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

      44.8k9118272




      44.8k9118272










      asked Mar 5 '15 at 0:04









      HK LeeHK Lee

      13.9k52159




      13.9k52159






















          1 Answer
          1






          active

          oldest

          votes


















          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











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1175922%2fexercise-in-hahn-banach-theorem-finding-linear-functional-p-x-leq-fx-leq%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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












              $begingroup$

              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

                44.8k9118272




                44.8k9118272






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1175922%2fexercise-in-hahn-banach-theorem-finding-linear-functional-p-x-leq-fx-leq%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Ellipse (mathématiques)

                    Image
                    Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                    Read more

                    Quarter-circle Tiles

                    Image
                    up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                    Read more

                    Mont Emei

                    Image
                    Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                    Read more