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By Thomas Craig

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Extra info for A Treatise on Linear Differential Equations, Volume I: Equations with Uniform Coefficients

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Of J on U. 2,5 one has v(f) = v(/) = inf V (fa) = inf V(Ja) ;::: v(l) ;::: v(f), a a and thus 2) ==:} 5). Let now v be maximal on U and let v -< '11, '11 a boundary probability measure on U. Note that v and '11 have the same barycenter in U. 6 implies £('11) = '11 0 ~. 10,2, '11 = v and so 1) ==:} 3) . L is a boundary probability measure on U. 2) Each v E 1'(U) with £(v) E U is dominated by a boundary probability measure onU. Proof. Part 1) is immediate and 2) follows from the observation that £('11) U for any boundary probability measure '11, v -< '11 on U.

8 Let 81 E ~ c P(X), let U = co(~) and let f be a bounded upper semicontinuous function on X. Assume that the following conditions hold: 1) 81 (J) > 8(J) for any 8 E ~, 8 ¥- 81 ; 2) for any 800 EE \P(X) there exists a function g E Cb(X), g ~ f, such that 800 (g) < 81 (J). Then 81 E 8e (U). Proof. 31]). l(J) is bounded, affine and upper semicontinuous on U. Let its maximum be attained at 8 E 8e (U). 59-60]), and 800 (J) ::; 800 (g) < 81 (J) for 800 EE \~, it follows that 8 = 81 . 2, to our noncompact case.

0. 0. 0. 5). 6 is called a subspace of the Khinchin space (X, U) and U1 is called a substructure of the Khinchin structure U or the restriction of U to Xl. Notice that U1 is the least fine Khinchin structure on Xl such that the canonical injection Xl --t X is a Khinchin morphism. 0. 6). 0. 0.. 0. 1)) is a Khinchin space. Proof. 0. 0. 5. 0. 5. c U, but not We now turn to the construction of a product of two Khinchin spaces. Let Xl and X 2 be completely regular spaces. For s = 1,2, let 'Irs be the projection from the product space X = Xl X X 2 onto XS.

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