A Note on a Brunn-Minkowski Inequality for the Gaussian Measure

We give the counter-examples related to a Gaussian Brunn-Minkowski inequality and the (B) conjecture.


Introduction and notation
Let γ n be the standard Gaussian distribution on R n , i.e. the measure with the density g n (x) = 1 (2π) n/2 e −|x| 2 /2 , where | · | stands for the standard Euclidean norm. A powerful tool in convex geometry is the Brunn-Minkowski inequality for Lebesgue measure (see [Sch] for more information). Concerning the Gaussian measure, the following question has recently been posed.
A counter-example is given in this note. However, we believe that this question has an affirmative answer in the case of o-symmetric convex sets, i.e. the sets satisfying K = −K.
In [CFM] it is proved that for an o-symmetric convex set K in R n the function is log-concave. This was conjectured by W. Banaszczyk and popularized by R. Lata la [Lat]. It turns out that the (B) conjecture cannot be extended to the class of sets which are not necessarily o-symmetric yet contain the origin, as one of the sets provided in our counter-example shows.
As for the notation, we frequently use the function

Counter-examples
Now we construct the convex sets A, B ⊂ R 2 containing the origin such that inequality (GBM) does not hold. Later on we show that for the set B the (B) conjecture is not true.
Fix α ∈ (0, π/2) and ε > 0. Take Clearly, A, B are convex and 0 ∈ A ∩ B. Moreover, from convexity of A we have λA + (1 − λ)A = A and therefore Observe that and that these expressions are analytic functions of ε. We will expand these functions in ε up to the order 2. Let for k = 0, 1, 2, where T (k) is the k-th derivative of T (we adopt the standard notation T (0) = T ). We get Taking ε(1 − λ) instead of ε we obtain we will have a counter-example if we find α ∈ (0, π/2) such that 2a 0 a 2 − a 2 1 > 0.
Recall that a 0 = 1 2 γ 2 (A) = 1 2 1 2 − α π . The integrals that define the a k 's can be calculated. Namely, Therefore, which is positive for α close to π/2. Now we turn our attention to the (B) conjecture. We are to check that for the set B = B ε the function R ∋ t → γ n (e t B) is not log-concave, provided that ε is sufficiently small. Since This produces the desired counter-example for sufficiently small ε as the function t → βe t , where is convex.
Remark. The set B ε which serves as a counter-example to the (B) conjecture in the nonsymmetric case works when the parameter α = 0 as well (and ε is sufficiently small). Since B ε is simply a halfspace in this case, it shows that symmetry of K is required for log-concavity of (1) even in the onedimensional case.