Multidimensionally Lipschitz continuous iff Lipschitz continuous in every coordinate











up vote
1
down vote

favorite













Definition. A function $f$ defined on a set $S subseteq mathbb R$ is said to be Lipschitz continuous on $S$ if there exists an $L geq 0$ such that $$|f(x_1) - f(x_2)| le L|x_1 - x_2|$$ for all $x_1$ and $x_2$ in $S$ such that $x_1 ne x_2$.




Consider some function $f: mathbb R^n to mathbb R^n$ taking several variables $x^{(k)}$ with $1 < k leq n$ as argument. Let $f_i: mathbb R^n to mathbb R$ be the function of $i$-th coordinate of the function value of $f$. Are the following statements true?




  • If $f$ is Lipschitz continuous, then $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$.

  • If $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$, then $f$ is Lipschitz continuous.










share|cite|improve this question


























    up vote
    1
    down vote

    favorite













    Definition. A function $f$ defined on a set $S subseteq mathbb R$ is said to be Lipschitz continuous on $S$ if there exists an $L geq 0$ such that $$|f(x_1) - f(x_2)| le L|x_1 - x_2|$$ for all $x_1$ and $x_2$ in $S$ such that $x_1 ne x_2$.




    Consider some function $f: mathbb R^n to mathbb R^n$ taking several variables $x^{(k)}$ with $1 < k leq n$ as argument. Let $f_i: mathbb R^n to mathbb R$ be the function of $i$-th coordinate of the function value of $f$. Are the following statements true?




    • If $f$ is Lipschitz continuous, then $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$.

    • If $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$, then $f$ is Lipschitz continuous.










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      Definition. A function $f$ defined on a set $S subseteq mathbb R$ is said to be Lipschitz continuous on $S$ if there exists an $L geq 0$ such that $$|f(x_1) - f(x_2)| le L|x_1 - x_2|$$ for all $x_1$ and $x_2$ in $S$ such that $x_1 ne x_2$.




      Consider some function $f: mathbb R^n to mathbb R^n$ taking several variables $x^{(k)}$ with $1 < k leq n$ as argument. Let $f_i: mathbb R^n to mathbb R$ be the function of $i$-th coordinate of the function value of $f$. Are the following statements true?




      • If $f$ is Lipschitz continuous, then $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$.

      • If $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$, then $f$ is Lipschitz continuous.










      share|cite|improve this question














      Definition. A function $f$ defined on a set $S subseteq mathbb R$ is said to be Lipschitz continuous on $S$ if there exists an $L geq 0$ such that $$|f(x_1) - f(x_2)| le L|x_1 - x_2|$$ for all $x_1$ and $x_2$ in $S$ such that $x_1 ne x_2$.




      Consider some function $f: mathbb R^n to mathbb R^n$ taking several variables $x^{(k)}$ with $1 < k leq n$ as argument. Let $f_i: mathbb R^n to mathbb R$ be the function of $i$-th coordinate of the function value of $f$. Are the following statements true?




      • If $f$ is Lipschitz continuous, then $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$.

      • If $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$, then $f$ is Lipschitz continuous.







      multivariable-calculus continuity lipschitz-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 25 at 10:03









      meisterluk

      14610




      14610






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          $f$ is K-Lipschitz continuous implies $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$:



          $left | f_i(x)-f_i(y) right |leq left | f(x)-f(y) right |leq Kleft | x-y right |$



          Each coordinate $f_i$ is $K_i$-Lipschitz continuous implies $f$ is Lipschitz-continuous:



          $f=sum_{i=1}^{n}f_imathbf{e_i}$ where $mathbf{e_i}=left (0,...,0,1,0,...,0 right )$ where the $1$ is at the $i$-th position.



          Note that $left | f_1(x)mathbf{e_1}+f_2(x)mathbf{e_2}-f_1(y)mathbf{e_1}-f_2(y)mathbf{e_2} right |leq left | f_1(x)-f_1(y) right |+left |f_2(x)-f_2(y) right |leq L_1left | x-y right |+L_2left | x-y right |leq (L_1+L_2)left | x-y right |$



          The result follows by induction.






          share|cite|improve this answer





















          • So it is. Thank you very much for showing it formally :)
            – meisterluk
            Nov 26 at 10:14











          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',
          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%2f3012643%2fmultidimensionally-lipschitz-continuous-iff-lipschitz-continuous-in-every-coordi%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








          up vote
          1
          down vote



          accepted










          $f$ is K-Lipschitz continuous implies $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$:



          $left | f_i(x)-f_i(y) right |leq left | f(x)-f(y) right |leq Kleft | x-y right |$



          Each coordinate $f_i$ is $K_i$-Lipschitz continuous implies $f$ is Lipschitz-continuous:



          $f=sum_{i=1}^{n}f_imathbf{e_i}$ where $mathbf{e_i}=left (0,...,0,1,0,...,0 right )$ where the $1$ is at the $i$-th position.



          Note that $left | f_1(x)mathbf{e_1}+f_2(x)mathbf{e_2}-f_1(y)mathbf{e_1}-f_2(y)mathbf{e_2} right |leq left | f_1(x)-f_1(y) right |+left |f_2(x)-f_2(y) right |leq L_1left | x-y right |+L_2left | x-y right |leq (L_1+L_2)left | x-y right |$



          The result follows by induction.






          share|cite|improve this answer





















          • So it is. Thank you very much for showing it formally :)
            – meisterluk
            Nov 26 at 10:14















          up vote
          1
          down vote



          accepted










          $f$ is K-Lipschitz continuous implies $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$:



          $left | f_i(x)-f_i(y) right |leq left | f(x)-f(y) right |leq Kleft | x-y right |$



          Each coordinate $f_i$ is $K_i$-Lipschitz continuous implies $f$ is Lipschitz-continuous:



          $f=sum_{i=1}^{n}f_imathbf{e_i}$ where $mathbf{e_i}=left (0,...,0,1,0,...,0 right )$ where the $1$ is at the $i$-th position.



          Note that $left | f_1(x)mathbf{e_1}+f_2(x)mathbf{e_2}-f_1(y)mathbf{e_1}-f_2(y)mathbf{e_2} right |leq left | f_1(x)-f_1(y) right |+left |f_2(x)-f_2(y) right |leq L_1left | x-y right |+L_2left | x-y right |leq (L_1+L_2)left | x-y right |$



          The result follows by induction.






          share|cite|improve this answer





















          • So it is. Thank you very much for showing it formally :)
            – meisterluk
            Nov 26 at 10:14













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          $f$ is K-Lipschitz continuous implies $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$:



          $left | f_i(x)-f_i(y) right |leq left | f(x)-f(y) right |leq Kleft | x-y right |$



          Each coordinate $f_i$ is $K_i$-Lipschitz continuous implies $f$ is Lipschitz-continuous:



          $f=sum_{i=1}^{n}f_imathbf{e_i}$ where $mathbf{e_i}=left (0,...,0,1,0,...,0 right )$ where the $1$ is at the $i$-th position.



          Note that $left | f_1(x)mathbf{e_1}+f_2(x)mathbf{e_2}-f_1(y)mathbf{e_1}-f_2(y)mathbf{e_2} right |leq left | f_1(x)-f_1(y) right |+left |f_2(x)-f_2(y) right |leq L_1left | x-y right |+L_2left | x-y right |leq (L_1+L_2)left | x-y right |$



          The result follows by induction.






          share|cite|improve this answer












          $f$ is K-Lipschitz continuous implies $f_i$ is Lipschitz continuous $forall i in {1, dots, n}$:



          $left | f_i(x)-f_i(y) right |leq left | f(x)-f(y) right |leq Kleft | x-y right |$



          Each coordinate $f_i$ is $K_i$-Lipschitz continuous implies $f$ is Lipschitz-continuous:



          $f=sum_{i=1}^{n}f_imathbf{e_i}$ where $mathbf{e_i}=left (0,...,0,1,0,...,0 right )$ where the $1$ is at the $i$-th position.



          Note that $left | f_1(x)mathbf{e_1}+f_2(x)mathbf{e_2}-f_1(y)mathbf{e_1}-f_2(y)mathbf{e_2} right |leq left | f_1(x)-f_1(y) right |+left |f_2(x)-f_2(y) right |leq L_1left | x-y right |+L_2left | x-y right |leq (L_1+L_2)left | x-y right |$



          The result follows by induction.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 10:27









          John11

          8861721




          8861721












          • So it is. Thank you very much for showing it formally :)
            – meisterluk
            Nov 26 at 10:14


















          • So it is. Thank you very much for showing it formally :)
            – meisterluk
            Nov 26 at 10:14
















          So it is. Thank you very much for showing it formally :)
          – meisterluk
          Nov 26 at 10:14




          So it is. Thank you very much for showing it formally :)
          – meisterluk
          Nov 26 at 10:14


















          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%2f3012643%2fmultidimensionally-lipschitz-continuous-iff-lipschitz-continuous-in-every-coordi%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

          Tonle Sap (See)

          I get strange results when I access the Sqlitedatabase with Unity C# via XAMPP

          Guatemaltekische Davis-Cup-Mannschaft