By Paul Koosis
The subject of this certain paintings, the logarithmic crucial, lies athwart a lot of 20th century research. it's a thread connecting many it sounds as if separate components of the topic, and is a usual element at which to start a significant learn of actual and complicated research. Professor Koosis' target is to teach how, from uncomplicated rules, one could building up an research that explains and clarifies many alternative, likely unrelated difficulties; to teach, in influence, how arithmetic grows. The presentation is easy, so this, the 1st of 2 volumes, is self-contained, yet extra importantly, via following the subject, Professor Koosis has produced a piece that may be learn as an entire. He has introduced jointly right here many effects, a few new and unpublished, making this a key reference for graduate scholars and researchers.
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Additional resources for The Logarithmic Integral: Volume 1
If we refer to the fourth theorem of §C (p. 28), we see that we can give the result just proved a more general formulation. The statement thus obtained, which we give as a corollary, also goes under the names of Paley and Wiener. Corollary. Let f(z) be entire and of (some) exponential type, with f(x)eL2(R), and let limsuploglf(iY)I = b, Y limsup log l f (iY) I Y- - OD = - a. IYI Then f(z) = 27r f ba e -'z'9(A) where cpeL2(a,b). Proof. , f is bounded on P if we only assume f eL2(P), provided that it is entire and of exponential type.
IYI Therefore Kph(),) = 0 outside [a - h, b + h], by Paley-Wiener. We're done. Remark. The same result holds (with almost the same proof) if we replace 9(A)d. 1), p being any finite signed measure of compact support. E. Introduction to the condition 1',,, (iog + I f(x) I /(1 + x 2)) dx < oo The entire functions of exponential type considered in the previous § certainly satisfy this condition, as do those arising in the study of many questions in analysis. We will meet repeatedly with such functions in the following chapters of this book, and the rest of the present chapter is mainly concerned with them.
For 2 > b'. The support of cp would thus be contained in [a, b'], so [a, b] would not be Bp's supporting interval, and we have a contradiction. If [a, b] is the supporting interval of cp, we must therefore have limsup log I f (iy) = b. Y _'O Y It is clear that this b can only come from the portions _ fb 1 27r e-iz'gp(2)d2 b- of the integral giving f (z), s > 0. e. on any interval of the form [b - e, b], but it is still quite conceivable that fbb-e eyzrp(1) dl could come out much smaller than eby for large y on account of cancellation.
The Logarithmic Integral: Volume 1 by Paul Koosis