Does $int_0^1 f(x)=int_0^1 xf(x)$ imply $int_0^x f(t)$ has a root












3












$begingroup$


My question is whether or not the following is true:



If $f:[0,1]to mathbb{R}$ is a continous function such that
$$int_0^1 f(x)dx=int_0^1 xf(x)dx$$
then there exist $cin(0,1)$ such that
$$int_0^c f(x)dx=0$$



It is quite clear that in the interval $(0,1)$ there must be some points $a$ such that $f(1)f(a)<0$ but I can't say from this if the statement above is true or not.






real-analysis definite-integrals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you have an example of a non-zero function that satisfies the first equality?
    $endgroup$
    – Yanko
    Jan 5 at 12:27












  • $begingroup$
    @Yanko $operatorname{sinc}(2pi (1-x) )$ works
    $endgroup$
    – Calvin Khor
    Jan 5 at 12:35








  • 1




    $begingroup$
    @Yanko Surley there are such functions. There exist $a,b$ such that $ax+b$ satisfies that equation.
    $endgroup$
    – Kavi Rama Murthy
    Jan 5 at 12:35










  • $begingroup$
    @KaviRamaMurthy Right, for $a=6,b=-2$. Thanks.
    $endgroup$
    – Yanko
    Jan 5 at 12:37
















3












$begingroup$


My question is whether or not the following is true:



If $f:[0,1]to mathbb{R}$ is a continous function such that
$$int_0^1 f(x)dx=int_0^1 xf(x)dx$$
then there exist $cin(0,1)$ such that
$$int_0^c f(x)dx=0$$



It is quite clear that in the interval $(0,1)$ there must be some points $a$ such that $f(1)f(a)<0$ but I can't say from this if the statement above is true or not.






real-analysis definite-integrals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you have an example of a non-zero function that satisfies the first equality?
    $endgroup$
    – Yanko
    Jan 5 at 12:27












  • $begingroup$
    @Yanko $operatorname{sinc}(2pi (1-x) )$ works
    $endgroup$
    – Calvin Khor
    Jan 5 at 12:35








  • 1




    $begingroup$
    @Yanko Surley there are such functions. There exist $a,b$ such that $ax+b$ satisfies that equation.
    $endgroup$
    – Kavi Rama Murthy
    Jan 5 at 12:35










  • $begingroup$
    @KaviRamaMurthy Right, for $a=6,b=-2$. Thanks.
    $endgroup$
    – Yanko
    Jan 5 at 12:37














3












3








3





$begingroup$


My question is whether or not the following is true:



If $f:[0,1]to mathbb{R}$ is a continous function such that
$$int_0^1 f(x)dx=int_0^1 xf(x)dx$$
then there exist $cin(0,1)$ such that
$$int_0^c f(x)dx=0$$



It is quite clear that in the interval $(0,1)$ there must be some points $a$ such that $f(1)f(a)<0$ but I can't say from this if the statement above is true or not.






real-analysis definite-integrals







share|cite|improve this question









$endgroup$




My question is whether or not the following is true:



If $f:[0,1]to mathbb{R}$ is a continous function such that
$$int_0^1 f(x)dx=int_0^1 xf(x)dx$$
then there exist $cin(0,1)$ such that
$$int_0^c f(x)dx=0$$



It is quite clear that in the interval $(0,1)$ there must be some points $a$ such that $f(1)f(a)<0$ but I can't say from this if the statement above is true or not.






real-analysis definite-integrals




real-analysis definite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 12:18









Andrew VAndrew V

31411




31411












  • $begingroup$
    Do you have an example of a non-zero function that satisfies the first equality?
    $endgroup$
    – Yanko
    Jan 5 at 12:27












  • $begingroup$
    @Yanko $operatorname{sinc}(2pi (1-x) )$ works
    $endgroup$
    – Calvin Khor
    Jan 5 at 12:35








  • 1




    $begingroup$
    @Yanko Surley there are such functions. There exist $a,b$ such that $ax+b$ satisfies that equation.
    $endgroup$
    – Kavi Rama Murthy
    Jan 5 at 12:35










  • $begingroup$
    @KaviRamaMurthy Right, for $a=6,b=-2$. Thanks.
    $endgroup$
    – Yanko
    Jan 5 at 12:37


















  • $begingroup$
    Do you have an example of a non-zero function that satisfies the first equality?
    $endgroup$
    – Yanko
    Jan 5 at 12:27












  • $begingroup$
    @Yanko $operatorname{sinc}(2pi (1-x) )$ works
    $endgroup$
    – Calvin Khor
    Jan 5 at 12:35








  • 1




    $begingroup$
    @Yanko Surley there are such functions. There exist $a,b$ such that $ax+b$ satisfies that equation.
    $endgroup$
    – Kavi Rama Murthy
    Jan 5 at 12:35










  • $begingroup$
    @KaviRamaMurthy Right, for $a=6,b=-2$. Thanks.
    $endgroup$
    – Yanko
    Jan 5 at 12:37
















$begingroup$
Do you have an example of a non-zero function that satisfies the first equality?
$endgroup$
– Yanko
Jan 5 at 12:27






$begingroup$
Do you have an example of a non-zero function that satisfies the first equality?
$endgroup$
– Yanko
Jan 5 at 12:27














$begingroup$
@Yanko $operatorname{sinc}(2pi (1-x) )$ works
$endgroup$
– Calvin Khor
Jan 5 at 12:35






$begingroup$
@Yanko $operatorname{sinc}(2pi (1-x) )$ works
$endgroup$
– Calvin Khor
Jan 5 at 12:35






1




1




$begingroup$
@Yanko Surley there are such functions. There exist $a,b$ such that $ax+b$ satisfies that equation.
$endgroup$
– Kavi Rama Murthy
Jan 5 at 12:35




$begingroup$
@Yanko Surley there are such functions. There exist $a,b$ such that $ax+b$ satisfies that equation.
$endgroup$
– Kavi Rama Murthy
Jan 5 at 12:35












$begingroup$
@KaviRamaMurthy Right, for $a=6,b=-2$. Thanks.
$endgroup$
– Yanko
Jan 5 at 12:37




$begingroup$
@KaviRamaMurthy Right, for $a=6,b=-2$. Thanks.
$endgroup$
– Yanko
Jan 5 at 12:37










2 Answers
2






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oldest

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3












$begingroup$

Let $F(x)=int_0^{x} f(t), dt$. Then $F(1)=int_0^{1} f(t), dt=int_0^{1} tf(t), dt=tF(t)|_0^{1}-int_0^{1}F(t), dt$. This gives $int_0^{1}F(t), dt=0$. If $F$ has no zeros then it is strictly positive or strictly negative throughout and its integral cannot be $0$. Hence $F(c)=0$ for some $c$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint: Note that
    $$
    int_0^1 F(x)dx =int_0^1left(int_0^x f(t)dtright)dx=int_0^1left(int_t^1dxright)f(t)dt= int_0^1 (1-t)f(t)dt
    $$     where $F(x) = int_0^x f(t)dt$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Let $F(x)=int_0^{x} f(t), dt$. Then $F(1)=int_0^{1} f(t), dt=int_0^{1} tf(t), dt=tF(t)|_0^{1}-int_0^{1}F(t), dt$. This gives $int_0^{1}F(t), dt=0$. If $F$ has no zeros then it is strictly positive or strictly negative throughout and its integral cannot be $0$. Hence $F(c)=0$ for some $c$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Let $F(x)=int_0^{x} f(t), dt$. Then $F(1)=int_0^{1} f(t), dt=int_0^{1} tf(t), dt=tF(t)|_0^{1}-int_0^{1}F(t), dt$. This gives $int_0^{1}F(t), dt=0$. If $F$ has no zeros then it is strictly positive or strictly negative throughout and its integral cannot be $0$. Hence $F(c)=0$ for some $c$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Let $F(x)=int_0^{x} f(t), dt$. Then $F(1)=int_0^{1} f(t), dt=int_0^{1} tf(t), dt=tF(t)|_0^{1}-int_0^{1}F(t), dt$. This gives $int_0^{1}F(t), dt=0$. If $F$ has no zeros then it is strictly positive or strictly negative throughout and its integral cannot be $0$. Hence $F(c)=0$ for some $c$.






          share|cite|improve this answer









          $endgroup$



          Let $F(x)=int_0^{x} f(t), dt$. Then $F(1)=int_0^{1} f(t), dt=int_0^{1} tf(t), dt=tF(t)|_0^{1}-int_0^{1}F(t), dt$. This gives $int_0^{1}F(t), dt=0$. If $F$ has no zeros then it is strictly positive or strictly negative throughout and its integral cannot be $0$. Hence $F(c)=0$ for some $c$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 at 12:30









          Kavi Rama MurthyKavi Rama Murthy

          65.1k42766




          65.1k42766























              1












              $begingroup$

              Hint: Note that
              $$
              int_0^1 F(x)dx =int_0^1left(int_0^x f(t)dtright)dx=int_0^1left(int_t^1dxright)f(t)dt= int_0^1 (1-t)f(t)dt
              $$               where $F(x) = int_0^x f(t)dt$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Hint: Note that
                $$
                int_0^1 F(x)dx =int_0^1left(int_0^x f(t)dtright)dx=int_0^1left(int_t^1dxright)f(t)dt= int_0^1 (1-t)f(t)dt
                $$                 where $F(x) = int_0^x f(t)dt$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Hint: Note that
                  $$
                  int_0^1 F(x)dx =int_0^1left(int_0^x f(t)dtright)dx=int_0^1left(int_t^1dxright)f(t)dt= int_0^1 (1-t)f(t)dt
                  $$                   where $F(x) = int_0^x f(t)dt$.






                  share|cite|improve this answer









                  $endgroup$



                  Hint: Note that
                  $$
                  int_0^1 F(x)dx =int_0^1left(int_0^x f(t)dtright)dx=int_0^1left(int_t^1dxright)f(t)dt= int_0^1 (1-t)f(t)dt
                  $$                   where $F(x) = int_0^x f(t)dt$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 12:30









                  SongSong

                  16.8k11042




                  16.8k11042






























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