Quantitative Steinitz theorem: A polynomial bound

The classical Steinitz theorem states that if the origin belongs to the interior of the convex hull of a set S⊂Rd$S \subset \mathbb {R}^d$ , then there are at most 2d$2d$ points of S$S$ whose convex hull contains the origin in the interior. Bárány, Katchalski, and Pach proved the following quantitative version of Steinitz's theorem. Let Q$Q$ be a convex polytope in Rd$\mathbb {R}^d$ containing the standard Euclidean unit ball Bd$\mathbf {B}^d$ . Then there exist at most 2d$2d$ vertices of Q$Q$ whose convex hull Q′$Q^\prime$ satisfies rBd⊂Q′$$\begin{equation*} \hspace*{100pt} r \mathbf {B}^d \subset Q^\prime \end{equation*}$$with r⩾d−2d$r\geqslant d^{-2d}$ . They conjectured that r⩾cd−1/2$r\geqslant c d^{-1/2}$ holds with a universal constant c>0$c>0$ . We prove r⩾15d2$r \geqslant \frac{1}{5d^2}$ , the first polynomial lower bound on r$r$ . Furthermore, we show that r$r$ is not greater than 2d$\frac{2}{\sqrt {d}}$ .


Introduction
The goal of this paper is to establish a quantitative version of the following classical result of E. Steinitz [Ste13].
Proposition 1.1 (Steinitz theorem).Let the origin belong to the interior of the convex hull of a set S ⊂ R d .Then there are at most 2d points of S whose convex hull contains the origin in the interior.
The first quantitative version of this result was obtained in [BKP82], where the following statement was proven.
Proposition 1.2 (Quantitative Steinitz theorem).There exists a constant r = r(d) > 0 such that for any subset Q of R d whose convex hull contains the Euclidean unit ball B d , there exists a subset F of Q of size at most 2d whose convex hull contains the ball rB d .
It was also shown that r(d) > d −2d .With the exception of the planar case d = 2 [KMY92, Bra97,BH94], no significant improvement on r(d) has been obtained (see also [DLLHRS17]).Now we state the main result of this paper in which we obtain a polynomial bound on r(d).
Theorem 1 (Q.S.T. with polynomial bound).Let Q be a subset of R d whose convex hull contains the Euclidean unit ball B d .Then there exist at most 2d points of Q whose convex hull We conjecture the following.
Conjecture 1.1.There is a constant c > 0 such that in any subset Q of R d whose convex hull contains the Euclidean unit ball B d , there are at most 2d points whose convex hull We provide an upper bound on r(d).
Theorem 2. Let u 1 , . . ., u n be unit vectors in R d .Then their absolute convex hull, that is the convex hull of ±u 1 , . . ., ±u n , does not contain the ball √ n d + ε B d for any positive ε.It follows that if u 1 , . . ., u m form a sufficiently dense subset of the unit sphere (with a large m), then their convex hull is almost the unit ball, while for any n of them with n ≤ 2d, we have that their convex hull does not contain the ball 2 √ d , which shows that the order of magnitude of r(d) in Conjecture 1.1 is sharp if the conjecture holds.
We mention the following conjecture which is closely related to Theorem 2. It can be found in a different formulation in [BJB + 04, p.194].
Conjecture 1.2.Let {u 1 , . . ., u 2d } be unit vectors in R d .Then there is a point in the set 2. The main steps in the proof of Theorem 1 Since r(1) = 1, we will assume that d ≥ 2 throughout the paper.First, we reduce the problem for the polytopal case.By the classical Carathéodory theorem [Car11, p.200], any point of a convex hull of a subset Q of R d can be represented as a convex combination of at most d + 1 points of Q.Thus, taking a sufficiently dense subset of the unit sphere, we observe that for any ǫ ∈ (0, 1) and any set  [BKP82] to prove certain quantitative versions of the Helly theorem.The connection between the quantitative Steinitz result and the quantitative Helly-type result is via polar duality.Recently, the authors of this paper [IN22] have proposed a new approach to quantitative Helly-type results via sparse approximation of polytopes.The connection between the sparse approximation of polytopes and the quantitative Helly-type result is via polar duality again.We state a refined version of the result on the sparse approximation of polytopes obtained by Almendra-Hernández, Ambrus, and Kendall in [AHAK22, Theorem 1].
Proposition 2.1 (Almendra-Hernández et.al.).Let λ > 0, and L ⊂ R d be a convex polytope such that L ⊂ −λL.Then there exist at most 2d vertices of L whose convex hull L ′ satisfies Choosing the origin smartly, one can achieve λ = d.For instance, the following statement holds.
Proposition 2.2.Let K be a convex body in R d .Then the inclusion (K −c) ⊂ −d(K −c) holds for some point c in the interior of K, for example, if c is the centroid of K or of a maximal volume simplex within K.
We recall that the polar of the set S ⊂ R d is defined by Our idea of the proof of Theorem 3 is to use duality twice: We will start with translating the assertion of the theorem in terms of the polar polytope Q • of Q.Then we will choose a point c "deep" in Q • and consider (Q • − c) • .Roughly speaking, by changing the center of polarity, we obtain a more well-structured convex polytope.Next, we use Proposition 2.1 to obtain a sufficiently reasonable bound on r(d), which is not destroyed on the way back to Q • and then to Q.
We use [n] to denote the sets {1, . . ., n}.The convex hull of a set S is denoted by conv S. For a non-zero vector v ∈ R d , H v denotes the half-space We use vert P to denote the vertex set of a polytope P.
For the sake of completeness, we provide a shortened original proof of Proposition 2.1.
Proof of Proposition 2.1.The condition L ⊆ −λL ensures that the origin belongs to the interior of L. Among all simplices with d vertices from the set of vertices of L and one vertex at the origin, consider a simplex S = conv{0, v 1 , . . ., v d } with maximal volume.The simplex S can be represented as It is easy to see that P is a paralletope that can be represented as (2) Since S is chosen maximally, equation (2) shows that for any vertex v of L, v ∈ P .By convexity, (3) L ⊂ P.
Let y be the intersection of the ray emanating from 0 in the direction Since L ⊂ −λL, we also have that Combining it with (3), (4), (5), we obtain (6) Completing the proof of Proposition 2.1.

Proof of Theorem 1
As was explained in the previous section, it suffices to prove Theorem 3, which we proceed to work with.
containing the origin in its interior.By duality, it suffices to show that there are at most 2d half-spaces Let c be a point in the interior of K such that the inclusion Now, we use Proposition 2.1 with λ = d.We obtain that there are w 1 , . . ., w m ∈ vert L for some integer m satisfying m ≤ 2d such that Considering the polar sets, we get Recall that c is an interior point of the polytope K.By (7), one has that for any w ∈ vert L, H w = H v − c for some v ∈ vert Q.It means that Since d ≥ 2, the desired bound for Q ′ = conv {v i : i ∈ [m]} follows.The proof of Theorem 3 is complete, which implies Theorem 1 as was discussed earlier.

Proof of Theorem 2
In this section, we prove Theorem 2, which is a dual version of [IN22, Theorem 1.4] and immediately follows from it.For the sake of completeness, we prove Theorem 2 here.We first state the main ingredient of the proof obtained by K. Ball and M. Prodromou.such that p, A −1 p ≥ trace A −1 .Denote q = A −1/2 p.Then, by the choice of p, That is, q ∈ K. On the other hand, |q| 2 = A −1/2 p, A −1/2 p = p, A −1 p ≥ trace A −1 .
Finally, since trace A = n and by the Cauchy-Schwarz inequality, one sees that trace A −1 is at least d 2 n .Thus, |q| ≥ d √ n .This completes the proof of Theorem 2.
Theorem 1 follows from the following polytopal version.Theorem 3. Let Q be a convex polytope in R d containing the Euclidean unit ball B d .Then there are at most 2d vertices of Q whose Proposition 4.1 ([BP09], Theorem 1.4).Let vectors {v 1 , . . ., v n } ⊂ R d satisfy n 1v i ⊗ v i = Id.Then for any positive semi-definite operator T : R d → R d , there is a point p in the intersection of the strips {x ∈ Rd : | x, v i | ≤ 1} satisfying p, T p ≥ trace T .Proof of Theorem 2. There is nothing to prove if the absolute convex hull conv {±u i : i ∈ [n]} does not contain the origin in its interior.So, assume that conv {±u i : i ∈ [n]} contains the origin in its interior.Set K = (conv{±u i : i ∈ [n]}) • .By duality, it suffices to show that K contains a point of Euclidean norm d √ n .Clearly, {u i : i ∈ [n]} spans R d .Consider A = i∈[n]u i ⊗ u i .Since the vectors span the space, A is positive definite.Using Proposition 4.1 with v i = A −1/2 u i , i ∈ [n], and T = A −1 , we find a point p in i∈[n]    {x : | v i , x | ≤ 1}.