On convex intersection bodies and unique determination problems for convex bodies

J. Math. Anal. Appl. 443 (2016) 295–312

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

On convex intersection bodies and unique determination problems for convex bodies Matthew Stephen Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, Alberta, T6G 2G1, Canada

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Article history: Received 9 December 2015 Available online 17 May 2016 Submitted by J. Bastero

We describe a general result which ensures counter-examples for certain problems of unique determination for convex bodies. Using this result, we show a convex body K ⊂ Rn is not uniquely determined by its convex intersection body CI(K), introduced by Meyer and Reisner in 2011. © 2016 Elsevier Inc. All rights reserved.

Keywords: Intersection body Cross-section body Convex intersection body Klee’s problem

1. Introduction Let K ⊂ Rn be a convex body, a convex and compact subset which includes the origin as an interior point. Many classic problems in convex geometry are concerned with the unique determination, possibly up to congruency, of a convex body within some collection. One well-known positive result is the Minkowski–Funk Section Theorem (see, for example, Corollary 3.9 in [9]), which may be stated in terms of intersection bodies. Recall that L ⊂ Rn is a star body if the closed line segment connecting the origin to every x ∈ L is contained in L, and if its radial function    ρL (ξ) = max a ≥ 0 aξ ∈ L ,

ξ ∈ S n−1 ,

is positive and continuous. Note that all convex bodies are also star bodies. We say L is origin-symmetric if L = −L. More generally, L is centrally-symmetric if one of its translates is origin-symmetric. The intersection body of L is the star body IL ⊂ Rn with radial function   ρIL (ξ) = voln−1 L ∩ ξ ⊥ ,

E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jmaa.2016.05.023 0022-247X/© 2016 Elsevier Inc. All rights reserved.

ξ ∈ S n−1 .

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Intersection bodies were first introduced by Lutwak in [10] in connection with the Busemann–Petty problem, and they continue to be an active area of research. Now, the aforementioned theorem is as follows: Theorem 1 (Minkowski–Funk Section Theorem). If K, L ⊂ Rn are origin-symmetric star bodies with IK = IL, then K = L. There are counter-examples for this result when K and L are not assumed to be origin-symmetric; see, for instance, [3]. It is natural to look for conditions for unique determination which hold in the absence of origin-symmetry. Klee proposed such conditions in [8]. We will state Klee’s problem in terms of cross-section bodies, which were introduced by Martini in [12]. Consider again a general convex body K ⊂ Rn . The cross-section body of K is the star body CK ⊂ Rn with   ρCK (ξ) = max voln−1 K ∩ {ξ ⊥ + tξ} , t∈R

ξ ∈ S n−1 .

It is obvious that IK ⊂ CK, and by Brunn’s Theorem, IK = CK when K is origin-symmetric. In [11], it was proven that IK = CK only if K is origin-symmetric. Now, Klee asked whether CK = CL implies K = L, for (not necessarily origin-symmetric) convex bodies K, L ⊂ Rn . This question was only recently proven in the negative in [4], where an explicit counter-example was constructed. Subsequently, [16] and [17] gave alternative constructions. In general, intersection and cross-section bodies are not convex bodies. Busemann’s Theorem implies that for a convex body K, IK is convex when K is origin-symmetric. If K is not origin-symmetric, then IK is not necessarily convex, nor is it necessary that CK is so when n > 3; see [2,13,1]. In [14], Meyer and Reisner introduced convex intersection bodies. Let K ⊂ Rn be a convex body. Let g = g(K) ∈ K be the centroid of K, and let K ∗y denote the polar body of K with respect to the point y ∈ int(K); that is,    K ∗y = x ∈ Rn  x − y, z − y ≤ 1 ∀ z ∈ K . The convex intersection body of K is the (a priori) star body CI(K) ⊂ Rn with 

   ∗y     y ∈ int K ∗g ξ ⊥ , ρCI(K) (ξ) = min voln−1 K ∗g ξ ⊥ ξ ∈ S n−1 ,  where · |ξ ⊥ is the orthogonal projection onto the hyperplane perpendicular to ξ. The main result in [14] is that CI(K) is always a convex body. Furthermore, when the centroid of K is at the origin, the relationship between CI(K) and IK parallels that between IK and CK. Indeed, with the assumption g(K) = 0, it was proved in [14] that CI(K) ⊂ IK and CI(K) = IK if and only if K is origin-symmetric. Is a convex body uniquely determined, up to congruency, by its convex intersection body? If it is centrallysymmetric, then yes, as follows from above. Recall that a convex body K is infinitely smooth if its radial function is infinitely smooth on S n−1 . We will say K is a convex or star body of rotation if its radial function is rotationally symmetric about the x1 -axis; i.e. ρK (ξ) = ρK (η) whenever ξ, η ∈ S n−1

and ξ, e1  = η, e1 ,

where e1 is the unit vector in the direction of the positive x1 -axis. We prove the following: Theorem 2. Let n ∈ N, n ≥ 2. There are infinitely smooth convex bodies of rotation K, L ⊂ Rn such that K is not centrally-symmetric, L is origin-symmetric, and CI(K) = CI(L).

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The convex bodies in the above theorem are necessarily non-congruent, so they provide a negative answer to our question of unique determination. In Section 3, we adapt the method of construction used in [4] so that it generates counter-examples to a more general question of unique determination for convex bodies. Roughly, if we “smoothly” associate  ⊂ Rn such that K  = IK whenever K every convex body K ⊂ Rn with an origin-symmetric star body K  does not uniquely determine K; for the precise statement, see Theorem 5. In is origin-symmetric, then K Section 4, we prove Theorem 2 by showing convex intersection bodies satisfy the hypotheses of Theorem 5. Finally, we offer some concluding remarks in Section 5. 2. Preliminaries and additional notation The constant ωn will be used for the surface area of S n−1 . Whenever we integrate over the sphere S n−1 , we use non-normalized spherical measure; that is, the appropriately scaled (n − 1)-dimensional Hausdorff measure on Rn . A Euclidean ball centered at the origin in Rn and with radius r > 0 will be represented by B n (r). The Legendre polynomial of dimension n ∈ N, n > 1, and degree three is given by

n+2 3 3 t − t; P3n (t) = n−1 n−1 equation (3.3.18) in [6] gives the general formula for Legendre polynomials. A function f : Rn \{0} → C is homogeneous of degree q ∈ R if x ∀ x ∈ Rn \{0}. f (x) = |x|q f |x| For f : S n−1 → C, we will let f˜ denote its homogeneous extension to Rn \{0} of degree zero. n We let Zn≥0 denote the collection of n-tuples of non-negative integers. For α ∈ Zn≥0 , we let |α| = j=1 αj , and let Dα denote the differential operator 1 ∂xα 1

∂ |α| . n · · · ∂xα n

When differentiating a function f = f (x, y), (x, y) ∈ Rm × Rn , the addition of a subscript Dyα f indicates we are differentiating with respect to y ∈ Rn . As usual, C k (S n−1 ), k ∈ N, is the space of complex-valued functions on the sphere which are k-times continuously differentiable, and  · C k (S n−1 ) will denote the standard norm. It is easy to see that f ∈ C k (S n−1 ) if and only if f˜ ∈ C k (Rn \{0}). Furthermore, limm→∞ fm − f C k (S n−1 ) = 0 for a sequence {fm } ⊂ C k (S n−1 ) if and only if     lim sup Dα f˜m (x) − Dα f˜ = 0 ∀ α ∈ Zn≥0 with |α| ≤ k; m→∞ x∈S n−1

this statement remains true if f˜m , f˜ are replaced with the homogeneous extensions of degree q ∈ R. We will say f ∈ C k (S m−1 × S n−1 ) if

  x y , ∈ C k Rm \{0} × Rn \{0} . f |x| |y| Let r = r(φ) ∈ C 2 (S 1 ) be a planar curve in polar coordinates. Its curvature is then given by the well-known formula

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2(r )2 − r · r + r2  3 ; (r )2 + r2 2

(1)

see, for example, formula (0.41) in [3]. The Laplacian operator iterated k-times is represented by Δk . When needed, the addition of a subscript Δz will indicate with respect to what variables the Laplacian is taken. The spherical Laplacian of a function f ∈ C 2 (S n−1 ) is defined by Δo f := Δf˜. For a homogeneous function f : Rn \{0} → C of degree m ∈ C, there is the well-known relation (Δf )(ξ) = (Δo f )(ξ) + m(m + n − 2)f (ξ),

ξ ∈ S n−1 ;

(2)

this appears as equation (1.2.9) in [6]. Let S = S (Rn ) be the Schwartz space of test functions f ∈ C ∞ (Rn ) whose derivatives decrease faster than any rational function. Recall that elements of the continuous dual S  are called distributions. The Fourier transform is defined on S by   φ ∈ S, φ(x) = φ(y)e−i x,y dy, Rn

and extended to f ∈ S  by  f, φ := f, φ

∀ φ ∈ S,

where f, · gives the action of f . Remember, any f ∈ C(Rn \{0}) which is homogeneous of degree greater than −n is a distribution which acts on S by integration. If, in addition, this f is infinitely smooth, then its distributional Fourier transform exists as a homogeneous function; see Theorem 3.1 in [5]. The spherical Radon transform R : C(S n−1 ) → C(S n−1 ) is defined by  Rf (u) = f (ξ) dξ, u ∈ S n−1 , f ∈ C(S n−1 ). S n−1 ∩u⊥

The following connection between the spherical Radon and Fourier transforms is well-known; for example, it appears as Lemma 3.7 in [9]. Lemma 3. Let f ∈ C(Rn \{0}) be an even homogeneous function of degree −n + 1. Then f ∈ C(Rn \{0}) is an even homogeneous function of degree −1 such that  1 ∀ ξ ∈ S n−1 . Rf (ξ) = f (η) dη = f(ξ) π S n−1 ∩ξ ⊥

The restriction R : Ce∞ (S n−1 ) → Ce∞ (S n−1 ) to the even and infinitely smooth functions on the sphere gives a bijection, and we may consider the inverse transform R−1 on this domain; see Theorem C.2.5 in [3]. The Santaló point s = s(K) ∈ Rn of a convex body K ⊂ Rn is the unique point such that 

  voln (K ∗s ) = min voln (K ∗y ) y ∈ int(K) ; see [18] or [19]. Let Kn and S n respectively denote the collections of convex bodies and star bodies in Rn . The support function for K ∈ Kn is

M. Stephen / J. Math. Anal. Appl. 443 (2016) 295–312

hK (ξ) = max x, ξ,

299

ξ ∈ S n−1 .

x∈K

The Hausdorff metric is then defined by   dH (K, L) = max hK (ξ) − hL (ξ), ξ∈S n−1

K, L ∈ Kn .

The following lemma appears as Proposition 1 in [7]. Lemma 4. The Santaló map s : (Kn , dH ) → Rn is continuous. Furthermore, for every convex body K, there exist positive constants C = C(K) and δ = δ(K) so that   s(K) − s(L) ≤ C dH (K, L) whenever L ∈ Kn and dH (K, L) ≤ δ. Using the Santaló point, the radial function of CI(K) may be rewritten as ρCI(K) (ξ) = voln−1



 ∗s(ξ) K ∗g ξ ⊥ ,

   where s(ξ) = s K ∗g ξ ⊥ . We let K ∗ denote K ∗0 . From the formula voln (K ∗y ) =

 K∗

1  n+1 dx, 1 − y, x

which is given by Lemma 3 of [15], we get voln−1



  ∗y K ∗g ξ ⊥ =

 

 ∗ K ∗g ξ ⊥

1 n dx. 1 − y, x

If, additionally, the centroid of K is at the origin, then  1  n dx. ρCI(K) (ξ) = 1 − s(ξ), x

(3)

K∩ξ ⊥

3. The general method of construction We can consider intersection bodies, cross-section bodies, and convex intersection bodies as maps from Kn to S n : K → IK,

K → CK,

K → CI(K) for K ∈ Kn .

The questions of unique determination described in the introduction are equivalent to asking whether these maps are injective. In [4], the constructed counter-example is specifically for Klee’s problem, and the methods  of construction are not stated in general terms. However, these methods can be applied to any map K → K n n from K to S which shares certain key properties with the maps above.  has the following properties: Suppose K → K  is always origin-symmetric. • K  for all origin-symmetric K ∈ Kn . • IK = K

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 m } are infinitely • There is a sequence {Km } ⊂ Kn of non-centrally-symmetric convex bodies such that {K smooth, with     lim ρK − a = 0 ∀ k ∈ N,  m m→∞

C k (S n−1 )

where a > 0 is a constant independent of k.  We can think of this last property as a type of smoothness for the map K → K.  The above three properties ensure K → K is not injective: Theorem 5. There exists a non-centrally-symmetric convex body K and an (infinitely smooth) origin = L.  Namely, take L = Lm defined by symmetric convex body L such that K 1  n−1 ρLm = (n − 1)R−1 ρK , m

and K = Km , for large enough m. At its core, Theorem 5 asserts ρLm is positive with non-negative curvature for large enough m. The main idea behind the proof is that smooth convergence of functions on S n−1 implies smooth convergence of the distributional Fourier transforms of their homogeneous extensions. This intuition, clarified in the following lemma and corollary, was used in [4]. For convenience, we present proofs of these auxiliary results. Lemma 6 is a simple generalization of Lemma 3.1 in [20]. Lemma 6. Let f ∈ Ce∞ (S n−1 ). For any k ∈ N ∪ {0} and q ∈ R with 0 < q < 2k + 1, we have

∧ y −n+q f |y| |x| (x) |y| ⎧ (−1)k+1π  ⎪ ⎪ ⎪ 2Γ(2k−q−1) sin π(2k−q) 2 ⎪    ⎪  ⎪ ⎪ y 2k−q k −n+q ⎪ × |x, ξ| Δ f |y| (ξ) dξ if q is not even, ⎨ |y| S n−1        = (−1) q2  k  x y 2k−q −n+q ⎪ Δ − log , ξ f |y| (ξ) dξ   ⎪ n−1 x, ξ (2k−q)! |x| |y| S ⎪ ⎪  ⎪    ⎪  ⎪ ⎪ y ⎩ + n−1 x, ξ2k−q Δk f |y| |y|−n+q log |y| (ξ) dξ if q is even, S 2k

for all x ∈ Rn \{0}. Proof. The formula for when q is not even is given by Lemma 3.16 in [9]. Suppose q is even. We will use the first formula in the lemma statement to calculate

∧ y 2k −n+p f |y| lim |x| (x). p→q |y|

(4)

Indeed, as p approaches q, both the numerator and denominator of     y (−1)k+1 π S n−1 |x, ξ|2k−p Δk f |y| |y|−n+p (ξ) dξ   2Γ(2k − p − 1) sin π(2k−p) 2 approach zero. This is clear for the denominator; we now prove this is also true for the numerator. That is, we need to show

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301

 gΔk h dξ = 0, S n−1

where g(ξ) = x, ξa , with a = 2k−q, and h is the homogeneous extension of f of degree −n +q. Equation (2) gives Δk h(ξ) = Δ(Δk−1 h)(ξ) = Δo (Δk−1 h)(ξ) + (−n − 2k + q + 2)(−2k + q)Δk−1 h(ξ),

ξ ∈ S n−1 ,

since Δk−1 h is homogeneous of degree −n − 2k + q + 2. Recalling that the spherical Laplacian is self-adjoint, we then have  gΔk h dξ S n−1



 (Δo g)Δk−1 h dξ + (−n − 2k + q + 2)(−2k + q)

= S n−1

gΔk−1 h dξ.

S n−1

But again by equation (2), Δo g(ξ) = Δg(ξ) − (2k − q)(n + 2k − q − 2)g(ξ), so

ξ ∈ S n−1 ,



 gΔk h dξ = S n−1

(Δg)Δk−1 h dξ. S n−1

It is clear that we can continue, repeatedly reducing the iterations of the Laplace transform on h by transferring them to g. Continuing a/2 times, we get   gΔk h dξ = (Δa/2+1 g) Δk−a/2−1 h dξ S n−1

S n−1



0 · Δk−a/2−1 h dξ = 0

= S n−1

because g is a polynomial in ξ of degree a. Using l’Hospital’s rule, we find that the limit (4) is equal to     q  (−1) 2 k k − g(ξ) log |x, ξ|Δ h(ξ) dξ + g(ξ)Δ h(y) log |y| (ξ) dξ . (2k − q)! S n−1

S n−1

This expression is equal to the formula given in the lemma statement for even q; use the identity log |x, ξ| = log |x/|x|, ξ| + log |x| and equation (5) to verify. Finally, we note that

∧

∧ y y |y|−n+q |y|−n+p |x|2k f (x) = lim |x|2k f (x); p→q |y| |y| this follows from Lemma 3.11 in [9].

2

(5)

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As a consequence of Lemma 6, we have the following result; it is a generalization of Corollary 3.17 in [9]. Corollary 7. Let {fm } ⊂ Ce∞ (S n−1 ) be a sequence of functions converging to f ∈ Ce∞ (S n−1 ) with respect to  · C k (S n−1 ) , for every k ∈ N. Then for every q > 0,  ∧  ∧    x x   −n+q −n+q |x| |x| lim  fm − f  m→∞   |x| |x|

=0

∀ k ∈ N.

C k (S n−1 )

Proof. Observe that y x, |y|

!2l−q and

y x, |y|

!2l−q

 !  x y   , log  |x| |y| 

extend to k-smooth functions on Rn \{0} × Rn \{0}, for large enough l ∈ N. For any α ∈ Zn≥0 with |α| ≤ k, we can then calculate  ∧  ∧ x x |x|−n+q |x|−n+q Dα fm and Dα f |x| |x| using the formulas from Lemma 6, where the derivatives will pass through the integrals. 2 −1 Proof of Theorem 5. We already know ρK ρK m is even and infinitely smooth, so R m is well-defined. Exn tending ρK to R \{0} with its natural homogeneity of degree −1, it then follows from Lemma 3, Lemma 6, m and the identity ∧∧ (ρK = (2π)n ρK m ) m

that R−1 ρK m (ξ) =

π (ρ  )∧ (ξ) ∀ ξ ∈ S n−1 . (2π)n Km

Since lim ρK m − aC k (S n−1 ) = 0

m→∞

for every k ∈ N, Corollary 7 implies    ∧ −1 ∧  lim (ρK ) − (a|x| )  m

m→∞

C 2 (S n−1 )

= 0.

∧ n−1 In particular, this shows (ρK . Therefore, ρLm defines m ) converges uniformly to a positive constant on S ∧ a star body for large enough m. We also see that the first and second order angular derivatives of (ρK m ) converge uniformly to zero; the same is true for ρLm . It then easily follows from formula (1) that the restriction of ρLm to any plane ξ ⊥ has positive curvature, which means Lm ∩ ξ ⊥ is convex for every ξ ∈ S n−1 ; see the proof of Lemma 8 for a similar and more explicit argument. We can conclude Lm is an origin-symmetric convex body, for large enough m. Finally, from the definition of Lm and the polar coordinate formula for volume, we have  1 n−1 ρL m (ξ) = ρILm (ξ) = ρn−1 . 2 m (ξ) ∀ ξ ∈ S Lm (η) dη = ρK n−1 S n−1 ∩ξ ⊥

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4. The construction for convex intersection bodies Recall that CI(K) is origin-symmetric for every K ∈ Kn , with CI(K) = IK whenever K is originsymmetric. To apply Theorem 5 to the map K → CI(K), we need to find a sequence {Km } ⊂ Kn of non-centrally-symmetric convex bodies such that   lim ρCI(K)m − aC k (S n−1 ) = 0 ∀ k ∈ N,

m→∞

where a > 0 is a constant independent of k. We do this through the following series of lemmas. Lemma 8. Define the function 1   n+1 ρKε (ξ) = 1 + εP (ξ, e1 )

for ξ ∈ S n−1 , where P = P3n and ε > 0. For sufficiently small ε > 0, ρKε is the radial function of an infinitely smooth convex body Kε which is a body of rotation about the x1 -axis, whose centroid is at the origin, and which is not centrally-symmetric. Proof. Clearly, the homogeneous extension of ρKε of degree −1 is positive and infinitely smooth on Rn \{0}, once ε > 0 is smaller than the maximum value of |P | on the interval [−1, 1]. So, ρKε defines a infinitely smooth star body, Kε . Given that  · , e1  is rotationally-symmetric about the x1 -axis, Kε is a star body of rotation about this axis. It is then necessary that the centroid of Kε lies on the x1 -axis, with its x1 coordinate given by 1 voln (Kε )



1 x1 dx = voln (Kε )



ρK ε (ξ)

rn ξ1 dr dξ S n−1

Kε

=

1 (n + 1)voln (Kε )

0





 1 + εP (ξ, e1 ) ξ1 dξ

S n−1

=

ε (n + 1)voln (Kε )



P (ξ, e1 ) ξ, e1  dξ. S n−1

This last integral is equal to zero, because P ( · , e1 ) and  · , e1  are spherical harmonics of different degrees. Therefore, Kε has its centroid at the origin. If we show that the restriction of Kε to the x1 , x2 -plane is convex, then this will imply Kε is convex, since it is a body of rotation. Letting θ be the angle with the positive x1 -axis, 1   n+1 r(θ) := 1 + εP (cos θ)

gives the boundary of Kε in the x1 , x2 -plane in polar coordinates. Recall formula (1). The restriction of Kε will be convex if the curvature of r is non-negative for all θ, so we need to show that the numerator of (1) is non-negative. This follows from the observations that r is bounded away from zero, every term of r · r is multiplied by a factor of ε, and so 2(r )2 − r · r + r2 ≥ r2 − r · r ≥ 0 for small enough ε > 0.

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If Kε has a center of symmetry, then it must lie on the x1 -axis, and r must have the same curvature at θ = 0 and θ = π. Let f (t) :=

(n − 1) + (n + 2)t n+2

(n − 1)(1 + t) n+1

.

Using the facts P (±1) = ±1 and P  (±1) = 3(n + 1)/(n − 1), it is easy to verify that the curvatures at θ = 0, π are, respectively, f (ε) and f (−ε). However, there is some open interval containing t = 0 on which f strictly increases, since f  (0) > 0. So f (ε) = f (−ε) for small enough ε > 0. 2 For φ ∈ S n−1 lying in the x1 , x2 -plane, let v = v(φ) ∈ S n−1

(6)

be the unit vector obtained by rotating φ counter-clockwise about the origin in the x1 , x2 -plane by π/2 radians. It is clear that Kε∗g φ⊥ = Kε∗ φ⊥ is a body of rotation about the axis in the direction v. The Santaló point of a convex body is affinely invariant, so we may uniquely define sε = sε (φ) ∈ R so that    s Kε∗ φ⊥ = sε v. (7) We will show sε is an infinitely smooth function on S 1 with its derivatives absolutely bounded by ε > 0. This is done with the help of the following lemmas and corollary.   Lemma 9. Let f ∈ C ∞ S n−1 , and let F be the homogeneous extension of Rf to Rn \{0} of degree zero. Then, for any k ∈ N and y ∈ Rn \{0},    C z 2k+1 k+1 −n+1 |z| (ξ) dξ; F (y) = 2k+1 |y, ξ| Δ f |y| |z| S n−1

C > 0 is a constant depending on k. Furthermore, F is infinitely smooth on its specified domain. Proof. Let f e denote the even part of f on S n−1 . For all y ∈ Rn \{0}, we have  y = F (y) = Rf f e (ξ) dξ |y| S n−1 ∩(y/|y|)⊥

∧  1 z y |z|−n+1 f e π |z| |y| ∧  |y| z |z|−n+1 f e (y) = π |z|

=

using Lemma 3, and ∧  |y| z |z|−n+1 f e (y) π |z|    z C 2k+1 k+1 −n+1 e |z| (ξ) dξ |y, ξ| Δ f = 2k+1 |y| |z| S n−1

=

using Lemma 6.

C |y|2k+1



|y, ξ|

2k+1

S n−1

Δ

k+1

  z −n+1 |z| (ξ) dξ f |z|

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Combining these equations, we get the alternate definition of F in the lemma statement. Noting that |y, z|2k+1 is at least k-smooth on R2n , and   z Δk+1 |z|−n+1 f |z| is infinitely smooth on Rn \{0}, it follows that F is k-smooth on Rn \{0}. However, k ∈ N is arbitrary, so F is infinitely smooth. 2 We will use Lemma 9 in the form of the following immediate corollary.   Corollary 10. Let f ∈ C ∞ S m−1 × S n−1 , and define

  y x ,· = F (x, y) := Rf |x| |y|



 f S n−1 ∩(y/|y|)⊥

x ,ξ |x|

dξ

for (x, y) ∈ Rm \{0} × Rn \{0}. Then, for any k ∈ N, F (x, y) =

C |y|2k+1

 S n−1



 x z −n+1 |z| , (ξ) dξ; |y, ξ|2k+1 Δk+1 f z |x| |z|

C > 0 is a constant depending on k. Furthermore, F is infinitely smooth on its specified domain. Lemma 11. Let m1 , m2 ∈ Z≥0 , let H be a k-dimensional subspace of Rn , let f : S n−1 → R be an odd function absolutely bounded by M > 0, and let g : S n−1 → R be absolutely bounded by ε > 0. If Kε is the convex body defined in Lemma 8, then      ρK ε (ξ)   m1 r   f (ξ) dr dξ   ≤ Cε   (1 − r g(ξ))m2 S n−1 ∩H  0 for small enough ε > 0. The constant C ≥ 0 depends on M , m1 , m2 , and n. Proof. Because f is odd, the even part of ρK ε (ξ)

h(ξ) := 0

rm1 dr (1 − r g(ξ))m2

vanishes when integrating against f over S n−1 ∩ H. Recall   1 ρKε (ξ) = 1 + εP (ξ, en ) n+1 , where P is an odd polynomial. Let N be the maximum value of |P | on the interval [−1, 1]. It follows that         M    h(ξ) − h(−ξ)| dξ f (ξ)h(ξ) dξ  ≤  2  n−1  S ∩H S n−1 ∩H

 M (1 − N ε)m1 +1 (1 + N ε)m1 +1 ≤ dξ − 2 (m1 + 1)(1 − 2ε)m2 (m1 + 1)(1 + 2ε)m2 S n−1 ∩H

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M. Stephen / J. Math. Anal. Appl. 443 (2016) 295–312

M ωk ≤ g(ε) := 2(m1 + 1)



(1 − N ε)m1 +1 (1 + N ε)m1 +1 − (1 − 2ε)m2 (1 + 2ε)m2

The function g is smooth at 0 ∈ R with g(0) = 0, giving the desired result.

. 2

We are now ready to prove the smoothness result for sε . Lemma 12. Let sε be the function on S 1 defined by equation (7), where S 1 is the intersection of S n−1 with the x1 , x2 -plane. For small enough ε > 0, sε ∈ C ∞ (S 1 ). Furthermore, for any α = (α1 , α2 ) ∈ Z2≥0 with α1 + α2 = k, |Dα s˜ε (ξ)| ≤ C(k, n)ε

∀ ξ ∈ S1.

(8)

The constant C(k, n) ≥ 0 depends only on k and the dimension n. Proof. Again let N be the maximum value of |P | on the interval [−1, 1]. Define the function g(t) = (1 + 1 N t) n+1 . We then have the containment B n (g(−ε)) ⊂ Kε ⊂ B n (g(ε)); therefore        dH Kε φ⊥ , B n−1 (1) = max hKε φ⊥ (ξ) − 1 ≤ max |hKε (ξ) − 1| ξ∈S n−1 ∩φ⊥

ξ∈S n−1

≤ max{|g(ε) − g(0)|, |g(−ε) − g(0)|} ≤ Cε ∀ φ ∈ S 1 , because g is smooth at 0 ∈ R. With the observation that the Santaló point of a Euclidean ball is always at its center, Lemma 4 implies       |sε (φ)| = s Kε∗ φ⊥ − s B n−1 (1)  ≤ Cε ∀ φ ∈ S 1 , for small enough ε > 0. Next, we show sε is infinitely smooth. The centroid of Kε is at the origin, so we may use equation (3) to get  1 ρCI(Kε ) (φ) = dx, φ ∈ S 1 ⊂ S n−1 , (9) (1 − sε v, x)n Kε ∩φ⊥

where v = v(φ) is the vector-valued function on S 1 defined by equation (6). Since Kε is contained in B n (2) for small enough ε > 0, the function  1  n dx G(φ, t) = 1 − v, xt Kε ∩φ⊥



ρK ε (ξ)

= S n−1 ∩φ⊥

0

rn−2  n dr dξ 1 − v, rξt

is defined for φ ∈ S 1 and −1/2 < t < 1/2. Because v is infinitely smooth on S 1 and ρKε is infinitely smooth on S n−1 , it follows from Corollary 10 that G is also so on its specified domain. For small enough ε > 0, we know that Kε contains B n (1/2), and −1/2 < sε (φ) < 1/2 for all φ. The uniqueness of the Santaló point then implies sε is the unique real number with −1/2 < sε < 1/2 and   ∂G  nv, x 0= = dx.  ∂t t=sε (1 − v, xsε )n+1 Kε ∩φ⊥

M. Stephen / J. Math. Anal. Appl. 443 (2016) 295–312

307

For all φ ∈ S 1 , we have  ∂ 2 G  = ∂t2 t=sε

 Kε ∩φ⊥

n(n + 1)v, x2 dx > 0, (1 − v, xsε )n+2

so the Implicit Function Theorem implies sε ∈ C ∞ (S 1 ). We will now prove inequality (8) for the first order partial derivatives of s˜ε . From now on, suppose that x ∈ Rn \{0} lies in the x1 , x2 -plane. From our application of the Implicit Function Theorem, we have

 F (x, x) :=

f S n−1 ∩(x/|x|)⊥



x ,ξ |x|

dξ

˜ v , y

=

n+1 dy

(1 − ˜ v , y s˜ε )

Kε ∩(x/|x|)⊥

=0

(10)

where ρK ε (ξ)

f (φ, ξ) =

rn−1 v, ξ

n+1 dr.

(1 − rv, ξ sε )

0

Here, v˜ and s˜ε are functions of x ∈ R2 , while v and sε are functions of φ ∈ S 1 . This function f : S 1 × S n−1 → C is infinitely smooth. Using the expression from Corollary 10 to calculate the partial derivative of F (x, x) with respect to xj (j = 1, 2), we find that it is equal to 

 (2k + 1)xj ∂ x f , ξ dξ − ∂xj |x| |x|2

 S n−1 ∩(x/|x|)⊥

(2k + 1)C + |x|2k+1

f S n−1 ∩(x/|x|)⊥



 |x, ξ|

2k

sgn(x, ξ)ξj Δk+1 z





−n+1

|z|



f

S n−1

x , ξ dξ |x|

x z , |x| |z|

(ξ)dξ.

(11)

We will expand these integrals into several more terms; one term will be equal to ∂˜ sε /∂xj multiplied by a factor bounded away from zero, and the remaining terms will be bounded by an appropriate constant times s˜ε or ε. With equation (10) and inequality (8) for s˜ε , this will imply inequality (8) for ∂˜ sε /∂xj . Any constants mentioned will be independent of ε, when ε > 0 is small enough, and of the variables x ∈ Rn and ξ ∈ S n−1 when present. Consider the first integral from expression (11). We have  S n−1 ∩(x/|x|)⊥



 x ∂ f ,ξ dξ ∂xj |x|



∂˜ v ,ξ ∂xj

= S n−1 ∩(x/|x|)⊥

! ρKε (ξ)

n+1

0

 + (n + 1) S n−1 ∩(x/|x|)⊥

rn−1

∂˜ v ,ξ ∂xj

(1 − r˜ v , ξ˜ sε ) ! ρKε (ξ)

dr dξ

rn ˜ v , ξ n+2

0

(1 − r˜ v , ξ˜ sε )

dr dξ s˜ε

M. Stephen / J. Math. Anal. Appl. 443 (2016) 295–312

308



ρK ε (ξ)

+ (n + 1)

rn ˜ v , ξ2 n+2

S n−1 ∩(x/|x|)⊥

(1 − r˜ v , ξ˜ sε )

0

dr dξ

∂˜ sε . ∂xj

(12)

Now, restrict x to S 1 . By Lemma 11, the first term above is absolutely bounded by a constant times ε > 0. The second term consists of an absolutely bounded integral multiplied by sε . For the final term, we have ∂˜ sε /∂xj multiplied by 

ρK ε (ξ)

(n + 1)

n+2

S n−1 ∩(x/|x|)⊥

(1 − r˜ v , ξ˜ sε )

0



n+1 ≥ n+2 2

dr dξ

ρK ε (ξ)

rn ˜ v , ξ2 dr dξ S n−1 ∩(x/|x|)⊥

=

rn ˜ v , ξ2

0



1

˜ v , ξ2 ρn+1 Kε (ξ) dξ

2n+2 S n−1 ∩(x/|x|)⊥

≥



1

ωn−2 ˜ v , ξ dξ = n+3 2

1 t2 (1 − t2 )

2

2n+3 S n−1 ∩(x/|x|)⊥

where the last equality comes from Lemma 1.3.1 in [6]. The second integral from expression (11),  (2k + 1)xj − |x|2

n−4 2

dt > 0,

−1

f

S n−1 ∩(x/|x|)⊥

x ,ξ |x|

dξ,

(13)

is absolutely bounded by a constant times ε > 0 by Lemma 11. Finally, consider the third integral from expression (11),

 (2k + 1)C x z 2k k+1 −n+1 |z| , (ξ) dξ. |x, ξ| sgn(x, ξ)ξ Δ f j z |x|2k+1 |x| |z|

(14)

S n−1

Observe that |x, ξ|2k sgn(x, ξ)ξj is even with respect to ξ; in fact, all further partial derivatives of |x, ξ|2k sgn(x, ξ)ξj with respect to xj (j = 1, 2) will be even with respect to ξ. Now, we need to determine the (k + 1)-iterated Laplacian of h(z) := |z|−n+1 f



z φ, |z|

v, z = |z|n

ρ˜K ε (z)

 0

rn−1 1 − rv, z/|z|sε

n+1 dr.

Letting i = 1, 2, . . . , n, we have ∂2h = ∂zi2



  ρ˜Kε (z) ∂ 2 v, z rn−1  n+1 dr 2 n ∂zi |z| 1 − rv, z/|z|sε 0

+2



∂ v, z ∂zi |z|n



⎛

⎜ ∂ ⎝ ∂zi

ρ˜K ε (z)

 0

r

n−1

1 − rv, z/|z|sε

⎞ ⎟ n+1 dr⎠

M. Stephen / J. Math. Anal. Appl. 443 (2016) 295–312

⎛ +

v, z ⎜ ∂ 2 ⎝ 2 |z|n ∂zi

ρ˜K ε (z)

0

309

⎞ rn−1 ⎟  n+1 dr⎠ . 1 − rv, z/|z|sε

Note that partial derivatives of v, z/|z|n of even orders are odd with respect to the variable z. Also, ∂ ∂zi

ρ˜K ε (z)

 0

rn−1 1 − rv, z/|z|sε

n+1 dr

(15)

n−1 ρ˜K ∂ ρ˜Kε ε = n+1 ∂zl 1 − ρ˜Kε v, z/|z|sε ⎛ ⎞ ρ˜K   ε (z) n r ⎜ ⎟ ∂ v, z sε . + (n + 1) ⎝ dr  n+2 ⎠ ∂zi |z| 1 − rv, z/|z|sε 0

Again, we know that sε and the derivatives of ρKε uniformly converge to zero with ε. Observe that all further derivatives of (15) (with respect to z) are similarly bounded. Given the even and odd symmetry mentioned above, it is clear that        2k k+1  |x, ξ| sgn(x, ξ)ξj Δ h(ξ) dξ  ≤ Cε.   n−1  S

We have finished bounding the integrals in expression (11); with equation (10), these bounds give inequality (8) for the first order partial derivatives of s˜ε . Recursion gives inequality (8) for the partial derivatives of higher order; let us make some additional comments on this. The first term in equation (12) is of a similar form as the right hand side of equation (10); differentiating this term with respect to xj proceeds in a similar way. Further derivatives of the remainder of equation (12) result in appropriately bounded terms multiplied by derivatives of s˜ε . Expression (13) is also of a similar form as the right hand side of equation (10); its derivatives are treated correspondingly. Derivatives of expression (14) may be bounded using the identity Dxα

 



 x z x z k+1 −n+1 k+1 −n+1 α Δz |z| |z| , = Δz , , f Dx f |x| |z| |x| |z|

where α ∈ Z2≥0 . Indeed, 

 x z , Dxα f |x| |z| will consist of terms which are multiplied by derivatives of s˜ε of degree not greater than |α|, and a term of the form Dα v˜, z hα (z) := |z|n We can then bound

ρ˜K ε (z)

 0

rn−1 1 − rv, z/|z|˜ sε

n+1 dr.

M. Stephen / J. Math. Anal. Appl. 443 (2016) 295–312

310

 |x, ξ|2k sgn(x, ξ)ξj Δk+1 hα (ξ) dξ S n−1

in the same way (14) was bounded. 2 Lemma 13. Let Kε be the convex body defined in Lemma 8. Define ωn−1 , n−1

gε (ξ) := ρCI(Kε ) (ξ) −

ξ ∈ S n−1 .

For small enough ε > 0, gε ∈ C ∞ (S n−1 ). Furthermore, for any α = (α1 , . . . , αn ) ∈ Zn≥0 with |α| = k, |Dα g˜ε (ξ)| ≤ C(k, n)ε

∀ ξ ∈ S n−1 .

The constants C(k, n) ≥ 0 depend only on k and the dimension n. Proof. It is clear that gε is rotationally symmetric about the x1 -axis, because ρCI(Kε ) is so. We will show that the restriction of g˜ε to the x1 , x2 -plane is infinitely smooth with bounded partial derivatives; the general result will then follow from the rotational symmetry. From now on, suppose x ∈ Rn \{0} lies in the x1 , x2 -plane. Using the representation of ρCI(Kε ) on the x1 , x2 -plane given by equation (9), we have ωn−1 = g˜ε (x) = ρ˜CI(Kε ) (x) − n−1



 f S n−1 ∩(x/|x|)⊥

x ,ξ |x|

dξ

where ρK ε (ξ) (

f (φ, ξ) =

) n−2 n ρn−1 r Kε (ξ) − (1 − rv, ξsε ) n ρn−1 Kε (ξ)(1 − rv, ξsε )

0

dr.

This function f : S 1 × S n−1 → C is infinitely smooth (for small enough ε > 0) because ρKε and sε are so. Observe that for all φ ∈ S 1 and ξ ∈ S n−1 , we have   f (φ, ξ) ≤

ρK ε (ξ) (

0

  ρn−1 (ξ) − (1 − rv, ξs )n ) rn−2  ε  Kε    dr n   ρn−1 (ξ)(1 − rv, ξs ) ε K ε

(1 + N ε)n−2 ≤ (1 − N ε)n−1 (1 − Cε(1 + εN )n

1+N  ε

 n−1  ρ (ξ) − (1 − rv, ξsε )n  dr Kε

0

≤ h(ε), where C = C(0, n) is the constant from Lemma 12, N is the maximum value of |P | on the interval [−1, 1], and  (1 + N ε)n−1 n−1 n − (1 − Cε(1 + N ε)) (1 + N ε) (1 − N ε)n−1 (1 − Cε(1 + εN )n  (1 + N ε)n−1 + (1 + Cε(1 + N ε))n − (1 − N ε)n−1 . n−1 n (1 − N ε) (1 − Cε(1 + εN )

h(ε) :=

M. Stephen / J. Math. Anal. Appl. 443 (2016) 295–312

311

The definition of h is independent of φ and ξ, and it is smooth in a neighborhood of 0 ∈ R with h(0) = 0. Therefore, there is another constant C > 0 such that |f (φ, ξ)| ≤ Cε for all (φ, ξ) ∈ S 1 × S n−1 , and which is independent of small enough ε > 0. For any α = (α1 , α2 ) ∈ Z2≥0 and β = (β1 , . . . , βn ) ∈ Zn≥0 , it is easily seen that every term of 

 x z , Dxα Dzβ f |x| |z| is multiplied by s˜ε or one of its derivatives, or by a derivative of ρ˜Kε . Using Lemma 12, such partial derivatives of f can be absolutely bounded by ε > 0 multiplied by a constant depending only on the dimension and the order of the derivative. Using Corollary 10 to calculate the partial derivative of g˜ε with respect to xj (j = 1, 2), we obtain the same expression as (11). Given our previous remarks, we can appropriately bound this derivative of g˜ε , as well as all partial derivatives of higher orders. 2 We can conclude from the previous lemma that    ωn−1   ρ lim  − = 0 ∀ k ∈ N. CI(K1/m ) m→∞  n − 1 C k (S n−1 ) Theorem 2 now follows immediately from Theorem 5. 5. Concluding remarks The authors of [17] proved that cross-section bodies do not uniquely determine the Euclidean ball in any dimension. It is natural to ask the corresponding question for convex intersection bodies. Our construction for Theorem 2 does not show that the origin-symmetric convex body L can be taken to be a Euclidean ball. However, this is possible with a simpler construction in two dimensions. Consider  K ∈ R2 . For any φ ∈ S 1 , K ∗g φ⊥ is a line segment with length hK ∗g (v(φ)) + hK ∗g (−v(φ)), where v ∈ SO(2) again denotes rotation about the origin by π/2 radians. With the simple observation that the Santaló point of a compact line segment is its center, we have ρCI(K) (φ) = vol1 =



 ∗s(φ) K ∗g φ⊥

4 , hK ∗g ((v(φ)) + hK ∗g (−v(φ))

φ ∈ S1.

Let L ⊂ R2 be a convex body of constant width which is not centrally-symmetric, and put K = L∗s . It is a property of the Santaló point that g(K) = s(L) (see [19], page 420), so CI(K) is necessarily a Euclidean disk. Finally, we have CI(K) = IB = CI(B) for some disk B ⊂ R2 with appropriate diameter. This counter-example should be compared with the two dimensional case for cross-section bodies. Indeed, the cross-section body of a convex body in R2 of constant width is always a disk (see, for example, Theorem 8.3.5 in [3]). Given the preceding comments, we ask the following: Question. Let n ∈ Z, n ≥ 3. Is there a convex body K ⊂ Rn which is not centrally-symmetric, and whose convex intersection body CI(K) is a Euclidean ball?

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Acknowledgments The author thanks Vladyslav Yaskin for his guidance and helpful suggestions. The author was partially supported by NSERC. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]

U. Brehm, Convex bodies with non-convex cross-section bodies, Mathematika 46 (1999) 127–129. H.T. Croft, Two problems on convex bodies, Math. Proc. Cambridge Philos. Soc. 58 (1962) 1–7. R. Gardner, Geometric Tomography, second edition, Cambridge University Press, New York, 2006. R.J. Gardner, D. Ryabogin, V. Yaskin, A. Zvavitch, A problem of Klee on inner section functions of convex bodies, J. Differential Geom. 91 (2012) 261–279. P. Goodey, V. Yaskin, M. Yaskina, Fourier transforms and the Funk–Hecke theorem in convex geometry, J. Lond. Math. Soc. 80 (2009) 388–404. H. Groemer, Geometric Applications of Fourier Series and Spherical Harmonics, Cambridge University Press, New York, 1996. J. Kim, S. Reisner, Local minimality of the volume-product at the simplex, Mathematika 57 (2011) 121–134. V. Klee, Research problems: is a body spherical if its HA-measurements are constant?, Amer. Math. Monthly 76 (1969) 539–542. A. Koldobsky, Fourier Analysis in Convex Geometry, American Mathematical Society, Providence, RI, 2005. E. Lutwak, Intersection bodies and dual mixed volumes, Adv. Math. 71 (1988) 232–261. E. Makai, H. Martini, T. Ódor, Maximal sections and centrally symmetric bodies, Mathematika 47 (2000) 19–30. H. Martini, Cross-sectional measures, in: Intuitive Geometry, Szeged, 1991, in: Colloq. Math. Soc. János Bolyai, vol. 63, North-Holland, Amsterdam, 1994, pp. 269–310. M. Meyer, Maximal hyperplane sections of convex bodies, Mathematika 46 (1999) 131–136. M. Meyer, S. Reisner, The convex intersection body of a convex body, Glasg. Math. J. 53 (2011) 523–534. M. Meyer, E. Werner, The Santaló-regions of a convex body, Trans. Amer. Math. Soc. 350 (1998) 4569–4591. F. Nazarov, D. Ryabogin, A. Zvavitch, Non-uniqueness of convex bodies with prescribed volumes of sections and projections, Mathematika 59 (2013) 213–221. F. Nazarov, D. Ryabogin, A. Zvavitch, An asymmetric convex body with maximal sections of constant volume, J. Amer. Math. Soc. 27 (2014) 43–68. L.A. Santaló, Un invariante afin para los cuerpos convexos del espacio de n dimensiones, Port. Math. 8 (1949) 155–161. R. Schneider, Convex Bodies: The Brunn–Minkowski Theory, Cambridge University Press, Cambridge, 2014. V. Yaskin, On strict inclusions in hierarchies of convex bodies, Proc. Amer. Math. Soc. 136 (2008) 3281–3291.