Is this statement true for an arbitrary real function? If $int_{a}^{b} f^2(x) = 0$ then $f(x) = 0$.












1












$begingroup$


Is this statement true for arbitrary real functions? If so, how to prove it?




If $int_{a}^{b} f^2(x) = 0$ then $f(x) = 0$.







integration







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    No take the dirac delta function
    $endgroup$
    – Tsemo Aristide
    Jan 22 at 15:14










  • $begingroup$
    No. take $f(x)=0$ for $x<1$ and $f(1)=1$. However for continuous functions your statement is true.
    $endgroup$
    – Maksim
    Jan 22 at 15:15












  • $begingroup$
    dirac function isn't a real function as title asks but not body...)
    $endgroup$
    – coffeemath
    Jan 22 at 15:16






  • 2




    $begingroup$
    @TsemoAristide: The Dirac delta function is not a function but a distribution.
    $endgroup$
    – gerw
    Jan 22 at 15:16






  • 2




    $begingroup$
    This is true if you add the condition that $f$ be continuous.
    $endgroup$
    – ncmathsadist
    Jan 22 at 15:30
















1












$begingroup$


Is this statement true for arbitrary real functions? If so, how to prove it?




If $int_{a}^{b} f^2(x) = 0$ then $f(x) = 0$.







integration







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    No take the dirac delta function
    $endgroup$
    – Tsemo Aristide
    Jan 22 at 15:14










  • $begingroup$
    No. take $f(x)=0$ for $x<1$ and $f(1)=1$. However for continuous functions your statement is true.
    $endgroup$
    – Maksim
    Jan 22 at 15:15












  • $begingroup$
    dirac function isn't a real function as title asks but not body...)
    $endgroup$
    – coffeemath
    Jan 22 at 15:16






  • 2




    $begingroup$
    @TsemoAristide: The Dirac delta function is not a function but a distribution.
    $endgroup$
    – gerw
    Jan 22 at 15:16






  • 2




    $begingroup$
    This is true if you add the condition that $f$ be continuous.
    $endgroup$
    – ncmathsadist
    Jan 22 at 15:30














1












1








1


0



$begingroup$


Is this statement true for arbitrary real functions? If so, how to prove it?




If $int_{a}^{b} f^2(x) = 0$ then $f(x) = 0$.







integration







share|cite|improve this question











$endgroup$




Is this statement true for arbitrary real functions? If so, how to prove it?




If $int_{a}^{b} f^2(x) = 0$ then $f(x) = 0$.







integration




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 15:16









Blue

48.6k870156




48.6k870156










asked Jan 22 at 15:13









fsociety_1729fsociety_1729

82




82








  • 1




    $begingroup$
    No take the dirac delta function
    $endgroup$
    – Tsemo Aristide
    Jan 22 at 15:14










  • $begingroup$
    No. take $f(x)=0$ for $x<1$ and $f(1)=1$. However for continuous functions your statement is true.
    $endgroup$
    – Maksim
    Jan 22 at 15:15












  • $begingroup$
    dirac function isn't a real function as title asks but not body...)
    $endgroup$
    – coffeemath
    Jan 22 at 15:16






  • 2




    $begingroup$
    @TsemoAristide: The Dirac delta function is not a function but a distribution.
    $endgroup$
    – gerw
    Jan 22 at 15:16






  • 2




    $begingroup$
    This is true if you add the condition that $f$ be continuous.
    $endgroup$
    – ncmathsadist
    Jan 22 at 15:30














  • 1




    $begingroup$
    No take the dirac delta function
    $endgroup$
    – Tsemo Aristide
    Jan 22 at 15:14










  • $begingroup$
    No. take $f(x)=0$ for $x<1$ and $f(1)=1$. However for continuous functions your statement is true.
    $endgroup$
    – Maksim
    Jan 22 at 15:15












  • $begingroup$
    dirac function isn't a real function as title asks but not body...)
    $endgroup$
    – coffeemath
    Jan 22 at 15:16






  • 2




    $begingroup$
    @TsemoAristide: The Dirac delta function is not a function but a distribution.
    $endgroup$
    – gerw
    Jan 22 at 15:16






  • 2




    $begingroup$
    This is true if you add the condition that $f$ be continuous.
    $endgroup$
    – ncmathsadist
    Jan 22 at 15:30








1




1




$begingroup$
No take the dirac delta function
$endgroup$
– Tsemo Aristide
Jan 22 at 15:14




$begingroup$
No take the dirac delta function
$endgroup$
– Tsemo Aristide
Jan 22 at 15:14












$begingroup$
No. take $f(x)=0$ for $x<1$ and $f(1)=1$. However for continuous functions your statement is true.
$endgroup$
– Maksim
Jan 22 at 15:15






$begingroup$
No. take $f(x)=0$ for $x<1$ and $f(1)=1$. However for continuous functions your statement is true.
$endgroup$
– Maksim
Jan 22 at 15:15














$begingroup$
dirac function isn't a real function as title asks but not body...)
$endgroup$
– coffeemath
Jan 22 at 15:16




$begingroup$
dirac function isn't a real function as title asks but not body...)
$endgroup$
– coffeemath
Jan 22 at 15:16




2




2




$begingroup$
@TsemoAristide: The Dirac delta function is not a function but a distribution.
$endgroup$
– gerw
Jan 22 at 15:16




$begingroup$
@TsemoAristide: The Dirac delta function is not a function but a distribution.
$endgroup$
– gerw
Jan 22 at 15:16




2




2




$begingroup$
This is true if you add the condition that $f$ be continuous.
$endgroup$
– ncmathsadist
Jan 22 at 15:30




$begingroup$
This is true if you add the condition that $f$ be continuous.
$endgroup$
– ncmathsadist
Jan 22 at 15:30










3 Answers
3






active

oldest

votes


















4












$begingroup$

It is false in general: take a function that equals $1$ on some point, and $0$ everywhere else.



For continuous functions it is true.



Proof: suppose $f:[a,b]to mathbb{R}$ is continuous and not identically zero. Let $x_0 in [a,b]$ such that $f(x_0)neq 0$, then $f(x_0)^2 > 0$. Since $f$ is continuous, $f^2$ is also continuous so there exists an interval with non-empty interior containing $x_0$ where $f(x)^2 > eta$ where $eta = f(x_0)^2/2$. Let $ell > 0$ be the length of this interval. By breaking the interval into three pieces, and noticing that $f(x)^2geq 0$ for all $x in [a,b]$ one obtains that $int_a^b f^2 geq ell cdot eta > 0$. In particular, $int_a^b f^2 neq 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you give me an example of where this statement can be true?
    $endgroup$
    – fsociety_1729
    Jan 22 at 15:28










  • $begingroup$
    @user637244 take $f:[0,1] to mathbb{R}$ such that $f(0)=1$ and $f(x) = 0$ for all $0<xleq 1$
    $endgroup$
    – user1892304
    Jan 22 at 15:30





















4












$begingroup$

No, this is not true. For example you can pick $f(x)=0$ except for a finite set of points for which you can assign arbitrary nonzero values. Then the integral of $f^2$ is still $0$. Intuitively speaking, the integral measures the area behind the graph of the function, and modifying a finite set of points of the graph does not modify the area.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is true, however, if you identify such functions in the usual way.
    $endgroup$
    – Klaus
    Jan 22 at 15:23






  • 1




    $begingroup$
    @Klaus After you've identified functions that differ only on a set of measure zero, they are not functions anymore (contrary to the popular abuse of language).
    $endgroup$
    – lisyarus
    Jan 22 at 15:29










  • $begingroup$
    Can you give me an example of where this statement can be true?
    $endgroup$
    – fsociety_1729
    Jan 22 at 15:31










  • $begingroup$
    @user637244 Do you mean an example in which your conjecture is true, or an specific example of my argument?
    $endgroup$
    – Jose Brox
    Jan 22 at 15:34










  • $begingroup$
    @lisyarus Well, technically you are right. What I was trying to say is that the statement is correct if you add "almost everywhere".
    $endgroup$
    – Klaus
    Jan 22 at 15:34



















0












$begingroup$

Your statement is true, if you change it like that



If $int_{a}^{b} f^2(x) = 0$ then $f(x) = 0$ a.e. (a.e. = almost everywhere).



And the opposite direction holds as well ... $ddotsmile$






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









    4












    $begingroup$

    It is false in general: take a function that equals $1$ on some point, and $0$ everywhere else.



    For continuous functions it is true.



    Proof: suppose $f:[a,b]to mathbb{R}$ is continuous and not identically zero. Let $x_0 in [a,b]$ such that $f(x_0)neq 0$, then $f(x_0)^2 > 0$. Since $f$ is continuous, $f^2$ is also continuous so there exists an interval with non-empty interior containing $x_0$ where $f(x)^2 > eta$ where $eta = f(x_0)^2/2$. Let $ell > 0$ be the length of this interval. By breaking the interval into three pieces, and noticing that $f(x)^2geq 0$ for all $x in [a,b]$ one obtains that $int_a^b f^2 geq ell cdot eta > 0$. In particular, $int_a^b f^2 neq 0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can you give me an example of where this statement can be true?
      $endgroup$
      – fsociety_1729
      Jan 22 at 15:28










    • $begingroup$
      @user637244 take $f:[0,1] to mathbb{R}$ such that $f(0)=1$ and $f(x) = 0$ for all $0<xleq 1$
      $endgroup$
      – user1892304
      Jan 22 at 15:30


















    4












    $begingroup$

    It is false in general: take a function that equals $1$ on some point, and $0$ everywhere else.



    For continuous functions it is true.



    Proof: suppose $f:[a,b]to mathbb{R}$ is continuous and not identically zero. Let $x_0 in [a,b]$ such that $f(x_0)neq 0$, then $f(x_0)^2 > 0$. Since $f$ is continuous, $f^2$ is also continuous so there exists an interval with non-empty interior containing $x_0$ where $f(x)^2 > eta$ where $eta = f(x_0)^2/2$. Let $ell > 0$ be the length of this interval. By breaking the interval into three pieces, and noticing that $f(x)^2geq 0$ for all $x in [a,b]$ one obtains that $int_a^b f^2 geq ell cdot eta > 0$. In particular, $int_a^b f^2 neq 0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can you give me an example of where this statement can be true?
      $endgroup$
      – fsociety_1729
      Jan 22 at 15:28










    • $begingroup$
      @user637244 take $f:[0,1] to mathbb{R}$ such that $f(0)=1$ and $f(x) = 0$ for all $0<xleq 1$
      $endgroup$
      – user1892304
      Jan 22 at 15:30
















    4












    4








    4





    $begingroup$

    It is false in general: take a function that equals $1$ on some point, and $0$ everywhere else.



    For continuous functions it is true.



    Proof: suppose $f:[a,b]to mathbb{R}$ is continuous and not identically zero. Let $x_0 in [a,b]$ such that $f(x_0)neq 0$, then $f(x_0)^2 > 0$. Since $f$ is continuous, $f^2$ is also continuous so there exists an interval with non-empty interior containing $x_0$ where $f(x)^2 > eta$ where $eta = f(x_0)^2/2$. Let $ell > 0$ be the length of this interval. By breaking the interval into three pieces, and noticing that $f(x)^2geq 0$ for all $x in [a,b]$ one obtains that $int_a^b f^2 geq ell cdot eta > 0$. In particular, $int_a^b f^2 neq 0$.






    share|cite|improve this answer









    $endgroup$



    It is false in general: take a function that equals $1$ on some point, and $0$ everywhere else.



    For continuous functions it is true.



    Proof: suppose $f:[a,b]to mathbb{R}$ is continuous and not identically zero. Let $x_0 in [a,b]$ such that $f(x_0)neq 0$, then $f(x_0)^2 > 0$. Since $f$ is continuous, $f^2$ is also continuous so there exists an interval with non-empty interior containing $x_0$ where $f(x)^2 > eta$ where $eta = f(x_0)^2/2$. Let $ell > 0$ be the length of this interval. By breaking the interval into three pieces, and noticing that $f(x)^2geq 0$ for all $x in [a,b]$ one obtains that $int_a^b f^2 geq ell cdot eta > 0$. In particular, $int_a^b f^2 neq 0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 22 at 15:26









    user1892304user1892304

    1,482917




    1,482917












    • $begingroup$
      Can you give me an example of where this statement can be true?
      $endgroup$
      – fsociety_1729
      Jan 22 at 15:28










    • $begingroup$
      @user637244 take $f:[0,1] to mathbb{R}$ such that $f(0)=1$ and $f(x) = 0$ for all $0<xleq 1$
      $endgroup$
      – user1892304
      Jan 22 at 15:30




















    • $begingroup$
      Can you give me an example of where this statement can be true?
      $endgroup$
      – fsociety_1729
      Jan 22 at 15:28










    • $begingroup$
      @user637244 take $f:[0,1] to mathbb{R}$ such that $f(0)=1$ and $f(x) = 0$ for all $0<xleq 1$
      $endgroup$
      – user1892304
      Jan 22 at 15:30


















    $begingroup$
    Can you give me an example of where this statement can be true?
    $endgroup$
    – fsociety_1729
    Jan 22 at 15:28




    $begingroup$
    Can you give me an example of where this statement can be true?
    $endgroup$
    – fsociety_1729
    Jan 22 at 15:28












    $begingroup$
    @user637244 take $f:[0,1] to mathbb{R}$ such that $f(0)=1$ and $f(x) = 0$ for all $0<xleq 1$
    $endgroup$
    – user1892304
    Jan 22 at 15:30






    $begingroup$
    @user637244 take $f:[0,1] to mathbb{R}$ such that $f(0)=1$ and $f(x) = 0$ for all $0<xleq 1$
    $endgroup$
    – user1892304
    Jan 22 at 15:30













    4












    $begingroup$

    No, this is not true. For example you can pick $f(x)=0$ except for a finite set of points for which you can assign arbitrary nonzero values. Then the integral of $f^2$ is still $0$. Intuitively speaking, the integral measures the area behind the graph of the function, and modifying a finite set of points of the graph does not modify the area.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It is true, however, if you identify such functions in the usual way.
      $endgroup$
      – Klaus
      Jan 22 at 15:23






    • 1




      $begingroup$
      @Klaus After you've identified functions that differ only on a set of measure zero, they are not functions anymore (contrary to the popular abuse of language).
      $endgroup$
      – lisyarus
      Jan 22 at 15:29










    • $begingroup$
      Can you give me an example of where this statement can be true?
      $endgroup$
      – fsociety_1729
      Jan 22 at 15:31










    • $begingroup$
      @user637244 Do you mean an example in which your conjecture is true, or an specific example of my argument?
      $endgroup$
      – Jose Brox
      Jan 22 at 15:34










    • $begingroup$
      @lisyarus Well, technically you are right. What I was trying to say is that the statement is correct if you add "almost everywhere".
      $endgroup$
      – Klaus
      Jan 22 at 15:34
















    4












    $begingroup$

    No, this is not true. For example you can pick $f(x)=0$ except for a finite set of points for which you can assign arbitrary nonzero values. Then the integral of $f^2$ is still $0$. Intuitively speaking, the integral measures the area behind the graph of the function, and modifying a finite set of points of the graph does not modify the area.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It is true, however, if you identify such functions in the usual way.
      $endgroup$
      – Klaus
      Jan 22 at 15:23






    • 1




      $begingroup$
      @Klaus After you've identified functions that differ only on a set of measure zero, they are not functions anymore (contrary to the popular abuse of language).
      $endgroup$
      – lisyarus
      Jan 22 at 15:29










    • $begingroup$
      Can you give me an example of where this statement can be true?
      $endgroup$
      – fsociety_1729
      Jan 22 at 15:31










    • $begingroup$
      @user637244 Do you mean an example in which your conjecture is true, or an specific example of my argument?
      $endgroup$
      – Jose Brox
      Jan 22 at 15:34










    • $begingroup$
      @lisyarus Well, technically you are right. What I was trying to say is that the statement is correct if you add "almost everywhere".
      $endgroup$
      – Klaus
      Jan 22 at 15:34














    4












    4








    4





    $begingroup$

    No, this is not true. For example you can pick $f(x)=0$ except for a finite set of points for which you can assign arbitrary nonzero values. Then the integral of $f^2$ is still $0$. Intuitively speaking, the integral measures the area behind the graph of the function, and modifying a finite set of points of the graph does not modify the area.






    share|cite|improve this answer









    $endgroup$



    No, this is not true. For example you can pick $f(x)=0$ except for a finite set of points for which you can assign arbitrary nonzero values. Then the integral of $f^2$ is still $0$. Intuitively speaking, the integral measures the area behind the graph of the function, and modifying a finite set of points of the graph does not modify the area.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 22 at 15:20









    Jose BroxJose Brox

    3,15711128




    3,15711128












    • $begingroup$
      It is true, however, if you identify such functions in the usual way.
      $endgroup$
      – Klaus
      Jan 22 at 15:23






    • 1




      $begingroup$
      @Klaus After you've identified functions that differ only on a set of measure zero, they are not functions anymore (contrary to the popular abuse of language).
      $endgroup$
      – lisyarus
      Jan 22 at 15:29










    • $begingroup$
      Can you give me an example of where this statement can be true?
      $endgroup$
      – fsociety_1729
      Jan 22 at 15:31










    • $begingroup$
      @user637244 Do you mean an example in which your conjecture is true, or an specific example of my argument?
      $endgroup$
      – Jose Brox
      Jan 22 at 15:34










    • $begingroup$
      @lisyarus Well, technically you are right. What I was trying to say is that the statement is correct if you add "almost everywhere".
      $endgroup$
      – Klaus
      Jan 22 at 15:34


















    • $begingroup$
      It is true, however, if you identify such functions in the usual way.
      $endgroup$
      – Klaus
      Jan 22 at 15:23






    • 1




      $begingroup$
      @Klaus After you've identified functions that differ only on a set of measure zero, they are not functions anymore (contrary to the popular abuse of language).
      $endgroup$
      – lisyarus
      Jan 22 at 15:29










    • $begingroup$
      Can you give me an example of where this statement can be true?
      $endgroup$
      – fsociety_1729
      Jan 22 at 15:31










    • $begingroup$
      @user637244 Do you mean an example in which your conjecture is true, or an specific example of my argument?
      $endgroup$
      – Jose Brox
      Jan 22 at 15:34










    • $begingroup$
      @lisyarus Well, technically you are right. What I was trying to say is that the statement is correct if you add "almost everywhere".
      $endgroup$
      – Klaus
      Jan 22 at 15:34
















    $begingroup$
    It is true, however, if you identify such functions in the usual way.
    $endgroup$
    – Klaus
    Jan 22 at 15:23




    $begingroup$
    It is true, however, if you identify such functions in the usual way.
    $endgroup$
    – Klaus
    Jan 22 at 15:23




    1




    1




    $begingroup$
    @Klaus After you've identified functions that differ only on a set of measure zero, they are not functions anymore (contrary to the popular abuse of language).
    $endgroup$
    – lisyarus
    Jan 22 at 15:29




    $begingroup$
    @Klaus After you've identified functions that differ only on a set of measure zero, they are not functions anymore (contrary to the popular abuse of language).
    $endgroup$
    – lisyarus
    Jan 22 at 15:29












    $begingroup$
    Can you give me an example of where this statement can be true?
    $endgroup$
    – fsociety_1729
    Jan 22 at 15:31




    $begingroup$
    Can you give me an example of where this statement can be true?
    $endgroup$
    – fsociety_1729
    Jan 22 at 15:31












    $begingroup$
    @user637244 Do you mean an example in which your conjecture is true, or an specific example of my argument?
    $endgroup$
    – Jose Brox
    Jan 22 at 15:34




    $begingroup$
    @user637244 Do you mean an example in which your conjecture is true, or an specific example of my argument?
    $endgroup$
    – Jose Brox
    Jan 22 at 15:34












    $begingroup$
    @lisyarus Well, technically you are right. What I was trying to say is that the statement is correct if you add "almost everywhere".
    $endgroup$
    – Klaus
    Jan 22 at 15:34




    $begingroup$
    @lisyarus Well, technically you are right. What I was trying to say is that the statement is correct if you add "almost everywhere".
    $endgroup$
    – Klaus
    Jan 22 at 15:34











    0












    $begingroup$

    Your statement is true, if you change it like that



    If $int_{a}^{b} f^2(x) = 0$ then $f(x) = 0$ a.e. (a.e. = almost everywhere).



    And the opposite direction holds as well ... $ddotsmile$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your statement is true, if you change it like that



      If $int_{a}^{b} f^2(x) = 0$ then $f(x) = 0$ a.e. (a.e. = almost everywhere).



      And the opposite direction holds as well ... $ddotsmile$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your statement is true, if you change it like that



        If $int_{a}^{b} f^2(x) = 0$ then $f(x) = 0$ a.e. (a.e. = almost everywhere).



        And the opposite direction holds as well ... $ddotsmile$






        share|cite|improve this answer









        $endgroup$



        Your statement is true, if you change it like that



        If $int_{a}^{b} f^2(x) = 0$ then $f(x) = 0$ a.e. (a.e. = almost everywhere).



        And the opposite direction holds as well ... $ddotsmile$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 22 at 15:51









        MaksimMaksim

        60718




        60718






























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