partition of unity

Hi, I have the follow question:

Construct a function $\displaystyle \phi\in C_c^\infty (\mathbb{R})$ such that $\displaystyle \phi\geq 0$, $\displaystyle supp\phi\subset (-1,1)$

$\displaystyle \psi_n(x)=\phi(x-n)/\displaystyle_{m\in\mathbb{Z}}\phi(x-m),\;\;\; n\in\mathbb{Z}$

are a partition of unity subordinated to the cover $\displaystyle \cup_{n\in\mathbb{Z}}(n-1,n+)$ of $\displaystyle \mathbb{R}$.

Thanks!!

Quote:

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Originally Posted by yemino

Hi, I have the follow question:

Construct a function $\displaystyle \phi\in C_c^\infty (\mathbb{R})$ such that $\displaystyle \phi\geq 0$, $\displaystyle supp\phi\subset (-1,1)$

$\displaystyle \psi_n(x)=\phi(x-n)\big/{\color{red}\textstyle\sum}_{m\in\mathbb{Z}}\phi(x-m),\;\;\; n\in\mathbb{Z}$

are a partition of unity subordinated to the cover $\displaystyle \textstyle\bigcup_{n\in\mathbb{Z}}(n-1,n+{\color{red}1})$ of $\displaystyle \mathbb{R}$.

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The standard example of such a function is $\displaystyle \phi(x) = \begin{cases}e^{\frac1{1-x^2}}&(|x|<1),\\ 0&\text{(otherwise).} \end{cases}$

Thanks Opalg,

but, how can I prove that

$\displaystyle \displaystyle\sum_{m\in\mathbb{Z}}\phi (x-m)<\infty$

Quote:

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Originally Posted by yemino

how can I prove that

$\displaystyle \displaystyle\sum_{m\in\mathbb{Z}}\phi (x-m)<\infty$

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The sum is not really an infinite sum. In fact, the support of $\displaystyle \phi$ is the interval [–1,1], so $\displaystyle \phi(x-m)$ can only be nonzero if |x-m|<1. For a given value of x, this will happen for only two values of m. Therefore $\displaystyle \Bigl|\sum_{m\in\mathbb{Z}}\phi (x-m)\Bigr| < 2\sup\{|\phi|\}$.

Thanks.. excelent answer..!! (Clapping) thanks again.