Linearity of functionals imply the inclusion of kernels












1












$begingroup$


I understand most of the following claim's proof except statement I've highlighted below:




Let $X$ be a linear space and $W$ be a subspace of $X^{text{#}}$ (linear functionals - not necessarily bounded). Then a linear functional $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous if and only if it belongs to $W$.




Proof:




The reverse directions follows from definition. In the forward direction, suppose $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous. By the continuity of $psi$ at $0$, there is a neighborhood $N$ of $0$ for which $|psi(x)| = |psi(x)-psi(0)| < 1$ if $x in N$. There is a neighborhood in the base for the $W$-weak toplogy (denoted below as $N_{epsilon, psi_1, ldots, psi_n}$) at $0$ that is contained in $N$. Choose $epsilon > 0$ and $psi_1,...psi_n$ in $W$ for which:
$$N_{epsilon, psi_1, ldots, psi_n}= {x' in X ,big| ,, |psi_k(x')-psi_k(0)| < epsilon quad 1 leq k leq n} subset N$$Thus:
$$|psi(x)|<1 text{ if } |psi_k(x)|<epsilon quad forall 1leq k leq n$$
By linearity of $psi$ and $psi_k$ we have the inclusion $bigcap ker psi_k subset ker psi$. By a previous theorem, this implies $psi$ is a linear combination of the $psi_k$'s.




I don't understand the bolded statement. Namely, suppose $A subset N_{epsilon, psi_1, ldots, psi_n}$ is the set of all $x$ for which all the $psi_k$ are zero -- i.e. $A = bigcap ker psi_k$. The implication shows $|psi(A)|<1$. This doesn't imply $|psi(A)|=0$. So how do we get $A subset ker psi$?






functional-analysis proof-explanation







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I understand most of the following claim's proof except statement I've highlighted below:




    Let $X$ be a linear space and $W$ be a subspace of $X^{text{#}}$ (linear functionals - not necessarily bounded). Then a linear functional $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous if and only if it belongs to $W$.




    Proof:




    The reverse directions follows from definition. In the forward direction, suppose $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous. By the continuity of $psi$ at $0$, there is a neighborhood $N$ of $0$ for which $|psi(x)| = |psi(x)-psi(0)| < 1$ if $x in N$. There is a neighborhood in the base for the $W$-weak toplogy (denoted below as $N_{epsilon, psi_1, ldots, psi_n}$) at $0$ that is contained in $N$. Choose $epsilon > 0$ and $psi_1,...psi_n$ in $W$ for which:
    $$N_{epsilon, psi_1, ldots, psi_n}= {x' in X ,big| ,, |psi_k(x')-psi_k(0)| < epsilon quad 1 leq k leq n} subset N$$Thus:
    $$|psi(x)|<1 text{ if } |psi_k(x)|<epsilon quad forall 1leq k leq n$$
    By linearity of $psi$ and $psi_k$ we have the inclusion $bigcap ker psi_k subset ker psi$. By a previous theorem, this implies $psi$ is a linear combination of the $psi_k$'s.




    I don't understand the bolded statement. Namely, suppose $A subset N_{epsilon, psi_1, ldots, psi_n}$ is the set of all $x$ for which all the $psi_k$ are zero -- i.e. $A = bigcap ker psi_k$. The implication shows $|psi(A)|<1$. This doesn't imply $|psi(A)|=0$. So how do we get $A subset ker psi$?






    functional-analysis proof-explanation







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I understand most of the following claim's proof except statement I've highlighted below:




      Let $X$ be a linear space and $W$ be a subspace of $X^{text{#}}$ (linear functionals - not necessarily bounded). Then a linear functional $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous if and only if it belongs to $W$.




      Proof:




      The reverse directions follows from definition. In the forward direction, suppose $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous. By the continuity of $psi$ at $0$, there is a neighborhood $N$ of $0$ for which $|psi(x)| = |psi(x)-psi(0)| < 1$ if $x in N$. There is a neighborhood in the base for the $W$-weak toplogy (denoted below as $N_{epsilon, psi_1, ldots, psi_n}$) at $0$ that is contained in $N$. Choose $epsilon > 0$ and $psi_1,...psi_n$ in $W$ for which:
      $$N_{epsilon, psi_1, ldots, psi_n}= {x' in X ,big| ,, |psi_k(x')-psi_k(0)| < epsilon quad 1 leq k leq n} subset N$$Thus:
      $$|psi(x)|<1 text{ if } |psi_k(x)|<epsilon quad forall 1leq k leq n$$
      By linearity of $psi$ and $psi_k$ we have the inclusion $bigcap ker psi_k subset ker psi$. By a previous theorem, this implies $psi$ is a linear combination of the $psi_k$'s.




      I don't understand the bolded statement. Namely, suppose $A subset N_{epsilon, psi_1, ldots, psi_n}$ is the set of all $x$ for which all the $psi_k$ are zero -- i.e. $A = bigcap ker psi_k$. The implication shows $|psi(A)|<1$. This doesn't imply $|psi(A)|=0$. So how do we get $A subset ker psi$?






      functional-analysis proof-explanation







      share|cite|improve this question











      $endgroup$




      I understand most of the following claim's proof except statement I've highlighted below:




      Let $X$ be a linear space and $W$ be a subspace of $X^{text{#}}$ (linear functionals - not necessarily bounded). Then a linear functional $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous if and only if it belongs to $W$.




      Proof:




      The reverse directions follows from definition. In the forward direction, suppose $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous. By the continuity of $psi$ at $0$, there is a neighborhood $N$ of $0$ for which $|psi(x)| = |psi(x)-psi(0)| < 1$ if $x in N$. There is a neighborhood in the base for the $W$-weak toplogy (denoted below as $N_{epsilon, psi_1, ldots, psi_n}$) at $0$ that is contained in $N$. Choose $epsilon > 0$ and $psi_1,...psi_n$ in $W$ for which:
      $$N_{epsilon, psi_1, ldots, psi_n}= {x' in X ,big| ,, |psi_k(x')-psi_k(0)| < epsilon quad 1 leq k leq n} subset N$$Thus:
      $$|psi(x)|<1 text{ if } |psi_k(x)|<epsilon quad forall 1leq k leq n$$
      By linearity of $psi$ and $psi_k$ we have the inclusion $bigcap ker psi_k subset ker psi$. By a previous theorem, this implies $psi$ is a linear combination of the $psi_k$'s.




      I don't understand the bolded statement. Namely, suppose $A subset N_{epsilon, psi_1, ldots, psi_n}$ is the set of all $x$ for which all the $psi_k$ are zero -- i.e. $A = bigcap ker psi_k$. The implication shows $|psi(A)|<1$. This doesn't imply $|psi(A)|=0$. So how do we get $A subset ker psi$?






      functional-analysis proof-explanation




      functional-analysis proof-explanation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 22:58









      mechanodroid

      28.9k62648




      28.9k62648










      asked Jan 8 at 21:56









      yoshiyoshi

      1,256917




      1,256917






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Assume that $x in bigcap_{k=1}^n ker psi_k$. Then $|psi_k(x)| = 0 < varepsilon$ for all $1 le k le n$ so it follows $|psi(x)| < 1$.



          But also $n x in bigcap_{k=1}^n ker psi_k$ for any $n in mathbb{N}$ so similarly $n|psi(x)| = |psi(nx)| < 1$. If $psi(x) ne 0$, the left hand side would be unbounded as $n to infty$ so it has to be $psi(x) = 0$.






          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%2f3066775%2flinearity-of-functionals-imply-the-inclusion-of-kernels%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









            1












            $begingroup$

            Assume that $x in bigcap_{k=1}^n ker psi_k$. Then $|psi_k(x)| = 0 < varepsilon$ for all $1 le k le n$ so it follows $|psi(x)| < 1$.



            But also $n x in bigcap_{k=1}^n ker psi_k$ for any $n in mathbb{N}$ so similarly $n|psi(x)| = |psi(nx)| < 1$. If $psi(x) ne 0$, the left hand side would be unbounded as $n to infty$ so it has to be $psi(x) = 0$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Assume that $x in bigcap_{k=1}^n ker psi_k$. Then $|psi_k(x)| = 0 < varepsilon$ for all $1 le k le n$ so it follows $|psi(x)| < 1$.



              But also $n x in bigcap_{k=1}^n ker psi_k$ for any $n in mathbb{N}$ so similarly $n|psi(x)| = |psi(nx)| < 1$. If $psi(x) ne 0$, the left hand side would be unbounded as $n to infty$ so it has to be $psi(x) = 0$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Assume that $x in bigcap_{k=1}^n ker psi_k$. Then $|psi_k(x)| = 0 < varepsilon$ for all $1 le k le n$ so it follows $|psi(x)| < 1$.



                But also $n x in bigcap_{k=1}^n ker psi_k$ for any $n in mathbb{N}$ so similarly $n|psi(x)| = |psi(nx)| < 1$. If $psi(x) ne 0$, the left hand side would be unbounded as $n to infty$ so it has to be $psi(x) = 0$.






                share|cite|improve this answer









                $endgroup$



                Assume that $x in bigcap_{k=1}^n ker psi_k$. Then $|psi_k(x)| = 0 < varepsilon$ for all $1 le k le n$ so it follows $|psi(x)| < 1$.



                But also $n x in bigcap_{k=1}^n ker psi_k$ for any $n in mathbb{N}$ so similarly $n|psi(x)| = |psi(nx)| < 1$. If $psi(x) ne 0$, the left hand side would be unbounded as $n to infty$ so it has to be $psi(x) = 0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 22:56









                mechanodroidmechanodroid

                28.9k62648




                28.9k62648






























                    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%2f3066775%2flinearity-of-functionals-imply-the-inclusion-of-kernels%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

                    Bressuire

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more

                    Cabo Verde

                    Image
                      Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                    Read more

                    Gyllenstierna

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more