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
            2 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
            2 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
            2 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 3 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
            2 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
            2 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
          2 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
          2 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

          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