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

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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









    $endgroup$













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      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






























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