نتایج جستجو برای: hilbert type inequality

تعداد نتایج: 1414224  

Journal: :Axioms 2022

We study the approximation capability of orthogonal super greedy algorithm (OSGA) with respect to μ-coherent dictionaries in Hilbert spaces. establish Lebesgue-type inequalities for OSGA, which show that OSGA provides an almost optimal on first [1/(18μs)] steps. Moreover, we improve asymptotic constant inequality OGA obtained by Livshitz E D.

2011
Charles Byrne Yair Censor Aviv Gibali Simeon Reich

We introduce and study the Split Common Null Point Problem (SCNPP) for set-valued maximal monotone mappings in Hilbert space. This problem generalizes our Split Variational Inequality Problem (SVIP) [Y. Censor, A. Gibali and S. Reich, Algorithms for the split variational inequality problem, Numerical Algorithms, accepted for publication, DOI 10.1007/s11075-011-9490-5]. The SCNPP with only two s...

ژورنال: پژوهش های ریاضی 2020

‎I‎n this paper‎, ‎a‎ forward-‎b‎ackward projection algorithm is considered for finding zero points of the sum of two operators‎ ‎in Hilbert spaces‎. ‎The sequence generated by algorithm converges strongly to the zero point of the sum of an $alpha$-inverse strongly‎ ‎monotone operator and a maximal monotone operator‎. ‎We apply the result for solving the variational inequality problem, fixed po...

Journal: :The American Mathematical Monthly 2008
Jonathan M. Borwein

We explore a variety of pleasing connections between analysis, number theory and operator theory, while revisiting a number of beautiful inequalities originating with Hilbert, Hardy and others. We shall first establish the aforementioned Hilbert inequality [?, ?] and then apply it to various multiple zeta values. In consequence we obtain the norm of the classical Hilbert matrix, while illustrat...

2012
HUSSIEN ALBADAWI

A singular value inequality for sums and products of Hilbert space operators is given. This inequality generalizes several recent singular value inequalities, and includes that if A, B, and X are positive operators on a complex Hilbert space H, then sj ( A 1/2 XB 1/2 ) ≤ 1 2 ‖X‖ sj (A+B) , j = 1, 2, · · · , which is equivalent to sj ( A 1/2 XA 1/2 −B 1/2 XB 1/2 ) ≤ ‖X‖ sj (A⊕B) , j = 1, 2, · · ...

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