Suppose that $f$ and $g$ are continuous functions on $[a,b]$, prove...












0












$begingroup$


Not sure what to use here, would like some help. Thank you. These are Riemann integrals also.



My finished proof:



Since $f$ and $g$ are both continuous on $[a,b]$ then we have that $f$ and $g$ are both integrable on $[a,b]$. Using the proofs that $f^2,g^2,-2fg$ are all integrable on $[a,b]$ we have that $(f-g)^2$ is integrable on $[a,b]$ and that $(f-g)^2ge0$ implies:



$int_a^b(f-g)^2=int_a^bf^2+g^2-2fg=int_a^bf^2+int_a^bg^2-int_a^b2fgge0$



therefore,



$int_a^bf^2+int_a^bg^2geint_a^b2fg$



hence,



$frac{1}{2}(int_a^bf^2+int_a^bg^2)geint_a^bfg$



∎






real-analysis integration







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$endgroup$








  • 1




    $begingroup$
    This is a version of what's called the "AM-GM" inequality, in case you didn't know.
    $endgroup$
    – rubikscube09
    Nov 28 '18 at 23:07
















0












$begingroup$


Not sure what to use here, would like some help. Thank you. These are Riemann integrals also.



My finished proof:



Since $f$ and $g$ are both continuous on $[a,b]$ then we have that $f$ and $g$ are both integrable on $[a,b]$. Using the proofs that $f^2,g^2,-2fg$ are all integrable on $[a,b]$ we have that $(f-g)^2$ is integrable on $[a,b]$ and that $(f-g)^2ge0$ implies:



$int_a^b(f-g)^2=int_a^bf^2+g^2-2fg=int_a^bf^2+int_a^bg^2-int_a^b2fgge0$



therefore,



$int_a^bf^2+int_a^bg^2geint_a^b2fg$



hence,



$frac{1}{2}(int_a^bf^2+int_a^bg^2)geint_a^bfg$



∎






real-analysis integration







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is a version of what's called the "AM-GM" inequality, in case you didn't know.
    $endgroup$
    – rubikscube09
    Nov 28 '18 at 23:07














0












0








0





$begingroup$


Not sure what to use here, would like some help. Thank you. These are Riemann integrals also.



My finished proof:



Since $f$ and $g$ are both continuous on $[a,b]$ then we have that $f$ and $g$ are both integrable on $[a,b]$. Using the proofs that $f^2,g^2,-2fg$ are all integrable on $[a,b]$ we have that $(f-g)^2$ is integrable on $[a,b]$ and that $(f-g)^2ge0$ implies:



$int_a^b(f-g)^2=int_a^bf^2+g^2-2fg=int_a^bf^2+int_a^bg^2-int_a^b2fgge0$



therefore,



$int_a^bf^2+int_a^bg^2geint_a^b2fg$



hence,



$frac{1}{2}(int_a^bf^2+int_a^bg^2)geint_a^bfg$



∎






real-analysis integration







share|cite|improve this question











$endgroup$




Not sure what to use here, would like some help. Thank you. These are Riemann integrals also.



My finished proof:



Since $f$ and $g$ are both continuous on $[a,b]$ then we have that $f$ and $g$ are both integrable on $[a,b]$. Using the proofs that $f^2,g^2,-2fg$ are all integrable on $[a,b]$ we have that $(f-g)^2$ is integrable on $[a,b]$ and that $(f-g)^2ge0$ implies:



$int_a^b(f-g)^2=int_a^bf^2+g^2-2fg=int_a^bf^2+int_a^bg^2-int_a^b2fgge0$



therefore,



$int_a^bf^2+int_a^bg^2geint_a^b2fg$



hence,



$frac{1}{2}(int_a^bf^2+int_a^bg^2)geint_a^bfg$



∎






real-analysis integration




real-analysis integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 3:13







Albert Diaz

















asked Nov 28 '18 at 21:46









Albert DiazAlbert Diaz

925




925








  • 1




    $begingroup$
    This is a version of what's called the "AM-GM" inequality, in case you didn't know.
    $endgroup$
    – rubikscube09
    Nov 28 '18 at 23:07














  • 1




    $begingroup$
    This is a version of what's called the "AM-GM" inequality, in case you didn't know.
    $endgroup$
    – rubikscube09
    Nov 28 '18 at 23:07








1




1




$begingroup$
This is a version of what's called the "AM-GM" inequality, in case you didn't know.
$endgroup$
– rubikscube09
Nov 28 '18 at 23:07




$begingroup$
This is a version of what's called the "AM-GM" inequality, in case you didn't know.
$endgroup$
– rubikscube09
Nov 28 '18 at 23:07










2 Answers
2






active

oldest

votes


















3












$begingroup$

Hint:



$$ (a - b)^2 ge 0 $$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    ah, would this be a hint towards the polynomial $f^2+g^2+2fg$? seems really obvious now
    $endgroup$
    – Albert Diaz
    Nov 28 '18 at 21:50












  • $begingroup$
    @Albert $-2fg$ but yes.
    $endgroup$
    – Trevor Gunn
    Nov 28 '18 at 22:04



















2












$begingroup$

Hint:



$(f(x) pm g(x))^2 =f^2(x)+g^2(x) pm 2f(x)g(x) ge 0.$



$f^2(x)+g^2(x) ge 2 |f(x)g(x)|$.



Hence



$f(x)g(x) le |f(x)g(x)| le$



$ 2|f(x)g(x)| le f^2(x)+g(x)^2.$






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Hint:



    $$ (a - b)^2 ge 0 $$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      ah, would this be a hint towards the polynomial $f^2+g^2+2fg$? seems really obvious now
      $endgroup$
      – Albert Diaz
      Nov 28 '18 at 21:50












    • $begingroup$
      @Albert $-2fg$ but yes.
      $endgroup$
      – Trevor Gunn
      Nov 28 '18 at 22:04
















    3












    $begingroup$

    Hint:



    $$ (a - b)^2 ge 0 $$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      ah, would this be a hint towards the polynomial $f^2+g^2+2fg$? seems really obvious now
      $endgroup$
      – Albert Diaz
      Nov 28 '18 at 21:50












    • $begingroup$
      @Albert $-2fg$ but yes.
      $endgroup$
      – Trevor Gunn
      Nov 28 '18 at 22:04














    3












    3








    3





    $begingroup$

    Hint:



    $$ (a - b)^2 ge 0 $$






    share|cite|improve this answer









    $endgroup$



    Hint:



    $$ (a - b)^2 ge 0 $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 28 '18 at 21:49









    Trevor GunnTrevor Gunn

    14.5k32046




    14.5k32046








    • 1




      $begingroup$
      ah, would this be a hint towards the polynomial $f^2+g^2+2fg$? seems really obvious now
      $endgroup$
      – Albert Diaz
      Nov 28 '18 at 21:50












    • $begingroup$
      @Albert $-2fg$ but yes.
      $endgroup$
      – Trevor Gunn
      Nov 28 '18 at 22:04














    • 1




      $begingroup$
      ah, would this be a hint towards the polynomial $f^2+g^2+2fg$? seems really obvious now
      $endgroup$
      – Albert Diaz
      Nov 28 '18 at 21:50












    • $begingroup$
      @Albert $-2fg$ but yes.
      $endgroup$
      – Trevor Gunn
      Nov 28 '18 at 22:04








    1




    1




    $begingroup$
    ah, would this be a hint towards the polynomial $f^2+g^2+2fg$? seems really obvious now
    $endgroup$
    – Albert Diaz
    Nov 28 '18 at 21:50






    $begingroup$
    ah, would this be a hint towards the polynomial $f^2+g^2+2fg$? seems really obvious now
    $endgroup$
    – Albert Diaz
    Nov 28 '18 at 21:50














    $begingroup$
    @Albert $-2fg$ but yes.
    $endgroup$
    – Trevor Gunn
    Nov 28 '18 at 22:04




    $begingroup$
    @Albert $-2fg$ but yes.
    $endgroup$
    – Trevor Gunn
    Nov 28 '18 at 22:04











    2












    $begingroup$

    Hint:



    $(f(x) pm g(x))^2 =f^2(x)+g^2(x) pm 2f(x)g(x) ge 0.$



    $f^2(x)+g^2(x) ge 2 |f(x)g(x)|$.



    Hence



    $f(x)g(x) le |f(x)g(x)| le$



    $ 2|f(x)g(x)| le f^2(x)+g(x)^2.$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint:



      $(f(x) pm g(x))^2 =f^2(x)+g^2(x) pm 2f(x)g(x) ge 0.$



      $f^2(x)+g^2(x) ge 2 |f(x)g(x)|$.



      Hence



      $f(x)g(x) le |f(x)g(x)| le$



      $ 2|f(x)g(x)| le f^2(x)+g(x)^2.$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint:



        $(f(x) pm g(x))^2 =f^2(x)+g^2(x) pm 2f(x)g(x) ge 0.$



        $f^2(x)+g^2(x) ge 2 |f(x)g(x)|$.



        Hence



        $f(x)g(x) le |f(x)g(x)| le$



        $ 2|f(x)g(x)| le f^2(x)+g(x)^2.$






        share|cite|improve this answer









        $endgroup$



        Hint:



        $(f(x) pm g(x))^2 =f^2(x)+g^2(x) pm 2f(x)g(x) ge 0.$



        $f^2(x)+g^2(x) ge 2 |f(x)g(x)|$.



        Hence



        $f(x)g(x) le |f(x)g(x)| le$



        $ 2|f(x)g(x)| le f^2(x)+g(x)^2.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 21:58









        Peter SzilasPeter Szilas

        11.1k2821




        11.1k2821






























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