# KARAMATA INEQUALITY PDF

Life[ edit ] Jovan Karamata was born in Zagreb on February 1, into a family descended from merchants based in the city of Zemun , which was then in Austria-Hungary , and now in Serbia. In Lausanne , , he finished primary school oriented towards mathematics and sciences. In the same year he enrolled at the Engineering faculty of Belgrade University and, after several years moved to the Philosophy and Mathematicians sector, where he graduated in He spent the years — in Paris , as a fellow of the Rockefeller Foundation , and in he became Assistant for Mathematics at the Faculty of Philosophy of Belgrade University. In he was elected Full Professor at the University of Geneva. Author: Nizshura Vudozuru Country: Italy Language: English (Spanish) Genre: Career Published (Last): 22 March 2008 Pages: 57 PDF File Size: 18.6 Mb ePub File Size: 10.28 Mb ISBN: 139-6-62957-659-7 Downloads: 37618 Price: Free* [*Free Regsitration Required] Uploader: Zoloran Statement of the inequality Let be an interval of the real line and let denote a real-valued, convex function defined on. Then majorizes the -tuple, since the arithmetic mean of the largest numbers of is at least as large as the arithmetic mean of all the numbers, for every.

Proof of the inequality We may assume that the numbers are in decreasing order as specified in. If for all, then the inequality holds with equality, hence we may assume in the following that for at least one. If for an, then the inequality and the majorization properties and are not affected if we remove and. Hence we may assume that for all. This implies that for all. Define and for all. By the majorization property, for all and by,. To discuss the case of equality in, note that by and our assumption for all.

Let be the smallest index such that, which exists due to. Hence there is a strictly positive term in the sum on the right hand side of and equality in cannot hold. The relaxed condition means that, which is enough to conclude that in the last step of. It only remains to discuss the case. However, then there is a strictly positive term on the right hand side of and equality in cannot hold.

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## Questions tagged [karamata-inequality] Just as the Jensen inequality is used to define convex functions, can the Karamata inequality be used instead to define convex functions? Hence there is a strictly positive term in the sum on the right hand side of 7 and equality in 1 cannot hold. Sign up or log in Sign up using Google. However, then there is a strictly positive term on the right hand side of 7 and equality in 1 cannot karamats.

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## Karamata's inequality Statement of the inequality Let be an interval of the real line and let denote a real-valued, convex function defined on. Then majorizes the -tuple, since the arithmetic mean of the largest numbers of is at least as large as the arithmetic mean of all the numbers, for every. Proof of the inequality We may assume that the numbers are in decreasing order as specified in. If for all, then the inequality holds with equality, hence we may assume in the following that for at least one. If for an, then the inequality and the majorization properties and are not affected if we remove and.

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