How to show non existence of an operator $A: V to V^*$











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1
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Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$






functional-analysis operator-theory banach-spaces







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  • any additional assumptions on $A$, such as linearity or continuity?
    – supinf
    7 hours ago










  • No, but it is $p in (1,2)$, edited it
    – Tesla
    7 hours ago










  • $Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
    – LeB
    7 hours ago










  • @LeB its not an inner product, its a duality product
    – supinf
    7 hours ago






  • 1




    @supinf I guess "operator" usually refers to linear operator?
    – Cave Johnson
    7 hours ago

















up vote
1
down vote

favorite












Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$






functional-analysis operator-theory banach-spaces







share|cite|improve this question
























  • any additional assumptions on $A$, such as linearity or continuity?
    – supinf
    7 hours ago










  • No, but it is $p in (1,2)$, edited it
    – Tesla
    7 hours ago










  • $Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
    – LeB
    7 hours ago










  • @LeB its not an inner product, its a duality product
    – supinf
    7 hours ago






  • 1




    @supinf I guess "operator" usually refers to linear operator?
    – Cave Johnson
    7 hours ago















up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$






functional-analysis operator-theory banach-spaces







share|cite|improve this question















Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$






functional-analysis operator-theory banach-spaces




functional-analysis operator-theory banach-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









user587192

97810




97810










asked 7 hours ago









Tesla

900426




900426












  • any additional assumptions on $A$, such as linearity or continuity?
    – supinf
    7 hours ago










  • No, but it is $p in (1,2)$, edited it
    – Tesla
    7 hours ago










  • $Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
    – LeB
    7 hours ago










  • @LeB its not an inner product, its a duality product
    – supinf
    7 hours ago






  • 1




    @supinf I guess "operator" usually refers to linear operator?
    – Cave Johnson
    7 hours ago




















  • any additional assumptions on $A$, such as linearity or continuity?
    – supinf
    7 hours ago










  • No, but it is $p in (1,2)$, edited it
    – Tesla
    7 hours ago










  • $Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
    – LeB
    7 hours ago










  • @LeB its not an inner product, its a duality product
    – supinf
    7 hours ago






  • 1




    @supinf I guess "operator" usually refers to linear operator?
    – Cave Johnson
    7 hours ago


















any additional assumptions on $A$, such as linearity or continuity?
– supinf
7 hours ago




any additional assumptions on $A$, such as linearity or continuity?
– supinf
7 hours ago












No, but it is $p in (1,2)$, edited it
– Tesla
7 hours ago




No, but it is $p in (1,2)$, edited it
– Tesla
7 hours ago












$Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
– LeB
7 hours ago




$Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
– LeB
7 hours ago












@LeB its not an inner product, its a duality product
– supinf
7 hours ago




@LeB its not an inner product, its a duality product
– supinf
7 hours ago




1




1




@supinf I guess "operator" usually refers to linear operator?
– Cave Johnson
7 hours ago






@supinf I guess "operator" usually refers to linear operator?
– Cave Johnson
7 hours ago












1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Pick an arbitrary point $vin V$ such that $|v|=1$.



Let $n$ be an arbitrary integer.
(Our goal is a proof by contradiction as $ntoinfty$).



We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
Then, by assumption we have
$$
langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
geq
mu | v_{k+1}-v_k|^p
$$
for every $k=0,ldots,n-1$.



Summing up those terms will lead to the inequality
$$
tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
$$



For sufficiently large $n$, this cannot be true.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Pick an arbitrary point $vin V$ such that $|v|=1$.



    Let $n$ be an arbitrary integer.
    (Our goal is a proof by contradiction as $ntoinfty$).



    We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
    Then, by assumption we have
    $$
    langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
    geq
    mu | v_{k+1}-v_k|^p
    $$
    for every $k=0,ldots,n-1$.



    Summing up those terms will lead to the inequality
    $$
    tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
    $$



    For sufficiently large $n$, this cannot be true.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      Pick an arbitrary point $vin V$ such that $|v|=1$.



      Let $n$ be an arbitrary integer.
      (Our goal is a proof by contradiction as $ntoinfty$).



      We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
      Then, by assumption we have
      $$
      langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
      geq
      mu | v_{k+1}-v_k|^p
      $$
      for every $k=0,ldots,n-1$.



      Summing up those terms will lead to the inequality
      $$
      tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
      $$



      For sufficiently large $n$, this cannot be true.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Pick an arbitrary point $vin V$ such that $|v|=1$.



        Let $n$ be an arbitrary integer.
        (Our goal is a proof by contradiction as $ntoinfty$).



        We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
        Then, by assumption we have
        $$
        langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
        geq
        mu | v_{k+1}-v_k|^p
        $$
        for every $k=0,ldots,n-1$.



        Summing up those terms will lead to the inequality
        $$
        tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
        $$



        For sufficiently large $n$, this cannot be true.






        share|cite|improve this answer












        Pick an arbitrary point $vin V$ such that $|v|=1$.



        Let $n$ be an arbitrary integer.
        (Our goal is a proof by contradiction as $ntoinfty$).



        We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
        Then, by assumption we have
        $$
        langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
        geq
        mu | v_{k+1}-v_k|^p
        $$
        for every $k=0,ldots,n-1$.



        Summing up those terms will lead to the inequality
        $$
        tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
        $$



        For sufficiently large $n$, this cannot be true.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 7 hours ago









        supinf

        5,490926




        5,490926






























             

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