How to show $f(x)$ is $0$ in following problem? [duplicate]












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This question already has an answer here:




  • How to show $alphaint_a^cf(x)dx+beta int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$

    3 answers





Let $f$ be a continuous function defined on $[a, b]$. Assume that there exist constants $α$ and $β$ with $(α ≠ β)$ such that
$$alphaint_a^x f(u)du + βint_x^bf(u)du = 0 $$ for all $x$ belonging to $[a,b]$. Show that $f(x) = 0 $ for all $x$ belonging to $[a,b]$.




My attempt: if we take $x = a$, then we get $int_a^bf(x)dx = 0$. However this does not imply $f(x)$ is $0$.






real-analysis riemann-integration







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marked as duplicate by Martin R, Paul Frost, GNUSupporter 8964民主女神 地下教會, José Carlos Santos real-analysis
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 Dec 2 '18 at 12:32


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 2




    $begingroup$
    Amit.Differentiating(FTC): $alpha f(x) -beta f(x)=0$, for x in the interval. Can you proceed?
    $endgroup$
    – Peter Szilas
    Dec 2 '18 at 7:44










  • $begingroup$
    @PeterSzilas Thanks . I got it now .It basically involves second fundamental theorem of calculus .
    $endgroup$
    – Bhowmick
    Dec 2 '18 at 7:46












  • $begingroup$
    So $f(x) = beta - alpha$. The desired conclusion contradicts the assumption.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 2 '18 at 7:53










  • $begingroup$
    See also: How to show $alphaint_a^cf(x)dx+beta int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$
    $endgroup$
    – Martin Sleziak
    Dec 2 '18 at 9:04
















4












$begingroup$



This question already has an answer here:




  • How to show $alphaint_a^cf(x)dx+beta int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$

    3 answers





Let $f$ be a continuous function defined on $[a, b]$. Assume that there exist constants $α$ and $β$ with $(α ≠ β)$ such that
$$alphaint_a^x f(u)du + βint_x^bf(u)du = 0 $$ for all $x$ belonging to $[a,b]$. Show that $f(x) = 0 $ for all $x$ belonging to $[a,b]$.




My attempt: if we take $x = a$, then we get $int_a^bf(x)dx = 0$. However this does not imply $f(x)$ is $0$.






real-analysis riemann-integration







share|cite|improve this question











$endgroup$



marked as duplicate by Martin R, Paul Frost, GNUSupporter 8964民主女神 地下教會, José Carlos Santos real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

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 Dec 2 '18 at 12:32


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 2




    $begingroup$
    Amit.Differentiating(FTC): $alpha f(x) -beta f(x)=0$, for x in the interval. Can you proceed?
    $endgroup$
    – Peter Szilas
    Dec 2 '18 at 7:44










  • $begingroup$
    @PeterSzilas Thanks . I got it now .It basically involves second fundamental theorem of calculus .
    $endgroup$
    – Bhowmick
    Dec 2 '18 at 7:46












  • $begingroup$
    So $f(x) = beta - alpha$. The desired conclusion contradicts the assumption.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 2 '18 at 7:53










  • $begingroup$
    See also: How to show $alphaint_a^cf(x)dx+beta int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$
    $endgroup$
    – Martin Sleziak
    Dec 2 '18 at 9:04














4












4








4


2



$begingroup$



This question already has an answer here:




  • How to show $alphaint_a^cf(x)dx+beta int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$

    3 answers





Let $f$ be a continuous function defined on $[a, b]$. Assume that there exist constants $α$ and $β$ with $(α ≠ β)$ such that
$$alphaint_a^x f(u)du + βint_x^bf(u)du = 0 $$ for all $x$ belonging to $[a,b]$. Show that $f(x) = 0 $ for all $x$ belonging to $[a,b]$.




My attempt: if we take $x = a$, then we get $int_a^bf(x)dx = 0$. However this does not imply $f(x)$ is $0$.






real-analysis riemann-integration







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • How to show $alphaint_a^cf(x)dx+beta int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$

    3 answers





Let $f$ be a continuous function defined on $[a, b]$. Assume that there exist constants $α$ and $β$ with $(α ≠ β)$ such that
$$alphaint_a^x f(u)du + βint_x^bf(u)du = 0 $$ for all $x$ belonging to $[a,b]$. Show that $f(x) = 0 $ for all $x$ belonging to $[a,b]$.




My attempt: if we take $x = a$, then we get $int_a^bf(x)dx = 0$. However this does not imply $f(x)$ is $0$.





This question already has an answer here:




  • How to show $alphaint_a^cf(x)dx+beta int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$

    3 answers







real-analysis riemann-integration




real-analysis riemann-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 7:40









Jimmy R.

33k42157




33k42157










asked Dec 2 '18 at 7:33









BhowmickBhowmick

1438




1438




marked as duplicate by Martin R, Paul Frost, GNUSupporter 8964民主女神 地下教會, José Carlos Santos real-analysis
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 Dec 2 '18 at 12:32


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Martin R, Paul Frost, GNUSupporter 8964民主女神 地下教會, José Carlos Santos real-analysis
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 Dec 2 '18 at 12:32


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 2




    $begingroup$
    Amit.Differentiating(FTC): $alpha f(x) -beta f(x)=0$, for x in the interval. Can you proceed?
    $endgroup$
    – Peter Szilas
    Dec 2 '18 at 7:44










  • $begingroup$
    @PeterSzilas Thanks . I got it now .It basically involves second fundamental theorem of calculus .
    $endgroup$
    – Bhowmick
    Dec 2 '18 at 7:46












  • $begingroup$
    So $f(x) = beta - alpha$. The desired conclusion contradicts the assumption.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 2 '18 at 7:53










  • $begingroup$
    See also: How to show $alphaint_a^cf(x)dx+beta int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$
    $endgroup$
    – Martin Sleziak
    Dec 2 '18 at 9:04














  • 2




    $begingroup$
    Amit.Differentiating(FTC): $alpha f(x) -beta f(x)=0$, for x in the interval. Can you proceed?
    $endgroup$
    – Peter Szilas
    Dec 2 '18 at 7:44










  • $begingroup$
    @PeterSzilas Thanks . I got it now .It basically involves second fundamental theorem of calculus .
    $endgroup$
    – Bhowmick
    Dec 2 '18 at 7:46












  • $begingroup$
    So $f(x) = beta - alpha$. The desired conclusion contradicts the assumption.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 2 '18 at 7:53










  • $begingroup$
    See also: How to show $alphaint_a^cf(x)dx+beta int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$
    $endgroup$
    – Martin Sleziak
    Dec 2 '18 at 9:04








2




2




$begingroup$
Amit.Differentiating(FTC): $alpha f(x) -beta f(x)=0$, for x in the interval. Can you proceed?
$endgroup$
– Peter Szilas
Dec 2 '18 at 7:44




$begingroup$
Amit.Differentiating(FTC): $alpha f(x) -beta f(x)=0$, for x in the interval. Can you proceed?
$endgroup$
– Peter Szilas
Dec 2 '18 at 7:44












$begingroup$
@PeterSzilas Thanks . I got it now .It basically involves second fundamental theorem of calculus .
$endgroup$
– Bhowmick
Dec 2 '18 at 7:46






$begingroup$
@PeterSzilas Thanks . I got it now .It basically involves second fundamental theorem of calculus .
$endgroup$
– Bhowmick
Dec 2 '18 at 7:46














$begingroup$
So $f(x) = beta - alpha$. The desired conclusion contradicts the assumption.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 2 '18 at 7:53




$begingroup$
So $f(x) = beta - alpha$. The desired conclusion contradicts the assumption.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 2 '18 at 7:53












$begingroup$
See also: How to show $alphaint_a^cf(x)dx+beta int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$
$endgroup$
– Martin Sleziak
Dec 2 '18 at 9:04




$begingroup$
See also: How to show $alphaint_a^cf(x)dx+beta int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$
$endgroup$
– Martin Sleziak
Dec 2 '18 at 9:04










2 Answers
2






active

oldest

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2












$begingroup$

Hint:begin{align}frac{d}{dx}left(alphaint_a^xf(u)du+betaint_x^bf(u)duright)=(alpha -beta)f(x)end{align}






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I got the hint .It involves second fundamental theorem of calculus .Thanks
    $endgroup$
    – Bhowmick
    Dec 2 '18 at 7:47



















0












$begingroup$

Note that
begin{align*}
alpha int_a^x f + beta int_x^b f = betaint_a^b f + (alpha - beta)int_a^x f.
end{align*}
In particular, setting $x= 0$ shows that $int_a^bf = 0$. Thus $int_a^x f$ for all $xin[a, b]$. Since $f$ is continuous, it must vanish everywhere.






share|cite|improve this answer









$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Hint:begin{align}frac{d}{dx}left(alphaint_a^xf(u)du+betaint_x^bf(u)duright)=(alpha -beta)f(x)end{align}






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I got the hint .It involves second fundamental theorem of calculus .Thanks
      $endgroup$
      – Bhowmick
      Dec 2 '18 at 7:47
















    2












    $begingroup$

    Hint:begin{align}frac{d}{dx}left(alphaint_a^xf(u)du+betaint_x^bf(u)duright)=(alpha -beta)f(x)end{align}






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I got the hint .It involves second fundamental theorem of calculus .Thanks
      $endgroup$
      – Bhowmick
      Dec 2 '18 at 7:47














    2












    2








    2





    $begingroup$

    Hint:begin{align}frac{d}{dx}left(alphaint_a^xf(u)du+betaint_x^bf(u)duright)=(alpha -beta)f(x)end{align}






    share|cite|improve this answer









    $endgroup$



    Hint:begin{align}frac{d}{dx}left(alphaint_a^xf(u)du+betaint_x^bf(u)duright)=(alpha -beta)f(x)end{align}







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 7:39









    Jimmy R.Jimmy R.

    33k42157




    33k42157








    • 1




      $begingroup$
      I got the hint .It involves second fundamental theorem of calculus .Thanks
      $endgroup$
      – Bhowmick
      Dec 2 '18 at 7:47














    • 1




      $begingroup$
      I got the hint .It involves second fundamental theorem of calculus .Thanks
      $endgroup$
      – Bhowmick
      Dec 2 '18 at 7:47








    1




    1




    $begingroup$
    I got the hint .It involves second fundamental theorem of calculus .Thanks
    $endgroup$
    – Bhowmick
    Dec 2 '18 at 7:47




    $begingroup$
    I got the hint .It involves second fundamental theorem of calculus .Thanks
    $endgroup$
    – Bhowmick
    Dec 2 '18 at 7:47











    0












    $begingroup$

    Note that
    begin{align*}
    alpha int_a^x f + beta int_x^b f = betaint_a^b f + (alpha - beta)int_a^x f.
    end{align*}
    In particular, setting $x= 0$ shows that $int_a^bf = 0$. Thus $int_a^x f$ for all $xin[a, b]$. Since $f$ is continuous, it must vanish everywhere.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Note that
      begin{align*}
      alpha int_a^x f + beta int_x^b f = betaint_a^b f + (alpha - beta)int_a^x f.
      end{align*}
      In particular, setting $x= 0$ shows that $int_a^bf = 0$. Thus $int_a^x f$ for all $xin[a, b]$. Since $f$ is continuous, it must vanish everywhere.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Note that
        begin{align*}
        alpha int_a^x f + beta int_x^b f = betaint_a^b f + (alpha - beta)int_a^x f.
        end{align*}
        In particular, setting $x= 0$ shows that $int_a^bf = 0$. Thus $int_a^x f$ for all $xin[a, b]$. Since $f$ is continuous, it must vanish everywhere.






        share|cite|improve this answer









        $endgroup$



        Note that
        begin{align*}
        alpha int_a^x f + beta int_x^b f = betaint_a^b f + (alpha - beta)int_a^x f.
        end{align*}
        In particular, setting $x= 0$ shows that $int_a^bf = 0$. Thus $int_a^x f$ for all $xin[a, b]$. Since $f$ is continuous, it must vanish everywhere.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 7:50









        anomalyanomaly

        17.5k42664




        17.5k42664















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