Prove generalised Hölder's inequality without calculus or analysis.

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It is the generalised Hölder's inequality.I saw many analytical proofs in this site but I don't know analysis. So I need a basic proof. I proved it by A.M.-G.M. for $m=n=3$. Please help me.
Proof when $m=n=3$



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  • "I need a basic proof" - why? My recommendation is to learn analysis!
    – Brevan Ellefsen
    Jul 23 at 3:29






  • 1




    as an aside, Corollary 1 in your image sounds horrifying. IMHO no proof should ever start by defining that many different variables without subscripts or something.... 0_0
    – Brevan Ellefsen
    Jul 23 at 3:30











  • Could you post your proof for $m = n = 3$? It will help us calibrate our answers, plus showing effort will help protect against down/close votes.
    – Theo Bendit
    Jul 23 at 3:37










  • @Theo Bendit The question is edited!
    – Sufaid Saleel
    Jul 23 at 4:18














up vote
1
down vote

favorite












enter image description here



It is the generalised Hölder's inequality.I saw many analytical proofs in this site but I don't know analysis. So I need a basic proof. I proved it by A.M.-G.M. for $m=n=3$. Please help me.
Proof when $m=n=3$



enter image description here




inequality holder-inequality




share|cite|improve this question





















  • "I need a basic proof" - why? My recommendation is to learn analysis!
    – Brevan Ellefsen
    Jul 23 at 3:29






  • 1




    as an aside, Corollary 1 in your image sounds horrifying. IMHO no proof should ever start by defining that many different variables without subscripts or something.... 0_0
    – Brevan Ellefsen
    Jul 23 at 3:30











  • Could you post your proof for $m = n = 3$? It will help us calibrate our answers, plus showing effort will help protect against down/close votes.
    – Theo Bendit
    Jul 23 at 3:37










  • @Theo Bendit The question is edited!
    – Sufaid Saleel
    Jul 23 at 4:18












up vote
1
down vote

favorite









up vote
1
down vote

favorite











enter image description here



It is the generalised Hölder's inequality.I saw many analytical proofs in this site but I don't know analysis. So I need a basic proof. I proved it by A.M.-G.M. for $m=n=3$. Please help me.
Proof when $m=n=3$



enter image description here




inequality holder-inequality




share|cite|improve this question













enter image description here



It is the generalised Hölder's inequality.I saw many analytical proofs in this site but I don't know analysis. So I need a basic proof. I proved it by A.M.-G.M. for $m=n=3$. Please help me.
Proof when $m=n=3$



enter image description here





inequality holder-inequality





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 23 at 8:11









Bernard

110k635103




110k635103









asked Jul 23 at 3:06









Sufaid Saleel

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  • "I need a basic proof" - why? My recommendation is to learn analysis!
    – Brevan Ellefsen
    Jul 23 at 3:29






  • 1




    as an aside, Corollary 1 in your image sounds horrifying. IMHO no proof should ever start by defining that many different variables without subscripts or something.... 0_0
    – Brevan Ellefsen
    Jul 23 at 3:30











  • Could you post your proof for $m = n = 3$? It will help us calibrate our answers, plus showing effort will help protect against down/close votes.
    – Theo Bendit
    Jul 23 at 3:37










  • @Theo Bendit The question is edited!
    – Sufaid Saleel
    Jul 23 at 4:18
















  • "I need a basic proof" - why? My recommendation is to learn analysis!
    – Brevan Ellefsen
    Jul 23 at 3:29






  • 1




    as an aside, Corollary 1 in your image sounds horrifying. IMHO no proof should ever start by defining that many different variables without subscripts or something.... 0_0
    – Brevan Ellefsen
    Jul 23 at 3:30











  • Could you post your proof for $m = n = 3$? It will help us calibrate our answers, plus showing effort will help protect against down/close votes.
    – Theo Bendit
    Jul 23 at 3:37










  • @Theo Bendit The question is edited!
    – Sufaid Saleel
    Jul 23 at 4:18















"I need a basic proof" - why? My recommendation is to learn analysis!
– Brevan Ellefsen
Jul 23 at 3:29




"I need a basic proof" - why? My recommendation is to learn analysis!
– Brevan Ellefsen
Jul 23 at 3:29




1




1




as an aside, Corollary 1 in your image sounds horrifying. IMHO no proof should ever start by defining that many different variables without subscripts or something.... 0_0
– Brevan Ellefsen
Jul 23 at 3:30





as an aside, Corollary 1 in your image sounds horrifying. IMHO no proof should ever start by defining that many different variables without subscripts or something.... 0_0
– Brevan Ellefsen
Jul 23 at 3:30













Could you post your proof for $m = n = 3$? It will help us calibrate our answers, plus showing effort will help protect against down/close votes.
– Theo Bendit
Jul 23 at 3:37




Could you post your proof for $m = n = 3$? It will help us calibrate our answers, plus showing effort will help protect against down/close votes.
– Theo Bendit
Jul 23 at 3:37












@Theo Bendit The question is edited!
– Sufaid Saleel
Jul 23 at 4:18




@Theo Bendit The question is edited!
– Sufaid Saleel
Jul 23 at 4:18










1 Answer
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We can extend your proof more generall. If we have $m$ sequences of $n$ entries, denoted $(a_i,1, a_i,2, ldots, a_i,n)$ for $i = 1, ldots, m$, then



beginalign*
m &= sum_i=1^m 1 \
&= sum_i=1^m sum_j=1^n fraca_i,jsum_k=1^n a_i,k \
1 &= sum_j=1^n frac1m sum_i=1^m fraca_i,jsum_k=1^n a_i,k \
&ge sum_j=1^n prod_i=1^m fracsqrt[m]a_i,jsqrt[m]sum_k=1^n a_i,k \
&= frac1prod_i=1^m sqrt[m]sum_k=1^n a_i,ksum_j=1^n sqrt[m]prod_i=1^m a_i,j,
endalign*
which implies
$$prod_i=1^m sum_k=1^n a_i,k ge left(sum_j=1^n sqrt[m]prod_i=1^m a_i,jright)^m$$
as required. The only inequality is that of the AGM; the rest are elementary manipulations of sums and products.



The inequality will be equal whenever we have
$$frac1m sum_i=1^m fraca_i,jsum_k=1^n a_i,k
= prod_i=1^m fracsqrt[m]a_i,jsqrt[m]sum_k=1^n a_i,k$$
for any $j = 1, ldots, n$. Equality occurs in the AGM whenever the sequence is constant, hence, there must be some $C_j$, constant with respect to $i$, such that
$$fraca_i,jsum_k=1^n a_i,k = C_j$$
for all $i, j$ over their respective ranges. Let $b_i = sum_k=1^n a_i,k$. Then
$$a_i, j = C_j b_i,$$
in other words, the sequences $(a_i, j)_i=1^m$, as $j = 1, ldots, n$, are just multiples of each other, also as required.






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    We can extend your proof more generall. If we have $m$ sequences of $n$ entries, denoted $(a_i,1, a_i,2, ldots, a_i,n)$ for $i = 1, ldots, m$, then



    beginalign*
    m &= sum_i=1^m 1 \
    &= sum_i=1^m sum_j=1^n fraca_i,jsum_k=1^n a_i,k \
    1 &= sum_j=1^n frac1m sum_i=1^m fraca_i,jsum_k=1^n a_i,k \
    &ge sum_j=1^n prod_i=1^m fracsqrt[m]a_i,jsqrt[m]sum_k=1^n a_i,k \
    &= frac1prod_i=1^m sqrt[m]sum_k=1^n a_i,ksum_j=1^n sqrt[m]prod_i=1^m a_i,j,
    endalign*
    which implies
    $$prod_i=1^m sum_k=1^n a_i,k ge left(sum_j=1^n sqrt[m]prod_i=1^m a_i,jright)^m$$
    as required. The only inequality is that of the AGM; the rest are elementary manipulations of sums and products.



    The inequality will be equal whenever we have
    $$frac1m sum_i=1^m fraca_i,jsum_k=1^n a_i,k
    = prod_i=1^m fracsqrt[m]a_i,jsqrt[m]sum_k=1^n a_i,k$$
    for any $j = 1, ldots, n$. Equality occurs in the AGM whenever the sequence is constant, hence, there must be some $C_j$, constant with respect to $i$, such that
    $$fraca_i,jsum_k=1^n a_i,k = C_j$$
    for all $i, j$ over their respective ranges. Let $b_i = sum_k=1^n a_i,k$. Then
    $$a_i, j = C_j b_i,$$
    in other words, the sequences $(a_i, j)_i=1^m$, as $j = 1, ldots, n$, are just multiples of each other, also as required.






    share|cite|improve this answer

























      up vote
      2
      down vote













      We can extend your proof more generall. If we have $m$ sequences of $n$ entries, denoted $(a_i,1, a_i,2, ldots, a_i,n)$ for $i = 1, ldots, m$, then



      beginalign*
      m &= sum_i=1^m 1 \
      &= sum_i=1^m sum_j=1^n fraca_i,jsum_k=1^n a_i,k \
      1 &= sum_j=1^n frac1m sum_i=1^m fraca_i,jsum_k=1^n a_i,k \
      &ge sum_j=1^n prod_i=1^m fracsqrt[m]a_i,jsqrt[m]sum_k=1^n a_i,k \
      &= frac1prod_i=1^m sqrt[m]sum_k=1^n a_i,ksum_j=1^n sqrt[m]prod_i=1^m a_i,j,
      endalign*
      which implies
      $$prod_i=1^m sum_k=1^n a_i,k ge left(sum_j=1^n sqrt[m]prod_i=1^m a_i,jright)^m$$
      as required. The only inequality is that of the AGM; the rest are elementary manipulations of sums and products.



      The inequality will be equal whenever we have
      $$frac1m sum_i=1^m fraca_i,jsum_k=1^n a_i,k
      = prod_i=1^m fracsqrt[m]a_i,jsqrt[m]sum_k=1^n a_i,k$$
      for any $j = 1, ldots, n$. Equality occurs in the AGM whenever the sequence is constant, hence, there must be some $C_j$, constant with respect to $i$, such that
      $$fraca_i,jsum_k=1^n a_i,k = C_j$$
      for all $i, j$ over their respective ranges. Let $b_i = sum_k=1^n a_i,k$. Then
      $$a_i, j = C_j b_i,$$
      in other words, the sequences $(a_i, j)_i=1^m$, as $j = 1, ldots, n$, are just multiples of each other, also as required.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        We can extend your proof more generall. If we have $m$ sequences of $n$ entries, denoted $(a_i,1, a_i,2, ldots, a_i,n)$ for $i = 1, ldots, m$, then



        beginalign*
        m &= sum_i=1^m 1 \
        &= sum_i=1^m sum_j=1^n fraca_i,jsum_k=1^n a_i,k \
        1 &= sum_j=1^n frac1m sum_i=1^m fraca_i,jsum_k=1^n a_i,k \
        &ge sum_j=1^n prod_i=1^m fracsqrt[m]a_i,jsqrt[m]sum_k=1^n a_i,k \
        &= frac1prod_i=1^m sqrt[m]sum_k=1^n a_i,ksum_j=1^n sqrt[m]prod_i=1^m a_i,j,
        endalign*
        which implies
        $$prod_i=1^m sum_k=1^n a_i,k ge left(sum_j=1^n sqrt[m]prod_i=1^m a_i,jright)^m$$
        as required. The only inequality is that of the AGM; the rest are elementary manipulations of sums and products.



        The inequality will be equal whenever we have
        $$frac1m sum_i=1^m fraca_i,jsum_k=1^n a_i,k
        = prod_i=1^m fracsqrt[m]a_i,jsqrt[m]sum_k=1^n a_i,k$$
        for any $j = 1, ldots, n$. Equality occurs in the AGM whenever the sequence is constant, hence, there must be some $C_j$, constant with respect to $i$, such that
        $$fraca_i,jsum_k=1^n a_i,k = C_j$$
        for all $i, j$ over their respective ranges. Let $b_i = sum_k=1^n a_i,k$. Then
        $$a_i, j = C_j b_i,$$
        in other words, the sequences $(a_i, j)_i=1^m$, as $j = 1, ldots, n$, are just multiples of each other, also as required.






        share|cite|improve this answer













        We can extend your proof more generall. If we have $m$ sequences of $n$ entries, denoted $(a_i,1, a_i,2, ldots, a_i,n)$ for $i = 1, ldots, m$, then



        beginalign*
        m &= sum_i=1^m 1 \
        &= sum_i=1^m sum_j=1^n fraca_i,jsum_k=1^n a_i,k \
        1 &= sum_j=1^n frac1m sum_i=1^m fraca_i,jsum_k=1^n a_i,k \
        &ge sum_j=1^n prod_i=1^m fracsqrt[m]a_i,jsqrt[m]sum_k=1^n a_i,k \
        &= frac1prod_i=1^m sqrt[m]sum_k=1^n a_i,ksum_j=1^n sqrt[m]prod_i=1^m a_i,j,
        endalign*
        which implies
        $$prod_i=1^m sum_k=1^n a_i,k ge left(sum_j=1^n sqrt[m]prod_i=1^m a_i,jright)^m$$
        as required. The only inequality is that of the AGM; the rest are elementary manipulations of sums and products.



        The inequality will be equal whenever we have
        $$frac1m sum_i=1^m fraca_i,jsum_k=1^n a_i,k
        = prod_i=1^m fracsqrt[m]a_i,jsqrt[m]sum_k=1^n a_i,k$$
        for any $j = 1, ldots, n$. Equality occurs in the AGM whenever the sequence is constant, hence, there must be some $C_j$, constant with respect to $i$, such that
        $$fraca_i,jsum_k=1^n a_i,k = C_j$$
        for all $i, j$ over their respective ranges. Let $b_i = sum_k=1^n a_i,k$. Then
        $$a_i, j = C_j b_i,$$
        in other words, the sequences $(a_i, j)_i=1^m$, as $j = 1, ldots, n$, are just multiples of each other, also as required.







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        answered Jul 23 at 5:26









        Theo Bendit

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