Proving that $|x , g(y) - y , g(x)| leq C$ for $x,y in [-1,1]$, where $|g''(t)| leq C$












1














Let $f : mathbb{R}^2 rightarrow mathbb{R}^2$ given by $f(x,y) = x , g(y) - y , g(x)$, where $g:mathbb{R} rightarrow mathbb{R}$ such that $g in C^2$, $g(0)=0$ and $|g''(x)| leq C $ $forall x in mathbb{R}$. I Want to prove:



a) If $(x,y) in [-1,1]times[-1,1]$ then $|f(x,y)| leq C$.



b) Evaluate $int^1_0 int^1_0 f(x,y) , dx , dy$.



For the b part i guess that a simple application of green theorem will solve it, the value will be in function of $g$.



But, for the a part I'm getting trouble.



I tried to use the mean value theorem for the derivative of $g$ and then obtain that $|g'(x)|leq$ sup $ g^{''}$ $|x|$ on $x in [-1,1]$ and then conclude $ |g'(x)| leq C$ either.



So, applying again to $f$ I would get $|f(x,y)|$ is less or equal to a multiple of $C$.






calculus derivatives absolute-value







share|cite|improve this question
























  • Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
    – PhoemueX
    Nov 23 '18 at 19:03










  • @PhoemueX, there is no change, but I still can't see for where it is going
    – Eduardo Silva
    Nov 23 '18 at 19:20










  • I now also added an answer for part b)
    – PhoemueX
    Nov 28 '18 at 10:48
















1














Let $f : mathbb{R}^2 rightarrow mathbb{R}^2$ given by $f(x,y) = x , g(y) - y , g(x)$, where $g:mathbb{R} rightarrow mathbb{R}$ such that $g in C^2$, $g(0)=0$ and $|g''(x)| leq C $ $forall x in mathbb{R}$. I Want to prove:



a) If $(x,y) in [-1,1]times[-1,1]$ then $|f(x,y)| leq C$.



b) Evaluate $int^1_0 int^1_0 f(x,y) , dx , dy$.



For the b part i guess that a simple application of green theorem will solve it, the value will be in function of $g$.



But, for the a part I'm getting trouble.



I tried to use the mean value theorem for the derivative of $g$ and then obtain that $|g'(x)|leq$ sup $ g^{''}$ $|x|$ on $x in [-1,1]$ and then conclude $ |g'(x)| leq C$ either.



So, applying again to $f$ I would get $|f(x,y)|$ is less or equal to a multiple of $C$.






calculus derivatives absolute-value







share|cite|improve this question
























  • Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
    – PhoemueX
    Nov 23 '18 at 19:03










  • @PhoemueX, there is no change, but I still can't see for where it is going
    – Eduardo Silva
    Nov 23 '18 at 19:20










  • I now also added an answer for part b)
    – PhoemueX
    Nov 28 '18 at 10:48














1












1








1







Let $f : mathbb{R}^2 rightarrow mathbb{R}^2$ given by $f(x,y) = x , g(y) - y , g(x)$, where $g:mathbb{R} rightarrow mathbb{R}$ such that $g in C^2$, $g(0)=0$ and $|g''(x)| leq C $ $forall x in mathbb{R}$. I Want to prove:



a) If $(x,y) in [-1,1]times[-1,1]$ then $|f(x,y)| leq C$.



b) Evaluate $int^1_0 int^1_0 f(x,y) , dx , dy$.



For the b part i guess that a simple application of green theorem will solve it, the value will be in function of $g$.



But, for the a part I'm getting trouble.



I tried to use the mean value theorem for the derivative of $g$ and then obtain that $|g'(x)|leq$ sup $ g^{''}$ $|x|$ on $x in [-1,1]$ and then conclude $ |g'(x)| leq C$ either.



So, applying again to $f$ I would get $|f(x,y)|$ is less or equal to a multiple of $C$.






calculus derivatives absolute-value







share|cite|improve this question















Let $f : mathbb{R}^2 rightarrow mathbb{R}^2$ given by $f(x,y) = x , g(y) - y , g(x)$, where $g:mathbb{R} rightarrow mathbb{R}$ such that $g in C^2$, $g(0)=0$ and $|g''(x)| leq C $ $forall x in mathbb{R}$. I Want to prove:



a) If $(x,y) in [-1,1]times[-1,1]$ then $|f(x,y)| leq C$.



b) Evaluate $int^1_0 int^1_0 f(x,y) , dx , dy$.



For the b part i guess that a simple application of green theorem will solve it, the value will be in function of $g$.



But, for the a part I'm getting trouble.



I tried to use the mean value theorem for the derivative of $g$ and then obtain that $|g'(x)|leq$ sup $ g^{''}$ $|x|$ on $x in [-1,1]$ and then conclude $ |g'(x)| leq C$ either.



So, applying again to $f$ I would get $|f(x,y)|$ is less or equal to a multiple of $C$.






calculus derivatives absolute-value




calculus derivatives absolute-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 10:48









PhoemueX

27.3k22456




27.3k22456










asked Nov 23 '18 at 18:08









Eduardo Silva

68239




68239












  • Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
    – PhoemueX
    Nov 23 '18 at 19:03










  • @PhoemueX, there is no change, but I still can't see for where it is going
    – Eduardo Silva
    Nov 23 '18 at 19:20










  • I now also added an answer for part b)
    – PhoemueX
    Nov 28 '18 at 10:48


















  • Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
    – PhoemueX
    Nov 23 '18 at 19:03










  • @PhoemueX, there is no change, but I still can't see for where it is going
    – Eduardo Silva
    Nov 23 '18 at 19:20










  • I now also added an answer for part b)
    – PhoemueX
    Nov 28 '18 at 10:48
















Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
– PhoemueX
Nov 23 '18 at 19:03




Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
– PhoemueX
Nov 23 '18 at 19:03












@PhoemueX, there is no change, but I still can't see for where it is going
– Eduardo Silva
Nov 23 '18 at 19:20




@PhoemueX, there is no change, but I still can't see for where it is going
– Eduardo Silva
Nov 23 '18 at 19:20












I now also added an answer for part b)
– PhoemueX
Nov 28 '18 at 10:48




I now also added an answer for part b)
– PhoemueX
Nov 28 '18 at 10:48










1 Answer
1






active

oldest

votes


















1














a) If we change from $g$ to $widetilde{g}(x) = g(x) + a x$ for any $a in Bbb{R}$, this does not change the function $f$. Also, we will still have $widetilde{g}(0) = 0$ and $|widetilde{g}''(x)| leq C$. Therefore, we can without loss of generality assume that $g'(0) = 0$.



Now, Taylor's theorem, with the Lagrange form of the remainder shows that for each $x in [-1,1]$, there is some $xi in [0,x] cup [x,0] subset [-1,1]$ satisfying
$$
g(x) = g(0) + g'(0) cdot x + frac{g''(xi)}{2} (x - 0)^2 , ,
$$
and hence $|g(x)| leq frac{C}{2} cdot x^2 leq frac{C}{2}$ for all $x in [-1,1]$.



Now, we easily see
$$
|f(x,y)| leq |x| cdot |g(y)| + |y| cdot |g(x)| leq frac{C}{2} + frac{C}{2} = C
$$
for all $x,y in [-1,1]$.



Final comment: This is more or less what you were also trying to do (using the estimate for $g''$ to estimate $g'$, and then $g$). The main points was to see that one can assume $g'(x) = 0$. Finally, instead of using the mean-value theorem twice, we used Taylor's theorem, which got rid of the factor two that you would otherwise get if you naively apply the mean-value theorem.



b) This part of the question is actually pretty easy; one does not need anything fancy like Green's theorem. Simply note that if we set $a := int_0^1 t , d t$ (so that $a = 1/2$) and $b := int_0^1 g(t) , dt$, then
begin{align*}
int_0^1 int_0^1 f(x,y) , d x , d y
& = int_0^1 int_0^1 x , g(y) , d x , d y
- int_0^1 int_0^1 y , g(x) , d x , d y \
& = int_0^1 g(y) cdot a , d y
- int_0^1 y , b , d y \
& = a cdot b - a cdot b
= 0 , .
end{align*}






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%2f3010661%2fproving-that-x-gy-y-gx-leq-c-for-x-y-in-1-1-where-g%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    a) If we change from $g$ to $widetilde{g}(x) = g(x) + a x$ for any $a in Bbb{R}$, this does not change the function $f$. Also, we will still have $widetilde{g}(0) = 0$ and $|widetilde{g}''(x)| leq C$. Therefore, we can without loss of generality assume that $g'(0) = 0$.



    Now, Taylor's theorem, with the Lagrange form of the remainder shows that for each $x in [-1,1]$, there is some $xi in [0,x] cup [x,0] subset [-1,1]$ satisfying
    $$
    g(x) = g(0) + g'(0) cdot x + frac{g''(xi)}{2} (x - 0)^2 , ,
    $$
    and hence $|g(x)| leq frac{C}{2} cdot x^2 leq frac{C}{2}$ for all $x in [-1,1]$.



    Now, we easily see
    $$
    |f(x,y)| leq |x| cdot |g(y)| + |y| cdot |g(x)| leq frac{C}{2} + frac{C}{2} = C
    $$
    for all $x,y in [-1,1]$.



    Final comment: This is more or less what you were also trying to do (using the estimate for $g''$ to estimate $g'$, and then $g$). The main points was to see that one can assume $g'(x) = 0$. Finally, instead of using the mean-value theorem twice, we used Taylor's theorem, which got rid of the factor two that you would otherwise get if you naively apply the mean-value theorem.



    b) This part of the question is actually pretty easy; one does not need anything fancy like Green's theorem. Simply note that if we set $a := int_0^1 t , d t$ (so that $a = 1/2$) and $b := int_0^1 g(t) , dt$, then
    begin{align*}
    int_0^1 int_0^1 f(x,y) , d x , d y
    & = int_0^1 int_0^1 x , g(y) , d x , d y
    - int_0^1 int_0^1 y , g(x) , d x , d y \
    & = int_0^1 g(y) cdot a , d y
    - int_0^1 y , b , d y \
    & = a cdot b - a cdot b
    = 0 , .
    end{align*}






    share|cite|improve this answer




























      1














      a) If we change from $g$ to $widetilde{g}(x) = g(x) + a x$ for any $a in Bbb{R}$, this does not change the function $f$. Also, we will still have $widetilde{g}(0) = 0$ and $|widetilde{g}''(x)| leq C$. Therefore, we can without loss of generality assume that $g'(0) = 0$.



      Now, Taylor's theorem, with the Lagrange form of the remainder shows that for each $x in [-1,1]$, there is some $xi in [0,x] cup [x,0] subset [-1,1]$ satisfying
      $$
      g(x) = g(0) + g'(0) cdot x + frac{g''(xi)}{2} (x - 0)^2 , ,
      $$
      and hence $|g(x)| leq frac{C}{2} cdot x^2 leq frac{C}{2}$ for all $x in [-1,1]$.



      Now, we easily see
      $$
      |f(x,y)| leq |x| cdot |g(y)| + |y| cdot |g(x)| leq frac{C}{2} + frac{C}{2} = C
      $$
      for all $x,y in [-1,1]$.



      Final comment: This is more or less what you were also trying to do (using the estimate for $g''$ to estimate $g'$, and then $g$). The main points was to see that one can assume $g'(x) = 0$. Finally, instead of using the mean-value theorem twice, we used Taylor's theorem, which got rid of the factor two that you would otherwise get if you naively apply the mean-value theorem.



      b) This part of the question is actually pretty easy; one does not need anything fancy like Green's theorem. Simply note that if we set $a := int_0^1 t , d t$ (so that $a = 1/2$) and $b := int_0^1 g(t) , dt$, then
      begin{align*}
      int_0^1 int_0^1 f(x,y) , d x , d y
      & = int_0^1 int_0^1 x , g(y) , d x , d y
      - int_0^1 int_0^1 y , g(x) , d x , d y \
      & = int_0^1 g(y) cdot a , d y
      - int_0^1 y , b , d y \
      & = a cdot b - a cdot b
      = 0 , .
      end{align*}






      share|cite|improve this answer


























        1












        1








        1






        a) If we change from $g$ to $widetilde{g}(x) = g(x) + a x$ for any $a in Bbb{R}$, this does not change the function $f$. Also, we will still have $widetilde{g}(0) = 0$ and $|widetilde{g}''(x)| leq C$. Therefore, we can without loss of generality assume that $g'(0) = 0$.



        Now, Taylor's theorem, with the Lagrange form of the remainder shows that for each $x in [-1,1]$, there is some $xi in [0,x] cup [x,0] subset [-1,1]$ satisfying
        $$
        g(x) = g(0) + g'(0) cdot x + frac{g''(xi)}{2} (x - 0)^2 , ,
        $$
        and hence $|g(x)| leq frac{C}{2} cdot x^2 leq frac{C}{2}$ for all $x in [-1,1]$.



        Now, we easily see
        $$
        |f(x,y)| leq |x| cdot |g(y)| + |y| cdot |g(x)| leq frac{C}{2} + frac{C}{2} = C
        $$
        for all $x,y in [-1,1]$.



        Final comment: This is more or less what you were also trying to do (using the estimate for $g''$ to estimate $g'$, and then $g$). The main points was to see that one can assume $g'(x) = 0$. Finally, instead of using the mean-value theorem twice, we used Taylor's theorem, which got rid of the factor two that you would otherwise get if you naively apply the mean-value theorem.



        b) This part of the question is actually pretty easy; one does not need anything fancy like Green's theorem. Simply note that if we set $a := int_0^1 t , d t$ (so that $a = 1/2$) and $b := int_0^1 g(t) , dt$, then
        begin{align*}
        int_0^1 int_0^1 f(x,y) , d x , d y
        & = int_0^1 int_0^1 x , g(y) , d x , d y
        - int_0^1 int_0^1 y , g(x) , d x , d y \
        & = int_0^1 g(y) cdot a , d y
        - int_0^1 y , b , d y \
        & = a cdot b - a cdot b
        = 0 , .
        end{align*}






        share|cite|improve this answer














        a) If we change from $g$ to $widetilde{g}(x) = g(x) + a x$ for any $a in Bbb{R}$, this does not change the function $f$. Also, we will still have $widetilde{g}(0) = 0$ and $|widetilde{g}''(x)| leq C$. Therefore, we can without loss of generality assume that $g'(0) = 0$.



        Now, Taylor's theorem, with the Lagrange form of the remainder shows that for each $x in [-1,1]$, there is some $xi in [0,x] cup [x,0] subset [-1,1]$ satisfying
        $$
        g(x) = g(0) + g'(0) cdot x + frac{g''(xi)}{2} (x - 0)^2 , ,
        $$
        and hence $|g(x)| leq frac{C}{2} cdot x^2 leq frac{C}{2}$ for all $x in [-1,1]$.



        Now, we easily see
        $$
        |f(x,y)| leq |x| cdot |g(y)| + |y| cdot |g(x)| leq frac{C}{2} + frac{C}{2} = C
        $$
        for all $x,y in [-1,1]$.



        Final comment: This is more or less what you were also trying to do (using the estimate for $g''$ to estimate $g'$, and then $g$). The main points was to see that one can assume $g'(x) = 0$. Finally, instead of using the mean-value theorem twice, we used Taylor's theorem, which got rid of the factor two that you would otherwise get if you naively apply the mean-value theorem.



        b) This part of the question is actually pretty easy; one does not need anything fancy like Green's theorem. Simply note that if we set $a := int_0^1 t , d t$ (so that $a = 1/2$) and $b := int_0^1 g(t) , dt$, then
        begin{align*}
        int_0^1 int_0^1 f(x,y) , d x , d y
        & = int_0^1 int_0^1 x , g(y) , d x , d y
        - int_0^1 int_0^1 y , g(x) , d x , d y \
        & = int_0^1 g(y) cdot a , d y
        - int_0^1 y , b , d y \
        & = a cdot b - a cdot b
        = 0 , .
        end{align*}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 28 '18 at 10:44

























        answered Nov 25 '18 at 18:17









        PhoemueX

        27.3k22456




        27.3k22456






























            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%2f3010661%2fproving-that-x-gy-y-gx-leq-c-for-x-y-in-1-1-where-g%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

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück ...
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
            Read more