Exercise in Hahn-Banach Theorem; Finding linear functional $-p(-x)leq f(x)leq p(x)$












4












$begingroup$


(The following exercises are in
Kreyszig's book 218 page; EXE 10) I want to solve the following
exercise : If $X=l^infty$, let $p(x)=limsup x_i $, which is
sublinear. Then find a linear functional $f(x)$ s.t. $$ -p(-x)leq
f(x) leq p(x)$$



Background : If $p$ is a sublinear function on
a real vector space, i.e.,
$$ p(x+y)leq p(x) + p(y), p(cx)=cp(x), cgeq 0 $$



then there exists linear functional $f$ s.t. $$ -p(-x)leq f(x) leq
p(x) $$



Proof : By Hahn-Banach theorem we have $f(x)leq p(x)
$ so that $$ f(-x)leq p(-x)$$



That is $$ -p(-x)leq f(x) leq f(x)$$



Now, we return to original question : $X=l^infty$, let
$p(x)=limsup x_i $. So
$$-p(-x)=-limsup (-x_i)=liminf x_i$$ Hence if such functional $f$ exists, then
$f(x)=lim x_i$ when $lim x_i$ exists. If $x_{2i+1}=2, x_{2i}=1$
then $v_{2i+1}=1, v_{2i}=2$ then $ f(x+v)=3 $. That is, the problem
is how determine $f$ on $x$ where $lim x_i$ does not exist.



Try : If we let $f(x)=frac{liminf x_i + limsup x_i}{2} $, then note
that $f(x+v)neq f(x)+f(v)$. Thank you in advance.






functional-analysis normed-spaces







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    (The following exercises are in
    Kreyszig's book 218 page; EXE 10) I want to solve the following
    exercise : If $X=l^infty$, let $p(x)=limsup x_i $, which is
    sublinear. Then find a linear functional $f(x)$ s.t. $$ -p(-x)leq
    f(x) leq p(x)$$



    Background : If $p$ is a sublinear function on
    a real vector space, i.e.,
    $$ p(x+y)leq p(x) + p(y), p(cx)=cp(x), cgeq 0 $$



    then there exists linear functional $f$ s.t. $$ -p(-x)leq f(x) leq
    p(x) $$



    Proof : By Hahn-Banach theorem we have $f(x)leq p(x)
    $     so that $$ f(-x)leq p(-x)$$



    That is $$ -p(-x)leq f(x) leq f(x)$$



    Now, we return to original question : $X=l^infty$, let
    $p(x)=limsup x_i $. So
    $$-p(-x)=-limsup (-x_i)=liminf x_i$$ Hence if such functional $f$ exists, then
    $f(x)=lim x_i$ when $lim x_i$ exists. If $x_{2i+1}=2, x_{2i}=1$
    then $v_{2i+1}=1, v_{2i}=2$ then $ f(x+v)=3 $. That is, the problem
    is how determine $f$ on $x$ where $lim x_i$ does not exist.



    Try : If we let $f(x)=frac{liminf x_i + limsup x_i}{2} $, then note
    that $f(x+v)neq f(x)+f(v)$. Thank you in advance.






    functional-analysis normed-spaces







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      2



      $begingroup$


      (The following exercises are in
      Kreyszig's book 218 page; EXE 10) I want to solve the following
      exercise : If $X=l^infty$, let $p(x)=limsup x_i $, which is
      sublinear. Then find a linear functional $f(x)$ s.t. $$ -p(-x)leq
      f(x) leq p(x)$$



      Background : If $p$ is a sublinear function on
      a real vector space, i.e.,
      $$ p(x+y)leq p(x) + p(y), p(cx)=cp(x), cgeq 0 $$



      then there exists linear functional $f$ s.t. $$ -p(-x)leq f(x) leq
      p(x) $$



      Proof : By Hahn-Banach theorem we have $f(x)leq p(x)
      $       so that $$ f(-x)leq p(-x)$$



      That is $$ -p(-x)leq f(x) leq f(x)$$



      Now, we return to original question : $X=l^infty$, let
      $p(x)=limsup x_i $. So
      $$-p(-x)=-limsup (-x_i)=liminf x_i$$ Hence if such functional $f$ exists, then
      $f(x)=lim x_i$ when $lim x_i$ exists. If $x_{2i+1}=2, x_{2i}=1$
      then $v_{2i+1}=1, v_{2i}=2$ then $ f(x+v)=3 $. That is, the problem
      is how determine $f$ on $x$ where $lim x_i$ does not exist.



      Try : If we let $f(x)=frac{liminf x_i + limsup x_i}{2} $, then note
      that $f(x+v)neq f(x)+f(v)$. Thank you in advance.






      functional-analysis normed-spaces







      share|cite|improve this question











      $endgroup$




      (The following exercises are in
      Kreyszig's book 218 page; EXE 10) I want to solve the following
      exercise : If $X=l^infty$, let $p(x)=limsup x_i $, which is
      sublinear. Then find a linear functional $f(x)$ s.t. $$ -p(-x)leq
      f(x) leq p(x)$$



      Background : If $p$ is a sublinear function on
      a real vector space, i.e.,
      $$ p(x+y)leq p(x) + p(y), p(cx)=cp(x), cgeq 0 $$



      then there exists linear functional $f$ s.t. $$ -p(-x)leq f(x) leq
      p(x) $$



      Proof : By Hahn-Banach theorem we have $f(x)leq p(x)
      $       so that $$ f(-x)leq p(-x)$$



      That is $$ -p(-x)leq f(x) leq f(x)$$



      Now, we return to original question : $X=l^infty$, let
      $p(x)=limsup x_i $. So
      $$-p(-x)=-limsup (-x_i)=liminf x_i$$ Hence if such functional $f$ exists, then
      $f(x)=lim x_i$ when $lim x_i$ exists. If $x_{2i+1}=2, x_{2i}=1$
      then $v_{2i+1}=1, v_{2i}=2$ then $ f(x+v)=3 $. That is, the problem
      is how determine $f$ on $x$ where $lim x_i$ does not exist.



      Try : If we let $f(x)=frac{liminf x_i + limsup x_i}{2} $, then note
      that $f(x+v)neq f(x)+f(v)$. Thank you in advance.






      functional-analysis normed-spaces




      functional-analysis normed-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 12:08









      Martin Sleziak

      44.7k9117272




      44.7k9117272










      asked Mar 5 '15 at 0:04









      HK LeeHK Lee

      13.9k52159




      13.9k52159






















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          $begingroup$

          I think that the exercise is expecting you to do exactly what you did - to prove that there exists a functional with the given properties. (Not to explicitly write down such $f$.)



          You can notice that in this way you get a functional $fin ell_infty^* setminus ell_1$. (A few proofs of the fact that $ell_infty^*neell_1$ are also collected here: Dual of $l^infty$ is not $l^1$.) Existence of such functional cannot be proved in ZF, so any proof has to use some non-constructive step at some point. Some posts where the fact that this is not provable in ZF is mentioned: Nonnegative linear functionals over $l^infty$ and $ell^1$ vs. continuous dual of $ell^{infty}$ in ZF+AD.






          share|cite|improve this answer











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            active

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            active

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            $begingroup$

            I think that the exercise is expecting you to do exactly what you did - to prove that there exists a functional with the given properties. (Not to explicitly write down such $f$.)



            You can notice that in this way you get a functional $fin ell_infty^* setminus ell_1$. (A few proofs of the fact that $ell_infty^*neell_1$ are also collected here: Dual of $l^infty$ is not $l^1$.) Existence of such functional cannot be proved in ZF, so any proof has to use some non-constructive step at some point. Some posts where the fact that this is not provable in ZF is mentioned: Nonnegative linear functionals over $l^infty$ and $ell^1$ vs. continuous dual of $ell^{infty}$ in ZF+AD.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              I think that the exercise is expecting you to do exactly what you did - to prove that there exists a functional with the given properties. (Not to explicitly write down such $f$.)



              You can notice that in this way you get a functional $fin ell_infty^* setminus ell_1$. (A few proofs of the fact that $ell_infty^*neell_1$ are also collected here: Dual of $l^infty$ is not $l^1$.) Existence of such functional cannot be proved in ZF, so any proof has to use some non-constructive step at some point. Some posts where the fact that this is not provable in ZF is mentioned: Nonnegative linear functionals over $l^infty$ and $ell^1$ vs. continuous dual of $ell^{infty}$ in ZF+AD.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                I think that the exercise is expecting you to do exactly what you did - to prove that there exists a functional with the given properties. (Not to explicitly write down such $f$.)



                You can notice that in this way you get a functional $fin ell_infty^* setminus ell_1$. (A few proofs of the fact that $ell_infty^*neell_1$ are also collected here: Dual of $l^infty$ is not $l^1$.) Existence of such functional cannot be proved in ZF, so any proof has to use some non-constructive step at some point. Some posts where the fact that this is not provable in ZF is mentioned: Nonnegative linear functionals over $l^infty$ and $ell^1$ vs. continuous dual of $ell^{infty}$ in ZF+AD.






                share|cite|improve this answer











                $endgroup$



                I think that the exercise is expecting you to do exactly what you did - to prove that there exists a functional with the given properties. (Not to explicitly write down such $f$.)



                You can notice that in this way you get a functional $fin ell_infty^* setminus ell_1$. (A few proofs of the fact that $ell_infty^*neell_1$ are also collected here: Dual of $l^infty$ is not $l^1$.) Existence of such functional cannot be proved in ZF, so any proof has to use some non-constructive step at some point. Some posts where the fact that this is not provable in ZF is mentioned: Nonnegative linear functionals over $l^infty$ and $ell^1$ vs. continuous dual of $ell^{infty}$ in ZF+AD.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 10 '18 at 17:47

























                answered Dec 10 '18 at 12:16









                Martin SleziakMartin Sleziak

                44.7k9117272




                44.7k9117272






























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