What is the operator norm of $Tf(x) = x^2f(x)$?












3












$begingroup$


Let $H = L^2([0,1],mathbb{R})$ and $T : H to H,, Tf(x) = x^2f(x) $.



$T$ is linear.



$$|Tf|_{L^2([0,1],mathbb{R})} = sqrt{int_0^1x^4f^2(x)dx} leqsqrt{int_0^1f^2(x)dx} = |f|_{L^2([0,1],mathbb{R})} $$



$T$ is linear and bounded therefore it's continuous. Also
$|T|| leq 1$.



I tried finding solution to $|Tf|_{L^2([0,1],mathbb{R})} = |f|_{L^2([0,1],mathbb{R})} $



I found $$f(x) = sqrt{frac{2x-1}{x^4-1}}$$



but it's not in $L^2$ so it doesn't work.



anyone knows an $f$ to reach $1$, I'm not even sure it's $1$.



any help will be greatly appreciated !






functional-analysis operator-theory hilbert-spaces norm lp-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
    $endgroup$
    – Mindlack
    Jan 7 at 18:16










  • $begingroup$
    The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
    $endgroup$
    – Dirk
    Jan 7 at 18:31
















3












$begingroup$


Let $H = L^2([0,1],mathbb{R})$ and $T : H to H,, Tf(x) = x^2f(x) $.



$T$ is linear.



$$|Tf|_{L^2([0,1],mathbb{R})} = sqrt{int_0^1x^4f^2(x)dx} leqsqrt{int_0^1f^2(x)dx} = |f|_{L^2([0,1],mathbb{R})} $$



$T$ is linear and bounded therefore it's continuous. Also
$|T|| leq 1$.



I tried finding solution to $|Tf|_{L^2([0,1],mathbb{R})} = |f|_{L^2([0,1],mathbb{R})} $



I found $$f(x) = sqrt{frac{2x-1}{x^4-1}}$$



but it's not in $L^2$ so it doesn't work.



anyone knows an $f$ to reach $1$, I'm not even sure it's $1$.



any help will be greatly appreciated !






functional-analysis operator-theory hilbert-spaces norm lp-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
    $endgroup$
    – Mindlack
    Jan 7 at 18:16










  • $begingroup$
    The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
    $endgroup$
    – Dirk
    Jan 7 at 18:31














3












3








3





$begingroup$


Let $H = L^2([0,1],mathbb{R})$ and $T : H to H,, Tf(x) = x^2f(x) $.



$T$ is linear.



$$|Tf|_{L^2([0,1],mathbb{R})} = sqrt{int_0^1x^4f^2(x)dx} leqsqrt{int_0^1f^2(x)dx} = |f|_{L^2([0,1],mathbb{R})} $$



$T$ is linear and bounded therefore it's continuous. Also
$|T|| leq 1$.



I tried finding solution to $|Tf|_{L^2([0,1],mathbb{R})} = |f|_{L^2([0,1],mathbb{R})} $



I found $$f(x) = sqrt{frac{2x-1}{x^4-1}}$$



but it's not in $L^2$ so it doesn't work.



anyone knows an $f$ to reach $1$, I'm not even sure it's $1$.



any help will be greatly appreciated !






functional-analysis operator-theory hilbert-spaces norm lp-spaces







share|cite|improve this question











$endgroup$




Let $H = L^2([0,1],mathbb{R})$ and $T : H to H,, Tf(x) = x^2f(x) $.



$T$ is linear.



$$|Tf|_{L^2([0,1],mathbb{R})} = sqrt{int_0^1x^4f^2(x)dx} leqsqrt{int_0^1f^2(x)dx} = |f|_{L^2([0,1],mathbb{R})} $$



$T$ is linear and bounded therefore it's continuous. Also
$|T|| leq 1$.



I tried finding solution to $|Tf|_{L^2([0,1],mathbb{R})} = |f|_{L^2([0,1],mathbb{R})} $



I found $$f(x) = sqrt{frac{2x-1}{x^4-1}}$$



but it's not in $L^2$ so it doesn't work.



anyone knows an $f$ to reach $1$, I'm not even sure it's $1$.



any help will be greatly appreciated !






functional-analysis operator-theory hilbert-spaces norm lp-spaces




functional-analysis operator-theory hilbert-spaces norm lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 9:30









Davide Giraudo

128k17154268




128k17154268










asked Jan 7 at 18:09









rapidracimrapidracim

1,7291419




1,7291419








  • 1




    $begingroup$
    Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
    $endgroup$
    – Mindlack
    Jan 7 at 18:16










  • $begingroup$
    The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
    $endgroup$
    – Dirk
    Jan 7 at 18:31














  • 1




    $begingroup$
    Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
    $endgroup$
    – Mindlack
    Jan 7 at 18:16










  • $begingroup$
    The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
    $endgroup$
    – Dirk
    Jan 7 at 18:31








1




1




$begingroup$
Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
$endgroup$
– Mindlack
Jan 7 at 18:16




$begingroup$
Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
$endgroup$
– Mindlack
Jan 7 at 18:16












$begingroup$
The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
$endgroup$
– Dirk
Jan 7 at 18:31




$begingroup$
The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
$endgroup$
– Dirk
Jan 7 at 18:31










2 Answers
2






active

oldest

votes


















5












$begingroup$

Consider $f_n(x)=x^n$ for $ninBbb N$.



Then $$|f_n|_2=sqrt{frac1{2n+1}}$$
and
$$|Tf_n|_2=sqrt{frac1{2n+5}}$$
Since
$$lim_{ntoinfty}frac{|Tf_n|_2}{|f_n|_2}=1$$
we have that $|T|ge1$.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Multiplication by $x^2$ on $[0,1]$. $T$ is a self-adjoint operator. The essential range of the multiplier $x^2$ is $[0,1]$, so the spectrum of $T$ is $[0,1]$. The spectral radius of $T$ is $1$, and (since $T$ is self-adjoint), the norm of $T$ is $1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
      $endgroup$
      – GEdgar
      Jan 8 at 14:24












    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%2f3065296%2fwhat-is-the-operator-norm-of-tfx-x2fx%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Consider $f_n(x)=x^n$ for $ninBbb N$.



    Then $$|f_n|_2=sqrt{frac1{2n+1}}$$
    and
    $$|Tf_n|_2=sqrt{frac1{2n+5}}$$
    Since
    $$lim_{ntoinfty}frac{|Tf_n|_2}{|f_n|_2}=1$$
    we have that $|T|ge1$.






    share|cite|improve this answer











    $endgroup$


















      5












      $begingroup$

      Consider $f_n(x)=x^n$ for $ninBbb N$.



      Then $$|f_n|_2=sqrt{frac1{2n+1}}$$
      and
      $$|Tf_n|_2=sqrt{frac1{2n+5}}$$
      Since
      $$lim_{ntoinfty}frac{|Tf_n|_2}{|f_n|_2}=1$$
      we have that $|T|ge1$.






      share|cite|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        Consider $f_n(x)=x^n$ for $ninBbb N$.



        Then $$|f_n|_2=sqrt{frac1{2n+1}}$$
        and
        $$|Tf_n|_2=sqrt{frac1{2n+5}}$$
        Since
        $$lim_{ntoinfty}frac{|Tf_n|_2}{|f_n|_2}=1$$
        we have that $|T|ge1$.






        share|cite|improve this answer











        $endgroup$



        Consider $f_n(x)=x^n$ for $ninBbb N$.



        Then $$|f_n|_2=sqrt{frac1{2n+1}}$$
        and
        $$|Tf_n|_2=sqrt{frac1{2n+5}}$$
        Since
        $$lim_{ntoinfty}frac{|Tf_n|_2}{|f_n|_2}=1$$
        we have that $|T|ge1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 7 at 20:05

























        answered Jan 7 at 18:16









        ajotatxeajotatxe

        54.1k24190




        54.1k24190























            3












            $begingroup$

            Multiplication by $x^2$ on $[0,1]$. $T$ is a self-adjoint operator. The essential range of the multiplier $x^2$ is $[0,1]$, so the spectrum of $T$ is $[0,1]$. The spectral radius of $T$ is $1$, and (since $T$ is self-adjoint), the norm of $T$ is $1$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
              $endgroup$
              – GEdgar
              Jan 8 at 14:24
















            3












            $begingroup$

            Multiplication by $x^2$ on $[0,1]$. $T$ is a self-adjoint operator. The essential range of the multiplier $x^2$ is $[0,1]$, so the spectrum of $T$ is $[0,1]$. The spectral radius of $T$ is $1$, and (since $T$ is self-adjoint), the norm of $T$ is $1$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
              $endgroup$
              – GEdgar
              Jan 8 at 14:24














            3












            3








            3





            $begingroup$

            Multiplication by $x^2$ on $[0,1]$. $T$ is a self-adjoint operator. The essential range of the multiplier $x^2$ is $[0,1]$, so the spectrum of $T$ is $[0,1]$. The spectral radius of $T$ is $1$, and (since $T$ is self-adjoint), the norm of $T$ is $1$.






            share|cite|improve this answer









            $endgroup$



            Multiplication by $x^2$ on $[0,1]$. $T$ is a self-adjoint operator. The essential range of the multiplier $x^2$ is $[0,1]$, so the spectrum of $T$ is $[0,1]$. The spectral radius of $T$ is $1$, and (since $T$ is self-adjoint), the norm of $T$ is $1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 7 at 18:17









            GEdgarGEdgar

            63.3k268172




            63.3k268172












            • $begingroup$
              As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
              $endgroup$
              – GEdgar
              Jan 8 at 14:24


















            • $begingroup$
              As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
              $endgroup$
              – GEdgar
              Jan 8 at 14:24
















            $begingroup$
            As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
            $endgroup$
            – GEdgar
            Jan 8 at 14:24




            $begingroup$
            As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
            $endgroup$
            – GEdgar
            Jan 8 at 14:24


















            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%2f3065296%2fwhat-is-the-operator-norm-of-tfx-x2fx%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