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




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      edited Jan 8 at 22:50







      Minato

















      asked Jan 8 at 17:31









      MinatoMinato

      445212




      445212






















          1 Answer
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          $begingroup$

          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





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          1 Answer
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          active

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          0












          $begingroup$

          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
















          0












          $begingroup$

          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














          0












          0








          0





          $begingroup$

          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

          27.2k54565




          27.2k54565












          • $begingroup$
            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


















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