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Let $\{U_s\}_{s\in S}$ be a family of open subsets of $\mathbb R$ with union $W$ with the following property: for each $x\in U_s$ we have $x+1, x-1\in U_s$.

Does there exist a sequence $\phi_1,\phi_2, \ldots$ of $1$-periodic, smooth, compactly supported nonnegative functions such that:

  1. for each $n\in \mathbb N$ there is a $s(n)\in S$ such that $supp \phi_n \subset U_{s(n)}$

  2. $\sum_{n=1}^\infty \phi_n(x)=1$ for $x\in \mathbb R$,

  3. for each compact $K\subset W$ there is an open set $G$ such that $G$ intersects only finite among sets $supp \phi_1, supp \phi_2,\ldots$.

A.B
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1 Answers1

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Since the closed interval $[0,1]$ is compact, finitely many of $U_s$'s, say $U_{s_1},\cdots, U_{s_m}$ , can cover it. Because of the assumption, these open sets cover whole $\mathbb{R}$ too. So they make an open cover for $\mathbb{R}$.

Because of your assumption the image of elements of these open cover under the quotient map $\mathbb{R}\to \mathbb{T}$, $t\mapsto e^{2\pi it}$ is an open cover for the circle $\mathbb{T}$.

Consider a partition of unity subordinate to these smaller open cover for $\mathbb{T}$, say $\psi_1,\cdots, \psi_m$. The pullback of this partition of unity under the above quotient map is the answer. By pullback of $\psi_i$, I mean the function defined by $\phi_i(x):=\psi_i(e^{2\pi ix})$ for all $i=1,\cdots,m$.

Remember such a partition of unity has to be finite and the last part of the question automatically follows from the construction.

Edit: The above answer is correct only if $W=\mathbb{R}$, and only in this case the smaller open cover is finite. When $W\neq \mathbb{R}$, you can check that the image of $W$ under the quotient map, say $\overline{W}$, is an open subset of $\mathbb{T}$ and instead of a finite cover for $\mathbb{T}$, you need to consider an infinite cover for $\overline{W}$ and find a subordinate partition of unity with respect to this infinite open cover, say $\{ \psi_i\}_{i=1}^\infty$ and then take the pullback of this partition of unity.

  • At first, I didn't pay attention to the requirement that $\phi_i$ should be periodic. But I think the current version is fine. –  Jul 21 '14 at 19:13
  • Two questions. 1. Have you assumed additionally that the union of $U_s$ is $R$? 2.Is there a theorem about smooth partition of unity for $T$? – A.B Jul 21 '14 at 19:18
  • It follows from your assumption that $x\in U_s$ implies that $x+1, x-1\in U_s$, and more generally $x+n, x-n\in U_s$ for all $n\in \mathbb{N}$. –  Jul 21 '14 at 19:23
  • If you take for example family consisting of one set $U_1=\bigcup_{k\in \mathbb Z}(-1/3+k,1/3+k)$ the sum this family is $U_1$ which is not $R$. – A.B Jul 21 '14 at 19:30