Unicity of the projection on a closed convex subset of $L^p$












1












$begingroup$


The original problem (not full) statement is the following :




Let $(Omega, mathfrak{B},mu)$ a measured space and $p>1$ and let $C$ be a closed convex subset of $L^p$.



For $uin L^p$ let $d_u:=inf{|phi-u |, phi in C}$



Through some questions, the problem shows the existence of a $phi in C$ such that $|phi-u |=d_u$.




The problem I have is with unicity. I read somewhere on MSE that the $L^p$ norm is strictly convex, which implies unicity.



As far as I know, no norm is strictly convex, so the previous affirmation might be wrong so how could we deal with unicity.






functional-analysis measure-theory lebesgue-measure lp-spaces projection







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Strictly_convex_space
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 11:52






  • 2




    $begingroup$
    is $p=infty$ possible? Because i think then the solution is not unique
    $endgroup$
    – supinf
    Dec 5 '18 at 11:59






  • 1




    $begingroup$
    Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
    $endgroup$
    – MaoWao
    Dec 5 '18 at 13:28
















1












$begingroup$


The original problem (not full) statement is the following :




Let $(Omega, mathfrak{B},mu)$ a measured space and $p>1$ and let $C$ be a closed convex subset of $L^p$.



For $uin L^p$ let $d_u:=inf{|phi-u |, phi in C}$



Through some questions, the problem shows the existence of a $phi in C$ such that $|phi-u |=d_u$.




The problem I have is with unicity. I read somewhere on MSE that the $L^p$ norm is strictly convex, which implies unicity.



As far as I know, no norm is strictly convex, so the previous affirmation might be wrong so how could we deal with unicity.






functional-analysis measure-theory lebesgue-measure lp-spaces projection







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Strictly_convex_space
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 11:52






  • 2




    $begingroup$
    is $p=infty$ possible? Because i think then the solution is not unique
    $endgroup$
    – supinf
    Dec 5 '18 at 11:59






  • 1




    $begingroup$
    Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
    $endgroup$
    – MaoWao
    Dec 5 '18 at 13:28














1












1








1





$begingroup$


The original problem (not full) statement is the following :




Let $(Omega, mathfrak{B},mu)$ a measured space and $p>1$ and let $C$ be a closed convex subset of $L^p$.



For $uin L^p$ let $d_u:=inf{|phi-u |, phi in C}$



Through some questions, the problem shows the existence of a $phi in C$ such that $|phi-u |=d_u$.




The problem I have is with unicity. I read somewhere on MSE that the $L^p$ norm is strictly convex, which implies unicity.



As far as I know, no norm is strictly convex, so the previous affirmation might be wrong so how could we deal with unicity.






functional-analysis measure-theory lebesgue-measure lp-spaces projection







share|cite|improve this question









$endgroup$




The original problem (not full) statement is the following :




Let $(Omega, mathfrak{B},mu)$ a measured space and $p>1$ and let $C$ be a closed convex subset of $L^p$.



For $uin L^p$ let $d_u:=inf{|phi-u |, phi in C}$



Through some questions, the problem shows the existence of a $phi in C$ such that $|phi-u |=d_u$.




The problem I have is with unicity. I read somewhere on MSE that the $L^p$ norm is strictly convex, which implies unicity.



As far as I know, no norm is strictly convex, so the previous affirmation might be wrong so how could we deal with unicity.






functional-analysis measure-theory lebesgue-measure lp-spaces projection




functional-analysis measure-theory lebesgue-measure lp-spaces projection






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 11:21









Oussama SarihOussama Sarih

47827




47827








  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Strictly_convex_space
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 11:52






  • 2




    $begingroup$
    is $p=infty$ possible? Because i think then the solution is not unique
    $endgroup$
    – supinf
    Dec 5 '18 at 11:59






  • 1




    $begingroup$
    Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
    $endgroup$
    – MaoWao
    Dec 5 '18 at 13:28














  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Strictly_convex_space
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 11:52






  • 2




    $begingroup$
    is $p=infty$ possible? Because i think then the solution is not unique
    $endgroup$
    – supinf
    Dec 5 '18 at 11:59






  • 1




    $begingroup$
    Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
    $endgroup$
    – MaoWao
    Dec 5 '18 at 13:28








1




1




$begingroup$
See en.wikipedia.org/wiki/Strictly_convex_space
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 11:52




$begingroup$
See en.wikipedia.org/wiki/Strictly_convex_space
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 11:52




2




2




$begingroup$
is $p=infty$ possible? Because i think then the solution is not unique
$endgroup$
– supinf
Dec 5 '18 at 11:59




$begingroup$
is $p=infty$ possible? Because i think then the solution is not unique
$endgroup$
– supinf
Dec 5 '18 at 11:59




1




1




$begingroup$
Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
$endgroup$
– MaoWao
Dec 5 '18 at 13:28




$begingroup$
Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
$endgroup$
– MaoWao
Dec 5 '18 at 13:28










1 Answer
1






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2












$begingroup$

Hint:



For $pin(1,infty)$, assume there are two different solutions $phi,psiin C$.
Then, by convexity of the norm, one can show that $(phi+psi)/2$ is closer to $u$ than $phi$ or $psi$.



In order to show the last part, you have to use the fact that for real numbers $a,b$ it holds that
$$
| (a+b)/2 | ^p = (|a|^p+|b|^p)/2
quadLeftrightarrowquad
a = b.
$$






share|cite|improve this answer









$endgroup$













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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Hint:



    For $pin(1,infty)$, assume there are two different solutions $phi,psiin C$.
    Then, by convexity of the norm, one can show that $(phi+psi)/2$ is closer to $u$ than $phi$ or $psi$.



    In order to show the last part, you have to use the fact that for real numbers $a,b$ it holds that
    $$
    | (a+b)/2 | ^p = (|a|^p+|b|^p)/2
    quadLeftrightarrowquad
    a = b.
    $$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint:



      For $pin(1,infty)$, assume there are two different solutions $phi,psiin C$.
      Then, by convexity of the norm, one can show that $(phi+psi)/2$ is closer to $u$ than $phi$ or $psi$.



      In order to show the last part, you have to use the fact that for real numbers $a,b$ it holds that
      $$
      | (a+b)/2 | ^p = (|a|^p+|b|^p)/2
      quadLeftrightarrowquad
      a = b.
      $$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint:



        For $pin(1,infty)$, assume there are two different solutions $phi,psiin C$.
        Then, by convexity of the norm, one can show that $(phi+psi)/2$ is closer to $u$ than $phi$ or $psi$.



        In order to show the last part, you have to use the fact that for real numbers $a,b$ it holds that
        $$
        | (a+b)/2 | ^p = (|a|^p+|b|^p)/2
        quadLeftrightarrowquad
        a = b.
        $$






        share|cite|improve this answer









        $endgroup$



        Hint:



        For $pin(1,infty)$, assume there are two different solutions $phi,psiin C$.
        Then, by convexity of the norm, one can show that $(phi+psi)/2$ is closer to $u$ than $phi$ or $psi$.



        In order to show the last part, you have to use the fact that for real numbers $a,b$ it holds that
        $$
        | (a+b)/2 | ^p = (|a|^p+|b|^p)/2
        quadLeftrightarrowquad
        a = b.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 12:11









        supinfsupinf

        6,3111028




        6,3111028






























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