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
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asked Nov 16 at 1...