Fake proof of differentiability












3














It's a theorem that if $fcolon UsubsetBbb R^nto Bbb R^m$ has the property that each of the partial derivatives $partial_if_j$ exist and are continuous $pin U$, then $f$ is differentiable at $p$. When I was trying to prove this, I came up with the following "proof" which doesn't use the continuity hypothesis. Can someone tell me what's wrong with this proof?



Since $f_j$ is differentiable at $p$, we can write
$$
f_j(p+v) = f_j(p) + sum_i partial_if_j(p)v_i + R_j(v),
$$
where $|R_j(v)|/|v| to 0$ as $vto 0$. Hence, we can write
begin{align*}
f(p+v) &= f(p) + big(sum_i partial_if_1(p)v_i + R_1(v),dots,sum_i partial_if_m(p)v_i + R_m(v)big) \
&= f(p) + sum_jbig(sum_ipartial_if_j(p)v_ibig)e_j + R_j(v)e_j \
&= f(p) + [Df_p][v] + (R_1,dots,R_m)(v),
end{align*}
where $[Df_p] = [partial_if_j(p)]$ is the usual Jacobian matrix, and $[v]$ is the column vector $[v_1 dotsb v_n]^T$. Now,
$$
frac{|(R_1,dots,R_m)(v)|^2}{|v|^2} = frac{R_1(v)^2 + dots + R_m(v)^2}{|v|^2} to 0,
$$
where the last expression goes to $0$ as $vto 0$ since it is a sum of finitely many terms, each of which goes to $0$. Hence we have written $f(p+v)$ as a sum of a constant term, a linear part, and a sublinear piece, so $f$ is differentiable at $p$. At no point did I explicitly use the continuity hypothesis, so what exactly is wrong with this proof? Best.






real-analysis multivariable-calculus derivatives proof-verification







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    3














    It's a theorem that if $fcolon UsubsetBbb R^nto Bbb R^m$ has the property that each of the partial derivatives $partial_if_j$ exist and are continuous $pin U$, then $f$ is differentiable at $p$. When I was trying to prove this, I came up with the following "proof" which doesn't use the continuity hypothesis. Can someone tell me what's wrong with this proof?



    Since $f_j$ is differentiable at $p$, we can write
    $$
    f_j(p+v) = f_j(p) + sum_i partial_if_j(p)v_i + R_j(v),
    $$
    where $|R_j(v)|/|v| to 0$ as $vto 0$. Hence, we can write
    begin{align*}
    f(p+v) &= f(p) + big(sum_i partial_if_1(p)v_i + R_1(v),dots,sum_i partial_if_m(p)v_i + R_m(v)big) \
    &= f(p) + sum_jbig(sum_ipartial_if_j(p)v_ibig)e_j + R_j(v)e_j \
    &= f(p) + [Df_p][v] + (R_1,dots,R_m)(v),
    end{align*}
    where $[Df_p] = [partial_if_j(p)]$ is the usual Jacobian matrix, and $[v]$ is the column vector $[v_1 dotsb v_n]^T$. Now,
    $$
    frac{|(R_1,dots,R_m)(v)|^2}{|v|^2} = frac{R_1(v)^2 + dots + R_m(v)^2}{|v|^2} to 0,
    $$
    where the last expression goes to $0$ as $vto 0$ since it is a sum of finitely many terms, each of which goes to $0$. Hence we have written $f(p+v)$ as a sum of a constant term, a linear part, and a sublinear piece, so $f$ is differentiable at $p$. At no point did I explicitly use the continuity hypothesis, so what exactly is wrong with this proof? Best.






    real-analysis multivariable-calculus derivatives proof-verification







    share|cite|improve this question

























      3












      3








      3







      It's a theorem that if $fcolon UsubsetBbb R^nto Bbb R^m$ has the property that each of the partial derivatives $partial_if_j$ exist and are continuous $pin U$, then $f$ is differentiable at $p$. When I was trying to prove this, I came up with the following "proof" which doesn't use the continuity hypothesis. Can someone tell me what's wrong with this proof?



      Since $f_j$ is differentiable at $p$, we can write
      $$
      f_j(p+v) = f_j(p) + sum_i partial_if_j(p)v_i + R_j(v),
      $$
      where $|R_j(v)|/|v| to 0$ as $vto 0$. Hence, we can write
      begin{align*}
      f(p+v) &= f(p) + big(sum_i partial_if_1(p)v_i + R_1(v),dots,sum_i partial_if_m(p)v_i + R_m(v)big) \
      &= f(p) + sum_jbig(sum_ipartial_if_j(p)v_ibig)e_j + R_j(v)e_j \
      &= f(p) + [Df_p][v] + (R_1,dots,R_m)(v),
      end{align*}
      where $[Df_p] = [partial_if_j(p)]$ is the usual Jacobian matrix, and $[v]$ is the column vector $[v_1 dotsb v_n]^T$. Now,
      $$
      frac{|(R_1,dots,R_m)(v)|^2}{|v|^2} = frac{R_1(v)^2 + dots + R_m(v)^2}{|v|^2} to 0,
      $$
      where the last expression goes to $0$ as $vto 0$ since it is a sum of finitely many terms, each of which goes to $0$. Hence we have written $f(p+v)$ as a sum of a constant term, a linear part, and a sublinear piece, so $f$ is differentiable at $p$. At no point did I explicitly use the continuity hypothesis, so what exactly is wrong with this proof? Best.






      real-analysis multivariable-calculus derivatives proof-verification







      share|cite|improve this question













      It's a theorem that if $fcolon UsubsetBbb R^nto Bbb R^m$ has the property that each of the partial derivatives $partial_if_j$ exist and are continuous $pin U$, then $f$ is differentiable at $p$. When I was trying to prove this, I came up with the following "proof" which doesn't use the continuity hypothesis. Can someone tell me what's wrong with this proof?



      Since $f_j$ is differentiable at $p$, we can write
      $$
      f_j(p+v) = f_j(p) + sum_i partial_if_j(p)v_i + R_j(v),
      $$
      where $|R_j(v)|/|v| to 0$ as $vto 0$. Hence, we can write
      begin{align*}
      f(p+v) &= f(p) + big(sum_i partial_if_1(p)v_i + R_1(v),dots,sum_i partial_if_m(p)v_i + R_m(v)big) \
      &= f(p) + sum_jbig(sum_ipartial_if_j(p)v_ibig)e_j + R_j(v)e_j \
      &= f(p) + [Df_p][v] + (R_1,dots,R_m)(v),
      end{align*}
      where $[Df_p] = [partial_if_j(p)]$ is the usual Jacobian matrix, and $[v]$ is the column vector $[v_1 dotsb v_n]^T$. Now,
      $$
      frac{|(R_1,dots,R_m)(v)|^2}{|v|^2} = frac{R_1(v)^2 + dots + R_m(v)^2}{|v|^2} to 0,
      $$
      where the last expression goes to $0$ as $vto 0$ since it is a sum of finitely many terms, each of which goes to $0$. Hence we have written $f(p+v)$ as a sum of a constant term, a linear part, and a sublinear piece, so $f$ is differentiable at $p$. At no point did I explicitly use the continuity hypothesis, so what exactly is wrong with this proof? Best.






      real-analysis multivariable-calculus derivatives proof-verification




      real-analysis multivariable-calculus derivatives proof-verification






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      asked 7 hours ago









      AOrtiz

      10.4k21341




      10.4k21341






















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          In short, if you don't assume that the $partial_i f_j$ are continuous then you can't assume that $f_j$ is differentiable at $p$.



          You only know that all partial derivatives of $f_j$ exist, but you need continuity to guarantee that $f_j$ is actually differentiable (that's the $m=1$ case of the theorem you talk about).






          share|cite|improve this answer





























            6














            The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $f_j$ is a differentiable function on $mathbb R^{n}$.






            share|cite|improve this answer





















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              2 Answers
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              active

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              4














              In short, if you don't assume that the $partial_i f_j$ are continuous then you can't assume that $f_j$ is differentiable at $p$.



              You only know that all partial derivatives of $f_j$ exist, but you need continuity to guarantee that $f_j$ is actually differentiable (that's the $m=1$ case of the theorem you talk about).






              share|cite|improve this answer


























                4














                In short, if you don't assume that the $partial_i f_j$ are continuous then you can't assume that $f_j$ is differentiable at $p$.



                You only know that all partial derivatives of $f_j$ exist, but you need continuity to guarantee that $f_j$ is actually differentiable (that's the $m=1$ case of the theorem you talk about).






                share|cite|improve this answer
























                  4












                  4








                  4






                  In short, if you don't assume that the $partial_i f_j$ are continuous then you can't assume that $f_j$ is differentiable at $p$.



                  You only know that all partial derivatives of $f_j$ exist, but you need continuity to guarantee that $f_j$ is actually differentiable (that's the $m=1$ case of the theorem you talk about).






                  share|cite|improve this answer












                  In short, if you don't assume that the $partial_i f_j$ are continuous then you can't assume that $f_j$ is differentiable at $p$.



                  You only know that all partial derivatives of $f_j$ exist, but you need continuity to guarantee that $f_j$ is actually differentiable (that's the $m=1$ case of the theorem you talk about).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 7 hours ago









                  0x539

                  599213




                  599213























                      6














                      The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $f_j$ is a differentiable function on $mathbb R^{n}$.






                      share|cite|improve this answer


























                        6














                        The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $f_j$ is a differentiable function on $mathbb R^{n}$.






                        share|cite|improve this answer
























                          6












                          6








                          6






                          The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $f_j$ is a differentiable function on $mathbb R^{n}$.






                          share|cite|improve this answer












                          The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $f_j$ is a differentiable function on $mathbb R^{n}$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 7 hours ago









                          Kavi Rama Murthy

                          49.2k31854




                          49.2k31854






























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