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












0














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




















  • 1




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
















0














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




















  • 1




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














0












0








0







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















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

























asked Nov 28 '18 at 21:46









Albert Diaz

925




925








  • 1




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














  • 1




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








1




1




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




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










2 Answers
2






active

oldest

votes


















3














Hint:



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






share|cite|improve this answer

















  • 1




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












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



















2














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





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017765%2fsuppose-that-f-and-g-are-continuous-functions-on-a-b-prove-int-abfg%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    Hint:



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






    share|cite|improve this answer

















    • 1




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












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
















    3














    Hint:



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






    share|cite|improve this answer

















    • 1




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












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














    3












    3








    3






    Hint:



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






    share|cite|improve this answer












    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 Gunn

    14.2k32046




    14.2k32046








    • 1




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












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














    • 1




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












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








    1




    1




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






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














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




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











    2














    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


























      2














      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
























        2












        2








        2






        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












        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 Szilas

        10.7k2720




        10.7k2720






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017765%2fsuppose-that-f-and-g-are-continuous-functions-on-a-b-prove-int-abfg%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Index of /

            This page is only for reference, If you need detailed information, please check here
            Read more

            Tribalistas

            Image
              Nota: Para o álbum homônimo, veja Tribalistas (álbum de 2002). Para o segundo álbum homônimo, veja Tribalistas (álbum de 2017). Para o tipo de organização tribal, veja Tribalismo. Tribalistas Informação geral Origem Salvador e Rio de Janeiro País   Brasil Gênero(s) MPB Pop Samba Período em atividade 2002–2004 2017–atualmente Gravadora(s) EMI Phonomotor Records Universal Music Afiliação(ões) Dadi Pretinho da Serrinha Margareth Menezes Integrantes Arnaldo Antunes Carlinhos Brown Marisa Monte Página oficial www2.uol.com.br/tribalistas/ ...
            Read more

            Listed building

            Image
            Listed building (literalmente: prédio listado ) é toda estrutura protegida por lei pelo complexo sistema de tombamento do Reino Unido. Um listed building é qualquer estrutura oficialmente designada como tendo especial interesse arquitetônico, histórico ou cultural. É um estatuto amplamente usado, aplicado a centenas de milhares de estruturas. Estruturas tombadas não podem ser alteradas sem permissão especial, e desfrutam de isenções fiscais e outros benefícios. Cada um dos países do Reino Unido trata tombamento de forma diferente: Índice 1 Inglaterra 2 País de Gales 3 Irlanda do Norte 4 Escócia 5 Referências Inglaterra | O Palácio de Buckingham, Londres. As estruturas tombadas são listadas no Statutory List of Buildings of Special Architectural or Historic Interest , e incluem campos, sinais de trânsito, piers e áreas geográficas. A principal legislação pertinente é o Ancient Monuments and Archaeological Areas Act 1979 , [ 1 ] embora legisla...
            Read more