Does $nabla u$ have null tangential component to $mathrm{supp} u$?












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Let $u: mathbb{R}^N to mathbb{R}$ be a smooth function. Does $nabla u$ have null tangential component to $mathrm{supp}, u $ ? That is, do we have
$$frac{partial u}{partial nu} = 0 Rightarrow |nabla u| = 0$$
on $mathrm{supp} u$, where $nu $ denotes the outer normal to $mathrm{supp} u$?






calculus multivariable-calculus derivatives differential-geometry







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    0












    $begingroup$


    Let $u: mathbb{R}^N to mathbb{R}$ be a smooth function. Does $nabla u$ have null tangential component to $mathrm{supp}, u $ ? That is, do we have
    $$frac{partial u}{partial nu} = 0 Rightarrow |nabla u| = 0$$
    on $mathrm{supp} u$, where $nu $ denotes the outer normal to $mathrm{supp} u$?






    calculus multivariable-calculus derivatives differential-geometry







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $u: mathbb{R}^N to mathbb{R}$ be a smooth function. Does $nabla u$ have null tangential component to $mathrm{supp}, u $ ? That is, do we have
      $$frac{partial u}{partial nu} = 0 Rightarrow |nabla u| = 0$$
      on $mathrm{supp} u$, where $nu $ denotes the outer normal to $mathrm{supp} u$?






      calculus multivariable-calculus derivatives differential-geometry







      share|cite|improve this question











      $endgroup$




      Let $u: mathbb{R}^N to mathbb{R}$ be a smooth function. Does $nabla u$ have null tangential component to $mathrm{supp}, u $ ? That is, do we have
      $$frac{partial u}{partial nu} = 0 Rightarrow |nabla u| = 0$$
      on $mathrm{supp} u$, where $nu $ denotes the outer normal to $mathrm{supp} u$?






      calculus multivariable-calculus derivatives differential-geometry




      calculus multivariable-calculus derivatives differential-geometry






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 13 at 1:01







      Dal

















      asked Jun 20 '18 at 12:53









      DalDal

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          $begingroup$

          $u$ is smooth on all of $Bbb R^n$, including on the boundary of $text{supp }u$. Outside of $text{supp }u$, we obviously have $nabla u = 0$. Therefore by continuity, we have $nabla u = 0$ on the boundary of $text{supp }u$ as well.






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            $begingroup$

            $u$ is smooth on all of $Bbb R^n$, including on the boundary of $text{supp }u$. Outside of $text{supp }u$, we obviously have $nabla u = 0$. Therefore by continuity, we have $nabla u = 0$ on the boundary of $text{supp }u$ as well.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              $u$ is smooth on all of $Bbb R^n$, including on the boundary of $text{supp }u$. Outside of $text{supp }u$, we obviously have $nabla u = 0$. Therefore by continuity, we have $nabla u = 0$ on the boundary of $text{supp }u$ as well.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                $u$ is smooth on all of $Bbb R^n$, including on the boundary of $text{supp }u$. Outside of $text{supp }u$, we obviously have $nabla u = 0$. Therefore by continuity, we have $nabla u = 0$ on the boundary of $text{supp }u$ as well.






                share|cite|improve this answer









                $endgroup$



                $u$ is smooth on all of $Bbb R^n$, including on the boundary of $text{supp }u$. Outside of $text{supp }u$, we obviously have $nabla u = 0$. Therefore by continuity, we have $nabla u = 0$ on the boundary of $text{supp }u$ as well.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jun 20 '18 at 23:53









                Paul SinclairPaul Sinclair

                19.7k21442




                19.7k21442






























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