Spherical sets avoiding a prescribed set of angles

Let $X$ be any subset of the interval $[-1,1]$. A subset $I$ of the unit sphere in $R^n$ will be called \emph{$X$-avoiding} if $\notin X$ for any $u,v \in I$. The problem of determining the maximum surface measure of a $\{ 0 \}$-avoiding set was first stated in a 1974 note by Witsenhausen; there the upper bound of $1/n$ times the surface measure of the sphere is derived from a simple averaging argument. A consequence of the Frankl-Wilson theorem is that this fraction decreases exponentially, but until now the $1/3$ upper bound for the case $n=3$ has not moved. We improve this bound to $0.313$ using an approach inspired by Delsarte's linear programming bounds for codes, combined with some combinatorial reasoning. In the second part of the paper, we use harmonic analysis to show that for $n\geq 3$ there always exists an $X$-avoiding set of maximum measure. We also show with an example that a maximiser need not exist when $n=2$.


Introduction
Witsenhausen [Wit74] in 1974 presented the following problem: Let S n−1 be the unit sphere in R n and suppose I ⊂ S n−1 is a Lebesgue measurable set such that no two vectors in I are orthogonal. What is the largest possible Lebesgue surface measure of I? Let α(n) denote the supremum of the measures of such sets I, divided by the total measure of S n−1 . The first upper bounds for α(n) appeared in [Wit74], where Witsenhausen deduced that α(n) ≤ 1/n. In [FW81] Frankl and Wilson proved their powerful combinatorial result on intersecting set systems, and as an application they gave the first exponentially decreasing upper bound α(n) ≤ (1 + o(1))(1.13) −n . Raigorodskii [Rai99] improved the bound to (1 + o(1))(1.225) −n using a refinement of the Frankl-Wilson method. Gil Kalai conjectured in his weblog [Kal09] that an extremal example is to take two opposite caps, each of geodesic radius π/4; if true, this implies that α(n) = ( √ 2 + o(1)) −n .
Besides being of independent interest, the above Double Cap Conjecture is also important because, if true, it would imply new lower bounds for the measurable chromatic number of Euclidean space, which we now discuss.
Let c(n) be the smallest integer k such that R n can be partitioned into sets X 1 , . . . , X k , with x − y 2 = 1 for each x, y ∈ X i , 1 ≤ i ≤ k. The number c(n) is called the chromatic number of R n , since the sets X 1 , . . . , X k can be thought of as colour classes for a proper colouring of the graph on the vertex set R n , in which we join two points with an edge when they have distance 1. Frankl and Wilson [FW81,Theorem 3] showed that c(n) ≥ (1 + o(1))(1.2) n , proving a conjecture of Erdős that c(n) grows exponentially. Raigorodskii in 2000 [Rai00] improved the lower bound to (1 + o(1))(1.239) n . Requiring the classes X 1 , . . . , X k to be Lebesgue measurable yields the measurable chromatic number c m (n). Clearly c m (n) ≥ c(n).
Remarkably, it is still open if the inequality is strict for at least one n, although one can prove better lower bounds on c m (n). In particular, the exponent in Raigorodskii's bound was recently beaten by Bachoc, Passuello and Thiery [BPT14] who showed that c m (n) ≥ (1.268 + o(1)) n . If the Double Cap Conjecture is true, then c m (n) ≥ ( √ 2 + o(1)) n because, as it is not hard to show, c m (n) ≥ 1/α(n) for every n ≥ 2. Note that the best known asymptotic upper bound on c m (n) (as well as on c(n)) is (3 + o(1)) n , by Larman and Rogers [LR72].
Let L be the σ-algebra of Lebesgue surface measurable subsets of S n−1 , and let λ be the surface measure, for simplicity normalised so that λ(S n−1 ) = 1. For X ⊂ [−1, 1], a subset I ⊂ S n−1 will be called X-avoiding if ξ, η / ∈ X for all ξ, η ∈ I, where ξ, η denotes the standard inner product of the vectors ξ, η. The corresponding extremal problem is to determine α X (n) := sup{λ(I) : I ∈ L, I is X-avoiding}. (1) For example, if t ∈ (−1, 1) and X = [−1, t), then I ⊂ S n−1 is X-avoiding if and only if its geodesic diameter is at most arccos(t). Thus Levy's isodiametric inequality [Lev51] shows that α X is given by a spherical cap of the appropriate size.
A priori, it is not clear that the value of α X (n) is actually attained by some measurable X-avoiding set I (so Witsenhausen [Wit74] had to use supremum to define α(n)). We prove in Theorem 16 that the supremum is attained as a maximum whenever n ≥ 3. Remarkably, this result holds under no additional assumptions whatsoever on the set X. However, in a sense only closed sets X matter: our Theorem 17 shows that α X (n) does not change if we replace X by its closure. When n = 2 the conclusion of Theorem 16 fails; that is, the supremum in (1) need not be a maximum: an example is given in Theorem 2.
Besides also answering a natural question, the importance of the attainment result can also be seen through the historical lens: In 1838 Jakob Steiner tried to prove that a circle maximizes the area among all plane figures having some given perimeter. He showed that any non-circle could be improved, but he was not able to rule out the possibility that a sequence of ever improving plane shapes of equal perimeter could have areas approaching some supremum which is not achieved as a maximum. Only 40 years later in 1879 was the proof completed, when Weierstrass showed that a maximizer must indeed exist.
The layout of the paper will be as follows. In Section 2 we make some general definitions and fix notation. In Section 3 we prove a simple and general proposition giving combinatorial upper bounds for α X (n); this is basically a formalisation of the method used by Witsenhausen in [Wit74] to obtain the α(n) ≤ 1/n bound. We then apply the proposition to calculate α X (2) when |X| = 1. In Section 5 we deduce linear programming upper bounds for α(n), in the spirit of the Delsarte bounds for binary [Del73] and spherical [DGS77] codes. We then strengthen the linear programming bound in the n = 3 case in Section 6 to obtain the first main result.
In Section 8 we prove that the supremum α X (n) is a maximum when n ≥ 3, and in Section 9 we show that α X (n) remains unchanged when X is replaced with its topological closure. In Section 10 we formulate a conjecture generalising the Double Cap Conjecture for the sphere in R 3 , in which other forbidden inner products are considered.

Preliminaries
If u, v ∈ R n are two vectors, their standard inner product will be denoted u, v . All vectors will be assumed to be column vectors. The transpose of a matrix A will be denoted A t . We denote by SO(n) the group of n × n matrices A over R having determinant 1, for which A t A is equal to the identity matrix. We will think of SO(n) as a compact topological group, and we will always assume its Haar measure is normalised so that SO(n) has measure 1. We denote by S n−1 the set of unit vectors in R n , S n−1 = {x ∈ R n : x, x = 1}, equipped with its usual topology. The Lebesgue measure λ on S n−1 is always taken to be normalised so that λ(S n−1 ) = 1. Recall that the standard surface measure of S n−1 is where Γ denotes Euler's gamma-function. The Lebesgue σ-algebra on S n−1 will be denoted by L. When (X, M, µ) is a measure space and 1 ≤ p < ∞, we use For f ∈ L p (X), we define f p := |f | p dµ 1/p . Identifying two functions when they agree µ-almost everywhere, we make L p (X) a Banach space with the norm · p .
We will use bold letters (for example X) for random variables. The expectation of a function f of a random variable X will be denoted E X [f (X)], or just E[f (X)]. The probability of an event E will be denoted P[E].
When X is a set, we use 1 X to denote its characteristic function; that is 1 X (x) = 1 if x ∈ X and 1 X (x) = 0 otherwise.
If G = (V, E) is a graph, a set I is called independent if {u, v} / ∈ E for any u, v ∈ I. The independence number α(G) of G is the cardinality of a largest independent set in G. We define α X (n) as in (1), and for brevity we let α(n) = α {0} (n).

Combinatorial upper bound
Let us begin by deriving a simple "combinatorial" upper bound for the quantity α X (n). Proof. Let I ⊂ S n−1 be an X-avoiding set, and take a uniform O ∈ SO(n). Let the random variable Y be the number of ξ ∈ V with Oξ ∈ I. Since Oξ ∈ S n−1 is uniformly distributed for every ξ ∈ V , we have by the linearity of expectation that E(Y ) = |V | λ(I). On the other hand, Y ≤ α(H) for every outcome O. Thus λ(I) ≤ α(H)/|V |.
We next use Proposition 1 to find the largest possible Lebesgue measure of a subset of the unit circle in R 2 in which no two points lie at some fixed forbidden angle.
Theorem 2. Let X = {x} and put t = arccos x 2π . If t is rational and t = p/q with p and q coprime integers, then α X (2) = 1/2, if q is even, In this case α X (2) is attained as a maximum. If t is irrational then α X (2) = 1/2, but there exists no measurable X-avoiding set I ⊂ S 1 with λ(I) = 1/2.
Proof. Write α = α X (2), and identify S 1 with the interval [0, 1) via the map (cos x, sin x) → x/2π. We regard [0, 1) as a group with the operation of addition modulo 1. Notice that I ⊂ [0, 1) is X-avoiding if and only if I ∩ (t + I) = ∅. This implies immediately that α ≤ 1/2 for all values of x.
Now suppose t = p/q with p and q coprime integers, and suppose that q is even. Let S be any open subinterval of [0, 1) of length 1/q, and define T : [0, 1) → [0, 1) by T x = x + t mod 1. Using the fact that p and q are coprime, one easily verifies that I = S ∪ T 2 S ∪ · · · ∪ T q−4 S ∪ T q−2 S has measure 1/2. Also S is X-avoiding since T S = T S ∪ T 3 S ∪ · · · ∪ T q−3 S ∪ T q−1 S is disjoint from S. Therefore α = 1/2.
Next suppose q is odd. With notation as before, a similar argument shows that I ∪T 2 I ∪· · ·∪T q−3 I is an X-avoiding set of measure (q −1)/(2q), and Proposition 1 shows that this is largest possible, since the points x, T x, T 2 x, . . . , T q−1 x induce a q-cycle.
Finally suppose that t is irrational. By Dirichlet's approximation theorem there exist infinitely many pairs of coprime integers p and q such that |t − p/q| < 1/q 2 . For each such pair, let ε = ε(q) = |t − p/q|. Using an open interval I of length 1 q − ε and applying the same construction as above with T defined by T x = x + p/q, one obtains an X-avoiding set of measure at least ( Alternatively, the lower bound α ≥ 1/2 follows from Rohlin's tower theorem (see e.g. [KM10, Theorem 169]) applied to the ergodic transformation T x = x + t. Therefore α = 1/2. However this supremum can never be attained. Indeed, if I ⊂ [0, 1) is an Xavoiding set with λ(I) = 1/2 and T is defined by T x = x + t, then I ∩ T I = ∅ and T I ∩ T 2 I = ∅. Since λ(I) = 1/2, this implies that I and T 2 I differ only on a nullset, contradicting the ergodicity of the irrational rotation T 2 .

Gegenbauer polynomials and Schoenberg's theorem
Before proving the first main result, we recall the Gegenbauer polynomials and Schoenberg's theorem from the theory of spherical harmonics. For ν > −1/2, define the Gegenbauer weight function To motivate this definition, observe that if we take a uniformly distributed vector ξ ∈ S n−1 , n ≥ 2, and project it to any given axis, then the density of the obtained random variable X ∈ [−1, 1] is proportional to r (n−2)/2 , with the coefficient Applying the Gram-Schmidt process to the polynomials 1, t, t 2 , . . . with respect to the inner product f, For a concise overview of these polynomials, see e.g. [DX13, Section B.2]. Here, we always use the normalisation Moreover, the convergence on the right-hand side is absolute and uniform for every positive definite function f . We are especially interested in the case n = 3. Then ν = 1/2, and the first few

Linear programming relaxation
Schoenberg's theorem allows us to set up a linear program whose value upper bounds α(n) for n ≥ 3. The same result appears in [BNdOFV09] and [dOF09]; we present a self-contained (and slightly simpler) proof for the reader's convenience.
In the next section we strengthen the linear program, obtaining a better bound for α(3).
Lemma 4. Suppose f, g ∈ L 2 (S n−1 ) and define k : where the expectation is taken over randomly chosen O ∈ SO(n), and ξ, η ∈ S n−1 are any two points satisfying ξ, η = t. Then k(t) exists for every −1 ≤ t ≤ 1, and k is continuous. If f = g, then k is positive definite.
Proof. The expectation in (3) clearly does not depend on the particular choice of ξ, η ∈ S n−1 . Fix any point ξ 0 ∈ S n−1 and let P : The being an inner product in L 2 (SO(n)), the expectation (4) therefore exists for every t ∈ [−1, 1].
We next show that k is continuous. For each O ∈ SO(n), let R O : L 2 (SO(n)) → L 2 (SO(n)) be the right translation operator defined by ( . Therefore the function t → R P (t) F is continuous from [−1, 1] to L 2 (SO(n)). Using F (O) = g(Oξ 0 ), the continuity of k follows.
Theorem 5. α(n) is no more than the value of the following infinite-dimensional linear program.
where each a i ≥ 0 and the convergence of the right-hand side is uniform on [−1, 1]. Moreover, for each fixed ξ 0 ∈ S n−1 , we have by Fubini's theorem and (3) that Note that whenever i ≥ 1 by the definition of the Gegenbauer polynomials. Putting (6) and (7) together and using that C (n−2)/2 0 ≡ 1, we conclude that a 0 = λ(I) 2 .
Unfortunately in the case n = 3, the value of (5) is at least 1/3, which is the same bound obtained when Witsenhausen first stated the problem in [Wit74]. This can be seen from the feasible solution x 0 = 1/3, x 2 = 2/3 and x i = 0 for all i = 0, 2.

Adding combinatorial constraints
For each ξ ∈ S n−1 and −1 < t < 1, let σ ξ,t be the unique probability measure on the Borel subsets of S n−1 whose support is equal to the set ξ t := {η ∈ S n−1 : η, ξ = t}, and which is invariant under all rotations fixing ξ. Now let n = 3. As before, let I ∈ L be a {0}-avoiding subset of S 2 and define k : [−1, 1] → R as in (3) with f = g = 1 I ; i.e.
Our aim now is to strengthen (5) for the case n = 3 by adding combinatorial inequalities coming from Proposition 1 applied to the sections of S 2 by affine planes. We proceed as follows. Let p and q be coprime integers with 1/4 ≤ p/q ≤ 1/2, and let Let ξ ∈ S n−1 be arbitrary. If we take two orthogonal unit vectors with endpoints in ξ tp,q and the centre ξ 0 = t p,q ξ of this circle, then we get an isosceles triangle with side lengths (1 − t 2 p,q ) 1/2 and base √ 2; by the Cosine Theorem, the angle at ξ 0 is 2πp/q. Let ξ 0 , η 0 ∈ S n−1 be arbitrary points satisfying ξ 0 , η 0 = t p,q . By Fubini's theorem we have But if q is odd, then ξ tp,q 0 1 I (Oη) dσ ξ 0 ,tp,q (η) ≤ q−1 2q for all O ∈ SO(n) by Proposition 1 applied to the circle (Oξ 0 ) tp,q ∼ = S 1 , since the subgraph it induces contains a cycle of length q. Therefore k(t p,q ) ≤ λ(I) q−1 2q . It follows that the inequalities are valid for the relaxation and can be added to (5). The same holds for the inequalities ∞ i=0 x i C 1/2 i (−t p,q ) ≤ (q − 1)/2q. So we have just proved the following result.
Theorem 6. α(3) is no more than the value of the following infinite-dimensional linear program.
Rather than attempting to find the exact value of the linear program (10), the idea will be to discard all but finitely many of the combinatorial constraints, and then to apply the weak duality theorem of linear programming. The dual linear program has only finitely many variables, and any feasible solution gives an upper bound for the value of program (10), and therefore also for α(3). At the heart of the proof is the verification of the feasibility of a particular dual solution which we give explicitly. While part of the verification has been carried out by computer in order to deal with the large numbers that appear, it can be done using only rational arithmetic and can therefore be considered rigorous. Proof. Consider the following linear program The linear programming dual of (11) is the following.
The rational arithmetic calculations required in the above proof were carried out with Mathematica. When verifying the upper bound for the right-hand side of (14), it is helpful to recall the identity Γ(i + 1/2) = (i − 1/2)(i − 3/2) · · · (1/2) √ π. When verifying the constraints (13) for i = 0, 1, . . . , 39, it can be helpful to observe that t 1,3 and ±t 2,5 are roots of the polynomials x 2 − 1/3 and x 4 − 1/5 respectively; this can be used to cut down the degree of the polynomials C 1/2 i (x) to at most 3 before evaluating them. The ancillary folder of the arxiv.org version of this paper contains a Mathematica notebook that verifies all calculations.
The combinatorial inequalities of the form (9) we chose to include in the strengthened linear program (11) were found as follows: Let L 0 denote the linear program (5). We first find an optimal solution σ 0 to L 0 . We then proceed recursively; having defined the linear program L i−1 and found an optimal solution σ i−1 , we search through the inequalities (9) until one is found which is violated by σ i−1 , and we strengthen L i−1 with that inequality to produce L i . At each stage, an optimal solution to L i is found by first solving the dual minimisation problem, and then applying the complementary slackness theorem from linear programming to reduce L i to a linear programming maximisation problem with just a finite number of variables.
Adding more inequalities of the form (9) appears to give no improvement on the upper bound. Also adding the constraints ∞ i=0 x i C 1/2 i (t) ≥ 0 for −1 ≤ t ≤ 1 appears to give no improvement. A small (basically insignificant) improvement can be achieved by allowing the odd cycles to embed into S 2 in more general ways, for instance with the points lying on two different latitudes rather than just one.

Adjacency operator
Let n ≥ 3. For ξ ∈ S n−1 and −1 < t < 1, we use the notations ξ t and σ ξ,t from Section 6. For f ∈ L 2 (S n−1 ) define A t f : S n−1 → R by Here we establish some basic properties of A t which will be helpful later. The operator A t can be thought of as an adjacency operator for the graph with vertex set S n−1 , in which we join two points with an edge when their inner product is t. Adjacency operators for infinite graphs are explored in greater detail and generality in [BDdOFV14].
Proof. The right-hand side of (15) involves integration over nullsets of a function f ∈ L 2 (S n−1 ) which is only defined almost everywhere, and so strictly speaking one should argue that (15) really makes sense. In other words, given a particular representative f from its L 2 -equivalence class, we need to check that the integral on the right-hand side of (15) is defined for almost all ξ ∈ S n−1 , and that the L 2 -equivalence class of A t f does not depend on the particular choice of representative f .
Let us formally check all the hypotheses of Minkowski's integral inequality applied to F , where SO(n) is equipped with the Haar measure, and where S is equipped with the normalised Lebesgue measure; this will show that the functioñ F : SO(n) → R defined byF (ρ) = S F (ρ, η) dη belongs to L 2 (SO(n)).
Clearly the function F is measurable. To see that the function ρ → F (ρ, η) belongs to L 2 (SO(n)) for each fixed η ∈ S, simply note that That the function η → F (·, η) 2 belongs to L 1 (S) then also follows easily (in fact, this function is constant): Minkowski's integral inequality now gives that the function η → F (ρ, η) is in L 1 (S) for a.e. ρ, the functionF is in L 2 (SO(n)), and its norm can be bounded as follows: Applying (16) to f −g where g is a.e. equal to f , we conclude that the L 2 -equivalence class ofF does not depend on the particular choice of representative f from its equivalence class.
Now (A t f )(ξ) is simplyF (ρ), where ρ ∈ SO(n) can be any rotation such that ρe n = ξ. This shows that the integral in (15) makes sense for almost all ξ ∈ S n−1 .
We have A t ≤ 1 since for any f ∈ L 2 (S n−1 ), Finallly, applying A t to the constant function 1 shows that A t = 1.
Lemma 9. Let f and g be functions in L 2 (S n−1 ), let ξ, η ∈ S n−1 be arbitrary points, and write t = ξ, η . If O ∈ SO(n) is chosen uniformly at random with respect to the Haar measure on SO(n), then which is exactly the definition of k(t) from (3).

Proof. We have
If H is the subgroup of all elements in SO(n) which fix ξ, then the above integral can be rewritten By Fubini's theorem, this integral is equal to where we use the right-translation invariance of the Haar integral on SO(n) at the first equality, and the second inequality follows by noting that the integrand is constant with respect to h.

Existence of a measurable maximum independent set
Let n ≥ 2 and X ⊂ [−1, 1]. From Theorem 2 we know that the supremum α X (n) is sometimes attained as a maximum, and sometimes not. It is therefore interesting to ask when a maximizer exists. The main positive result in this direction is Theorem 16, which says that a largest measurable X-avoiding set always exists when n ≥ 3. Remarkably, this result holds under no additional restrictions (not even Lebesgue measurability) on the set X of forbidden inner products. Before arriving at this theorem, we shall need to establish a number of technical results. For the remainder of this section we suppose n ≥ 3.
For d ≥ 0, let H n d be the vector space of homogeneous polynomials p(x 1 , . . . , x n ) of degree d in n variables belonging to the kernel of the Laplace operator; that is Note that each H n d is finite-dimensional. The restrictions of the elements of H n d to the surface of the unit sphere are called the spherical harmonics. For fixed n, we have L 2 (S n−1 ) = ⊕ ∞ d=0 H n d ([DX13, Theorem 2.2.2]); that is, each function in L 2 (S n−1 ) can be written uniquely as an infinite sum of elements from H n d , d = 0, 1, 2, . . . , with convergence in the L 2 -norm.
Recall the definition (15) of the adjacency operator from Section 7: The next lemma states that each spherical harmonic is an eigenfunction of the operator A t . It extends the Funk-Hecke formula ([DX13, Theorem 1.2.9]) to the Dirac measures, obtaining the eigenvalues of A t explicitly. The proof relies on the fact that integral kernel operators K having the form (Kf )(ξ) = f (ζ)k( ζ, ξ ) dζ for some function k : [−1, 1] → R are diagonalised by the spherical harmonics, and moreover that the eigenvalue of a specific spherical harmonic depends only on its degree.
Proposition 11. Let t ∈ (−1, 1). Then for every spherical harmonic Y d of degree d, By taking limits, we finish the proof.
The next lemma is a general fact about weakly convergent sequences in a Hilbert space.
Lemma 12. Let H be a Hilbert space and let K : H → H be a compact operator.
Proof. Let C be the maximum of x and sup i≥1 x i , which is finite by the Principle of Uniform Boundedness. Let {K m } ∞ m=1 be a sequence of finite rank operators such that K m → K in the operator norm as m → ∞. Clearly for each m = 1, 2, . . . . Let ε > 0 be given and choose m 0 so that and the lemma follows.
The next corollary is also a result stated in [BDdOFV14].
Proof. The operator A t is diagonalisable by Proposition 11, since the spherical harmonics form an orthonormal basis for L 2 (S n−1 ). It therefore suffices to show that its eigenvalues cluster only at 0. By Theorem 8.21.8 of [Sze92] and Proposition 11, the eigenvalues µ d (t) tend to zero as d → ∞. The eigenspace corresponding to the eigenvalue µ d (t) is precisely the vector space of spherical harmonics of degree d, which is finite dimensional. Therefore A t is compact.
From now on we may therefore assume −1 < t < 1. Let h > 0 be a small number which will be determined later. Suppose x ∈ C h (ξ 1 ). The intersection x t ∩ C h (ξ 2 ) is a spherical cap in the (n − 2)-dimensional sphere x t having height proportional to h; this is because C h (ξ 2 ) is the intersection of S n−1 with a certain halfspace H, and x t ∩ C h (ξ 2 ) = x t ∩ H. We have σ x,t (x t ∩ C h (ξ 2 )) = Θ(h (n−2)/2 ) by Lemma 14, and it follows that there exists D > 0 such that σ x,t (x t ∩ C h (ξ 2 )) ≤ Dh (n−2)/2 for sufficiently small h > 0.
If x ∈ C h/2 (ξ 1 ), then x t ∩ C h/2 (ξ 2 ) = ∅ since x t is just a rotation of the hyperplane ξ t 1 through an angle equal to the angle between x and ξ 1 . Therefore x t ∩ C h (ξ 2 ) is a spherical cap in x t having height at least h/2.
We are now in a position to prove the second main result of this paper.
Theorem 16. Suppose n ≥ 3 and let X be any subset of [−1, 1]. Then there exists an X-avoiding set I ∈ L such that λ(I) = α X (n).
Proof. We may suppose that 1 ∈ X for otherwise every X-avoiding set is empty and the theorem holds with I = ∅.
be a sequence of measurable X-avoiding sets such that lim i→∞ λ(I i ) = α X (n). Passing to a subsequence if necessary, we may suppose that the sequence {1 I i } of characteristic functions converges weakly in L 2 (S n−1 ); let h be its limit.
Denote by I the set h −1 ((0, 1]), and let I be the set of Lebesgue density points of I . We claim that I is X-avoiding.
For all t ∈ X \ {−1}, the operator A t : L 2 (S n−1 ) → L 2 (S n−1 ) is self-adjoint and compact by Lemma 10 and Corollary 13. Since A t 1 I i , 1 I i = 0 for each i, Lemma 12 implies A t h, h = 0. Since h ≥ 0, it follows from the definition of A t that A t 1 I , 1 I = 0, and therefore also that A t 1 I , 1 I = 0. But if there exist points ξ, η ∈ I with t 0 = ξ, η ∈ X \ {−1}, then A t 0 1 I , 1 I > 0 by Lemma 15.
Thus, in order to show that I is X-avoiding, it remains to derive a contradiction from assuming that −1 ∈ X and −ξ, ξ ∈ I for some ξ ∈ S n−1 . Since ξ and −ξ are Lebesgue density points of I, there is a spherical cap C centred at ξ such that λ(I ∩ C) > 2 3 λ(C) and λ(I ∩ (−C)) > 2 3 λ(C). The same applies to I i for all large i (since a cap is a continuity set). But this contradicts the fact that I i and its reflection −I i are disjoint for every i. Thus I is X-avoiding.
9 Invariance of α X (n) under taking the closure of X Again let n ≥ 2 and X ⊂ [−1, 1]. We will use X to denote the toplogical closure of X in [−1, 1]. In general it is false that X-avoiding sets are X-avoiding. In spite of this, we have the following result.
Proof. Clearly α X (n) ≥ α X (n) For the reverse inequality, let I ⊂ S n−1 be any measurable X-avoiding set. Let I ⊂ I be the set of Lebesgue density points of I , and define k : [−1, 1] → R by k(t) = S n−1 1 I (ζ)(A t 1 I )(ζ) dζ. Then k is continuous by Lemma 4 and Lemma 9, and since k(t) = 0 for every t ∈ X, it follows that k(t) = 0 for every t ∈ X. Lemma 15 now implies that I is X-avoiding. The theorem now follows since I was arbitrary, and λ(I) = λ(I ) by the Lebesgue density theorem.

Single forbidden inner product
An interesting case to consider is when |X| = 1, motivated by the fact that 1/α {t} (n) is a lower bound on the measurable chromatic number of R n for any t ∈ (−1, 1) and this freedom of choosing t may lead to better bounds.
Let us restrict ourselves to the special case when n = 3 (that is, we look at the 2-dimensional sphere). For a range of t ∈ [−1, cos 2π 5 ], the best construction that we could find consists of one or two spherical caps as follows. Given t, let h be the maximum height of an open spherical cap which is {t}-avoiding. A simple calculation shows that h = 1 − (t + 1)/2. If t ≤ −1/2, then we just take a single cap C of height h, which gives that α {t} (3) ≥ h/2 then. When −1/2 < t ≤ 0, we can add another cap C whose centre is opposite to that of C. When t reaches 0, the caps C and C have the same height (and we get the two-cap construction from Kalai's conjecture). When 0 < t ≤ 2π 5 , we can form a {t}-avoiding set by taking two caps of the same height h. (Note that the last construction cannot be optimal for t > 2π 5 , as then the two caps can be arranged so that a set of positive measure can be added, see the third picture of Figure 1.) Calculations show that the above construction gives the following lower bound (where h = 1 − (t + 1)/2): 0 ≤ t ≤ cos 2π 5 . We conjecture.that the bounds in (18) are all equalities. In particular, our conjecture states that, for t ≤ −1/2, we can strengthen Levy's isodiametric inequality by forbidding a single inner product t instead of the whole interval [−1, t].
As in Section 6, one can write an infinite linear program that gives an upper bound on α {t} (3). Although our numerical experiments indicate that the upper bound given by the LP exceeds the lower bound in (18) by at most 0.062 for all −1 ≤ t ≤ 0.3, we were not able to determine the exact value of α {t} (3) for any single t ∈ (0, cos 2π 5 ].