Setup

Let n0 be the following parameter (that depends on both n and ε):

n0 :=n + 4pn ln(4/ε)1/2.

LetC0

convex denote the set of convex bodies inRn that are contained in Ball(n0), equivalently, Cconvex0 =C ∩ Ball(n0) : C ∈ Cconvex .

We prove the following claim that helps us focus on testing of C0

convex instead Cconvex. Claim 2.11. Suppose that there is a one-sided sample-based ε-testing algorithm A0 which, given any Lebesgue measurable target set S contained in Ball(n0_{), uses (n/ε)}O(n) _samples drawn from _{N (0, 1)}n to test whether S _{∈ C}0

convex versus S is ε-far from Cconvex0 . Then this implies Theorem 2.2.

Proof. Given A0 for C0

convex, we consider an algorithm A which works as follows to test whether an arbitrary Lebesgue measurable subset S of Rn _{is convex or ε-far from C}

convex: algorithm A runs A0 with parameter ε/2, but with the following modification: each time A0 receives from the oracle a labeled sample (x, b) with x /∈ Ball(n0_{), it replaces the label b with} 0 and gives the modified labeled sample to A0_{. When the run of A}0 _{is complete A returns}

If S _{⊆ R}n _{is the target set, then it is clear that the above modification results in running} A0 on S∩ Ball(n0_{). If S is convex, then S ∩ Ball(n}0_{) is also convex. As A}0 _{commits only} one-sided error, it will always output “accept,” and hence so will A. On the other hand, suppose that S is ε-far from _Cconvex. We claim that Vol(Ball(n0)) ≥ 1 − ε/4 (this will be shown below); given this claim, it must be the case that S∩ Ball(n0_{) is at least (3ε/4)-far} from_Cconvex and at least (3ε/4)-far fromCconvex0 as well. Consequently A0 will output “reject” with probability at least 2/3, and hence so will A.

To bound Vol(Ball(n0)), observe that it is the probability that an x← N (0, 1)n _has

kxk2 ≤ n + 4pn ln(4/ε).

It follows from Lemma 2.5 that the probability is at least 1_{− ε/4 as claimed.}

Given Claim 2.11, it suffices to prove the following slight variant of Theorem 2.2:

Theorem 2.12. There is a one-sided sample-based ε-testing algorithm A0 _{which, given any} Lebesgue measurable target setS contained in Ball(n0), uses (n/ε)O(n) _{samples fromN (0, 1)}n to test whether S _{∈ C}0

convex versus S is ε-far from Cconvex0 .

In the rest of this section we prove Theorem 2.12. We start with some terminology and concepts that we use in the description and analysis of our algorithm. Some of the notions that we introduce below, such as the notions of “boundary” cubes and “internal” cubes, are inspired by related notions that arise in earlier works such as [Ker92, Ras03].

Fix ` := ε3_/n4 _{in the rest of the section, and let Cube}

0 denote the following set

of side length ` that is centered at the origin. We say that a cube is a subset of Rn <sub>of the</sub> form Cube0+ `·(i1, . . . , in), where each ij ∈ Z, which contains at least one point of Ball(2n0). We use CubeSet to denote the set of all such cubes.

It is easy to see that

Ball(n0)⊂[CubeSet⊂ Ball(2n0+ `√n)⊂ Ball(3n0).

Fix an S _{⊆ Ball(n}0_{) as the target set being tested for membership inC}0

convex. Additionally fix a finite set T = _{(x1_{, S(x}1_{)), . . . , (x}M_{, S(x}M_{))} of labeled samples according to S, for} some positive integer M . (The set T will correspond to the set of labeled samples that the testing algorithm receives.) We classify cubes in the CubeSet based on T in the following way:

• A cube Cube is said to be an external cube if Cube ∩ T+ _{=∅ (i.e., no positive point} of T lies in Cube). We let EC denote the union of all the external cubes.

• Any cube which is not an external cube (equivalently, any cube that contains at least one positive point of T ) is said to be a positive cube.

• We say that two cubes Cube, Cube0 are adjacent if for any κ > 0 there exist x _{∈ Cube} and y∈ Cube0 that have Euclidean distance at most κ (in other words, two cubes are adjacent if their closure “touch anywhere, even only at a vertex;” note that each cube is adjacent to itself). If a cube is both (i) a positive cube and (ii) is adjacent to a cube (including itself) that contains at least one negative point of T , then we call it a boundary cube. We use BC to denote the union of all boundary cubes.

Figure 1: A 2D example of the different types of cubes induced by a set of labeled sam- ples. The target set S is a disk, and the solid and hollow dots are positive and negative samples, respectively. The hollow, hatched, and shaded boxes are external, boundary, and internal cubes, respectively.

• We say that a positive cube which is not a boundary cube is an internal cube. (Equivalently, a cube is internal if and only if it contains at least one positive point and all the points in T that are contained in any of its adjacent cubes, including itself, are positive.) We use IC to denote the union of all internal cubes.

We note that since each cube is either external, internal, or boundary, the set Ball(n0) is contained in the (disjoint) union of EC, BC and IC. Figure 1 illustrates the different types of cubes.

We will use the following useful property of internal cubes:

Lemma 2.13. Suppose a finite set of labeled samples T is such that every cube in CubeSet contains at least one point of T . Then every internal cube is contained in Conv(T+_).

The lemma is a direct consequence of the following claim by setting H = T+_:

Claim 2.14. Let H ⊆ Rn _{be any set that contains at least one point in each cube that is} adjacent to Cube0. Then Cube0 is contained in Conv(H).

Proof. We prove the claim by induction on the dimension n. When n = 1 the claim is trivial since Cube0 is simply the interval [−`/2, `/2) and by assumption, there is at least one point of H in [_{−3`/2, −`/2) and at least one point of H in [`/2, 3`/2).}

For n > 1, let P =_{{p ∈ H | p}n ≥ `/2} and P0 ={p0 ∈ H | p0n ≤ −`/2} be two subsets of H. Intuitively, the convex hulls of P and P0 “cover” Cube0 on both sides (by induction), so the convex hull of their union will contain the whole Cube0. More formally, let x be any point in Cube0. By projecting P, P0 and x onto the first n− 1 dimensions and using the inductive hypothesis2_{, we can find points y ∈ Conv(P ) and y}0 _{∈ Conv(P}0_{) such that y}

i = yi0 = xi for all i _{∈ [n − 1]. Since we have p}n ≥ 1/2 and p0n ≤ −1/2 for all p ∈ P and p0 ∈ P0, respectively, it follows directly that yn ≥ 1/2 and y0n ≤ −1/2. As x ∈ Cube0, x is on the line segment between y and y0 and thus is in the convex hull of H. Hence all of Cube0 is contained in Conv(H).