Showing that the function $f(x,y)=xsin y+ycos x$ is Lipschitz












1












$begingroup$


I wanted to show that $f(x,y)=xsin y+ycos x$ sastisfy Lipschitz conditions. but I can't separate it to $L|y_1-y_2|$.




According to my lecturer, the Lipschitz condition should be $$|f(x,y_1)-f(x,y_2)|le L|y2-y1|$$
I was able show that $x^2+y^2$ in the rectangle $|x|le a$, $|y|le b$ satisfies the Lipschitz condition, with my $L=2b$. But I had problem showing this for $f(x,y)=xsin y+ycos x $.






real-analysis lipschitz-functions







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Do you want Lipschitz in $y$ or Lipschitz in both variables?
    $endgroup$
    – Jose27
    Apr 4 '15 at 5:51










  • $begingroup$
    according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
    $endgroup$
    – James Anyanwu
    Apr 4 '15 at 6:52
















1












$begingroup$


I wanted to show that $f(x,y)=xsin y+ycos x$ sastisfy Lipschitz conditions. but I can't separate it to $L|y_1-y_2|$.




According to my lecturer, the Lipschitz condition should be $$|f(x,y_1)-f(x,y_2)|le L|y2-y1|$$
I was able show that $x^2+y^2$ in the rectangle $|x|le a$, $|y|le b$ satisfies the Lipschitz condition, with my $L=2b$. But I had problem showing this for $f(x,y)=xsin y+ycos x $.






real-analysis lipschitz-functions







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Do you want Lipschitz in $y$ or Lipschitz in both variables?
    $endgroup$
    – Jose27
    Apr 4 '15 at 5:51










  • $begingroup$
    according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
    $endgroup$
    – James Anyanwu
    Apr 4 '15 at 6:52














1












1








1





$begingroup$


I wanted to show that $f(x,y)=xsin y+ycos x$ sastisfy Lipschitz conditions. but I can't separate it to $L|y_1-y_2|$.




According to my lecturer, the Lipschitz condition should be $$|f(x,y_1)-f(x,y_2)|le L|y2-y1|$$
I was able show that $x^2+y^2$ in the rectangle $|x|le a$, $|y|le b$ satisfies the Lipschitz condition, with my $L=2b$. But I had problem showing this for $f(x,y)=xsin y+ycos x $.






real-analysis lipschitz-functions







share|cite|improve this question











$endgroup$




I wanted to show that $f(x,y)=xsin y+ycos x$ sastisfy Lipschitz conditions. but I can't separate it to $L|y_1-y_2|$.




According to my lecturer, the Lipschitz condition should be $$|f(x,y_1)-f(x,y_2)|le L|y2-y1|$$
I was able show that $x^2+y^2$ in the rectangle $|x|le a$, $|y|le b$ satisfies the Lipschitz condition, with my $L=2b$. But I had problem showing this for $f(x,y)=xsin y+ycos x $.






real-analysis lipschitz-functions




real-analysis lipschitz-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 5 '15 at 2:44







user147263

















asked Apr 4 '15 at 5:44









James AnyanwuJames Anyanwu

61




61








  • 3




    $begingroup$
    Do you want Lipschitz in $y$ or Lipschitz in both variables?
    $endgroup$
    – Jose27
    Apr 4 '15 at 5:51










  • $begingroup$
    according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
    $endgroup$
    – James Anyanwu
    Apr 4 '15 at 6:52














  • 3




    $begingroup$
    Do you want Lipschitz in $y$ or Lipschitz in both variables?
    $endgroup$
    – Jose27
    Apr 4 '15 at 5:51










  • $begingroup$
    according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
    $endgroup$
    – James Anyanwu
    Apr 4 '15 at 6:52








3




3




$begingroup$
Do you want Lipschitz in $y$ or Lipschitz in both variables?
$endgroup$
– Jose27
Apr 4 '15 at 5:51




$begingroup$
Do you want Lipschitz in $y$ or Lipschitz in both variables?
$endgroup$
– Jose27
Apr 4 '15 at 5:51












$begingroup$
according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
$endgroup$
– James Anyanwu
Apr 4 '15 at 6:52




$begingroup$
according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
$endgroup$
– James Anyanwu
Apr 4 '15 at 6:52










1 Answer
1






active

oldest

votes


















0












$begingroup$

It seems that you want to prove that $f$ is Lipschitz with respect to $y$. When $f$ is viewed as a global function $f:>{mathbb R}^2to{mathbb R}$ the statement is wrong, because
$$left|{fbigl(x,{piover 2}bigr)-f(x,0)over {piover2}}right|geq |x|-{piover2}$$
assumes arbitrarily large values. The function $f$ is, however, locally Lipschitz with respect to $y$ in ${mathbb R}^2$. This means that any point $(x_0,y_0)$ has a neighborhood $W$ whithin which the Lipschitz condition is fulfilled. In order to see this it is sufficient to note that $fin C^1({mathbb R}^2)$, but maybe you want a selfcontained proof.



From
$${partial f(x,y)overpartial y}= xcos y+cos x$$
it follows that $bigl|{partial f(x,y)overpartial y}bigr|leq |x|+1$. Therefore any point $(x_0,y_0)$ is the center of a quite large window $W$ such that for a suitable $M$ one has
$$left|{partial f(x,y)overpartial y}right|leq Mqquadforall (x,y)in W .$$
By means of the MVT we then conclude that
$$bigl|f(x,y_1)-f(x,y_2)bigr|leq M>|y_1-y_2|$$
for all $(x,y_1)$, $(x,y_2)in W$.






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%2f1219535%2fshowing-that-the-function-fx-y-x-sin-yy-cos-x-is-lipschitz%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









    0












    $begingroup$

    It seems that you want to prove that $f$ is Lipschitz with respect to $y$. When $f$ is viewed as a global function $f:>{mathbb R}^2to{mathbb R}$ the statement is wrong, because
    $$left|{fbigl(x,{piover 2}bigr)-f(x,0)over {piover2}}right|geq |x|-{piover2}$$
    assumes arbitrarily large values. The function $f$ is, however, locally Lipschitz with respect to $y$ in ${mathbb R}^2$. This means that any point $(x_0,y_0)$ has a neighborhood $W$ whithin which the Lipschitz condition is fulfilled. In order to see this it is sufficient to note that $fin C^1({mathbb R}^2)$, but maybe you want a selfcontained proof.



    From
    $${partial f(x,y)overpartial y}= xcos y+cos x$$
    it follows that $bigl|{partial f(x,y)overpartial y}bigr|leq |x|+1$. Therefore any point $(x_0,y_0)$ is the center of a quite large window $W$ such that for a suitable $M$ one has
    $$left|{partial f(x,y)overpartial y}right|leq Mqquadforall (x,y)in W .$$
    By means of the MVT we then conclude that
    $$bigl|f(x,y_1)-f(x,y_2)bigr|leq M>|y_1-y_2|$$
    for all $(x,y_1)$, $(x,y_2)in W$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It seems that you want to prove that $f$ is Lipschitz with respect to $y$. When $f$ is viewed as a global function $f:>{mathbb R}^2to{mathbb R}$ the statement is wrong, because
      $$left|{fbigl(x,{piover 2}bigr)-f(x,0)over {piover2}}right|geq |x|-{piover2}$$
      assumes arbitrarily large values. The function $f$ is, however, locally Lipschitz with respect to $y$ in ${mathbb R}^2$. This means that any point $(x_0,y_0)$ has a neighborhood $W$ whithin which the Lipschitz condition is fulfilled. In order to see this it is sufficient to note that $fin C^1({mathbb R}^2)$, but maybe you want a selfcontained proof.



      From
      $${partial f(x,y)overpartial y}= xcos y+cos x$$
      it follows that $bigl|{partial f(x,y)overpartial y}bigr|leq |x|+1$. Therefore any point $(x_0,y_0)$ is the center of a quite large window $W$ such that for a suitable $M$ one has
      $$left|{partial f(x,y)overpartial y}right|leq Mqquadforall (x,y)in W .$$
      By means of the MVT we then conclude that
      $$bigl|f(x,y_1)-f(x,y_2)bigr|leq M>|y_1-y_2|$$
      for all $(x,y_1)$, $(x,y_2)in W$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        It seems that you want to prove that $f$ is Lipschitz with respect to $y$. When $f$ is viewed as a global function $f:>{mathbb R}^2to{mathbb R}$ the statement is wrong, because
        $$left|{fbigl(x,{piover 2}bigr)-f(x,0)over {piover2}}right|geq |x|-{piover2}$$
        assumes arbitrarily large values. The function $f$ is, however, locally Lipschitz with respect to $y$ in ${mathbb R}^2$. This means that any point $(x_0,y_0)$ has a neighborhood $W$ whithin which the Lipschitz condition is fulfilled. In order to see this it is sufficient to note that $fin C^1({mathbb R}^2)$, but maybe you want a selfcontained proof.



        From
        $${partial f(x,y)overpartial y}= xcos y+cos x$$
        it follows that $bigl|{partial f(x,y)overpartial y}bigr|leq |x|+1$. Therefore any point $(x_0,y_0)$ is the center of a quite large window $W$ such that for a suitable $M$ one has
        $$left|{partial f(x,y)overpartial y}right|leq Mqquadforall (x,y)in W .$$
        By means of the MVT we then conclude that
        $$bigl|f(x,y_1)-f(x,y_2)bigr|leq M>|y_1-y_2|$$
        for all $(x,y_1)$, $(x,y_2)in W$.






        share|cite|improve this answer









        $endgroup$



        It seems that you want to prove that $f$ is Lipschitz with respect to $y$. When $f$ is viewed as a global function $f:>{mathbb R}^2to{mathbb R}$ the statement is wrong, because
        $$left|{fbigl(x,{piover 2}bigr)-f(x,0)over {piover2}}right|geq |x|-{piover2}$$
        assumes arbitrarily large values. The function $f$ is, however, locally Lipschitz with respect to $y$ in ${mathbb R}^2$. This means that any point $(x_0,y_0)$ has a neighborhood $W$ whithin which the Lipschitz condition is fulfilled. In order to see this it is sufficient to note that $fin C^1({mathbb R}^2)$, but maybe you want a selfcontained proof.



        From
        $${partial f(x,y)overpartial y}= xcos y+cos x$$
        it follows that $bigl|{partial f(x,y)overpartial y}bigr|leq |x|+1$. Therefore any point $(x_0,y_0)$ is the center of a quite large window $W$ such that for a suitable $M$ one has
        $$left|{partial f(x,y)overpartial y}right|leq Mqquadforall (x,y)in W .$$
        By means of the MVT we then conclude that
        $$bigl|f(x,y_1)-f(x,y_2)bigr|leq M>|y_1-y_2|$$
        for all $(x,y_1)$, $(x,y_2)in W$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 4 '15 at 15:22









        Christian BlatterChristian Blatter

        172k7113326




        172k7113326






























            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%2f1219535%2fshowing-that-the-function-fx-y-x-sin-yy-cos-x-is-lipschitz%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

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
            Read more

            Antonio Litta Visconti Arese

            Image
            Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
            Read more