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By Percival Frost Joseph Wolstenholme
The Authors of the next Treatise have endeavoured to give prior to scholars as complete a view of the topic as sure obstacles have allowed them to do. the need of those boundaries has built itself during getting ready the paintings in the course of a interval of 4 years. The examine of innumerable papers, by way of the main celebrated mathematicians of all nations, has confident the authors that the topic is sort of inexhaustible, and that, to have taken care of all components of it with whatever impending to the fulness with which they've got handled the 1st component, could have swelled their paintings in a anxious share to what it has already attained.
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1) with the constant n2 replaced by 47r2. 4 Carlson's Inequality for Finite Sums Suppose that m is a positive integer. e. if the series are replaced by finite sums with m terms. If, however, we restrict attention to finite sums with a fixed number of terms, the constant 7r2 need no longer be (and is not) sharp. We have 18 Multiplicative Inequalities of Carlson Type and Interpolation the following preliminary result by L. Larsson, Z. -E. Persson [52]. 1 Let a i , . . , a m be non-negative numbers, not all zero.
9. Let m m a = J2k S 'ark+1 and ^2ka2ark+1, T = k=l fe=l and let A be any positive number. TT . (AfcQl + jka*)^iak, we get by the Holder-Rogers inequality / m a E0 \r+\ / m A a, + A lfcaa <(D * \ r / _m_ ~ )~M E( ^(Xk^+x^k^yr) Afcai+A_lfca2 +1 \ K {XS + X^T). 21) can be estimated by an integral. ; A M B a 2 -ai A2 + m a 2 - a i is decreasing. 22), it follows that +1 (±aX <2\(*2 (-^—B( \t^[ J — ai a >~\, ( \{a2-oii)r r(ai+\-> ^ ^ - a\)r l + m a2_ai 2 «2- f —ai+r •S a2-<»l J 1 " 2 - i l . 20). 9 is not sharp for finite m.
1). 1). Here, however, we include a proof of the discrete version based on Hilbert's double sum inequality. 1), it still gives the same qualitative information. Proof by Hilbert 's Inequality. that fc=l 1=1 The well-known Hilbert inequality states p \k=l / \l=l / for all sequences {cik}k and {bi}i of non-negative numbers, with equality if and only if either all the a,k or all the 6; are zero. 7) that oo ak oo = J2 Y^ai k=l 1=1 SS' k+i 00 °° (k + l)akai oo oo *=i 1=1 /oo . K+ oo oo . + k=u=iK \ 3 / oo \ I 2„2 l <2* E°0 I>< fc=l / \(=1 provided not all Ofe are zero.