Proving that $f(x)=0$ in all points of continuity if $f$ is orthogonal to all polynomials












4












$begingroup$


Suppose that the function $f$ is:



1) Riemann integrable (not necessarily continuous) function on $big[a,b big]$;



2) $forall n geq 0$ $int_{a}^{b}{f(x) x^n} = 0$ (in particular, it means that the function is orthogonal to all polynomials).



Prove that $f(x) = 0$ in all points of continuity $f$.






real-analysis functional-analysis riemann-integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    I'm not sure if it would work, but have you tried using repeated integration by parts?
    $endgroup$
    – Seth
    Nov 25 '18 at 22:54










  • $begingroup$
    For integration by parts $f$ have to be differentiable but it is not.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:15










  • $begingroup$
    No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
    $endgroup$
    – Seth
    Nov 25 '18 at 23:16










  • $begingroup$
    It will not lead to anything.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:19










  • $begingroup$
    One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:22
















4












$begingroup$


Suppose that the function $f$ is:



1) Riemann integrable (not necessarily continuous) function on $big[a,b big]$;



2) $forall n geq 0$ $int_{a}^{b}{f(x) x^n} = 0$ (in particular, it means that the function is orthogonal to all polynomials).



Prove that $f(x) = 0$ in all points of continuity $f$.






real-analysis functional-analysis riemann-integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    I'm not sure if it would work, but have you tried using repeated integration by parts?
    $endgroup$
    – Seth
    Nov 25 '18 at 22:54










  • $begingroup$
    For integration by parts $f$ have to be differentiable but it is not.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:15










  • $begingroup$
    No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
    $endgroup$
    – Seth
    Nov 25 '18 at 23:16










  • $begingroup$
    It will not lead to anything.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:19










  • $begingroup$
    One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:22














4












4








4





$begingroup$


Suppose that the function $f$ is:



1) Riemann integrable (not necessarily continuous) function on $big[a,b big]$;



2) $forall n geq 0$ $int_{a}^{b}{f(x) x^n} = 0$ (in particular, it means that the function is orthogonal to all polynomials).



Prove that $f(x) = 0$ in all points of continuity $f$.






real-analysis functional-analysis riemann-integration







share|cite|improve this question









$endgroup$




Suppose that the function $f$ is:



1) Riemann integrable (not necessarily continuous) function on $big[a,b big]$;



2) $forall n geq 0$ $int_{a}^{b}{f(x) x^n} = 0$ (in particular, it means that the function is orthogonal to all polynomials).



Prove that $f(x) = 0$ in all points of continuity $f$.






real-analysis functional-analysis riemann-integration




real-analysis functional-analysis riemann-integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 25 '18 at 22:51









ModeGenModeGen

333




333












  • $begingroup$
    I'm not sure if it would work, but have you tried using repeated integration by parts?
    $endgroup$
    – Seth
    Nov 25 '18 at 22:54










  • $begingroup$
    For integration by parts $f$ have to be differentiable but it is not.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:15










  • $begingroup$
    No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
    $endgroup$
    – Seth
    Nov 25 '18 at 23:16










  • $begingroup$
    It will not lead to anything.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:19










  • $begingroup$
    One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:22


















  • $begingroup$
    I'm not sure if it would work, but have you tried using repeated integration by parts?
    $endgroup$
    – Seth
    Nov 25 '18 at 22:54










  • $begingroup$
    For integration by parts $f$ have to be differentiable but it is not.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:15










  • $begingroup$
    No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
    $endgroup$
    – Seth
    Nov 25 '18 at 23:16










  • $begingroup$
    It will not lead to anything.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:19










  • $begingroup$
    One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
    $endgroup$
    – ModeGen
    Nov 25 '18 at 23:22
















$begingroup$
I'm not sure if it would work, but have you tried using repeated integration by parts?
$endgroup$
– Seth
Nov 25 '18 at 22:54




$begingroup$
I'm not sure if it would work, but have you tried using repeated integration by parts?
$endgroup$
– Seth
Nov 25 '18 at 22:54












$begingroup$
For integration by parts $f$ have to be differentiable but it is not.
$endgroup$
– ModeGen
Nov 25 '18 at 23:15




$begingroup$
For integration by parts $f$ have to be differentiable but it is not.
$endgroup$
– ModeGen
Nov 25 '18 at 23:15












$begingroup$
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
$endgroup$
– Seth
Nov 25 '18 at 23:16




$begingroup$
No, because you are differentiating $x^n$, so you would be integrating $f(x)$, which is okay
$endgroup$
– Seth
Nov 25 '18 at 23:16












$begingroup$
It will not lead to anything.
$endgroup$
– ModeGen
Nov 25 '18 at 23:19




$begingroup$
It will not lead to anything.
$endgroup$
– ModeGen
Nov 25 '18 at 23:19












$begingroup$
One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
$endgroup$
– ModeGen
Nov 25 '18 at 23:22




$begingroup$
One way I see is to prove that $int_a^b{f(x)^2} = 0$. Then from here we will be able to easily derive the statement.
$endgroup$
– ModeGen
Nov 25 '18 at 23:22










3 Answers
3






active

oldest

votes


















2












$begingroup$

To keep the notation simple, suppose $f$ is Riemann integrable on $[-1,1],$ $f$ is continuous at $0,$ and $ int_{-1}^1 p(x)f(x), dx =0$ for all polynomials $p.$ We want to show $f(0)=0.$



Suppose, to reach a contradiction, that this fails. Then WLOG $f(0)>0.$ By the continuity of $f$ at $0,$ there exists $0<delta < 1$ such that $f>f(0)/2$ in $[-delta,delta].$



Define $p_n(x) = sqrt n(1-x^2)^n.$ Then



$$|int_{delta}^1 fp_n|le Msqrt n(1-delta^2)^n.$$



The right hand side $to 0$ as $nto infty.$ Same thing for the integral over $[-1,-delta].$



On the other hand, for large $n$ we have



$$int_{-delta}^{delta} fp_n ge (f(0)/2)int_{-delta}^{delta} sqrt n(1-x^2)^n, dx ge (f(0)/2)sqrt nint_{0}^{1/sqrt n} (1-x^2)^n, dx$$ $$ = int_0^1 (1-y^2/n)^n,dy to int_0^1 e^{-y^2},dy >0.$$



This proves that $int_{-1}^1 p_n(x)f(x), dx >0 $ for large $n,$ and we have our contradiction.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Because $f$ is a 2-norm limit of polynomials, you can deduce that $int_a^bf(x)^2=0$.



    Now suppose that $f(x_0)ne0$ for some $x_0$ where $f$ is continuous. Take $varepsilon=|f(x_0)|/2$; by continuity at $x_0$, there exists $delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $xin (x_0-delta,x_0+delta)$. From the reverse triangle inequality we have
    $$
    |f(x_0)|-|f(x)|<|f(x_0)|/2,
    $$
    so $|f(x)|>|f(x_0)|/2$. Then
    $$
    int_a^b f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x_0)^2/4=delta f(x_0)^2/2>0,
    $$
    a contradiction. So $f(x_0)=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
      $endgroup$
      – ModeGen
      Nov 26 '18 at 7:47










    • $begingroup$
      Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
      $endgroup$
      – Martin Argerami
      Nov 26 '18 at 14:23












    • $begingroup$
      I know the basics of the measure theory.
      $endgroup$
      – ModeGen
      Nov 26 '18 at 14:39










    • $begingroup$
      In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
      $endgroup$
      – Martin Argerami
      Nov 26 '18 at 14:59



















    2












    $begingroup$

    If you know something about Fourier analysis, you can use the Fejer kernel and the following to conclude that $f=0$ at all points of continuity:
    $$
    int_{a}^{b}f(x)e^{-isx}dx = sum_{n=0}^{infty}frac{(-is)^n}{n!}int_{a}^{b} f(x)x^n dx = 0,;;; sinmathbb{R}.
    $$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      To keep the notation simple, suppose $f$ is Riemann integrable on $[-1,1],$ $f$ is continuous at $0,$ and $ int_{-1}^1 p(x)f(x), dx =0$ for all polynomials $p.$ We want to show $f(0)=0.$



      Suppose, to reach a contradiction, that this fails. Then WLOG $f(0)>0.$ By the continuity of $f$ at $0,$ there exists $0<delta < 1$ such that $f>f(0)/2$ in $[-delta,delta].$



      Define $p_n(x) = sqrt n(1-x^2)^n.$ Then



      $$|int_{delta}^1 fp_n|le Msqrt n(1-delta^2)^n.$$



      The right hand side $to 0$ as $nto infty.$ Same thing for the integral over $[-1,-delta].$



      On the other hand, for large $n$ we have



      $$int_{-delta}^{delta} fp_n ge (f(0)/2)int_{-delta}^{delta} sqrt n(1-x^2)^n, dx ge (f(0)/2)sqrt nint_{0}^{1/sqrt n} (1-x^2)^n, dx$$ $$ = int_0^1 (1-y^2/n)^n,dy to int_0^1 e^{-y^2},dy >0.$$



      This proves that $int_{-1}^1 p_n(x)f(x), dx >0 $ for large $n,$ and we have our contradiction.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        To keep the notation simple, suppose $f$ is Riemann integrable on $[-1,1],$ $f$ is continuous at $0,$ and $ int_{-1}^1 p(x)f(x), dx =0$ for all polynomials $p.$ We want to show $f(0)=0.$



        Suppose, to reach a contradiction, that this fails. Then WLOG $f(0)>0.$ By the continuity of $f$ at $0,$ there exists $0<delta < 1$ such that $f>f(0)/2$ in $[-delta,delta].$



        Define $p_n(x) = sqrt n(1-x^2)^n.$ Then



        $$|int_{delta}^1 fp_n|le Msqrt n(1-delta^2)^n.$$



        The right hand side $to 0$ as $nto infty.$ Same thing for the integral over $[-1,-delta].$



        On the other hand, for large $n$ we have



        $$int_{-delta}^{delta} fp_n ge (f(0)/2)int_{-delta}^{delta} sqrt n(1-x^2)^n, dx ge (f(0)/2)sqrt nint_{0}^{1/sqrt n} (1-x^2)^n, dx$$ $$ = int_0^1 (1-y^2/n)^n,dy to int_0^1 e^{-y^2},dy >0.$$



        This proves that $int_{-1}^1 p_n(x)f(x), dx >0 $ for large $n,$ and we have our contradiction.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          To keep the notation simple, suppose $f$ is Riemann integrable on $[-1,1],$ $f$ is continuous at $0,$ and $ int_{-1}^1 p(x)f(x), dx =0$ for all polynomials $p.$ We want to show $f(0)=0.$



          Suppose, to reach a contradiction, that this fails. Then WLOG $f(0)>0.$ By the continuity of $f$ at $0,$ there exists $0<delta < 1$ such that $f>f(0)/2$ in $[-delta,delta].$



          Define $p_n(x) = sqrt n(1-x^2)^n.$ Then



          $$|int_{delta}^1 fp_n|le Msqrt n(1-delta^2)^n.$$



          The right hand side $to 0$ as $nto infty.$ Same thing for the integral over $[-1,-delta].$



          On the other hand, for large $n$ we have



          $$int_{-delta}^{delta} fp_n ge (f(0)/2)int_{-delta}^{delta} sqrt n(1-x^2)^n, dx ge (f(0)/2)sqrt nint_{0}^{1/sqrt n} (1-x^2)^n, dx$$ $$ = int_0^1 (1-y^2/n)^n,dy to int_0^1 e^{-y^2},dy >0.$$



          This proves that $int_{-1}^1 p_n(x)f(x), dx >0 $ for large $n,$ and we have our contradiction.






          share|cite|improve this answer









          $endgroup$



          To keep the notation simple, suppose $f$ is Riemann integrable on $[-1,1],$ $f$ is continuous at $0,$ and $ int_{-1}^1 p(x)f(x), dx =0$ for all polynomials $p.$ We want to show $f(0)=0.$



          Suppose, to reach a contradiction, that this fails. Then WLOG $f(0)>0.$ By the continuity of $f$ at $0,$ there exists $0<delta < 1$ such that $f>f(0)/2$ in $[-delta,delta].$



          Define $p_n(x) = sqrt n(1-x^2)^n.$ Then



          $$|int_{delta}^1 fp_n|le Msqrt n(1-delta^2)^n.$$



          The right hand side $to 0$ as $nto infty.$ Same thing for the integral over $[-1,-delta].$



          On the other hand, for large $n$ we have



          $$int_{-delta}^{delta} fp_n ge (f(0)/2)int_{-delta}^{delta} sqrt n(1-x^2)^n, dx ge (f(0)/2)sqrt nint_{0}^{1/sqrt n} (1-x^2)^n, dx$$ $$ = int_0^1 (1-y^2/n)^n,dy to int_0^1 e^{-y^2},dy >0.$$



          This proves that $int_{-1}^1 p_n(x)f(x), dx >0 $ for large $n,$ and we have our contradiction.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 18:48









          zhw.zhw.

          71.9k43075




          71.9k43075























              4












              $begingroup$

              Because $f$ is a 2-norm limit of polynomials, you can deduce that $int_a^bf(x)^2=0$.



              Now suppose that $f(x_0)ne0$ for some $x_0$ where $f$ is continuous. Take $varepsilon=|f(x_0)|/2$; by continuity at $x_0$, there exists $delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $xin (x_0-delta,x_0+delta)$. From the reverse triangle inequality we have
              $$
              |f(x_0)|-|f(x)|<|f(x_0)|/2,
              $$
              so $|f(x)|>|f(x_0)|/2$. Then
              $$
              int_a^b f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x_0)^2/4=delta f(x_0)^2/2>0,
              $$
              a contradiction. So $f(x_0)=0$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
                $endgroup$
                – ModeGen
                Nov 26 '18 at 7:47










              • $begingroup$
                Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
                $endgroup$
                – Martin Argerami
                Nov 26 '18 at 14:23












              • $begingroup$
                I know the basics of the measure theory.
                $endgroup$
                – ModeGen
                Nov 26 '18 at 14:39










              • $begingroup$
                In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
                $endgroup$
                – Martin Argerami
                Nov 26 '18 at 14:59
















              4












              $begingroup$

              Because $f$ is a 2-norm limit of polynomials, you can deduce that $int_a^bf(x)^2=0$.



              Now suppose that $f(x_0)ne0$ for some $x_0$ where $f$ is continuous. Take $varepsilon=|f(x_0)|/2$; by continuity at $x_0$, there exists $delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $xin (x_0-delta,x_0+delta)$. From the reverse triangle inequality we have
              $$
              |f(x_0)|-|f(x)|<|f(x_0)|/2,
              $$
              so $|f(x)|>|f(x_0)|/2$. Then
              $$
              int_a^b f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x_0)^2/4=delta f(x_0)^2/2>0,
              $$
              a contradiction. So $f(x_0)=0$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
                $endgroup$
                – ModeGen
                Nov 26 '18 at 7:47










              • $begingroup$
                Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
                $endgroup$
                – Martin Argerami
                Nov 26 '18 at 14:23












              • $begingroup$
                I know the basics of the measure theory.
                $endgroup$
                – ModeGen
                Nov 26 '18 at 14:39










              • $begingroup$
                In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
                $endgroup$
                – Martin Argerami
                Nov 26 '18 at 14:59














              4












              4








              4





              $begingroup$

              Because $f$ is a 2-norm limit of polynomials, you can deduce that $int_a^bf(x)^2=0$.



              Now suppose that $f(x_0)ne0$ for some $x_0$ where $f$ is continuous. Take $varepsilon=|f(x_0)|/2$; by continuity at $x_0$, there exists $delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $xin (x_0-delta,x_0+delta)$. From the reverse triangle inequality we have
              $$
              |f(x_0)|-|f(x)|<|f(x_0)|/2,
              $$
              so $|f(x)|>|f(x_0)|/2$. Then
              $$
              int_a^b f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x_0)^2/4=delta f(x_0)^2/2>0,
              $$
              a contradiction. So $f(x_0)=0$.






              share|cite|improve this answer











              $endgroup$



              Because $f$ is a 2-norm limit of polynomials, you can deduce that $int_a^bf(x)^2=0$.



              Now suppose that $f(x_0)ne0$ for some $x_0$ where $f$ is continuous. Take $varepsilon=|f(x_0)|/2$; by continuity at $x_0$, there exists $delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $xin (x_0-delta,x_0+delta)$. From the reverse triangle inequality we have
              $$
              |f(x_0)|-|f(x)|<|f(x_0)|/2,
              $$
              so $|f(x)|>|f(x_0)|/2$. Then
              $$
              int_a^b f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x)^2geqint_{x_0-delta}^{x_0+delta}f(x_0)^2/4=delta f(x_0)^2/2>0,
              $$
              a contradiction. So $f(x_0)=0$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 26 '18 at 22:29

























              answered Nov 26 '18 at 2:42









              Martin ArgeramiMartin Argerami

              125k1178178




              125k1178178












              • $begingroup$
                We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
                $endgroup$
                – ModeGen
                Nov 26 '18 at 7:47










              • $begingroup$
                Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
                $endgroup$
                – Martin Argerami
                Nov 26 '18 at 14:23












              • $begingroup$
                I know the basics of the measure theory.
                $endgroup$
                – ModeGen
                Nov 26 '18 at 14:39










              • $begingroup$
                In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
                $endgroup$
                – Martin Argerami
                Nov 26 '18 at 14:59


















              • $begingroup$
                We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
                $endgroup$
                – ModeGen
                Nov 26 '18 at 7:47










              • $begingroup$
                Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
                $endgroup$
                – Martin Argerami
                Nov 26 '18 at 14:23












              • $begingroup$
                I know the basics of the measure theory.
                $endgroup$
                – ModeGen
                Nov 26 '18 at 14:39










              • $begingroup$
                In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
                $endgroup$
                – Martin Argerami
                Nov 26 '18 at 14:59
















              $begingroup$
              We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
              $endgroup$
              – ModeGen
              Nov 26 '18 at 7:47




              $begingroup$
              We can not use Weierstrass Approximation Theorem because $f(x)$ is not continuous but only Riemann integrable.
              $endgroup$
              – ModeGen
              Nov 26 '18 at 7:47












              $begingroup$
              Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
              $endgroup$
              – Martin Argerami
              Nov 26 '18 at 14:23






              $begingroup$
              Indeed. I was trying to avoid using something more sophisticated, like measure theory or convolutions, to simplify the argument. Do you know any of those two?
              $endgroup$
              – Martin Argerami
              Nov 26 '18 at 14:23














              $begingroup$
              I know the basics of the measure theory.
              $endgroup$
              – ModeGen
              Nov 26 '18 at 14:39




              $begingroup$
              I know the basics of the measure theory.
              $endgroup$
              – ModeGen
              Nov 26 '18 at 14:39












              $begingroup$
              In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
              $endgroup$
              – Martin Argerami
              Nov 26 '18 at 14:59




              $begingroup$
              In measure theory, one proves that the polynomials are dense in the continuous functions when you when you use the 2-norm.
              $endgroup$
              – Martin Argerami
              Nov 26 '18 at 14:59











              2












              $begingroup$

              If you know something about Fourier analysis, you can use the Fejer kernel and the following to conclude that $f=0$ at all points of continuity:
              $$
              int_{a}^{b}f(x)e^{-isx}dx = sum_{n=0}^{infty}frac{(-is)^n}{n!}int_{a}^{b} f(x)x^n dx = 0,;;; sinmathbb{R}.
              $$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                If you know something about Fourier analysis, you can use the Fejer kernel and the following to conclude that $f=0$ at all points of continuity:
                $$
                int_{a}^{b}f(x)e^{-isx}dx = sum_{n=0}^{infty}frac{(-is)^n}{n!}int_{a}^{b} f(x)x^n dx = 0,;;; sinmathbb{R}.
                $$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  If you know something about Fourier analysis, you can use the Fejer kernel and the following to conclude that $f=0$ at all points of continuity:
                  $$
                  int_{a}^{b}f(x)e^{-isx}dx = sum_{n=0}^{infty}frac{(-is)^n}{n!}int_{a}^{b} f(x)x^n dx = 0,;;; sinmathbb{R}.
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  If you know something about Fourier analysis, you can use the Fejer kernel and the following to conclude that $f=0$ at all points of continuity:
                  $$
                  int_{a}^{b}f(x)e^{-isx}dx = sum_{n=0}^{infty}frac{(-is)^n}{n!}int_{a}^{b} f(x)x^n dx = 0,;;; sinmathbb{R}.
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 26 '18 at 15:16









                  DisintegratingByPartsDisintegratingByParts

                  58.9k42580




                  58.9k42580






























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