Definition of smooth functions on arbitrary subsets of $mathbb{R}^n$ and partial derivatives












0












$begingroup$


Let $A$ be an arbitrary subset of $mathbb{R}^n$, and let $f:Ato mathbb{R}$ be a function.



We say that $f$ is smooth if for each point $p$ in $A$ there exists an open subset $U$ of $mathbb{R}^n$ containing $p$ and a smooth map $g:Uto mathbb{R}$ such that $f$ and $g$ agrees on $A cap U$.



Now let $U$ be an open subset of $mathbb{H}^n$, with $mathbb{H}^n={xin mathbb{R}^n:x^nge0 }$



Let $f:Uto mathbb{R}$ be a smooth function in the above sense.
Let $pin U cap partialmathbb{H}^n$. Thus there are $tilde U$ open subset of $mathbb{R}^n$ and a smooth map $g:tilde Uto mathbb{R}$ such that $g$ and $f$ agrees on $U cap tilde U$.




I want to show that the partial derivatives of $f$ at $p$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension i.e. of the choice of $g$.




Here is my argument.



$$lim_{tto0^+}frac{f(p+te_i)-f(p)}{t}=lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}qquad [1]$$ beacuse $f(p)=g(p)$ and for sufficient positive small $t$ we have $p+te_i in U cap tilde U$ and so $f(p+te_i)=g(p+te_i)$.



Now, since $g$ is smooth at $p$ by hypothesis, we have that the right hand side of the above equation exists, and we have



$$lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}=lim_{tto0^-}frac{g(p+te_i)-g(p)}{t}=:partial_i|_p g$$



So I can define $partial_i|_p f:=partial_i|_p g$ and this value is indipendent of the choice of $g$ by $[1]$.



So I have shown the claim in the above yellow box.




Is my argument right? In John M. Lee Textbook Introduction to smooth manifolds, on page 27, he says that By continuity, all partial derivatives of $f$ at points of $Ucap partialmathbb{H}^n$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension.




But it seems I dont'use the continuity of any function, so I'm not convinced of having understood the matter.






multivariable-calculus differential-geometry smooth-manifolds smooth-functions







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $A$ be an arbitrary subset of $mathbb{R}^n$, and let $f:Ato mathbb{R}$ be a function.



    We say that $f$ is smooth if for each point $p$ in $A$ there exists an open subset $U$ of $mathbb{R}^n$ containing $p$ and a smooth map $g:Uto mathbb{R}$ such that $f$ and $g$ agrees on $A cap U$.



    Now let $U$ be an open subset of $mathbb{H}^n$, with $mathbb{H}^n={xin mathbb{R}^n:x^nge0 }$



    Let $f:Uto mathbb{R}$ be a smooth function in the above sense.
    Let $pin U cap partialmathbb{H}^n$. Thus there are $tilde U$ open subset of $mathbb{R}^n$ and a smooth map $g:tilde Uto mathbb{R}$ such that $g$ and $f$ agrees on $U cap tilde U$.




    I want to show that the partial derivatives of $f$ at $p$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension i.e. of the choice of $g$.




    Here is my argument.



    $$lim_{tto0^+}frac{f(p+te_i)-f(p)}{t}=lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}qquad [1]$$ beacuse $f(p)=g(p)$ and for sufficient positive small $t$ we have $p+te_i in U cap tilde U$ and so $f(p+te_i)=g(p+te_i)$.



    Now, since $g$ is smooth at $p$ by hypothesis, we have that the right hand side of the above equation exists, and we have



    $$lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}=lim_{tto0^-}frac{g(p+te_i)-g(p)}{t}=:partial_i|_p g$$



    So I can define $partial_i|_p f:=partial_i|_p g$ and this value is indipendent of the choice of $g$ by $[1]$.



    So I have shown the claim in the above yellow box.




    Is my argument right? In John M. Lee Textbook Introduction to smooth manifolds, on page 27, he says that By continuity, all partial derivatives of $f$ at points of $Ucap partialmathbb{H}^n$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension.




    But it seems I dont'use the continuity of any function, so I'm not convinced of having understood the matter.






    multivariable-calculus differential-geometry smooth-manifolds smooth-functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $A$ be an arbitrary subset of $mathbb{R}^n$, and let $f:Ato mathbb{R}$ be a function.



      We say that $f$ is smooth if for each point $p$ in $A$ there exists an open subset $U$ of $mathbb{R}^n$ containing $p$ and a smooth map $g:Uto mathbb{R}$ such that $f$ and $g$ agrees on $A cap U$.



      Now let $U$ be an open subset of $mathbb{H}^n$, with $mathbb{H}^n={xin mathbb{R}^n:x^nge0 }$



      Let $f:Uto mathbb{R}$ be a smooth function in the above sense.
      Let $pin U cap partialmathbb{H}^n$. Thus there are $tilde U$ open subset of $mathbb{R}^n$ and a smooth map $g:tilde Uto mathbb{R}$ such that $g$ and $f$ agrees on $U cap tilde U$.




      I want to show that the partial derivatives of $f$ at $p$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension i.e. of the choice of $g$.




      Here is my argument.



      $$lim_{tto0^+}frac{f(p+te_i)-f(p)}{t}=lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}qquad [1]$$ beacuse $f(p)=g(p)$ and for sufficient positive small $t$ we have $p+te_i in U cap tilde U$ and so $f(p+te_i)=g(p+te_i)$.



      Now, since $g$ is smooth at $p$ by hypothesis, we have that the right hand side of the above equation exists, and we have



      $$lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}=lim_{tto0^-}frac{g(p+te_i)-g(p)}{t}=:partial_i|_p g$$



      So I can define $partial_i|_p f:=partial_i|_p g$ and this value is indipendent of the choice of $g$ by $[1]$.



      So I have shown the claim in the above yellow box.




      Is my argument right? In John M. Lee Textbook Introduction to smooth manifolds, on page 27, he says that By continuity, all partial derivatives of $f$ at points of $Ucap partialmathbb{H}^n$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension.




      But it seems I dont'use the continuity of any function, so I'm not convinced of having understood the matter.






      multivariable-calculus differential-geometry smooth-manifolds smooth-functions







      share|cite|improve this question











      $endgroup$




      Let $A$ be an arbitrary subset of $mathbb{R}^n$, and let $f:Ato mathbb{R}$ be a function.



      We say that $f$ is smooth if for each point $p$ in $A$ there exists an open subset $U$ of $mathbb{R}^n$ containing $p$ and a smooth map $g:Uto mathbb{R}$ such that $f$ and $g$ agrees on $A cap U$.



      Now let $U$ be an open subset of $mathbb{H}^n$, with $mathbb{H}^n={xin mathbb{R}^n:x^nge0 }$



      Let $f:Uto mathbb{R}$ be a smooth function in the above sense.
      Let $pin U cap partialmathbb{H}^n$. Thus there are $tilde U$ open subset of $mathbb{R}^n$ and a smooth map $g:tilde Uto mathbb{R}$ such that $g$ and $f$ agrees on $U cap tilde U$.




      I want to show that the partial derivatives of $f$ at $p$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension i.e. of the choice of $g$.




      Here is my argument.



      $$lim_{tto0^+}frac{f(p+te_i)-f(p)}{t}=lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}qquad [1]$$ beacuse $f(p)=g(p)$ and for sufficient positive small $t$ we have $p+te_i in U cap tilde U$ and so $f(p+te_i)=g(p+te_i)$.



      Now, since $g$ is smooth at $p$ by hypothesis, we have that the right hand side of the above equation exists, and we have



      $$lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}=lim_{tto0^-}frac{g(p+te_i)-g(p)}{t}=:partial_i|_p g$$



      So I can define $partial_i|_p f:=partial_i|_p g$ and this value is indipendent of the choice of $g$ by $[1]$.



      So I have shown the claim in the above yellow box.




      Is my argument right? In John M. Lee Textbook Introduction to smooth manifolds, on page 27, he says that By continuity, all partial derivatives of $f$ at points of $Ucap partialmathbb{H}^n$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension.




      But it seems I dont'use the continuity of any function, so I'm not convinced of having understood the matter.






      multivariable-calculus differential-geometry smooth-manifolds smooth-functions




      multivariable-calculus differential-geometry smooth-manifolds smooth-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 22:50







      Minato

















      asked Jan 8 at 17:31









      MinatoMinato

      595314




      595314






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Yes, your argument works. The argument I had in mind was this: Given $pin partial mathbb H^n$, there's a sequence of points $q_jin text{Int} mathbb H^n$ such that $q_jto p$. Because $partial g/partial x^i$ is continuous,
          $$
          frac{partial g}{partial x^i}(p)
          =
          lim_{jto infty} frac{partial g}{partial x^i}(q_j)
          =
          lim_{jto infty} frac{partial f}{partial x^i}(q_j),
          $$
          and this last expression manifestly depends only on the original function $f$.



          This argument might be a little simpler, and it has the advantage that it applies to any subset of $mathbb R^n$ that is the closure of its interior, not just to $mathbb H^n$. But there's nothing wrong with your argument.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much prof. Lee :)
            $endgroup$
            – Minato
            Jan 9 at 17:55










          • $begingroup$
            @Minato — you’re welcome!
            $endgroup$
            – Jack Lee
            Jan 10 at 0:19












          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: "69"
          };
          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%2fmath.stackexchange.com%2fquestions%2f3066478%2fdefinition-of-smooth-functions-on-arbitrary-subsets-of-mathbbrn-and-partia%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









          1












          $begingroup$

          Yes, your argument works. The argument I had in mind was this: Given $pin partial mathbb H^n$, there's a sequence of points $q_jin text{Int} mathbb H^n$ such that $q_jto p$. Because $partial g/partial x^i$ is continuous,
          $$
          frac{partial g}{partial x^i}(p)
          =
          lim_{jto infty} frac{partial g}{partial x^i}(q_j)
          =
          lim_{jto infty} frac{partial f}{partial x^i}(q_j),
          $$
          and this last expression manifestly depends only on the original function $f$.



          This argument might be a little simpler, and it has the advantage that it applies to any subset of $mathbb R^n$ that is the closure of its interior, not just to $mathbb H^n$. But there's nothing wrong with your argument.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much prof. Lee :)
            $endgroup$
            – Minato
            Jan 9 at 17:55










          • $begingroup$
            @Minato — you’re welcome!
            $endgroup$
            – Jack Lee
            Jan 10 at 0:19
















          1












          $begingroup$

          Yes, your argument works. The argument I had in mind was this: Given $pin partial mathbb H^n$, there's a sequence of points $q_jin text{Int} mathbb H^n$ such that $q_jto p$. Because $partial g/partial x^i$ is continuous,
          $$
          frac{partial g}{partial x^i}(p)
          =
          lim_{jto infty} frac{partial g}{partial x^i}(q_j)
          =
          lim_{jto infty} frac{partial f}{partial x^i}(q_j),
          $$
          and this last expression manifestly depends only on the original function $f$.



          This argument might be a little simpler, and it has the advantage that it applies to any subset of $mathbb R^n$ that is the closure of its interior, not just to $mathbb H^n$. But there's nothing wrong with your argument.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much prof. Lee :)
            $endgroup$
            – Minato
            Jan 9 at 17:55










          • $begingroup$
            @Minato — you’re welcome!
            $endgroup$
            – Jack Lee
            Jan 10 at 0:19














          1












          1








          1





          $begingroup$

          Yes, your argument works. The argument I had in mind was this: Given $pin partial mathbb H^n$, there's a sequence of points $q_jin text{Int} mathbb H^n$ such that $q_jto p$. Because $partial g/partial x^i$ is continuous,
          $$
          frac{partial g}{partial x^i}(p)
          =
          lim_{jto infty} frac{partial g}{partial x^i}(q_j)
          =
          lim_{jto infty} frac{partial f}{partial x^i}(q_j),
          $$
          and this last expression manifestly depends only on the original function $f$.



          This argument might be a little simpler, and it has the advantage that it applies to any subset of $mathbb R^n$ that is the closure of its interior, not just to $mathbb H^n$. But there's nothing wrong with your argument.






          share|cite|improve this answer









          $endgroup$



          Yes, your argument works. The argument I had in mind was this: Given $pin partial mathbb H^n$, there's a sequence of points $q_jin text{Int} mathbb H^n$ such that $q_jto p$. Because $partial g/partial x^i$ is continuous,
          $$
          frac{partial g}{partial x^i}(p)
          =
          lim_{jto infty} frac{partial g}{partial x^i}(q_j)
          =
          lim_{jto infty} frac{partial f}{partial x^i}(q_j),
          $$
          and this last expression manifestly depends only on the original function $f$.



          This argument might be a little simpler, and it has the advantage that it applies to any subset of $mathbb R^n$ that is the closure of its interior, not just to $mathbb H^n$. But there's nothing wrong with your argument.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 5:39









          Jack LeeJack Lee

          27.7k54868




          27.7k54868












          • $begingroup$
            Thank you very much prof. Lee :)
            $endgroup$
            – Minato
            Jan 9 at 17:55










          • $begingroup$
            @Minato — you’re welcome!
            $endgroup$
            – Jack Lee
            Jan 10 at 0:19


















          • $begingroup$
            Thank you very much prof. Lee :)
            $endgroup$
            – Minato
            Jan 9 at 17:55










          • $begingroup$
            @Minato — you’re welcome!
            $endgroup$
            – Jack Lee
            Jan 10 at 0:19
















          $begingroup$
          Thank you very much prof. Lee :)
          $endgroup$
          – Minato
          Jan 9 at 17:55




          $begingroup$
          Thank you very much prof. Lee :)
          $endgroup$
          – Minato
          Jan 9 at 17:55












          $begingroup$
          @Minato — you’re welcome!
          $endgroup$
          – Jack Lee
          Jan 10 at 0:19




          $begingroup$
          @Minato — you’re welcome!
          $endgroup$
          – Jack Lee
          Jan 10 at 0:19


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • 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%2fmath.stackexchange.com%2fquestions%2f3066478%2fdefinition-of-smooth-functions-on-arbitrary-subsets-of-mathbbrn-and-partia%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

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more