Continuous functions of three variables as superpositions of two variable functions












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$begingroup$


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







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    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






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      share|cite|improve this question











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      asked 4 hours ago









      KhashFKhashF

      986




      986






















          1 Answer
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          $begingroup$


          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











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          $begingroup$


          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 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
            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


















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          濃尾地震

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          濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸
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