Function $f$ from $L^2[0,1]$ such that $int limits_0^1 f(x)g(x)dx = g(0)$, where $g$ is polynomial.












2












$begingroup$



Is there function $f$ from $L^2[0,1]$ such that $int limits_0^1 f(x)g(x)dx = g(0)$, where $g$ is:



a) Any polynomial of degree $leq n$?



b) Any polynomial of any degree?




I know that the answer for b) is no: if $f$ is such function, then $xf(x)$ is orthoganal to any polynomial and thus orthogonal to any element of $L^2[0, 1]$. Therefore $f = [0]$.



I also understand that we can get a polynomial, which is orthoganal to ${x, dots, x^n}$, using Gram–Schmidt process from $x$ to $x^{n+1}$. The problem is that I can't prove that such element is not orthoganal to $1$.



Thanks!






integration functional-analysis polynomials hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, the trivial case is if $gequiv 0$ of course, which is a polynomial of degree $-infty leq n$ by convention.
    $endgroup$
    – Thomas Lang
    Dec 4 '18 at 8:47












  • $begingroup$
    @ThomasLang I meant that $g$ is every polynomial with degree $leq n$. I'm sorry for poorly stated problem.
    $endgroup$
    – Gleb Chili
    Dec 4 '18 at 8:55


















2












$begingroup$



Is there function $f$ from $L^2[0,1]$ such that $int limits_0^1 f(x)g(x)dx = g(0)$, where $g$ is:



a) Any polynomial of degree $leq n$?



b) Any polynomial of any degree?




I know that the answer for b) is no: if $f$ is such function, then $xf(x)$ is orthoganal to any polynomial and thus orthogonal to any element of $L^2[0, 1]$. Therefore $f = [0]$.



I also understand that we can get a polynomial, which is orthoganal to ${x, dots, x^n}$, using Gram–Schmidt process from $x$ to $x^{n+1}$. The problem is that I can't prove that such element is not orthoganal to $1$.



Thanks!






integration functional-analysis polynomials hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, the trivial case is if $gequiv 0$ of course, which is a polynomial of degree $-infty leq n$ by convention.
    $endgroup$
    – Thomas Lang
    Dec 4 '18 at 8:47












  • $begingroup$
    @ThomasLang I meant that $g$ is every polynomial with degree $leq n$. I'm sorry for poorly stated problem.
    $endgroup$
    – Gleb Chili
    Dec 4 '18 at 8:55
















2












2








2


2



$begingroup$



Is there function $f$ from $L^2[0,1]$ such that $int limits_0^1 f(x)g(x)dx = g(0)$, where $g$ is:



a) Any polynomial of degree $leq n$?



b) Any polynomial of any degree?




I know that the answer for b) is no: if $f$ is such function, then $xf(x)$ is orthoganal to any polynomial and thus orthogonal to any element of $L^2[0, 1]$. Therefore $f = [0]$.



I also understand that we can get a polynomial, which is orthoganal to ${x, dots, x^n}$, using Gram–Schmidt process from $x$ to $x^{n+1}$. The problem is that I can't prove that such element is not orthoganal to $1$.



Thanks!






integration functional-analysis polynomials hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$





Is there function $f$ from $L^2[0,1]$ such that $int limits_0^1 f(x)g(x)dx = g(0)$, where $g$ is:



a) Any polynomial of degree $leq n$?



b) Any polynomial of any degree?




I know that the answer for b) is no: if $f$ is such function, then $xf(x)$ is orthoganal to any polynomial and thus orthogonal to any element of $L^2[0, 1]$. Therefore $f = [0]$.



I also understand that we can get a polynomial, which is orthoganal to ${x, dots, x^n}$, using Gram–Schmidt process from $x$ to $x^{n+1}$. The problem is that I can't prove that such element is not orthoganal to $1$.



Thanks!






integration functional-analysis polynomials hilbert-spaces lp-spaces




integration functional-analysis polynomials hilbert-spaces lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 9:26









Batominovski

1




1










asked Dec 4 '18 at 8:36









Gleb ChiliGleb Chili

44428




44428












  • $begingroup$
    Well, the trivial case is if $gequiv 0$ of course, which is a polynomial of degree $-infty leq n$ by convention.
    $endgroup$
    – Thomas Lang
    Dec 4 '18 at 8:47












  • $begingroup$
    @ThomasLang I meant that $g$ is every polynomial with degree $leq n$. I'm sorry for poorly stated problem.
    $endgroup$
    – Gleb Chili
    Dec 4 '18 at 8:55




















  • $begingroup$
    Well, the trivial case is if $gequiv 0$ of course, which is a polynomial of degree $-infty leq n$ by convention.
    $endgroup$
    – Thomas Lang
    Dec 4 '18 at 8:47












  • $begingroup$
    @ThomasLang I meant that $g$ is every polynomial with degree $leq n$. I'm sorry for poorly stated problem.
    $endgroup$
    – Gleb Chili
    Dec 4 '18 at 8:55


















$begingroup$
Well, the trivial case is if $gequiv 0$ of course, which is a polynomial of degree $-infty leq n$ by convention.
$endgroup$
– Thomas Lang
Dec 4 '18 at 8:47






$begingroup$
Well, the trivial case is if $gequiv 0$ of course, which is a polynomial of degree $-infty leq n$ by convention.
$endgroup$
– Thomas Lang
Dec 4 '18 at 8:47














$begingroup$
@ThomasLang I meant that $g$ is every polynomial with degree $leq n$. I'm sorry for poorly stated problem.
$endgroup$
– Gleb Chili
Dec 4 '18 at 8:55






$begingroup$
@ThomasLang I meant that $g$ is every polynomial with degree $leq n$. I'm sorry for poorly stated problem.
$endgroup$
– Gleb Chili
Dec 4 '18 at 8:55












2 Answers
2






active

oldest

votes


















3












$begingroup$

This is easy if you are allowed to use some Functional Analysis. Let $M_n$ be the subspace of $L^{2}$ consisting of polynomials with degree at most $n$. Then $p to p(0)$ is a linear map, hence continuous too: any linear map on a finite dimensional space is continuous. By Hahn Banach Theorem there is a continuous linear functional on $L^{2}$ which extends this and any continuous linear functional on $L^{2}$ is of the type $g to int fg$ for some $f in L^{2}$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    In addition to Kavi Rama Murthy's great answer, I am offering an explicit construction of $finmathcal{L}^2big([0,1]big)$ for each given $ninmathbb{Z}_{geq 0}$. This function $f$ is given by
    $$f(x)=f_n(x):=sum_{k=0}^n,(-1)^k,(k+1),binom{n+1}{k+1},binom{n+k+1}{k+1},x^k$$
    for all $xin[0,1]$. For example, $f_0(x)=1$, $f_1(x)=4-6x$, and $f_2(x)=9-36x+30x^2$. The coefficients of the polynomial $f_n$ are obtained by taking the first row of the inverse of the $(n+1)$-by-$(n+1)$ Hilbert matrix. See here.






    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%2f3025297%2ffunction-f-from-l20-1-such-that-int-limits-01-fxgxdx-g0-whe%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









      3












      $begingroup$

      This is easy if you are allowed to use some Functional Analysis. Let $M_n$ be the subspace of $L^{2}$ consisting of polynomials with degree at most $n$. Then $p to p(0)$ is a linear map, hence continuous too: any linear map on a finite dimensional space is continuous. By Hahn Banach Theorem there is a continuous linear functional on $L^{2}$ which extends this and any continuous linear functional on $L^{2}$ is of the type $g to int fg$ for some $f in L^{2}$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        This is easy if you are allowed to use some Functional Analysis. Let $M_n$ be the subspace of $L^{2}$ consisting of polynomials with degree at most $n$. Then $p to p(0)$ is a linear map, hence continuous too: any linear map on a finite dimensional space is continuous. By Hahn Banach Theorem there is a continuous linear functional on $L^{2}$ which extends this and any continuous linear functional on $L^{2}$ is of the type $g to int fg$ for some $f in L^{2}$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          This is easy if you are allowed to use some Functional Analysis. Let $M_n$ be the subspace of $L^{2}$ consisting of polynomials with degree at most $n$. Then $p to p(0)$ is a linear map, hence continuous too: any linear map on a finite dimensional space is continuous. By Hahn Banach Theorem there is a continuous linear functional on $L^{2}$ which extends this and any continuous linear functional on $L^{2}$ is of the type $g to int fg$ for some $f in L^{2}$.






          share|cite|improve this answer









          $endgroup$



          This is easy if you are allowed to use some Functional Analysis. Let $M_n$ be the subspace of $L^{2}$ consisting of polynomials with degree at most $n$. Then $p to p(0)$ is a linear map, hence continuous too: any linear map on a finite dimensional space is continuous. By Hahn Banach Theorem there is a continuous linear functional on $L^{2}$ which extends this and any continuous linear functional on $L^{2}$ is of the type $g to int fg$ for some $f in L^{2}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 8:53









          Kavi Rama MurthyKavi Rama Murthy

          55.9k42158




          55.9k42158























              2












              $begingroup$

              In addition to Kavi Rama Murthy's great answer, I am offering an explicit construction of $finmathcal{L}^2big([0,1]big)$ for each given $ninmathbb{Z}_{geq 0}$. This function $f$ is given by
              $$f(x)=f_n(x):=sum_{k=0}^n,(-1)^k,(k+1),binom{n+1}{k+1},binom{n+k+1}{k+1},x^k$$
              for all $xin[0,1]$. For example, $f_0(x)=1$, $f_1(x)=4-6x$, and $f_2(x)=9-36x+30x^2$. The coefficients of the polynomial $f_n$ are obtained by taking the first row of the inverse of the $(n+1)$-by-$(n+1)$ Hilbert matrix. See here.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                In addition to Kavi Rama Murthy's great answer, I am offering an explicit construction of $finmathcal{L}^2big([0,1]big)$ for each given $ninmathbb{Z}_{geq 0}$. This function $f$ is given by
                $$f(x)=f_n(x):=sum_{k=0}^n,(-1)^k,(k+1),binom{n+1}{k+1},binom{n+k+1}{k+1},x^k$$
                for all $xin[0,1]$. For example, $f_0(x)=1$, $f_1(x)=4-6x$, and $f_2(x)=9-36x+30x^2$. The coefficients of the polynomial $f_n$ are obtained by taking the first row of the inverse of the $(n+1)$-by-$(n+1)$ Hilbert matrix. See here.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  In addition to Kavi Rama Murthy's great answer, I am offering an explicit construction of $finmathcal{L}^2big([0,1]big)$ for each given $ninmathbb{Z}_{geq 0}$. This function $f$ is given by
                  $$f(x)=f_n(x):=sum_{k=0}^n,(-1)^k,(k+1),binom{n+1}{k+1},binom{n+k+1}{k+1},x^k$$
                  for all $xin[0,1]$. For example, $f_0(x)=1$, $f_1(x)=4-6x$, and $f_2(x)=9-36x+30x^2$. The coefficients of the polynomial $f_n$ are obtained by taking the first row of the inverse of the $(n+1)$-by-$(n+1)$ Hilbert matrix. See here.






                  share|cite|improve this answer









                  $endgroup$



                  In addition to Kavi Rama Murthy's great answer, I am offering an explicit construction of $finmathcal{L}^2big([0,1]big)$ for each given $ninmathbb{Z}_{geq 0}$. This function $f$ is given by
                  $$f(x)=f_n(x):=sum_{k=0}^n,(-1)^k,(k+1),binom{n+1}{k+1},binom{n+k+1}{k+1},x^k$$
                  for all $xin[0,1]$. For example, $f_0(x)=1$, $f_1(x)=4-6x$, and $f_2(x)=9-36x+30x^2$. The coefficients of the polynomial $f_n$ are obtained by taking the first row of the inverse of the $(n+1)$-by-$(n+1)$ Hilbert matrix. See here.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 9:22









                  BatominovskiBatominovski

                  1




                  1






























                      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%2f3025297%2ffunction-f-from-l20-1-such-that-int-limits-01-fxgxdx-g0-whe%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

                      Le Mesnil-Réaume

                      Image
                      Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
                      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

                      web3.py web3.isConnected() returns false always

                      Image
                      1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
                      Read more