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Difference between revisions of "Abel inequality"

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An estimate for the sum of products of two numbers. If sets of numbers <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a0101301.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a0101302.png" /> are given such that the absolute values of all sums <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a0101303.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a0101304.png" />, are bounded by a number <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a0101305.png" />, i.e. <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a0101306.png" />, and if either <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a0101307.png" /> or <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a0101308.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a0101309.png" />, then
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An estimate for the sum of products of two numbers. If sets of numbers $(a_k)$ and $(b_k)$ are given such that the absolute values of all sums $B_k = b_1 + \cdots + b_k$, $k=1,\ldots,n$, are bounded by a number $B$, i.e. $|B_k| \le B$, and if either $a_i \ge a_{i+1}$ or $a_i \le a_{i+1}$, $i = 1,2,\ldots,n-1$, then
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$$
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\left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B(|a_1| + 2|a_n|)
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$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a01013010.png" /></td> </tr></table>
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If the $a_k$ are non-increasing and non-negative, one has the simpler estimate:
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$$
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\left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B a_1 \ .
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$$
  
If the <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a01013011.png" /> are non-increasing and non-negative, one has the simpler estimate:
 
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a010/a010130/a01013012.png" /></td> </tr></table>
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Abel's inequality is proved by means of the [[Abel transformation]].
  
Abel's inequality is proved by means of the [[Abel transformation|Abel transformation]].
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{{TEX|done}}
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[[Category:Real functions]]

Latest revision as of 12:13, 9 November 2014

An estimate for the sum of products of two numbers. If sets of numbers $(a_k)$ and $(b_k)$ are given such that the absolute values of all sums $B_k = b_1 + \cdots + b_k$, $k=1,\ldots,n$, are bounded by a number $B$, i.e. $|B_k| \le B$, and if either $a_i \ge a_{i+1}$ or $a_i \le a_{i+1}$, $i = 1,2,\ldots,n-1$, then $$ \left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B(|a_1| + 2|a_n|) $$

If the $a_k$ are non-increasing and non-negative, one has the simpler estimate: $$ \left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B a_1 \ . $$


Abel's inequality is proved by means of the Abel transformation.

How to Cite This Entry:
Abel inequality. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Abel_inequality&oldid=18342
This article was adapted from an original article by L.D. Kudryavtsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article