Log-sum inequality












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$begingroup$


Let $(a^n)_{n in mathbb{N}} subset mathbb{R}^k$ be a sequence of vectors with non-negative components and with $sum_{i=1}^k a^n_i=1$ for all $ninmathbb{N}$. Then we have for each $ninmathbb{N}$ by the log-sum inequality:



$sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} geq left(sum_{i=1}^k a_i^{n+1}right) logdfrac{sum_{i=1}^k a_i^{n+1}}{sum_{i=1}^k a_i^{n}}=0,$



where the equality holds if and only if $a_i^{n+1}/a_i^n$ is constant for any $i=1,...,k$.



Now assume that



$limlimits_{nrightarrow infty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} = 0.$



Can we conclude that $limlimits_{nrightarrow infty} dfrac{a_i^{n+1}}{a_i^n} = 1$? Intuitively yes, but how can this be proven rigorously?



Thanks in advance!






real-analysis analysis inequality







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  • $begingroup$
    I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
    $endgroup$
    – marty cohen
    Dec 17 '18 at 18:18
















1












$begingroup$


Let $(a^n)_{n in mathbb{N}} subset mathbb{R}^k$ be a sequence of vectors with non-negative components and with $sum_{i=1}^k a^n_i=1$ for all $ninmathbb{N}$. Then we have for each $ninmathbb{N}$ by the log-sum inequality:



$sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} geq left(sum_{i=1}^k a_i^{n+1}right) logdfrac{sum_{i=1}^k a_i^{n+1}}{sum_{i=1}^k a_i^{n}}=0,$



where the equality holds if and only if $a_i^{n+1}/a_i^n$ is constant for any $i=1,...,k$.



Now assume that



$limlimits_{nrightarrow infty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} = 0.$



Can we conclude that $limlimits_{nrightarrow infty} dfrac{a_i^{n+1}}{a_i^n} = 1$? Intuitively yes, but how can this be proven rigorously?



Thanks in advance!






real-analysis analysis inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
    $endgroup$
    – marty cohen
    Dec 17 '18 at 18:18














1












1








1





$begingroup$


Let $(a^n)_{n in mathbb{N}} subset mathbb{R}^k$ be a sequence of vectors with non-negative components and with $sum_{i=1}^k a^n_i=1$ for all $ninmathbb{N}$. Then we have for each $ninmathbb{N}$ by the log-sum inequality:



$sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} geq left(sum_{i=1}^k a_i^{n+1}right) logdfrac{sum_{i=1}^k a_i^{n+1}}{sum_{i=1}^k a_i^{n}}=0,$



where the equality holds if and only if $a_i^{n+1}/a_i^n$ is constant for any $i=1,...,k$.



Now assume that



$limlimits_{nrightarrow infty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} = 0.$



Can we conclude that $limlimits_{nrightarrow infty} dfrac{a_i^{n+1}}{a_i^n} = 1$? Intuitively yes, but how can this be proven rigorously?



Thanks in advance!






real-analysis analysis inequality







share|cite|improve this question









$endgroup$




Let $(a^n)_{n in mathbb{N}} subset mathbb{R}^k$ be a sequence of vectors with non-negative components and with $sum_{i=1}^k a^n_i=1$ for all $ninmathbb{N}$. Then we have for each $ninmathbb{N}$ by the log-sum inequality:



$sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} geq left(sum_{i=1}^k a_i^{n+1}right) logdfrac{sum_{i=1}^k a_i^{n+1}}{sum_{i=1}^k a_i^{n}}=0,$



where the equality holds if and only if $a_i^{n+1}/a_i^n$ is constant for any $i=1,...,k$.



Now assume that



$limlimits_{nrightarrow infty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n} = 0.$



Can we conclude that $limlimits_{nrightarrow infty} dfrac{a_i^{n+1}}{a_i^n} = 1$? Intuitively yes, but how can this be proven rigorously?



Thanks in advance!






real-analysis analysis inequality




real-analysis analysis inequality






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asked Dec 17 '18 at 16:14









Max93Max93

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31529












  • $begingroup$
    I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
    $endgroup$
    – marty cohen
    Dec 17 '18 at 18:18


















  • $begingroup$
    I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
    $endgroup$
    – marty cohen
    Dec 17 '18 at 18:18
















$begingroup$
I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
$endgroup$
– marty cohen
Dec 17 '18 at 18:18




$begingroup$
I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$.
$endgroup$
– marty cohen
Dec 17 '18 at 18:18










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$begingroup$

Not necessarily. For instance, put $k=2$ and for each $ninBbb N$ put $a^n_1=frac 1{2^n}$ and $a^n_2=1-frac 1{2^n}$. Then



$$lim_{ntoinfty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n}=
lim_{ntoinfty}frac 1{2^{n+1}}logdfrac{frac 1{2^{n+1}}}{frac 1{2^{n}}}+
left(1-frac 1{2^{n+1}}right)logdfrac{1-frac 1{2^{n+1}}}{1-frac 1{2^{n}}}=0,$$



but $$limlimits_{nrightarrow infty} dfrac{a_1^{n+1}}{a_1^n} = frac 12.$$






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    $begingroup$

    Not necessarily. For instance, put $k=2$ and for each $ninBbb N$ put $a^n_1=frac 1{2^n}$ and $a^n_2=1-frac 1{2^n}$. Then



    $$lim_{ntoinfty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n}=
    lim_{ntoinfty}frac 1{2^{n+1}}logdfrac{frac 1{2^{n+1}}}{frac 1{2^{n}}}+
    left(1-frac 1{2^{n+1}}right)logdfrac{1-frac 1{2^{n+1}}}{1-frac 1{2^{n}}}=0,$$



    but $$limlimits_{nrightarrow infty} dfrac{a_1^{n+1}}{a_1^n} = frac 12.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Not necessarily. For instance, put $k=2$ and for each $ninBbb N$ put $a^n_1=frac 1{2^n}$ and $a^n_2=1-frac 1{2^n}$. Then



      $$lim_{ntoinfty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n}=
      lim_{ntoinfty}frac 1{2^{n+1}}logdfrac{frac 1{2^{n+1}}}{frac 1{2^{n}}}+
      left(1-frac 1{2^{n+1}}right)logdfrac{1-frac 1{2^{n+1}}}{1-frac 1{2^{n}}}=0,$$



      but $$limlimits_{nrightarrow infty} dfrac{a_1^{n+1}}{a_1^n} = frac 12.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Not necessarily. For instance, put $k=2$ and for each $ninBbb N$ put $a^n_1=frac 1{2^n}$ and $a^n_2=1-frac 1{2^n}$. Then



        $$lim_{ntoinfty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n}=
        lim_{ntoinfty}frac 1{2^{n+1}}logdfrac{frac 1{2^{n+1}}}{frac 1{2^{n}}}+
        left(1-frac 1{2^{n+1}}right)logdfrac{1-frac 1{2^{n+1}}}{1-frac 1{2^{n}}}=0,$$



        but $$limlimits_{nrightarrow infty} dfrac{a_1^{n+1}}{a_1^n} = frac 12.$$






        share|cite|improve this answer









        $endgroup$



        Not necessarily. For instance, put $k=2$ and for each $ninBbb N$ put $a^n_1=frac 1{2^n}$ and $a^n_2=1-frac 1{2^n}$. Then



        $$lim_{ntoinfty}sum_{i=1}^k a_i^{n+1} logdfrac{a_i^{n+1}}{a_i^n}=
        lim_{ntoinfty}frac 1{2^{n+1}}logdfrac{frac 1{2^{n+1}}}{frac 1{2^{n}}}+
        left(1-frac 1{2^{n+1}}right)logdfrac{1-frac 1{2^{n+1}}}{1-frac 1{2^{n}}}=0,$$



        but $$limlimits_{nrightarrow infty} dfrac{a_1^{n+1}}{a_1^n} = frac 12.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 18:05









        Alex RavskyAlex Ravsky

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        41.4k32282






























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