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## TOPICS IN MATH

by: Joanne Bergnaum

41

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# TOPICS IN MATH MATH 4900

Joanne Bergnaum
UGA
GPA 3.56

Staff

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KARMA
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## Popular in Mathematics (M)

This 1 page Class Notes was uploaded by Joanne Bergnaum on Saturday September 12, 2015. The Class Notes belongs to MATH 4900 at University of Georgia taught by Staff in Fall. Since its upload, it has received 41 views. For similar materials see /class/202080/math-4900-university-of-georgia in Mathematics (M) at University of Georgia.

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Date Created: 09/12/15
Math 4900 6900 Spring 2007 ROTH S THEOREM In this note we give a selfcontained proof of the second most famous theorem of Klaus Roth7 namely the very special case of Szemer di s theorem the density version of van der Waerdenls theorem where the arithmetic progressions are of length threel Roth s Theorem Let 6 gt 0 There exists an absolute constant C gt 0 such that ifN 2 exp expC6 1 and A C 17 N with lAl 6N then A necessarily contains a nontrivial arithmetic progression of length three 11 FOURIER ANALYSIS ON Z If f Z A C and E lt 00 then we de ne its Fourier transform by nez a Z fne 2quoti nez Our absolute summability assumption on f ensures that the in nite series de ning fconverges uniformly and hence that f is a continuous function on the circle In this setting the Fourier inversion formula and Plancherells identity are essentially immediate consequences of the familiar orthogonality relation 1 e2m39nada 1 1f n 0 0 0 if n f 0 Indeed it is an easy exercise7 using the uniform convergence of the in nite series de ning f to then estalish i Fourier inversion formula 1 1 mammal 0 ii Plancherells identity 1 lfa12da Z W 0 neZ 21 PROOF OF ROTH S THEOREM We introduce the trilinear form 1 mow Z Z fnyn dgthltn 2d fa 2aha da nez 0162 0 where the second identity can be easily veri ed using the Fourier inversion formula The signi cance of the trilinear form is that A31A1A1A in fact equals the exact number of threeterm arithmetic progressions in A including the lAl trivial progressions where d 0 In order to prove Rothls theorem it therefore suf ces to show that A31A1A1A gt 5N1 For technical reasons it shall be convenient to consider functions of mean value zerol De nition 1 Balanced function We de ne the balanced function of A to be fA 1A 511NA Since A has density 6 on 17 N it is a simple exercise to verify that indeed E fAn 01 Writing the second 1A as 1A 611N fA we obtain 1 A31A71A71A 6 1An1An2d A31A7fA71A7 nEZ 0162 and note by considering the even and odd elements of A that A2 ZZ1Altngt1Altn2dgt 1A1 nEZdEZ The leading term in identity 1 is therefore approximately SSN2 which is instructive as this is also approximately the number of threeterm arithmetic progressions that we would expect A to contain if it where random7 obtained by selecting each natural number from 1 to N independently with probability 61 1

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