A Spherical Surface Measure Inequality for Convex Sets

Abstract

Let the set C in the Euclidean space of.ndimensions be closed, symmetric under reflection in the origin, and convex.The portion of the surface of the unit ball lying in C is shown to decrease in (the uniform) surface measure when C is replaced by AC, the image of C under any linear transformation A with norm no greater than one.Some cases of equality are discussed, and an application is given.1. Introduction.We prove that the hypersurface area of the intersection of a convex symmetric set Cs£" (C= -C) with the unit sphere S decreases when we shrink C. Precisely, let /* be the uniform surface measure on S and let A .En-*En be a linear transformation of norm <1.(The norm of A, denoted \\A\\, is given by \\A\\ =sup{Ux| :xeS},and |_y|2=2 }'«•) Theorem 1.If C is closed, convex, and symmetric about 0, then H(ACr\S)<ii{Cr\S).A simple geometric argument proves Theorem 1 in the two-dimensional case.First let C be an infinite strip L meeting S in /U(-/), the union of a symmetric pair of arcs (see Figure 1).Since the width of AL is no greater than that of L, fi(ALnS)^fi(LnS).Next assume C is a polygon.Then CnS may be expressed as the union of a finite collection of symmetric pairs of arcs fjt -1}.For each j let L, be the symmetric strip such that LjrtS=IjKj( -lj) (see Figure 2).Since || A || ^ 1, the part of C contained in the open unit ball B plays no role (that is, ACr\S-A(C^~B)r\S).The remainder of C is contained in the union

Locations

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