Proving an inequality related to Riemann integration












0












$begingroup$


Problem




Let $f$ be a function such that $|f(u)-f(v)| leq |u-v|$.
Assume that $f$ is integrable on $[a,b]$.
Prove that
$$left|int_{a}^{b} f(x)dx - (b-a)f(c)right| leq (b-a)^2/2$$
where $cin [a,b]$




Attempt



It is easy to prove that $f$ is continuous on $[a,b]$ (uniformly).



Now since $f$ is continuous on a compact set, it attains its extremum in that set. So it is bounded. Say $|f(x)| leq M$ where $M=f(c')$ for some $cin [a,b]$.
Now,
$$left|int_{a}^{b} f(x)dx - (b-a)f(c)right| leq |(f(c')-f(c))(b-a)| leq (b-a)^2.$$
Why I am missing $1/2$?



Thanks in advance.






real-analysis calculus







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Problem




    Let $f$ be a function such that $|f(u)-f(v)| leq |u-v|$.
    Assume that $f$ is integrable on $[a,b]$.
    Prove that
    $$left|int_{a}^{b} f(x)dx - (b-a)f(c)right| leq (b-a)^2/2$$
    where $cin [a,b]$




    Attempt



    It is easy to prove that $f$ is continuous on $[a,b]$ (uniformly).



    Now since $f$ is continuous on a compact set, it attains its extremum in that set. So it is bounded. Say $|f(x)| leq M$ where $M=f(c')$ for some $cin [a,b]$.
    Now,
    $$left|int_{a}^{b} f(x)dx - (b-a)f(c)right| leq |(f(c')-f(c))(b-a)| leq (b-a)^2.$$
    Why I am missing $1/2$?



    Thanks in advance.






    real-analysis calculus







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Problem




      Let $f$ be a function such that $|f(u)-f(v)| leq |u-v|$.
      Assume that $f$ is integrable on $[a,b]$.
      Prove that
      $$left|int_{a}^{b} f(x)dx - (b-a)f(c)right| leq (b-a)^2/2$$
      where $cin [a,b]$




      Attempt



      It is easy to prove that $f$ is continuous on $[a,b]$ (uniformly).



      Now since $f$ is continuous on a compact set, it attains its extremum in that set. So it is bounded. Say $|f(x)| leq M$ where $M=f(c')$ for some $cin [a,b]$.
      Now,
      $$left|int_{a}^{b} f(x)dx - (b-a)f(c)right| leq |(f(c')-f(c))(b-a)| leq (b-a)^2.$$
      Why I am missing $1/2$?



      Thanks in advance.






      real-analysis calculus







      share|cite|improve this question











      $endgroup$




      Problem




      Let $f$ be a function such that $|f(u)-f(v)| leq |u-v|$.
      Assume that $f$ is integrable on $[a,b]$.
      Prove that
      $$left|int_{a}^{b} f(x)dx - (b-a)f(c)right| leq (b-a)^2/2$$
      where $cin [a,b]$




      Attempt



      It is easy to prove that $f$ is continuous on $[a,b]$ (uniformly).



      Now since $f$ is continuous on a compact set, it attains its extremum in that set. So it is bounded. Say $|f(x)| leq M$ where $M=f(c')$ for some $cin [a,b]$.
      Now,
      $$left|int_{a}^{b} f(x)dx - (b-a)f(c)right| leq |(f(c')-f(c))(b-a)| leq (b-a)^2.$$
      Why I am missing $1/2$?



      Thanks in advance.






      real-analysis calculus




      real-analysis calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 14 '18 at 10:06









      Robert Z

      94.5k1063134




      94.5k1063134










      asked Dec 14 '18 at 9:51









      blue boyblue boy

      1,229613




      1,229613






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          You don't need the maximum of $f$. For all $c in [a, b]$
          $$
          left|int_{a}^{b} f(x) , dx - (b-a)f(c) right| =
          left|int_{a}^{b} (f(x)-f(c)) , dx right| \le
          int_{a}^{b} | f(x) - f(c) | , dx le
          int_{a}^{b} | x-c | , dx \
          = int_{a}^{c} (c-x) , dx + int_{c}^{b} (x-c) , dx \
          = frac{(c-a)^2}{2} + frac{(b-c)^2}{2} \ le frac{(b-a)^2}{2}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How did you get the last inequality?
            $endgroup$
            – blue boy
            Dec 14 '18 at 10:13






          • 1




            $begingroup$
            @blueboy: $x^2 + y^2 le (x+y)^2$ for nonnegative numbers. You can also see it geometrically if you draw the triangles corresponding to these areas.
            $endgroup$
            – Martin R
            Dec 14 '18 at 10:15










          • $begingroup$
            Oh. Thanks. When i used the maximum it wasn't using the required bound.
            $endgroup$
            – blue boy
            Dec 14 '18 at 10:17



















          0












          $begingroup$

          Let $c=frac {a+b} 2$. Note that $int_a^{b} [f(x)-f(c)]dx=int_a^{b} f(x)dx-f(c)(b-a)$. Hence $|int_a^{b} f(x)dx-f(c)(b-a)| leq int_a^{b} |f(x)-f(c)|dx leq int_a^{b} |x-c|dx$. calculate this last integal explicitly by making the change of variable $y=x-c$ and you will get the desired inequality.






          share|cite|improve this answer









          $endgroup$













            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%2f3039166%2fproving-an-inequality-related-to-riemann-integration%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









            2












            $begingroup$

            You don't need the maximum of $f$. For all $c in [a, b]$
            $$
            left|int_{a}^{b} f(x) , dx - (b-a)f(c) right| =
            left|int_{a}^{b} (f(x)-f(c)) , dx right| \le
            int_{a}^{b} | f(x) - f(c) | , dx le
            int_{a}^{b} | x-c | , dx \
            = int_{a}^{c} (c-x) , dx + int_{c}^{b} (x-c) , dx \
            = frac{(c-a)^2}{2} + frac{(b-c)^2}{2} \ le frac{(b-a)^2}{2}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How did you get the last inequality?
              $endgroup$
              – blue boy
              Dec 14 '18 at 10:13






            • 1




              $begingroup$
              @blueboy: $x^2 + y^2 le (x+y)^2$ for nonnegative numbers. You can also see it geometrically if you draw the triangles corresponding to these areas.
              $endgroup$
              – Martin R
              Dec 14 '18 at 10:15










            • $begingroup$
              Oh. Thanks. When i used the maximum it wasn't using the required bound.
              $endgroup$
              – blue boy
              Dec 14 '18 at 10:17
















            2












            $begingroup$

            You don't need the maximum of $f$. For all $c in [a, b]$
            $$
            left|int_{a}^{b} f(x) , dx - (b-a)f(c) right| =
            left|int_{a}^{b} (f(x)-f(c)) , dx right| \le
            int_{a}^{b} | f(x) - f(c) | , dx le
            int_{a}^{b} | x-c | , dx \
            = int_{a}^{c} (c-x) , dx + int_{c}^{b} (x-c) , dx \
            = frac{(c-a)^2}{2} + frac{(b-c)^2}{2} \ le frac{(b-a)^2}{2}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How did you get the last inequality?
              $endgroup$
              – blue boy
              Dec 14 '18 at 10:13






            • 1




              $begingroup$
              @blueboy: $x^2 + y^2 le (x+y)^2$ for nonnegative numbers. You can also see it geometrically if you draw the triangles corresponding to these areas.
              $endgroup$
              – Martin R
              Dec 14 '18 at 10:15










            • $begingroup$
              Oh. Thanks. When i used the maximum it wasn't using the required bound.
              $endgroup$
              – blue boy
              Dec 14 '18 at 10:17














            2












            2








            2





            $begingroup$

            You don't need the maximum of $f$. For all $c in [a, b]$
            $$
            left|int_{a}^{b} f(x) , dx - (b-a)f(c) right| =
            left|int_{a}^{b} (f(x)-f(c)) , dx right| \le
            int_{a}^{b} | f(x) - f(c) | , dx le
            int_{a}^{b} | x-c | , dx \
            = int_{a}^{c} (c-x) , dx + int_{c}^{b} (x-c) , dx \
            = frac{(c-a)^2}{2} + frac{(b-c)^2}{2} \ le frac{(b-a)^2}{2}
            $$






            share|cite|improve this answer









            $endgroup$



            You don't need the maximum of $f$. For all $c in [a, b]$
            $$
            left|int_{a}^{b} f(x) , dx - (b-a)f(c) right| =
            left|int_{a}^{b} (f(x)-f(c)) , dx right| \le
            int_{a}^{b} | f(x) - f(c) | , dx le
            int_{a}^{b} | x-c | , dx \
            = int_{a}^{c} (c-x) , dx + int_{c}^{b} (x-c) , dx \
            = frac{(c-a)^2}{2} + frac{(b-c)^2}{2} \ le frac{(b-a)^2}{2}
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 14 '18 at 10:03









            Martin RMartin R

            27.5k33255




            27.5k33255












            • $begingroup$
              How did you get the last inequality?
              $endgroup$
              – blue boy
              Dec 14 '18 at 10:13






            • 1




              $begingroup$
              @blueboy: $x^2 + y^2 le (x+y)^2$ for nonnegative numbers. You can also see it geometrically if you draw the triangles corresponding to these areas.
              $endgroup$
              – Martin R
              Dec 14 '18 at 10:15










            • $begingroup$
              Oh. Thanks. When i used the maximum it wasn't using the required bound.
              $endgroup$
              – blue boy
              Dec 14 '18 at 10:17


















            • $begingroup$
              How did you get the last inequality?
              $endgroup$
              – blue boy
              Dec 14 '18 at 10:13






            • 1




              $begingroup$
              @blueboy: $x^2 + y^2 le (x+y)^2$ for nonnegative numbers. You can also see it geometrically if you draw the triangles corresponding to these areas.
              $endgroup$
              – Martin R
              Dec 14 '18 at 10:15










            • $begingroup$
              Oh. Thanks. When i used the maximum it wasn't using the required bound.
              $endgroup$
              – blue boy
              Dec 14 '18 at 10:17
















            $begingroup$
            How did you get the last inequality?
            $endgroup$
            – blue boy
            Dec 14 '18 at 10:13




            $begingroup$
            How did you get the last inequality?
            $endgroup$
            – blue boy
            Dec 14 '18 at 10:13




            1




            1




            $begingroup$
            @blueboy: $x^2 + y^2 le (x+y)^2$ for nonnegative numbers. You can also see it geometrically if you draw the triangles corresponding to these areas.
            $endgroup$
            – Martin R
            Dec 14 '18 at 10:15




            $begingroup$
            @blueboy: $x^2 + y^2 le (x+y)^2$ for nonnegative numbers. You can also see it geometrically if you draw the triangles corresponding to these areas.
            $endgroup$
            – Martin R
            Dec 14 '18 at 10:15












            $begingroup$
            Oh. Thanks. When i used the maximum it wasn't using the required bound.
            $endgroup$
            – blue boy
            Dec 14 '18 at 10:17




            $begingroup$
            Oh. Thanks. When i used the maximum it wasn't using the required bound.
            $endgroup$
            – blue boy
            Dec 14 '18 at 10:17











            0












            $begingroup$

            Let $c=frac {a+b} 2$. Note that $int_a^{b} [f(x)-f(c)]dx=int_a^{b} f(x)dx-f(c)(b-a)$. Hence $|int_a^{b} f(x)dx-f(c)(b-a)| leq int_a^{b} |f(x)-f(c)|dx leq int_a^{b} |x-c|dx$. calculate this last integal explicitly by making the change of variable $y=x-c$ and you will get the desired inequality.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let $c=frac {a+b} 2$. Note that $int_a^{b} [f(x)-f(c)]dx=int_a^{b} f(x)dx-f(c)(b-a)$. Hence $|int_a^{b} f(x)dx-f(c)(b-a)| leq int_a^{b} |f(x)-f(c)|dx leq int_a^{b} |x-c|dx$. calculate this last integal explicitly by making the change of variable $y=x-c$ and you will get the desired inequality.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let $c=frac {a+b} 2$. Note that $int_a^{b} [f(x)-f(c)]dx=int_a^{b} f(x)dx-f(c)(b-a)$. Hence $|int_a^{b} f(x)dx-f(c)(b-a)| leq int_a^{b} |f(x)-f(c)|dx leq int_a^{b} |x-c|dx$. calculate this last integal explicitly by making the change of variable $y=x-c$ and you will get the desired inequality.






                share|cite|improve this answer









                $endgroup$



                Let $c=frac {a+b} 2$. Note that $int_a^{b} [f(x)-f(c)]dx=int_a^{b} f(x)dx-f(c)(b-a)$. Hence $|int_a^{b} f(x)dx-f(c)(b-a)| leq int_a^{b} |f(x)-f(c)|dx leq int_a^{b} |x-c|dx$. calculate this last integal explicitly by making the change of variable $y=x-c$ and you will get the desired inequality.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 14 '18 at 9:59









                Kavi Rama MurthyKavi Rama Murthy

                52.9k32055




                52.9k32055






























                    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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039166%2fproving-an-inequality-related-to-riemann-integration%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

                    Måne

                    Image
                    För jordens måne, se Månen. Några månar i solsystemet i skala till vår måne En måne , naturlig satellit eller drabant är en himlakropp som kretsar kring en planet eller en asteroid i ett solsystem. Ordet i bestämd form – månen – avser planeten jordens måne. Det finns inte i något fall belagt att någon måne har egna månar. De flesta stora månar i solsystemet är bundna i synkron rotation med en och samma sida vänd mot planeten. Ett exempel på en måne som inte beter sig på detta vis, är Hyperion som kaotiskt roterar kring Saturnus. En asteroidmåne är en måne som kretsar kring en asteroid. En måne kan ha bildats samtidigt som moderplaneten eller senare infångats av planetens gravitationskraft. I vårt eget solsystem känner vi i nuläget till 173 månar runt sex av de åtta planeterna [ 1 ] : Jorden – månen Mars – Phobos och Deimos Jupiter – Ganymedes (som är solsystemets största måne), Io, Europa, Callisto och 59 andra kända månar Saturnus – Titan (den enda måne...
                    Read more

                    Storängen

                    Image
                    För stadsdelen i Mariehamn, se Storängen, Åland. För ett naturreservat i Norrtälje kommun, se Storängens naturreservat. Storängen från luften i september 1967, vy mot väst. Till vänster syns Järlasjön och Saltsjöbanans spår. Till höger löper Värmdövägen och längst ut till höger syns anläggningsarbetena för Värmdöleden. Storängen är ett villaområde i centrala Nacka i Stockholms län. Området är beläget strax sydost om Nacka Forum vid Saltsjöbanan mellan Lillängen och Saltsjö-Duvnäs. Storängens villasamhälle grundades 1904 av "Tjänstemännens Egnahemsförening vid Storängen" och de första villorna började byggas samma år. Redan 1909 var 107 tomter bebyggda av totalt runt 160. [ 1 ] År 2013 fanns drygt 150 villor med lika många familjer i området. Egnahemsföreningen existerar fortfarande och fungerar idag som intresseförening för de boende i Storängen. Storängen anses vara ett av Sveriges bäst bevarade villasamhällen från 1900-talets början, tack vare sina tidstypis...
                    Read more

                    VLT Carioca

                    Image
                    VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
                    Read more