Continuous embeddings of $C(a,b)$ into itself with $L^p$ norms.












1












$begingroup$


Let $1leq p<qleqinfty$



Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.



Proof attempt:



Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
$|x|_Yleq C|x|_X$



We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.



Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
for $1leq p<qleq infty$
$|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$



So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.



However, $Y$ is not continuously embedded into $X$.



we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.






real-analysis functional-analysis operator-theory lp-spaces







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $1leq p<qleqinfty$



    Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.



    Proof attempt:



    Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
    $|x|_Yleq C|x|_X$



    We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.



    Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
    for $1leq p<qleq infty$
    $|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$



    So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.



    However, $Y$ is not continuously embedded into $X$.



    we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.






    real-analysis functional-analysis operator-theory lp-spaces







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $1leq p<qleqinfty$



      Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.



      Proof attempt:



      Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
      $|x|_Yleq C|x|_X$



      We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.



      Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
      for $1leq p<qleq infty$
      $|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$



      So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.



      However, $Y$ is not continuously embedded into $X$.



      we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.






      real-analysis functional-analysis operator-theory lp-spaces







      share|cite|improve this question











      $endgroup$




      Let $1leq p<qleqinfty$



      Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.



      Proof attempt:



      Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
      $|x|_Yleq C|x|_X$



      We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.



      Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
      for $1leq p<qleq infty$
      $|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$



      So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.



      However, $Y$ is not continuously embedded into $X$.



      we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.






      real-analysis functional-analysis operator-theory lp-spaces




      real-analysis functional-analysis operator-theory lp-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 15 '18 at 7:49









      Yiorgos S. Smyrlis

      63.4k1385163




      63.4k1385163










      asked Sep 21 '17 at 2:22









      Kernel_DirichletKernel_Dirichlet

      1,147416




      1,147416






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Consider $(a,b)=(0,1)$, $rin (p,q)$ and
          $$
          f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
          $$
          Then
          $$
          int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
          $$
          and hence
          $$
          |,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
          $$
          Meanwhile
          $$
          lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
          $$






          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%2f2438323%2fcontinuous-embeddings-of-ca-b-into-itself-with-lp-norms%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












            $begingroup$

            Consider $(a,b)=(0,1)$, $rin (p,q)$ and
            $$
            f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
            $$
            Then
            $$
            int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
            $$
            and hence
            $$
            |,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
            $$
            Meanwhile
            $$
            lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
            $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Consider $(a,b)=(0,1)$, $rin (p,q)$ and
              $$
              f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
              $$
              Then
              $$
              int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
              $$
              and hence
              $$
              |,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
              $$
              Meanwhile
              $$
              lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
              $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Consider $(a,b)=(0,1)$, $rin (p,q)$ and
                $$
                f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
                $$
                Then
                $$
                int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
                $$
                and hence
                $$
                |,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
                $$
                Meanwhile
                $$
                lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
                $$






                share|cite|improve this answer









                $endgroup$



                Consider $(a,b)=(0,1)$, $rin (p,q)$ and
                $$
                f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
                $$
                Then
                $$
                int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
                $$
                and hence
                $$
                |,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
                $$
                Meanwhile
                $$
                lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 21 '17 at 9:27









                Yiorgos S. SmyrlisYiorgos S. Smyrlis

                63.4k1385163




                63.4k1385163






























                    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%2f2438323%2fcontinuous-embeddings-of-ca-b-into-itself-with-lp-norms%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

                    Bundesstraße 106

                    Image
                    Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                    Read more

                    Verónica Boquete

                    Image
                    Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                    Read more

                    Ida-Boy-Ed-Garten

                    Image
                    Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                    Read more