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Math Help - partition of unity

  1. #1
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    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!!
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  2. #2
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    Quote Originally Posted by yemino View Post
    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}
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  3. #3
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    Thanks Opalg,

    but, how can I prove that

    \displaystyle\sum_{m\in\mathbb{Z}}\phi (x-m)<\infty
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  4. #4
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    Quote Originally Posted by yemino View Post
    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|\}.
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  5. #5
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    Thanks.. excelent answer..!! thanks again.
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