By Eisenhart L. P.
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32) implies lim − αj2 ) (the limit always exists) is strictly positive ∞ 2 and thus, j =0 αj < ∞. 20), lim a1 . . an exists in (0, ∞). We have thus proven that ∞ αj2 < ∞ ⇒ lim a1 . . 39) imply lim sup(a1 . . 33) has a limit. 2 Since bn+1 and an+1 − 1 are functions of α2n+j (j = −2, −1, 0, 1), we see that ∞ 2 bn2 < ∞ and (an2 − 1)2 < ∞. 34) holds. 2 2 1 Finally, when ∞ j =0 αj < ∞, an+1 − 1 and bn+1 are the sum of an L sequence 2 and a telescoping sequence, so an+1 − 1 and bn+1 are summable.
We will eventually keep track of this subtlety associated with zeros of w(θ ), but for the rest of this section we will ignore it. 15), is 1. 18). 2) that it has a simple direct proof. 14). dµN strips N α’s off the “bottom” while dµ(N) leaves the bottom N α’s and sets the others to zero. 2). 11). 11), but I know no direct proof. All that one gets from general principles is a semicontinuity. 3). Let dµ , dµ be nontrivial probability measures on ∂D so that dµ → dµ weakly (in the dual topology defined by C(∂D)).
18) When z = eiθ , pn (2 cos θ ) and qn−1 (2 cos θ ) are real, but (z − z −1 )/2 = i sin θ is pure imaginary, so the absolute value square has no cross-term. 19) where we used ([ 12 (1 + α2n−1 )]1/2 )2 + ([ 12 (1 − α2n−1 )]1/2 )2 = 1 to miraculously have α2n−1 drop out! 2 (Direct Geronimus Relations). Let dρ = Sz(dµ) for nontrivial probability measures on [−2, 2] and ∂D. Let {an , bn }∞ n=1 be the Jacobi parameters the Verblunsky coefficients for dµ. 22) (i) (a1 . . an )2 = 2(1 + α2n−1 ) j =0 Remark.