Continuous functions of three variables as superpositions of two variable functions












3












$begingroup$


Could we always locally represent a continuous function $F(x,y,z)$ in the form of $gleft(f(x,y),zright)$ for suitable continuous functions $f$, $g$ of two variables? I am aware of Vladimir Arnold's work on this problem, but it seems that in that context $F(x,y,z)$ is written as a sum of several expressions of this form. Can one reduce it to just a single superposition $gleft(f(x,y),zright)$; or does anyone know a counter example?






real-analysis







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Could we always locally represent a continuous function $F(x,y,z)$ in the form of $gleft(f(x,y),zright)$ for suitable continuous functions $f$, $g$ of two variables? I am aware of Vladimir Arnold's work on this problem, but it seems that in that context $F(x,y,z)$ is written as a sum of several expressions of this form. Can one reduce it to just a single superposition $gleft(f(x,y),zright)$; or does anyone know a counter example?






    real-analysis







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Could we always locally represent a continuous function $F(x,y,z)$ in the form of $gleft(f(x,y),zright)$ for suitable continuous functions $f$, $g$ of two variables? I am aware of Vladimir Arnold's work on this problem, but it seems that in that context $F(x,y,z)$ is written as a sum of several expressions of this form. Can one reduce it to just a single superposition $gleft(f(x,y),zright)$; or does anyone know a counter example?






      real-analysis







      share|cite|improve this question









      $endgroup$




      Could we always locally represent a continuous function $F(x,y,z)$ in the form of $gleft(f(x,y),zright)$ for suitable continuous functions $f$, $g$ of two variables? I am aware of Vladimir Arnold's work on this problem, but it seems that in that context $F(x,y,z)$ is written as a sum of several expressions of this form. Can one reduce it to just a single superposition $gleft(f(x,y),zright)$; or does anyone know a counter example?






      real-analysis




      real-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      KhashFKhashF

      986




      986






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$


          Example. The function $F(x,y,z)=x(1-z)+yz$ cannot be represented as $F(x,y,z)=g(f(x,y),z)$.




          Proof. Suppose to the contrary that we have such a representation. Let $g_1(t)=g(t,0)$. Then $g_1(f(x,y))=g(f(x,y),0)=F(x,y,0)=x$. That implies that for every $y$, $g_1$ is the inverse function of $xmapsto f(x,y)$ so
          $f(x,y)=g_1^{-1}(x)$.



          Let $g_2(t)=g(t,1)$. Then $g_2(f(x,y))=g(f(x,y),1)=F(x,y,1)=y$. That implies that for every $x$, $g_2$ is the inverse function of $ymapsto f(x,y)$ so $f(x,y)=g_2^{-1}(y)$.



          Let $c=f(0,0)$. Then
          $$
          f(x,y)=g_1^{-1}(x)=f(x,0)=g_2^{-1}(0)=f(0,0)=c.
          $$
          Therefore
          $$
          x(1-z)+yz=F(x,y,z)=g(f(x,y),z)=g(c,z)
          $$
          which is a contradiction, because the function on the left hand side depends on $x$ and $y$, while the function on the right hand side does not. $Box$



          Edit: I modified my proof using a suggestion of user44191.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks Piotr! But how did you deduce that $f$ has derivative? Does your example work in the continuous setting as well?
            $endgroup$
            – KhashF
            3 hours ago










          • $begingroup$
            Partial derivatives aren't necessary for this proof; if you let $c = f(0, 0)$, you get that $f(x, y) = g_1^{-1}(x) = f(x, 0) = g_2^{-1}(0) = f(0, 0) = c$.
            $endgroup$
            – user44191
            3 hours ago










          • $begingroup$
            @user44191 Thank you. I modified my answer.
            $endgroup$
            – Piotr Hajlasz
            3 hours ago










          • $begingroup$
            @KhashF I modified my answer.
            $endgroup$
            – Piotr Hajlasz
            3 hours ago






          • 1




            $begingroup$
            I’m sorry but where you used continuity of $f$ and $g$? Because without it answer to question in OP is that every function can be represented in such a manner (via bijection between $mathbb{R}$ and $mathbb{R}^2$). And also what is $g_1^{-1}(x)$ really? I see no reasons for $g_1$ to be injective.
            $endgroup$
            – Aleksei Kulikov
            3 hours ago











          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: "504"
          };
          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%2fmathoverflow.net%2fquestions%2f322184%2fcontinuous-functions-of-three-variables-as-superpositions-of-two-variable-functi%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









          4












          $begingroup$


          Example. The function $F(x,y,z)=x(1-z)+yz$ cannot be represented as $F(x,y,z)=g(f(x,y),z)$.




          Proof. Suppose to the contrary that we have such a representation. Let $g_1(t)=g(t,0)$. Then $g_1(f(x,y))=g(f(x,y),0)=F(x,y,0)=x$. That implies that for every $y$, $g_1$ is the inverse function of $xmapsto f(x,y)$ so
          $f(x,y)=g_1^{-1}(x)$.



          Let $g_2(t)=g(t,1)$. Then $g_2(f(x,y))=g(f(x,y),1)=F(x,y,1)=y$. That implies that for every $x$, $g_2$ is the inverse function of $ymapsto f(x,y)$ so $f(x,y)=g_2^{-1}(y)$.



          Let $c=f(0,0)$. Then
          $$
          f(x,y)=g_1^{-1}(x)=f(x,0)=g_2^{-1}(0)=f(0,0)=c.
          $$
          Therefore
          $$
          x(1-z)+yz=F(x,y,z)=g(f(x,y),z)=g(c,z)
          $$
          which is a contradiction, because the function on the left hand side depends on $x$ and $y$, while the function on the right hand side does not. $Box$



          Edit: I modified my proof using a suggestion of user44191.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks Piotr! But how did you deduce that $f$ has derivative? Does your example work in the continuous setting as well?
            $endgroup$
            – KhashF
            3 hours ago










          • $begingroup$
            Partial derivatives aren't necessary for this proof; if you let $c = f(0, 0)$, you get that $f(x, y) = g_1^{-1}(x) = f(x, 0) = g_2^{-1}(0) = f(0, 0) = c$.
            $endgroup$
            – user44191
            3 hours ago










          • $begingroup$
            @user44191 Thank you. I modified my answer.
            $endgroup$
            – Piotr Hajlasz
            3 hours ago










          • $begingroup$
            @KhashF I modified my answer.
            $endgroup$
            – Piotr Hajlasz
            3 hours ago






          • 1




            $begingroup$
            I’m sorry but where you used continuity of $f$ and $g$? Because without it answer to question in OP is that every function can be represented in such a manner (via bijection between $mathbb{R}$ and $mathbb{R}^2$). And also what is $g_1^{-1}(x)$ really? I see no reasons for $g_1$ to be injective.
            $endgroup$
            – Aleksei Kulikov
            3 hours ago
















          4












          $begingroup$


          Example. The function $F(x,y,z)=x(1-z)+yz$ cannot be represented as $F(x,y,z)=g(f(x,y),z)$.




          Proof. Suppose to the contrary that we have such a representation. Let $g_1(t)=g(t,0)$. Then $g_1(f(x,y))=g(f(x,y),0)=F(x,y,0)=x$. That implies that for every $y$, $g_1$ is the inverse function of $xmapsto f(x,y)$ so
          $f(x,y)=g_1^{-1}(x)$.



          Let $g_2(t)=g(t,1)$. Then $g_2(f(x,y))=g(f(x,y),1)=F(x,y,1)=y$. That implies that for every $x$, $g_2$ is the inverse function of $ymapsto f(x,y)$ so $f(x,y)=g_2^{-1}(y)$.



          Let $c=f(0,0)$. Then
          $$
          f(x,y)=g_1^{-1}(x)=f(x,0)=g_2^{-1}(0)=f(0,0)=c.
          $$
          Therefore
          $$
          x(1-z)+yz=F(x,y,z)=g(f(x,y),z)=g(c,z)
          $$
          which is a contradiction, because the function on the left hand side depends on $x$ and $y$, while the function on the right hand side does not. $Box$



          Edit: I modified my proof using a suggestion of user44191.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks Piotr! But how did you deduce that $f$ has derivative? Does your example work in the continuous setting as well?
            $endgroup$
            – KhashF
            3 hours ago










          • $begingroup$
            Partial derivatives aren't necessary for this proof; if you let $c = f(0, 0)$, you get that $f(x, y) = g_1^{-1}(x) = f(x, 0) = g_2^{-1}(0) = f(0, 0) = c$.
            $endgroup$
            – user44191
            3 hours ago










          • $begingroup$
            @user44191 Thank you. I modified my answer.
            $endgroup$
            – Piotr Hajlasz
            3 hours ago










          • $begingroup$
            @KhashF I modified my answer.
            $endgroup$
            – Piotr Hajlasz
            3 hours ago






          • 1




            $begingroup$
            I’m sorry but where you used continuity of $f$ and $g$? Because without it answer to question in OP is that every function can be represented in such a manner (via bijection between $mathbb{R}$ and $mathbb{R}^2$). And also what is $g_1^{-1}(x)$ really? I see no reasons for $g_1$ to be injective.
            $endgroup$
            – Aleksei Kulikov
            3 hours ago














          4












          4








          4





          $begingroup$


          Example. The function $F(x,y,z)=x(1-z)+yz$ cannot be represented as $F(x,y,z)=g(f(x,y),z)$.




          Proof. Suppose to the contrary that we have such a representation. Let $g_1(t)=g(t,0)$. Then $g_1(f(x,y))=g(f(x,y),0)=F(x,y,0)=x$. That implies that for every $y$, $g_1$ is the inverse function of $xmapsto f(x,y)$ so
          $f(x,y)=g_1^{-1}(x)$.



          Let $g_2(t)=g(t,1)$. Then $g_2(f(x,y))=g(f(x,y),1)=F(x,y,1)=y$. That implies that for every $x$, $g_2$ is the inverse function of $ymapsto f(x,y)$ so $f(x,y)=g_2^{-1}(y)$.



          Let $c=f(0,0)$. Then
          $$
          f(x,y)=g_1^{-1}(x)=f(x,0)=g_2^{-1}(0)=f(0,0)=c.
          $$
          Therefore
          $$
          x(1-z)+yz=F(x,y,z)=g(f(x,y),z)=g(c,z)
          $$
          which is a contradiction, because the function on the left hand side depends on $x$ and $y$, while the function on the right hand side does not. $Box$



          Edit: I modified my proof using a suggestion of user44191.






          share|cite|improve this answer











          $endgroup$




          Example. The function $F(x,y,z)=x(1-z)+yz$ cannot be represented as $F(x,y,z)=g(f(x,y),z)$.




          Proof. Suppose to the contrary that we have such a representation. Let $g_1(t)=g(t,0)$. Then $g_1(f(x,y))=g(f(x,y),0)=F(x,y,0)=x$. That implies that for every $y$, $g_1$ is the inverse function of $xmapsto f(x,y)$ so
          $f(x,y)=g_1^{-1}(x)$.



          Let $g_2(t)=g(t,1)$. Then $g_2(f(x,y))=g(f(x,y),1)=F(x,y,1)=y$. That implies that for every $x$, $g_2$ is the inverse function of $ymapsto f(x,y)$ so $f(x,y)=g_2^{-1}(y)$.



          Let $c=f(0,0)$. Then
          $$
          f(x,y)=g_1^{-1}(x)=f(x,0)=g_2^{-1}(0)=f(0,0)=c.
          $$
          Therefore
          $$
          x(1-z)+yz=F(x,y,z)=g(f(x,y),z)=g(c,z)
          $$
          which is a contradiction, because the function on the left hand side depends on $x$ and $y$, while the function on the right hand side does not. $Box$



          Edit: I modified my proof using a suggestion of user44191.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 4 hours ago









          Piotr HajlaszPiotr Hajlasz

          8,18942862




          8,18942862












          • $begingroup$
            Thanks Piotr! But how did you deduce that $f$ has derivative? Does your example work in the continuous setting as well?
            $endgroup$
            – KhashF
            3 hours ago










          • $begingroup$
            Partial derivatives aren't necessary for this proof; if you let $c = f(0, 0)$, you get that $f(x, y) = g_1^{-1}(x) = f(x, 0) = g_2^{-1}(0) = f(0, 0) = c$.
            $endgroup$
            – user44191
            3 hours ago










          • $begingroup$
            @user44191 Thank you. I modified my answer.
            $endgroup$
            – Piotr Hajlasz
            3 hours ago










          • $begingroup$
            @KhashF I modified my answer.
            $endgroup$
            – Piotr Hajlasz
            3 hours ago






          • 1




            $begingroup$
            I’m sorry but where you used continuity of $f$ and $g$? Because without it answer to question in OP is that every function can be represented in such a manner (via bijection between $mathbb{R}$ and $mathbb{R}^2$). And also what is $g_1^{-1}(x)$ really? I see no reasons for $g_1$ to be injective.
            $endgroup$
            – Aleksei Kulikov
            3 hours ago


















          • $begingroup$
            Thanks Piotr! But how did you deduce that $f$ has derivative? Does your example work in the continuous setting as well?
            $endgroup$
            – KhashF
            3 hours ago










          • $begingroup$
            Partial derivatives aren't necessary for this proof; if you let $c = f(0, 0)$, you get that $f(x, y) = g_1^{-1}(x) = f(x, 0) = g_2^{-1}(0) = f(0, 0) = c$.
            $endgroup$
            – user44191
            3 hours ago










          • $begingroup$
            @user44191 Thank you. I modified my answer.
            $endgroup$
            – Piotr Hajlasz
            3 hours ago










          • $begingroup$
            @KhashF I modified my answer.
            $endgroup$
            – Piotr Hajlasz
            3 hours ago






          • 1




            $begingroup$
            I’m sorry but where you used continuity of $f$ and $g$? Because without it answer to question in OP is that every function can be represented in such a manner (via bijection between $mathbb{R}$ and $mathbb{R}^2$). And also what is $g_1^{-1}(x)$ really? I see no reasons for $g_1$ to be injective.
            $endgroup$
            – Aleksei Kulikov
            3 hours ago
















          $begingroup$
          Thanks Piotr! But how did you deduce that $f$ has derivative? Does your example work in the continuous setting as well?
          $endgroup$
          – KhashF
          3 hours ago




          $begingroup$
          Thanks Piotr! But how did you deduce that $f$ has derivative? Does your example work in the continuous setting as well?
          $endgroup$
          – KhashF
          3 hours ago












          $begingroup$
          Partial derivatives aren't necessary for this proof; if you let $c = f(0, 0)$, you get that $f(x, y) = g_1^{-1}(x) = f(x, 0) = g_2^{-1}(0) = f(0, 0) = c$.
          $endgroup$
          – user44191
          3 hours ago




          $begingroup$
          Partial derivatives aren't necessary for this proof; if you let $c = f(0, 0)$, you get that $f(x, y) = g_1^{-1}(x) = f(x, 0) = g_2^{-1}(0) = f(0, 0) = c$.
          $endgroup$
          – user44191
          3 hours ago












          $begingroup$
          @user44191 Thank you. I modified my answer.
          $endgroup$
          – Piotr Hajlasz
          3 hours ago




          $begingroup$
          @user44191 Thank you. I modified my answer.
          $endgroup$
          – Piotr Hajlasz
          3 hours ago












          $begingroup$
          @KhashF I modified my answer.
          $endgroup$
          – Piotr Hajlasz
          3 hours ago




          $begingroup$
          @KhashF I modified my answer.
          $endgroup$
          – Piotr Hajlasz
          3 hours ago




          1




          1




          $begingroup$
          I’m sorry but where you used continuity of $f$ and $g$? Because without it answer to question in OP is that every function can be represented in such a manner (via bijection between $mathbb{R}$ and $mathbb{R}^2$). And also what is $g_1^{-1}(x)$ really? I see no reasons for $g_1$ to be injective.
          $endgroup$
          – Aleksei Kulikov
          3 hours ago




          $begingroup$
          I’m sorry but where you used continuity of $f$ and $g$? Because without it answer to question in OP is that every function can be represented in such a manner (via bijection between $mathbb{R}$ and $mathbb{R}^2$). And also what is $g_1^{-1}(x)$ really? I see no reasons for $g_1$ to be injective.
          $endgroup$
          – Aleksei Kulikov
          3 hours ago


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to MathOverflow!


          • 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%2fmathoverflow.net%2fquestions%2f322184%2fcontinuous-functions-of-three-variables-as-superpositions-of-two-variable-functi%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

          Сан-Квентин

          Image
          Сан-Квентин Местоположение Сан-Квентин, Калифорния Координаты 37°56′13″ с. ш. 122°29′20″ з. д. H G Я O L Текущий статус Действует Режим безопасности минимальный и максимальный Количество мест 3 302 Открытие Июнь 1852 Находится в ведомстве California Department of Corrections and Rehabilitation Начальник Mike Martel, начальник  Сан-Квентин на Викискладе Сан-Квентин (англ.  San Quentin ) — тюрьма, располагающаяся на 432 акрах (около 1,7 км²), на мысе Сан-Квентин, в округе Марин, штат Калифорния, США. Тюрьма штата Калифорния Сан-Квентин была открыта в июле 1852 года и является старейшей в штате. Она была построена заключёнными, которые проживали в течение строительства на тюремном судне Вабан. В Сан-Квентине отбывали наказание и мужчины, и женщины до 1934 года, когда построили женскую тюрьму в Техачапи и туда перевели всех заключённых женщин. В Сан-Квентине приводятся в исполнение смертные казни, для этого имеется газовая камера, но
          Read more

          Алькесар

          Image
          Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове
          Read more

          Josef Freinademetz

          Image
          San Ujöp (Giuseppe/Josef) Freinademetz Josef Freinademetz in abiti cinesi   Religioso e missionario   Nascita 15 aprile 1852 Morte 28 gennaio 1908 Venerato da Chiesa cattolica Beatificazione 1975 Canonizzazione 5 ottobre 2003 Ricorrenza 28 gennaio Manuale Ujöp Freinademetz noto anche come Giuseppe o Josef Freinademetz (Oies, 15 aprile 1852 – Taikia, 28 gennaio 1908) è stato un presbitero austriaco, membro della Società del Verbo Divino, fu missionario in Cina; è venerato come santo dalla Chiesa cattolica. .mw-parser-output .citazione-table{margin-bottom:.5em;font-size:95%}.mw-parser-output .citazione-table td{padding:0 1.2em 0 2.4em}.mw-parser-output .citazione-lang{vertical-align:top}.mw-parser-output .citazione-lang td{width:50%}.mw-parser-output .citazione-lang td:first-child{padding:0 0 0 2.4em}.mw-parser-output .citazione-lang td:nth-child(2){padding:0 1.2em} «Io amo la Cina e i cinesi; voglio morire in mezzo a loro, e tra loro
          Read more