By Ball K.
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Press, 1993. [Szarek 1978] S. J. Szarek, “On Kashin’s almost Euclidean orthogonal decomposition of n er. Sci. Math. Astronom. Phys. 26 (1978), 691–694. 1 ”, Bull. Acad. Polon. Sci. S´ [Talagrand 1988] M. Talagrand, “An isoperimetric inequality on the cube and the Khintchine-Kahane inequalities”, Proc. Amer. Math. Soc. 104 (1988), 905–909. [Talagrand 1990] M. Talagrand, “Embedding subspaces of L1 into Math. Soc. 108 (1990), 363–369. N 1 ”, Proc. Amer. [Talagrand 1991a] M. Talagrand, “A new isoperimetric inequality”, Geom.
We have seen how the second can be deduced from the first. The reverse implication also holds (apart from the precise constants involved). ) In many applications, exact solutions of the isoperimetric problem are not as important as deviation estimates of the kind we are discussing. In some cases where the exact solutions are known, the two properties above are a good deal easier to prove than the solutions: and in a great many situations, an exact isoperimetric inequality is not known, but the two properties are.
Vol(Bt )λ . Let f , g, and m be the functions on the line, given by f (x) = vol(Ax ), g(x) = vol(Bx ), m(x) = vol(Cx ). Then, for r, s, and t as above, m(s) ≥ f (r)1−λ g(t)λ . By the one-dimensional Pr´ekopa–Leindler inequality, 1−λ m≥ f λ g . But this is exactly the statement vol(C) ≥ vol(A)1−λ vol(B)λ , so the inductive step is complete. The proof illustrates clearly why the Pr´ekopa–Leindler inequality makes things go smoothly. Although we only carried out the induction for sets, we required the one-dimensional result for the functions we get by scanning sets in Rn .