A Treatise On the Theory of Functions by James. Harkness

By James. Harkness

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Thus, xo(k, 6) is obtained from So(k) by deleting the (k-n+2-~)-neighborhood of its selfintersections, which, in fact, are described by the von Laue diffraction conditions. 1. It contains two steps. 1 are satisfied and for all t in the complex (k-n+l-2~)-neighborhood of xo(k, 6) there exists a unique j E Z " such that I p~t ( t ) - k 2t 1< k 2t-~-6 and the second condition of Geometric L e m m a is satisfied. Secondly, we prove that the nonsingular set xo(k,6) is nonempty; moreover it is a subset of an asymptotically full measure on So(k).

In fact, they are expansions of eigenvalue ,~(t) and its spectral projection E(t) in powers of l). Therefore, )~(t) and E(t) are analytic functions of I7, but not, generally speaking, of V. Hence, it would be natural now to place a multiplier a before I7 (just as in the preceding considerations it was helpful to place a before V). However, we omit it to make formulae tess cumbersome. The results for the singular set were proved in [K121. First of all we describe more in detail operator Hq(t). As noted above, operator [-Iq(t) has only two nondiagonal terms not equal to zero: [7q(t)j,j_q = Hq(t)j_q,j = Vq.

If p,~(O) <_ lOpq(O), then considering that vectors pro(O), pq(O) are linearly independent, we obtain p~(0)• ~p~(0). 19), we get pq(0)• > c. 18), because pq(0) < k ~/2 by the hypothesis. 18). 9). The lemma is proved. 11 . " pro(O) < c, c ~s c(k). Then the following s(Sq(k, 1 + ~) N Ao(k, 6)) < k-~s(Sq(k, 1 + 6)). 2o) Proof. TEO It is easy to see that Sq(k, 1 + 6) N Sin(k, 1 + 6~f), when pro(0), pq(0) are linearly dependent. 10). Let Q1 = {m :pq,~ > 1}, q2 = { m : pq~ < 1, p,,(o) > 2k - 1}, q3 = {rn: pq~ < I p~(0) _< 2k - 1}.

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