1− ζ 2n+2n2n
V(∆◦n), thus
δvol(L◦,∆◦n)≥ ζ
2n+2n2nV(∆◦n).
On the other hand, (91) and(1 + 2nη)n <4/3yield
δvol(S1,∆◦n)≤((1 + 2nη)n−(1−nη)n)V(∆◦n)<4n2η V(∆◦n).
Therefore the triangle inequality implies that V(S1\L◦) = δvol(S1, L◦)≥
ζ
2n+2n2n −4n2η
V(∆◦n).
SinceV(∆◦n)>1by Lemma 5.7 (c), we deduce from (92) that n23nε14 V(∆◦n)≥V(S1\Z∞(µ)◦)≥V(S1\L◦)>
ζ
2n+2n2n −4n2η
V(∆◦n).
It follows fromη=n15nε14 that
ζ <2n+2n2n(n23nε14 + 4n2η)< n28nε14,
which proves Theorem 1.6 in the case where`(Z∞(µ)◦)≤(1 +ε)`(∆◦n)andε < n−100n. Since`(Z∞(µ)◦)≤(1 +ε)`(∆◦n)andW(Z∞(µ))≤(1 +ε)W(∆n)are equivalent according to (1), we have completed the proof of Theorem 1.6 ifε < n−100n.
However, ifε ≥ n−100n, then Theorem 1.6 trivially holds as for anyx∈ Sn−1 there exists a vertexwof∆nwithkx−wk ≤√
2. 2
11 Proof of Corollary 1.5
For the proof of Corollary 1.5, we need the following observation.
Lemma 11.1 If 1nBn ⊂K, C ⊂nBnfor convex bodiesK andC inRn, then
1
n2 δH(K, C)≤δH(K◦, C◦)≤n2δH(K, C).
Proof. We also have n1 Bn⊂K◦, C◦ ⊂nBn. First, we show
δH(K◦, C◦)≤n2δH(K, C). (98) SinceK ⊂C+δH(K, C)Bn⊂C+nδH(K, C)C= (1 +nδH(K, C))C, we have
C◦ ⊂(1 +nδH(K, C))K◦ ⊂K◦+n2δH(K, C)Bn.
By symmetry, we also haveK◦ ⊂C◦+n2δH(K, C)Bn, and thus we have verified (98).
Changing the roles ofK, C and their polarsK◦, C◦ in (98) (and using the bipolar theorem), we also deduce the inequalityδH(K, C)≤n2δH(K◦, C◦). 2 SinceW(K) = `(B2n)`(K◦)according to (1), we conclude Corollary 1.5 by combining The-orem 1.3 (ii), TheThe-orem 1.4 (ii) and Lemma 11.1. 2
Remark The factorn2in Lemma 11.1 is optimal.
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Authors’ addresses:
K´aroly J. B¨or¨oczky, MTA Alfr´ed R´enyi Institute of Mathematics, Hungarian Academy of Sci-ences, Re´altanoda u. 13-15, 1053 Budapest, Hungary. E-mail: carlos@renyi.hu
Ferenc Fodor, Department of Geometry, Bolyai Institute, University of Szeged, Aradi v´ertan´uk tere 1, 6720 Szeged, Hungary. E-mail: fodorf@math.u-szeged.hu
Daniel Hug, Karlsruhe Institute of Technology (KIT), D-76128 Karlsruhe, Germany. E-mail:
daniel.hug@kit.edu