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




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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 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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          body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} GameSpot 戰地風雲：惡名昭彰2在Gamespot的評論 网站类型 新聞 持有者 CBS 创始人 Pete Deemer Vince Broady Jon Epstein 网站 http://www.gamespot.com/ 注册 Optional (free and paid) 推出时间 1996年5月1日 [1] GameSpot （中国大陆：游戏基地），於1996年5月由Pete Deemer和Vince Broady創立，是一個提供新聞、評論、預告片、下載及其他的相關資訊的電子遊戲網站。GameSpot被一間後來被CNET收購的企業ZDNet所收購。根據Alexa，GameSpot.com是200個網路擁擠最嚴重的網站之一。 除了由GameSpot員工創作的內容，網站還允許用戶寫評論、網誌、之後在網路論壇分享。一些在CNET旗下的GameFAQs分享。 2004年， GameSpot被Spike TV的觀眾選上「電子遊戲賞節目」贏得「最傑出遊戲網站。 [2] 其他的遊戲網站還有IGN、1UP.com、GameSpy是它最大的競爭對手。2008年，根據Compete.com的統計，「gamespot.com」吸引了最少6000萬人的點擊率。 [3] GameSpot的主頁鏈結了到最近新聞、評論、預告、和一些有關遊戲機的入口：Wii、任天堂DS、電腦遊戲、Xbox 360、PSP、PlayStation 2、PlayStation 3。它還有一列「最受歡迎遊戲名單」，還有給用戶快速獲得遊戲資訊的搜尋器。GameSpot 還包括一些小範圍的遊戲機：任天堂64、GameCube、Game Boy Color、Game Boy Advance、Xbox、PlayStation、SEGA Saturn、Dreamcast、Neo Geo Pocket Color、N-Gage、手機遊戲。 目录 1 歷史 1.1 國際歷史 1.2 著名的員工 2 評論和分
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          Tu-95轟炸機

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          body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 图-95 Ту–95 北约代号：熊（Bear） 一架在安格斯空军基地的图-95MS 概觀 類型 战略轰炸机 代號 北约代号： Bear （熊） 乘員 7名 駕駛員2名，機尾炮手1名，其他人員4名 首飛 1952年11月12日 服役 1956年 設計 圖波列夫設計局 產量 超过500架 現況 现役 主要用戶   蘇聯   俄羅斯 衍生機型 图-114、图-119、Tu-95 技术数据 長度 49.50米（162呎5吋） 翼展 51.10米（167呎8吋） 高度 12.12米（39呎9吋） 翼面積 310平方米（3,330平方呎） 空重 90,000公斤（198,000磅） 最大起飛重量 188,000公斤（414,500磅） 發動機 4具库兹涅佐夫NK-12MV型渦輪螺旋槳發動機 功率 4×11,000千瓦（14,800馬力） 性能數據 最大速度 925公里／時（500節，575哩／時） 爬升率 10米／秒（2,000呎／分） 最大升限 12,000米（39,000呎） 最大航程 15,000公里（8,100海浬，9,400哩） 翼負荷 606公斤／平方米（124磅／平方呎） 推重比 235 W／公斤（0.143馬力／磅） 武器装备 機炮 1具/2具AM-23型23公釐雷達控制機砲（機尾） 飛彈 空对地导弹： Kh-20、Kh-22 Kh-26、Kh-55 炸彈 60枚FAB-250炸彈，或 30枚FAB-500炸彈。 其他 载弹量（包括飞弹）：最多15,000公斤（33,000磅） Tu-95熊式MR型機 图-95 （俄语： Ту–95 ， 英语： Tu-95 ），北約代號： Bear （ 熊 ），是蘇聯圖波列夫設計局所研製，是全世界唯一服役的大型四渦輪螺旋槳發動機
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