Smooth extension of function on $mathbb{R}^p$











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I have the following specific problem:



For an application I use the following function $g : Bbb R^{Nd} to [0,infty)$,



$$(x^1, ldots, x^N) mapsto begin{cases} prodlimits_{substack{(a,b)in{1,ldots,N }^2 \ a< b}} exp left( frac{1}{(Vert x^aVert - Vert x^b Vert)^2 -1} right) &: bigvert Vert x^aVert - Vert x^b Vert bigvert < 1 ;forall a,b=1,ldots,N \ 0 &: text{else} end{cases}$$
where $x^i inBbb R^d$ and $VertcdotVert$ denotes the euclidean norm on $Bbb R ^d$. This special form of function is just important on the complement $K^c$ of a compact set. Moreover, I need that the function is $C^2$, i.e. twice continuously differentiable, on $K^c$.



For this purpose, let $$B_R := left{xinBbb R^{Nd} : maxlimits_{a=1,ldots,N} Vert x^a Vert leq R right}$$ for a radius $R>0$. Note that $g$ restricted to $B_1^c$ is already $C^2$ on $Bbb R^{Nd} setminus B_1^c$.



So here my question:



Is there a way to get the existence of a function $G:Bbb R^{Nd} to [0,infty)$ which is $C^2$ and all derivatives of second order at most match with the derivatives of $g$ on $Bbb R^{Nd} setminus B_1^c$ ?



I found extensions theorems for sobolev spaces on domains with a nice boundary. Is there something similar applying here?










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    up vote
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    down vote

    favorite
    1












    I have the following specific problem:



    For an application I use the following function $g : Bbb R^{Nd} to [0,infty)$,



    $$(x^1, ldots, x^N) mapsto begin{cases} prodlimits_{substack{(a,b)in{1,ldots,N }^2 \ a< b}} exp left( frac{1}{(Vert x^aVert - Vert x^b Vert)^2 -1} right) &: bigvert Vert x^aVert - Vert x^b Vert bigvert < 1 ;forall a,b=1,ldots,N \ 0 &: text{else} end{cases}$$
    where $x^i inBbb R^d$ and $VertcdotVert$ denotes the euclidean norm on $Bbb R ^d$. This special form of function is just important on the complement $K^c$ of a compact set. Moreover, I need that the function is $C^2$, i.e. twice continuously differentiable, on $K^c$.



    For this purpose, let $$B_R := left{xinBbb R^{Nd} : maxlimits_{a=1,ldots,N} Vert x^a Vert leq R right}$$ for a radius $R>0$. Note that $g$ restricted to $B_1^c$ is already $C^2$ on $Bbb R^{Nd} setminus B_1^c$.



    So here my question:



    Is there a way to get the existence of a function $G:Bbb R^{Nd} to [0,infty)$ which is $C^2$ and all derivatives of second order at most match with the derivatives of $g$ on $Bbb R^{Nd} setminus B_1^c$ ?



    I found extensions theorems for sobolev spaces on domains with a nice boundary. Is there something similar applying here?










    share|cite|improve this question


























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      I have the following specific problem:



      For an application I use the following function $g : Bbb R^{Nd} to [0,infty)$,



      $$(x^1, ldots, x^N) mapsto begin{cases} prodlimits_{substack{(a,b)in{1,ldots,N }^2 \ a< b}} exp left( frac{1}{(Vert x^aVert - Vert x^b Vert)^2 -1} right) &: bigvert Vert x^aVert - Vert x^b Vert bigvert < 1 ;forall a,b=1,ldots,N \ 0 &: text{else} end{cases}$$
      where $x^i inBbb R^d$ and $VertcdotVert$ denotes the euclidean norm on $Bbb R ^d$. This special form of function is just important on the complement $K^c$ of a compact set. Moreover, I need that the function is $C^2$, i.e. twice continuously differentiable, on $K^c$.



      For this purpose, let $$B_R := left{xinBbb R^{Nd} : maxlimits_{a=1,ldots,N} Vert x^a Vert leq R right}$$ for a radius $R>0$. Note that $g$ restricted to $B_1^c$ is already $C^2$ on $Bbb R^{Nd} setminus B_1^c$.



      So here my question:



      Is there a way to get the existence of a function $G:Bbb R^{Nd} to [0,infty)$ which is $C^2$ and all derivatives of second order at most match with the derivatives of $g$ on $Bbb R^{Nd} setminus B_1^c$ ?



      I found extensions theorems for sobolev spaces on domains with a nice boundary. Is there something similar applying here?










      share|cite|improve this question















      I have the following specific problem:



      For an application I use the following function $g : Bbb R^{Nd} to [0,infty)$,



      $$(x^1, ldots, x^N) mapsto begin{cases} prodlimits_{substack{(a,b)in{1,ldots,N }^2 \ a< b}} exp left( frac{1}{(Vert x^aVert - Vert x^b Vert)^2 -1} right) &: bigvert Vert x^aVert - Vert x^b Vert bigvert < 1 ;forall a,b=1,ldots,N \ 0 &: text{else} end{cases}$$
      where $x^i inBbb R^d$ and $VertcdotVert$ denotes the euclidean norm on $Bbb R ^d$. This special form of function is just important on the complement $K^c$ of a compact set. Moreover, I need that the function is $C^2$, i.e. twice continuously differentiable, on $K^c$.



      For this purpose, let $$B_R := left{xinBbb R^{Nd} : maxlimits_{a=1,ldots,N} Vert x^a Vert leq R right}$$ for a radius $R>0$. Note that $g$ restricted to $B_1^c$ is already $C^2$ on $Bbb R^{Nd} setminus B_1^c$.



      So here my question:



      Is there a way to get the existence of a function $G:Bbb R^{Nd} to [0,infty)$ which is $C^2$ and all derivatives of second order at most match with the derivatives of $g$ on $Bbb R^{Nd} setminus B_1^c$ ?



      I found extensions theorems for sobolev spaces on domains with a nice boundary. Is there something similar applying here?







      real-analysis functions differential-topology smooth-functions






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      edited 9 hours ago

























      asked yesterday









      Falrach

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