Product of a special $C^infty (Bbb R^d)$ and $C^2 (overline{Bbb R^d setminus B(0,R)})$ is $C^2 (Bbb R^d)$
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Let $r > 0$. Suppose $fin C^2 (overline{Bbb R^d setminus B(0,r)})$, where $B(0,r) := { xin Bbb R^d : Vert x Vert leq r }$. This means $fin C^2 (Bbb R^d setminus B(0,r))$ and has a continuous extension to the boundary of $B(0,r)$. Take $R > r$. Now define
$$varphi (x) := begin{cases} exp left( frac{R^2 - Vert x Vert^2}{r^2 - Vert x Vert^2} right) &: r < Vert x Vert < R\ 1 &: Vert x Vert geq R \ 0 &: Vert x Vert leq r end{cases}$$
Does it hold that $varphi in C^infty(Bbb R^d)$ and $varphi f in C^2(Bbb R^d)$? And if so, how do I show it?
derivatives smooth-functions
asked Nov 16 at 19:06
Falrach
1,494223
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Let $r > 0$. Suppose $fin C^2 (overline{Bbb R^d setminus B(0,r)})$, where $B(0,r) := { xin Bbb R^d : Vert x Vert leq r }$. This means $fin C^2 (Bbb R^d setminus B(0,r))$ and has a continuous extension to the boundary of $B(0,r)$. Take $R > r$. Now define
$$varphi (x) := begin{cases} exp left( frac{R^2 - Vert x Vert^2}{r^2 - Vert x Vert^2} right) &: r < Vert x Vert < R\ 1 &: Vert x Vert geq R \ 0 &: Vert x Vert leq r end{cases}$$
Does it hold that $varphi in C^infty(Bbb R^d)$ and $varphi f in C^2(Bbb R^d)$? And if so, how do I show it?
derivatives smooth-functions
asked Nov 16 at 19:06
Falrach
1,494223
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $r > 0$. Suppose $fin C^2 (overline{Bbb R^d setminus B(0,r)})$, where $B(0,r) := { xin Bbb R^d : Vert x Vert leq r }$. This means $fin C^2 (Bbb R^d setminus B(0,r))$ and has a continuous extension to the boundary of $B(0,r)$. Take $R > r$. Now define
$$varphi (x) := begin{cases} exp left( frac{R^2 - Vert x Vert^2}{r^2 - Vert x Vert^2} right) &: r < Vert x Vert < R\ 1 &: Vert x Vert geq R \ 0 &: Vert x Vert leq r end{cases}$$
Does it hold that $varphi in C^infty(Bbb R^d)$ and $varphi f in C^2(Bbb R^d)$? And if so, how do I show it?
derivatives smooth-functions
asked Nov 16 at 19:06
Falrach
1,494223
Let $r > 0$. Suppose $fin C^2 (overline{Bbb R^d setminus B(0,r)})$, where $B(0,r) := { xin Bbb R^d : Vert x Vert leq r }$. This means $fin C^2 (Bbb R^d setminus B(0,r))$ and has a continuous extension to the boundary of $B(0,r)$. Take $R > r$. Now define
$$varphi (x) := begin{cases} exp left( frac{R^2 - Vert x Vert^2}{r^2 - Vert x Vert^2} right) &: r < Vert x Vert < R\ 1 &: Vert x Vert geq R \ 0 &: Vert x Vert leq r end{cases}$$
Does it hold that $varphi in C^infty(Bbb R^d)$ and $varphi f in C^2(Bbb R^d)$? And if so, how do I show it?
derivatives smooth-functions
derivatives smooth-functions
asked Nov 16 at 19:06
Falrach
1,494223
asked Nov 16 at 19:06
Falrach
1,494223
asked Nov 16 at 19:06
Falrach
1,494223
asked Nov 16 at 19:06
Falrach
1,494223
asked Nov 16 at 19:06
Falrach
1,494223
1,494223
add a comment |
add a comment |
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