By Sergey V. Ludkovsky

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1(c)[VTC85]) and using the normalization of a measure µ by 1 we consider the case y = 0. For each r > 0 we have: |Re(θ(0) − θ(x))| = 1/2 ≤ B(X,0,r) X (1 − [χe (x(u)) + χe (−x(u))]/2)µ(du) 1/2 2[χe (x(u)) − χe (−x(u))/(2i)]2 µ(du) + 2 ≤ 2π2 B(X,0,r) η(x(u))2 µ(dx) + 2µ([x : µ(du) X\B(X,0,r) x > r]). 5 for each b > 0 and δ > 0 there are a finite-dimensional over K subspace L in X and a compact subset W ⊂ X such that W ⊂ Lδ , µ|(X\W ) < b, hence µ|(X\Lδ ) < b. We consider the following expression: J( j, l) := 2π2 B(X,0,r) η(e j (u))η(el (u))µ(du), where (e j ) is the orthonormal basis in X which contains the orthonormal basis of L = Kn , n = dimK L.

42. Theorem. 40(4) exists. Proof. We shall say that a sequence α1 , . . , α2 , . . crosses infinite times the segment [β1 , β2 ] with β1 < β2 , if it is possible to choose k1 < k2 < · · · so that αk1 ≥ β2 , αk2 ≤ β1 ,. . , αk2n−1 ≥ β2 , αk2n ≤ β1 , . . Consider the set Bβ1 ,β2 consisting of all x ∈ X for which the sequence {φn (x) : n} crosses infinite number of times the segment [β1 , β2 ]. Denote by B− the set of all x ∈ X for which infn φn (x) = −∞, while B+ is the set of all x ∈ X with supn φn (x) = ∞.

Put ρ1n (x) := dν1Ln (PLn x)/dµ1Ln (PLn x). 1 there exists the limit lim ρ1 (x) n→∞ n = dν1 (x)/dµ1 (x) (mod µ1 ). But βρ1n (x) ≤ ρn (x), consequently, limn→∞ ρ1n (x) ≤ limn→∞ ρn (x)/b = 0, that is, dν1 (x)/dµ1 (x) = 0 (mod µ1 ) contradicting the supposition about the absolute continuity of ν1 relative to µ1 , hence µ1 ⊥ µ2 . 1. Theorem. 1 (ii) is accomplished in all cases. Proof. Let µ2 = βν1 + (1 − β)ν2 with ν1 ≪ µ1 and ν2 ⊥ µ1 and 0 ≤ β < 1. Denote by 1 νLn and ν2Ln projections of measures ν1 and ν2 on Ln , also ρ1n (x) = dν1Ln (PLn x)/dµ1Ln (PLn x), ρ2n (x) = dν2 (PLn x)/dµ1Ln (PLn x).