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Suppose that $E$ is a compact subset of $\mathbb R$. If for every $x\in E$ there exist a nonnegative function $f=f_x$ and an $r=r(x)>0$ such that $f$ is $C^\infty$ on $\mathbb R$, $f(t)=1$ for $t\in (x-r, x+r)$, and $f(t)=0$ for $t\notin (x-2r,x+2r)$, prove that there exist a differentiable function $f$, a nonzero constant $M$, and a bounded, open set $V$ which contains $E$ such that $1\leq f(x)\leq M$ for all $x\in E$ and $f(x)=0$ for $x\notin V$.

Robert Israel
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JFK
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2 Answers2

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If $f_x$ is such a function, so is $g(t) = f_x(x + a (t - x))$ with $r(x)$ replaced by $r(x)/a$. Take $f = g$ where $a > 0$ is small enough that $E \subseteq (x - r(x)/a, x + r(x)/a)$, and $V = (x - 2 r(x)/a, x + 2 r(x)/a)$. All we need about $E$ is that compact sets are bounded.

Robert Israel
  • 448,999
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Clearly, $$E\subseteq\bigcup_{x\in E}(x-r(x),x+r(x)).$$ Since $E$ is compact, there is a finite subcover $$E\subseteq\bigcup_{1\le i\le M}(x_i-r(x_i),x_i+r(x_i)).$$ Let $$f(x)=\sum_{i=1}^M f_{x_i}(x)$$ and $$V = \bigcup_{1\le i\le M}(x_i-2r(x_i),x_i+2r(x_i)).$$

Can you see that the required properties hold for this choice of $M,V,f$?

  • More simply, note that $E\subseteq F$ with $F=[-K,+K]$ for some $K$, choose $V=(-K-1,+K+1)$, $M=1$, and one of the usual smooth functions $f$ such that $\mathbb 1_F\leqslant f\leqslant\mathbf 1_V$. Or am I missing something? – Did Feb 17 '13 at 08:42