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Let $g$ be a $C^r$function defined in a neighborhood $U$ of $\mathbf{x_0}$ in $\mathbb{R}^n$. Show that if $\phi:\mathbb{R}^n\to \mathbb{R} $ is a $C^r$ function whose support lies in $U$, then the function:$$h(\mathbf{x})=\begin{cases}\phi(\mathbf{x})g(\mathbf{x})&\mbox{for }\mathbf{x}\in U\\0\qquad &\mbox{for }\mathbf{x}\notin \mbox{Support }\phi\end{cases}$$
is well-defined and of class $C^r$ on $\mathbb{R}^n$.

My approach: outside Support $\phi$, $h$ is clearly well-defined (if $\mathbf{x}\in U$ then $h$ seems to give two values, however $\phi$ is zero in a neighborhood of $\mathbf{x}$, hence $h=0$) and of class $C^r$ since it's identically zero (note that $\mathbb{R}^n-$ Support $\phi$ is open, since Support $\phi$ is closed). If $\mathbf{x}\in \operatorname{Int}$ (Support $\phi$), then $h$ equals the product of $\phi$ and $g$; therefore $h$ is of class $C^r$. Now what happens if $\mathbf{x}\in \operatorname{Bd}$ (Support $\phi$)? I'm not sure about my approach, any help would be greatly appreciated.

user239726
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  • Instead of looking at the exterior and interior of the support, look at the exterior of the support and $U$. – Daniel Fischer May 11 '15 at 17:43
  • What do you mean, am I making a mistake? – user239726 May 11 '15 at 17:46
  • Tactically, yes. It's much simpler to avoid the technicalities you face when explicitly looking at points in the boundary of the support. You have two open sets, $U$ and $\mathbb{R}^n \setminus \operatorname{supp} \phi$ (I assume that in your course, "neighbourhood" is defined as an open set, for if it's not, and neighbourhoods can be compact for example, the assertion need not hold). On each of the two open sets, the behaviour is simple. – Daniel Fischer May 11 '15 at 17:52
  • Yes, of course by neighbourhood I meant an open set – user239726 May 11 '15 at 17:55
  • Would you be so kind as to give me an answer, cause I'm new in calculus? :) – user239726 May 11 '15 at 17:59

2 Answers2

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I am assuming you mean that $\text{supp}(\phi)=\overline{\{x \in \mathbb{R}^n \ : \ \phi(x)\neq 0\}}.$ The key is that smoothness is a purely local phenomenon.

Let $x_0 \in V = \mathbb{R^n} - \text{supp}(\phi).$ Then we can find a ball $B(x_0) \subseteq V$ for which $h(x)=0$ on $B(x_0).$ Thus $h$ is $C^r$ on $B(x_0).$

Similary, $h(x)$ is $C^r$ on any ball $B(x_0)\subseteq U.$ Since $U \cup V = \mathbb{R}^n$, we have $h \in C^r(\mathbb{R}^n,\mathbb{R}).$

matt biesecker
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A point of the boundary of the support is the limit of some sequence of points out of the support. Since $\phi$ is continuous, $\phi$ vanishes at the boundary.

ajotatxe
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