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






      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
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          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





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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 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


















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          格利澤436b

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          格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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