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Fractional Planks
Authors:Aharoni  Holzman  Krivelevich  Meshulam
Affiliation:(1) Department of Mathematics, Technion, 32000 Haifa, Israel ra@tx.technion.ac.il holzman@tx.technion.ac.il meshulam@tx.technion.ac.il , IL;(2) Department of Mathematics, Sackler Faculty of Exact Sciences, Tel Aviv University, 69978 Tel Aviv, Israel krivelev@math.tau.ac.il, IL
Abstract:Abstract. In 1950 Bang proposed a conjecture which became known as ``the plank conjecture': Suppose that a convex set S contained in the unit cube of R n and touching all its sides is covered by planks. (A plank is a set of the form {(x 1 , ..., x n ): x j ∈ I} for some j ∈ {1, ...,n} and a measurable subset I of [0, 1]. Its width is defined as |I| .) Then the sum of the widths of the planks is at least 1 . We consider a version of the conjecture in which the planks are fractional. Namely, we look at n -tuples f 1 , ..., f n of nonnegative-valued measurable functions on [0,1] which cover the set S in the sense that ∑ f j (x j ) ≥ 1 for all (x 1 , ..., x n )∈ S . The width of a function f j is defined as ∈t 0 1 f j (x) dx . In particular, we are interested in conditions on a convex subset of the unit cube in R n which ensure that it cannot be covered by fractional planks (functions) whose sum of widths (integrals) is less than 1 . We prove that this (and, a fortiori, the plank conjecture) is true for sets which touch all edges incident with two antipodal points in the cube. For general convex bodies inscribed in the unit cube in R n we prove that the sum of widths must be at least 1/n (the true bound is conjectured to be 2/n ).
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