neighborhood of the ball B_{d}. This will immediately follow from the following

Theorem 1.

I^{m}L ⟶^{𝕊_{d}} B_{d} as m → ∞,

for all L sufficiently close to B_{d} in Banach-Mazur distance.

More information on this and analogous questions can be found in Chapter 8 of [Ga] (see Problems 8.6 and 8.7 page 337 and note 8.6 on page 341) and [Lu2], [GZ].

We also note that a similar question for projection bodies (see [Ga], [K]) is much better understood. It is quite easy to observe that the projection body of a cube is again (a dilation of) a cube. W. Weil (see [W]) described the polytopes that are stable under the projection body operation. Still the general question of the description of all fixed points remains open. Notation: For a convex body K ⊂ ℝ^{d}, consider the following two quantities:

d_{∞}(K) = inf{∥1 − ρ_{T K}∥_{∞} : T ∈ GL(d)},
d_{2}(K) = inf{||1 − ρ_{T K}∥_{2} : T ∈ GL(d)}.

Note that in the small neighborhood of B_{d}, the quantity d_{∞}(K) is comparable with log d_{BM}(K, B_{d}).

In this paper we will denote by |u| the Euclidean norm of a vector u ∈ ℝ^{d}. We will denote by C, c constants depending on d (dimension) only, which may change from line to line.

2. Plan of the proof of the Theorem.

To avoid writing irrelevant normalization constants in formulae, from now on, we shall denote by R the normalized Radon transform on S^{d−1} that differs from the usual one by the factor \frac{1}{vol_{d−2}(S^{d−2})}, so R1 = 1. It doesn’t change anything because homotheties have already been factored out in the definition of 𝕊_{d}.

Our main tool in the proof of Theorem 1 is spherical harmonics. We refer the reader to [Gr] for more information and definitions. We denote by H_{k} the space of spherical harmonics of degree k. We shall denote by H_{k}^{f} the projection of f to H_{k}, so

f ∼ ∑_{k≥0} H_{k}^{f}.

The following formula for computation of the Radon transform of a spherical harmonic H_{k} ∈ H_{k} for even order k is especially useful for our calculations (see Lemma 3.4.7 in [Gr]):

(2.1)  R H_{k} = (−1)^{k/2}v_{d,k}H_{k},

where

v_{d,k} = \frac{1 · 3 · · · · · (k − 1)}{(d − 1)(d + 1) … (d + k − 3)} ≈ k^{−(d−2)}.

Let K ∈ 𝕊_{d} be close to B_{d}. Our main goal is to show the following two things:

(1) I^{m}K is smooth for all large m.

(2) If K is sufficiently smooth and close to B_{d}, then d_{2}(IK) ≤ λd_{2}(K) with some λ < 1.

The first claim will follow from the smoothing properties of R. Since f : S^{d−1} → ℝ is C^{m}-smooth essentially if the norms of H_{k}^{f} decay as k^{−m} and since Rf ∼ ∑_{k≥0}(−1)^{k/2}v_{d,k}H_{k}^{f}, we conclude that the order of smoothness of Rf exceeds the

order of smoothness of f by roughly speaking d − 2 ≥ 1.

Raising f to the power d − 1 does not change its smoothness class but can drastically increase the norm of f in that class, so we shall need some accurate computation to show that the smoothing effect still prevails if f is close to constant.

To prove the second claim, we write ρ_{K} = 1+φ, where φ is a small, even function belonging to L^{∞} and ∫_{S^{n−1}} φ = 0. Then

ρ_{IK} = 1 + (d − 1)Rφ + R O(φ^{2}).

The main idea is to try to show that ∥(d − 1)Rφ∥_{L^{2}} ≤ λ∥φ∥_{L^{2}} with some λ < 1. Since ∥φ^{2}∥_{L^{2}} = O(∥φ∥_{L^{∞}}∥φ∥_{L^{2}}), and ∥R∥_{L^{2}→L^{2}} ≤ 1, we get ∥RO(φ^{2})∥_{L^{2}} ≤ C∥φ∥_{L^{∞}}∥φ∥_{L^{2}}. Thus,

∥RO(φ^{2})∥_{L^{2}} ≤ \frac{1 − λ}{2}∥φ∥_{L^{2}}, provided ∥φ∥_{L^{∞}} ≤ \frac{1 − λ}{2},

so the last term won’t give us any trouble.

Note that φ ∼ ∑_{l≥1} H_{2l}^{φ} and the terms H_{2l}^{φ} are orthogonal. If all the products

v_{d,2l}(d−1) were less than 1, our task would be trivial. Unfortunately, v_{d,2}(d−1) = 1 (but v_{d,2l}(d − 1) ≤ \frac{3}{d+1} ≤ \frac{3}{4}, for l > 1). Thus, we need to kill H_{2}^{φ} somehow. It turns out that it can be done by first applying a suitable linear transformation to K.

Remark 1. The proof below can be noticeably shortened in the case of convex bodies. Then we may use the Busemann theorem (see [Bu] or [MP], Theorem 3.9; [Ga], Theorem 8.10) to claim that I^{m}L is convex, for all m ≥ 1. Then one can compare L^{∞} and L^{2} norms of radial functions of convex bodies to avoid the smoothening procedure described below.

3. Auxiliary Lemmata.

For a function f : S^{d−1} → ℝ we define its homogeneous extension f̃ of degree 0 by

f̃(x) = f (\frac{x}{|x|}) ,

so if f is a smooth function on S^{d−1}, then f̃ is a smooth function on ℝ^{d} \ {0}. By Df and D^{2}f, we mean the restrictions to the units sphere S^{d−1} of the first and the second differentials of f̃. Note that Df̃ and D^{2}f̃ are homogeneous functions on ℝ^{d} \ {0} of degree −1 and −2 respectively, so the norms ∥Df̃∥_{L^{∞}} and ∥D^{2}f̃∥_{L^{∞}} do not bound the differentials Df̃ and D^{2}f̃ on the entire space ℝ^{d} \ {0}. Still they bound them (up to a constant factor) outside any ball of positive radius centered at the origin, which is enough to transfer to the sphere all usual estimates coming from the second order Taylor formula in ℝ^{d}.

Lemma 1. Suppose that f : S^{d−1} → ℝ satisfies ∥D^{2}f∥_{L^{∞}} ≤ 1, ∥f∥_{L^{2}} < ε, for some ε ∈ (0, 1). Then ∥f∥_{L^{∞}} ≤ Cε^{\frac{4}{d+3}} and ∥Df∥_{L^{∞}} ≤ Cε^{\frac{2}{d+3}} .

Proof. Replacing f by −f, if necessary, we may assume that

∥f∥_{L^{∞}} = max_{S^{d−1}} f = f(x_{0}) = M > 0.

Since D_{x_{0}}f = 0, we can use the second order Taylor formula to conclude that

f(x) ≥ M − C∥D^{2}f∥_{L^{∞}}|x − x_{0}|^{2} ≥ M − C|x − x_{0}|^{2}.

Thus, in the ball of radius c\sqrt{M}, centered at x_{0}, we have

f(x) ≥ M − Cc^{2}M ≥ \frac{1}{2}M, provided that c^{2}C < \frac{1}{2}.

Hence,

ε^{2} ≥ ∫_{S^{d−1}} f^{2} ≥ c\frac{M^{2}}{4} (\sqrt{M})^{d−1} = cM^{\frac{d+3}{2}}

and the first inequality follows immediately.

The second inequality can now easily be derived from the classical Landau-Kolmogorov inequality (see [HLP])

∥Df∥_{L^{∞}} ≤ C∥f∥L^{∞}^{\frac{1}{2}}∥D^{2}f∥L^{∞}^{\frac{1}{2}}.

□

Let T ∈ GL(d). We would like to define the action of T on bounded functions on S^{d−1} in such a way that, for a radial function ρ_{K}(x) = ∥x∥_{K}^{−1} of a star-shaped body K, the image T ρ_{K} would coincide with the radial function of T^{−1}K. To this end, note that

ρ_{T^{−1}K}(x) = ∥T x∥_{K}^{−1} = ∥\frac{T x}{|T x|}∥_{K}^{−1}|T x|^{−1} = ρ_{K}(\frac{T x}{|T x|})|T x|^{−1}.

Thus for an arbitrary bounded function f : S^{d−1} → ℝ, it is natural to put

(3.1)  T f(x) := f (ω(x)) |T x|^{−1},

where ω_{T} : S^{d−1} → S^{d−1} is given by ω_{T} (x) = \frac{T x}{|T x|}.

Lemma 2. Let T = I + Q, where Q is self-adjoint and ∥Q∥ is small. Then

|ω_{T} (x) − x| ≤ C∥Q∥.

Proof.

|ω_{T} (x) − x| = \frac{1}{|T x|}|T x − |T x|x| ≤ ∥T^{−1}∥|(T x − x) − (|T x| − 1)x|

≤ ∥T^{−1}∥[|T x − x| + ||T x| − 1| ] ≤ 2∥T^{−1}∥∥Q∥ ≤ \frac{2}{1 − ∥Q∥}∥Q∥.

□

4. Classes U_{α}

Let α ≥ 0. For a bounded function f on S^{d−1}, define ∥f∥_{U_{α}} to be the least constant M such that ∥f∥_{L^{∞}} ≤ M and for every n ≥ 1, there exists a polynomial p_{n} of degree n satisfying ∥f − p_{n}∥_{L^{2}} ≤ Mn^{−α}. We will say that f ∈ U_{α} if ∥f∥_{U_{α}} < ∞.

Fix an infinitely smooth function Θ on [0, +∞) such that Θ = 1 on [0, 1], Θ = 0 on [2, +∞) and 0 ≤ Θ ≤ 1 everywhere. Consider the multiplier operator

(4.1)  M_{n}f = M_{n}^{Θ}f = ∑_{k≥0} Θ (\frac{k}{n}) H_{k}^{f}.

We will use the following property: ∥M_{n}∥_{L^{p}→L^{p}} ≤ C(Θ), for all 1 ≤ p ≤ ∞. This result is well know to experts but, for the sake of competence, we will present a proof in Appendix.

Note that M_{n}f is a polynomial of degree 2n. Also M_{n}p_{n} = p_{n} for all polynomials p_{n} of degree n.

Suppose now that f ∈ U_{α}. Let q_{n} = M_{n}f. We have

∥f − q_{n}∥_{L^{2}} = ∥(f − p_{n}) − M_{n}(f − p_{n})∥_{L^{2}} ≤ C∥f − p_{n}∥_{L^{2}} ≤ C∥f∥_{U_{α}}n^{−α},

and

∥q_{n}∥_{L^{∞}} ≤ C∥f∥_{L^{∞}} ≤ C∥f∥_{U_{α}}.

Now we use the polynomials q_{n} to prove the following lemma describing the properties of the classes U_{α}.

Lemma 3.

(1) If f, g ∈ U_{α}, then fg ∈ U_{α} and ∥fg∥_{U_{α}} ≤ C∥f∥_{U_{α}}∥g∥_{U_{α}}.

(2) Let T ∈ GL(d) with ∥T ∥, ∥T^{−1}∥ ≤ 2. Then, for every δ > 0, f ∈ U_{α}, we

have T f ∈ U_{α−δ} and ∥T f∥_{U_{α−δ}} ≤ C_{δ}∥f∥_{U_{α}}.
(3) If f ∈ U_{α}, then Rf ∈ U_{α+d−2} and ∥Rf∥_{U_{α+d−2}} ≤ C∥f∥_{U_{α}}.

Proof. (1) We obviously have

∥fg∥_{L^{∞}} ≤ ∥f∥_{L^{∞}}∥g∥_{L^{∞}} ≤ ∥f∥_{U_{α}}∥g∥_{U_{α}}.

Now notice that

∥f − M_{n}f∥_{L^{2}} ≤ C∥f∥_{U_{α}}n^{−α} and ∥g − M_{n}g∥_{L^{2}} ≤ C∥g∥_{U_{α}}n^{−α},
∥M_{n}f∥_{L^{∞}} ≤ C∥f∥_{L^{∞}} ≤ C∥f∥_{U_{α}}
∥M_{n}g∥_{L^{∞}} ≤ C∥g∥_{L^{∞}} ≤ C∥g∥_{U_{α}},

and that p_{n} = M_{n}f · M_{n}g is a polynomial of degree 4n. Hence

∥fg − p_{n}∥_{L^{2}} = ∥(f − M_{n}f)g + M_{n}f(g − M_{n}g)∥_{L^{2}}
≤ ∥(f − M_{n}f)∥_{L_{2}}∥g∥_{L^{∞}} + ∥M_{n}f∥_{L^{∞}}∥(g − M_{n}g)∥_{L^{2}}
≤ C∥f∥_{U_{α}}∥g∥_{U_{α}}n^{−α}.

(2) Write f = p_{n} + g where p_{n} = M_{n}f and ∥g∥_{L^{2}} ≤ C∥f∥_{U_{α}}n^{−α}.

(T f) (x) = |T x|^{−1}f(ω_{T} x) = |T x|^{−1}p_{n}(ω_{T} x) + |T x|^{−1}g(ω_{T} x).

Since |T x|^{−1} ≤ ∥T^{−1}∥ ≤ 2 on S^{d−1} and ω_{T} is a diffeomorphism of the unit sphere, the L^{2}-norm of the second term does not exceed C∥g∥_{L^{2}} ≤ C∥f∥_{U_{α}}n^{−α}. Note now that x → |T x|^{−1} is a C^{∞}-function and ω_{T} is a C^{∞}-mapping on S^{d−1}. Moreover, their derivatives of all orders are bounded by some constants depending on the dimension and the order, but not on T (as long as ∥T ∥, ∥T^{−1}∥ ≤ 2).

We need the following approximation lemma (see for example [R], Theorem 3.3):

Lemma 4. If m ∈ ℕ, h ∈ C^{m}(S^{d−1}), then for every N, there exists a polynomial P_{N} of degree N such that ∥h − P_{N}∥_{L^{2}} ≤ C_{m}∥h∥_{C^{m}}N^{−m}.

Since both a multiplication by a C^{∞}-function and a C^{∞} change of variable are bounded operators in C^{m}, the function h(x) = |T x|^{−1}p_{n}(ω_{T} (x)) ∈ C^{m} and ∥h∥_{C^{m}} ≤ C_{m}∥p_{n}∥_{C^{m}}. By Bernstein inequality (see Theorem 3.2.6 in [S]),

∥p_{n}∥_{C^{m}} ≤ C_{m}∥p_{n}∥_{L^{∞}}n^{m} ≤ C_{m}∥f∥_{L^{∞}}n^{m} ≤ C_{m}∥f∥_{U_{α}}n^{m}.

Thus we can find a polynomial P_{N} of degree N = n^{1+ε} such that

∥h − P_{N}∥_{L^{2}} ≤ C_{m}∥f∥_{U_{α}}n^{m}N^{−m} = C_{m}∥f∥_{U_{α}}N^{− \frac{ε}{1+ε} m}.

Consider some δ > 0 and choose ε so small that \frac{α}{1+ε} > α − δ and m so large that \frac{ε}{1+ε}m > α − δ. Then we shall get

∥T f − P_{N}∥_{L^{2}} ≤ C_{m} (N^{−(α−δ)} + n^{−α}) ∥f∥_{U_{α}} ≤ C_{m} (N^{−α−δ} + N^{− \frac{α}{1+ε}} ) ∥f∥_{U_{α}}
≤ C_{m}N^{−(α−δ)}∥f∥_{U_{α}}.

(3) Obviously, ∥f∥_{L^{∞}} ≤ ∥f∥_{U_{α}}. Let Ψ = 1 − Θ. Note that R M_{n}f is a polynomial of degree 2n and

∥Rf − RM_{n}f∥_{L^{2}}^{2} = ∑_{k≥n} v_{d,k}^{2}Ψ (\frac{k}{n})^{2} ∥H_{k}^{f}∥_{L^{2}}^{2}
≤ Cn^{−2(d−2)} ∑_{k≥n} Ψ (\frac{k}{n})^{2} ∥H_{k}^{f}∥_{L^{2}}^{2}
= Cn^{−2(d−2)}∥f − M_{n}f∥_{L^{2}}^{2}
≤ C∥f∥_{U_{α}}n^{−2(d−2+α)}.
□

Lemma 5. Let β > α. Then for every σ > 0, there exists C = C_{σ,α,β} > 0 such that ∥f∥_{U_{α}} ≤ C∥f∥_{L^{∞}} + σ∥f∥_{U_{β}}.

Proof. We have ∥f∥_{L^{∞}} ≤ C∥f∥_{L^{∞}} as soon as C ≥ 1. Now take n ≥ 1. If n^{−(β−α)} > σ, take p_{n} = 0. Then,

∥f − p_{n}∥_{L^{2}} ≤ ∥f∥_{L^{∞}} ≤ C∥f∥_{L^{∞}}n^{−α},

provided that C > σ^{− \frac{α}{β−α}}. If n^{−(β−α)} ≤ σ, choose p_{n} so that

∥f − p_{n}∥_{L^{2}} ≤ ∥f∥_{U_{β}}n^{−β} = n^{−(β−α)}∥f∥_{U_{β}}n^{−α} ≤ σ∥f∥_{U_{β}}n^{−α}.
□

5. Iteration Lemma

Lemma 6. Fix α so large that U_{α} ⊂ C^{2}. Let L > 0 be a constant such that ∥ · ∥_{C^{2}} ≤ L∥ · ∥_{U_{α}}. There exist ε_{d} > 0 and λ_{d} < 1 with the following property. For every ε ∈ (0, ε_{d}) and every function f such that f = 1 + φ, ∫ φ = 0, ∥φ∥_{L^{2}} ≤ ε, ∥φ∥_{U_{α}} ≤ L^{−1}, there exists a linear operator T ∈ GL(d) and a positive number γ such that f̃ = γR(T f)^{d−1} can be written as 1 + φ̃ where ∫ φ̃ = 0, ∥φ̃∥_{L^{2}} ≤ λ_{d}ε, ∥φ̃∥_{U_{α}} ≤ L^{−1}.

Proof. Step 1: We show first that there exists an operator T, such that T f = 1+ψ, where ∥ψ∥_{2} ≤ ε + Cε^{\frac{d+5}{d+3}} and ∥H_{2}^{ψ}∥_{2} ≤ Cε^{\frac{d+5}{d+3}} .

We shall seek T in the form T = I + Q as in Lemma 2. We have

|T x| = \sqrt{1 + 2(Qx, x) + ∥Q∥^{2}} = 1 + (Qx, x) + O(∥Q∥^{2}).

Hence,

|T x|^{−1} = 1 + (Qx, x) + O(∥Qx∥^{2}).

Further, since ∥φ∥_{C^{2}} ≤ L||φ∥_{U_{α}} ≤ 1, Lemmata 1, 2 yield

|φ(ω_{T} (x)) − φ(x)| ≤ Cε^{\frac{2}{d+3}} |ω_{T} (x) − x| ≤ Cε^{\frac{2}{d+3}} ∥Q∥.

We also have

T f(x) = |T x|^{−1}(1 + φ(ω_{T} (x)))
(5.1)  = (1 − (Qx, x) + O(∥Q∥^{2}))(1 + φ(x) + O(ε^{\frac{2}{d+3}} ∥Q∥))
= 1 − (Qx, x) + φ(x) + O(∥Q∥ε^{\frac{2}{d+3}} + ∥Q∥^{2}).

Now we choose Q so that (Qx, x) = H_{2}^{φ}(x). Since ∥H_{2}^{φ}∥_{L^{2}} ≤ ∥φ∥_{L^{2}} ≤ ε, and H_{2}^{φ} is a quadratic polynomial, we can conclude that all its coefficients do not exceed Cε and thereby ∥Q∥ = O(ε). Also, applying Lemma 1 we get ∥φ∥_{L^{∞}} ≤ Cε^{\frac{4}{d+3}} . Thus, by (5.1), T f = 1 + ψ, where ψ = φ − H_{2}^{φ} + O (ε^{\frac{d+5}{d+3}} ). Note now that

∥φ − H_{2}^{φ}∥_{L^{2}} ≤ ∥φ∥_{L^{2}} ≤ ε,

so ∥ψ∥_{L^{2}} ≤ ε + O(ε^{\frac{d+5}{d+3}} ), and that φ − H_{2}^{φ} has no spherical harmonics of degree 2 in its decomposition, so ∥H_{2}^{ψ}∥_{L^{2}} = O (ε^{\frac{d+5}{d+3}} ). Also

(5.2)  ∥ψ∥_{L^{∞}} ≤ Cε^{\frac{4}{d+3}} .

Step 2: Now we compute (T f)^{d−1}. We have

(T f)^{d−1} = (1 + ψ)^{d−1} = 1 + (d − 1)ψ + η,

and (5.2) yields

∥η∥_{L^{2}} ≤ Cε^{\frac{4}{d+3}} ∥ψ∥_{L^{2}} ≤ Cε^{\frac{d+7}{d+3}} .

Applying the Radon transform, we get

R(T f)^{d−1} = 1+(d−1)H_{0}^{ψ} +H_{0}^{η} +R(d−1)H_{2}^{ψ} +(d−1)R(ψ−H_{0}^{ψ} −H_{2}^{ψ})+R(η−H_{0}^{η}).

Note that (d − 1)H_{0}^{ψ} + H_{0}^{η} is a constant function whose value ζ satisfies |ζ| ≤ ∥ψ∥_{L^{2}} ≤ Cε. We also have

(d − 1)∥RH_{2}^{ψ}∥_{L^{2}} = ∥H_{2}^{ψ}∥_{L^{2}} ≤ Cε^{\frac{d+5}{d+3}} ,
(d − 1)∥R(ψ − H_{0}^{ψ} − H_{2}^{ψ})∥_{L^{2}} ≤ λ_{d} ∥ψ∥_{L^{2}},

and

∥R(η − H_{0}^{η})∥_{L^{2}} ≤ ∥η∥_{L^{2}} ≤ Cε^{\frac{d+7}{d+3}} .

Now take γ = (1 + ζ)^{−1} = 1 + O(ε) and put

φ̃ = γ(RH_{2}^{ψ} + (d − 1)R(ψ − H_{0}^{ψ} − H_{2}^{ψ}) + R(η − H_{0}^{η})).

Note that

∥φ̃∥_{L^{2}} ≤ (1 + O(ε))(λ_{d}ε + O(ε^{\frac{d+5}{d+3}} )) = λ_{d}ε + O(ε^{\frac{d+5}{d+3}} ) < λ′_{d}ε,

provided that ε is small enough. Also ∫ φ̃ = 0, and γR(T f)^{d−1} = 1 + φ̃. At last

∥φ̃∥_{L^{∞}} ≤ C (∥ψ∥_{L^{∞}} + ∥η∥_{L^{∞}}) ≤ Cε^{\frac{4}{d+3}} .

Step 3: It remains to estimate ∥φ̃∥_{U_{α}}. Note that ∥f∥_{U_{α}} ≤ 2, so applying Lemma 3, with δ = 1/2, we get

∥T f∥_{U_{α−1/2}} ≤ C ⇒ ∥(T f)^{d−1}∥_{U_{α−1/2}} ≤ C′ ⇒ ∥R(T f)^{d−1}∥_{U_{β}} ≤ C′′ ⇒ ∥φ̃∥_{U_{β}} ≤ C′′′,

where β = α − 1/2 + d − 2 > α. Now choose σ > 0 so that C′′′σ ≤ \frac{1}{2L}. Then, by Lemma 5,

∥φ̃∥_{U_{α}} ≤ σC′′′ + C_{σ}C′ε^{\frac{4}{d+3}} ≤ \frac{1}{L},

provided that ε is small enough.

□

6. Smoothing

Fix α > 0, β > 0. Let f = 1 + φ, ∥φ∥_{L^{∞}} < ε < 1/2. Define the sequence f_{k} recursively by f_{0} = f, f_{k+1} = Rf_{k}^{d−1}. Using Lemma 3, we can conclude that f_{k} ∈ U_{β} for sufficiently large k and ∥f_{k}∥_{U_{β}} ≤ C(k). Also, it is easy to show by induction that

(1 − ε)^{(d−1)^{k}} ≤ f_{k} ≤ (1 + ε)^{(d−1)^{k}}.

Let μ = ∫ f_{k}. If ε > 0 is sufficiently small, then |μ − 1| is small and μ^{−1}f = 1 + ψ where ∫ ψ = 0 and ∥ψ∥_{L^{∞}} is small. Note that

∥ψ∥_{U_{β}} ≤ 1 + μ^{−1}∥f_{k}∥_{U_{α}} ≤ C′(k),

and, thereby, by Lemma 3, ∥ψ∥_{U_{α}} is also small, (∥ψ∥_{U_{β}} is bounded by a fixed constant and ∥ψ∥_{L^{∞}} → 0 as ε → 0). Applying this observation to the function ρ_{K}, we conclude that if K is sufficiently close to B_{d}, then, after proper normalization, ρ_{I^{k}K} can be written as 1 + φ with ∥φ∥_{U_{α}} as small as we want.

7. The end of the proof

Now we choose ε so small that the smoothing part results in a body K for which ρ_{K} satisfies the assumptions of Lemma 6. Then ρ_{K_{1}}, where K_{1} = γIT K satisfies the assumptions of Lemma 6 with λε instead of ε. Note that K_{1} =^{𝕊_{d}} IK. Applying Lemma 6 again, we get a body K_{2} =^{𝕊_{d}} I^{2}K such that ρ_{K_{2}} satisfies the assumption of Lemma 6 with λ^{2}ε instead of ε and so on.

In particular, it means that

∥ρ_{K_{m}} − 1∥_{L^{2}} ≤ λ^{m}ε → 0 as m → ∞

and ∥ρ_{K_{m}}∥_{C^{2}} ≤ 2.

This is enough to conclude that

d_{BM}(K_{m}, B_{d}) = d_{BM}(I^{m}K, B_{d}) → 0 as m → ∞.

8. Appendix

Theorem 2. Consider Θ ∈ C_{0}^{∞}(ℝ). Then the operator M_{n}^{Θ} defined in (4.1) is bounded in L^{p}, for all 1 ≤ p ≤ ∞, i.e.

(8.1)  ∥M_{n}^{Θ}f∥_{L^{p}(S^{d−1})} ≤ C ∥f∥_{L^{p}(S^{d−1})}.

We start the proof of Theorem 2 with some auxiliary lemmata.

For every z ∈ ℂ such that |z| < 1, define the function P_{z}(x, y) : S^{d−1}×S^{d−1} → ℂ buy

(8.2)  P_{z}(x, y) := \frac{1 − z^{2}}{(1 + z^{2} − 2z(x · y))^{d/2}} ,  z ∈ ℂ, |z| < 1,

where for odd d we pick the branch of an analytic function

z → g(z) = (1 + z^{2} − 2z(x · y))^{d/2}

in such a way that g(ℝ_{+}) ⊂ ℝ_{+}.

Lemma 7. For all x, y ∈ S^{d−1}, and |z| < 1

|P_{z}(x, y)| ≤ 2 · 3^{d}(\frac{|1 − z|}{1 − |z|})^{d+1}P_{|z|}(x, y).

Proof. For β ∈ ℂ, |β| = 1, we have

\frac{||z| − β|}{|z − β|} ≤ 1 + \frac{|z − |z||}{|z − β|} ≤ 1 + \frac{|z − |z||}{||z| − 1|} ≤ \frac{|z − |z|| + ||z| − 1|}{1 − |z|} ≤ 3 \frac{|1 − z|}{1 − |z|}.

We also have

\frac{|1 − z^{2}|}{1 − |z|^{2}} ≤ 2 \frac{|1 − z|}{1 − |z|}.

Since

1 + z^{2} − 2z(x · y) = (z − α)(z − ᾱ), for α = x · y + i\sqrt{1 − (x · y)^{2}},

we conclude

\frac{|P_{z}(x, y)|}{P_{|z|}(x, y)} = \frac{|1 − z^{2}| |(|z| − α)(|z| − ᾱ)|^{d/2}}{|1 − |z|^{2}| |(z − α)(z − ᾱ)|^{d/2}} ≤ 2 · 3^{d}(\frac{|1 − z|}{1 − |z|})^{d+1}.

□

Lemma 8. Let z ∈ ℂ, 0 < Im z < 2, and let n ∈ ℕ. Then,

∥P_{e^{iz/n}}∥_{L^{1}(S^{d−1})} ≤ 2^{d+2} · 3^{d} \frac{|z|}{Im z} .

Proof. Put ξ = iz/n. Then,

\frac{|1 − e^{ξ}|}{1 − e^{ξ}} ≤ 1 + \frac{e^{ξ} − e^{Re ξ}}{1 − e^{Re ξ}} ≤ 1 + \frac{e^{Re ξ}|Im ξ|}{1 − e^{Re ξ}} = 1 + \frac{|Im ξ|}{e^{−Re ξ} − 1} ≤ 1 + \frac{|Im ξ|}{|Re ξ|}
≤ \frac{2|ξ|}{|Re ξ|} = 2\frac{|z|}{Im z} .

Now by Lemma 7,

|P_{e^{iz/n}}(x, y)| ≤ 2 · 3^{d}(\frac{|1 − e^{iz/n}|}{1 − |e^{iz/n}|})^{d+1}P_{|e^{iz/n}|}(x, y) ≤ 2^{d+2} · 3^{d} \frac{|z|}{Im z} P_{|e^{iz/n}|}(x, y).

It remains to use ∥P_{|e^{iz/n}|}∥_{L^{1}(S^{d−1})} = 1. □

Let S(ℝ) be the Schwartz space. To prove (8.1), write

(8.3)  Θ (\frac{k}{n}) = ∫_{ℝ} ψ(x)e^{ikx/n}dx,

where ψ ∈ S(ℝ) is the Fourier transform of some C_{0}^{∞} extension of Θ to the entire real line.

Using the Stokes formula, we can rewrite the last integral as

2i ∫_{Im z>0} ¯∂Ψ(z)e^{ikz/n}dA(z),

where Ψ is any reasonable extension of ψ to the upper half-plane. To make this representation useful, we shall need the following lemma:

Lemma 9. For any ψ ∈ S(ℝ) there exists an extension Ψ(z), Im z ≥ 0, Ψ_{|_{ℝ}}(x) = ψ(x), such that

∫_{Imz>0} | ¯∂Φ(z)|(\frac{|z|}{Im z} )^{d+1}dA(z) < ∞.

Let us first show that Lemma 9 gives ∥M_{n}^{Θ}∥_{L_{p}→L_{p}} < ∞. Indeed, using (8.3), we can calculate the kernel K_{n} of the operator M_{n}^{Θ},

K_{n} := ∑_{k=0}^{∞} ϕ(k/n)H_{k} = ∫_{ℝ} ψ(x) ∑_{k=0}^{∞} e^{ikx/n}H_{k}dx =

2i ∫_{Im z>0} ¯∂Ψ(z) ∑_{k=0}^{∞} e^{ikz/n}H_{k}dA(z) = 2i ∫_{Im z>0} ¯∂Ψ(z)P_{e^{iz/n}}dA(z).

Now (8.1) follows by Lemma 9, since ∥K_{n}∥_{L^{1}(S^{d−1})} ≤ C.

Let us now prove Lemma 9 :

Proof. We define

Ψ(x + iy) = η(y)Ψ_{0}(x + iy), Ψ_{0}(x + iy) = ∑_{k=0}^{d+1} ψ^{(k)}(x)(iy)^{k}/k!,

where η : [0, ∞) → [0, 1] is infinitely differentiable, η(y) = 1 for 0 ≤ y ≤ 1, and η(y) = 0 for y ≥ 2. Observe that

|2¯∂Ψ(x + iy)| = |(∂/∂x + i∂/∂y)Ψ(x + iy)| ≤
|Ψ_{0}(x + iy)||η′(y)| + |η(y)||ψ^{(d+2)}(x)(iy)^{d+1}/(d + 1)!|.

Hence,

∫_{Im z>0} | ¯∂Φ(z)|(\frac{|z|}{Im z} )^{d+1}dA(z) ≤ 2 ∫_{Im z≤2} |Ψ_{0}(x + iy)||z|^{d+1}dA(z)+

1/(d + 1)! ∫_{Im z≤2} |ψ^{(d+2)}(x)||z|^{d+1}dA(z) ≤ C,

and we are done, since ψ ∈ S(ℝ). □

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Department of Mathematics, University of Wisconsin, Madison 480 Lincoln Drive Madison, WI 53706

   E-mail address: afish@math.wisc.edu

Department of Mathematics, University of Wisconsin, Madison 480 Lincoln Drive Madison, WI 53706

   E-mail address: nazarov@math.wisc.edu

Department of Mathematics, Kent State University, Kent, OH 44242, USA

   E-mail address: ryabogin@math.kent.edu

Department of Mathematics, Kent State University, Kent, OH 44242, USA

   E-mail address: zvavitch@math.kent.edu