# partition of unity

• Sep 11th 2009, 05:36 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!!
• Sep 11th 2009, 10:41 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 $\phi(x) = \begin{cases}e^{\frac1{1-x^2}}&(|x|<1),\\ 0&\text{(otherwise).} \end{cases}$
• Sep 12th 2009, 10:26 PM
yemino
Thanks Opalg,

but, how can I prove that

$\displaystyle\sum_{m\in\mathbb{Z}}\phi (x-m)<\infty$
• Sep 13th 2009, 12:55 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 $\Bigl|\sum_{m\in\mathbb{Z}}\phi (x-m)\Bigr| < 2\sup\{|\phi|\}$.
• Sep 13th 2009, 08:28 AM
yemino
Thanks.. excelent answer..!! (Clapping) thanks again.