partition of unity

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September 11th 2009, 06:36 AM
yemino

partition of unity

Hi, I have the follow question:

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

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

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

Thanks!!
September 11th 2009, 11:41 AM
Opalg

Quote:

Originally Posted by yemino

Hi, I have the follow question:

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

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

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

The standard example of such a function is $\phi(x) = \begin{cases}e^{\frac1{1-x^2}}&(|x|<1),\\ 0&\text{(otherwise).} \end{cases}$
September 12th 2009, 11:26 PM
yemino

Thanks Opalg,

but, how can I prove that

$\displaystyle\sum_{m\in\mathbb{Z}}\phi (x-m)<\infty$
September 13th 2009, 01:55 AM
Opalg

Quote:

Originally Posted by yemino

how can I prove that

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

The sum is not really an infinite sum. In fact, the support of $\phi$ is the interval [–1,1], so $\phi(x-m)$ can only be nonzero if |x-m|<1. For a given value of x, this will happen for only two values of m. Therefore $\Bigl|\sum_{m\in\mathbb{Z}}\phi (x-m)\Bigr| < 2\sup\{|\phi|\}$ .
September 13th 2009, 09:28 AM
yemino

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