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Proof of Theorem 1. Fix K G In order to show (4) we introduce the function g(z) = 1 — 1 k(z)- We adopt the standard convention that OlnO = 0, hence the desired ineąuality is equivalent to (6). Thus the application of Lemma 2 for the function g finishes the proof.    □

Theorem 1 immediately implies that the Cartesian products of cylinders sup-port SC-ineąuality. As a conseąuence, SC-inequality possesses a tensorization property.

Corollary 1. Assume sets K\ C Cⁿⁱ,...,Kg C C^ne support SC-inequality. Then the set K\ x ... x Ki also supports SC-inequality.

Another consequence of the main theorem concerns the standard exponential measure A_n on Rⁿ, i.e.

dA_n(x) = — e~\^x\^ldx, x G Rⁿ, ^v ’ 2ⁿ

where we denote |(aq,..., a^_n)|i = KI- It turns out that certain subsets of Rⁿ support the 5-inequality for A_n with strips as the optimal sets. To State the result a few definitions will be useful. We say that a set K C (R+)ⁿ is an ideał if along with any its point x G K it contains the cube [0,xi] x ... x [0, x_n]. A set K C Rⁿ is called unconditional if (eixi,..., e_nx_n) G K whenever (a;i,..., x_n) G K and ei,... ,e_n G {—1,1}- By an unconditional ideał K in Rⁿ we mean the unconditional set K such that the set K n (M+)ⁿ is an ideał. For instance, any unconditional convex set is also an unconditional ideał.

Theorem 2. For any closed unconditional ideał K c Mⁿ and for any strip P = {x G Rⁿ | |a?i | < p], p>0, we have

A_n(K) = A_n(P)    =4> Vt > 1 A_n(tK) > A_n(tP),    (7)

and, equivalently,

A_n(K) = A_n(P)    < 1 A_n{tK) < A_n(tP).    (8)

Proof. The equivalence between (7) and (8) is straightforward. For instance, assume the latter does not hołd. Then, there is to < 1 ^such that X_n(toK) > X_n{toP). We can find so < 1 for which X_n(sotoK) = X_n(toP). Using (7) we get a contradiction

A n(K) > A„(s₀ K) = A„ (J- (sotoK)) > A„ fi (t₀P) J = A„(P) = A „(if). Consider the mapping F : Cⁿ —> (R₊)ⁿ given by the formula F(z\,..., z_n) =    KI)-

Observe that for an ideał A C (R+)ⁿ, the set F~¹(A) is Reinhardt complete and integrating using the polar coordinates we find that