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Math Help - Del Operator

  1. #1
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    Del Operator

    Let U be an open bounded subset of \mathbb{R}^{n}, assume we have a finite open covering of U.
    Consider the partition of unity subordinated: \theta_{0},  \theta_{1}...\theta_{k} \in C^{\infty}(\mathbb{R}^{n}) and \sum_{i=0}^{k}\theta_{i} = 1 where 0 \leq \theta_{i} \leq 1.

    How does it follow that \nabla(\theta_{0}) = - \sum_{i=1}^{k}\nabla(\theta_{i}) ?
    Thanks!
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  2. #2
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    Re: Del Operator

    \theta_0=1-\sum_{i=1}^k \theta_i

    then as \nabla 1=0 and \nabla(\alpha A+\beta B)=\alpha \nabla(A)+\beta \nabla(B) the result follows.

    Though what your first sentence has to do with the rest of the question escapes me.

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    Last edited by zzephod; September 10th 2013 at 11:53 PM.
    Thanks from topsquark
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  3. #3
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    Re: Del Operator

    Thanks for response, for every open cover there exists a partition of unity.
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  4. #4
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    Re: Del Operator

    Quote Originally Posted by Johnyboy View Post
    Thanks for response, for every open cover there exists a partition of unity.
    I know what it means, but it has nothing to do with the second part, nor is it a question.

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  5. #5
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    Re: Del Operator

    Okay fair enough. I have another short question:

    Since \theta_{i} are real-valued functions, \nabla(\theta_{i}) are vector fields according to my understanding of how the \nabla operator works.
    In the book I am working from I am given that for i = 1,...,k it follows that \theta_{i} has compact support and it follows then that
    each \nabla(\theta_{i}) has compact support? But \nabla(\theta_{i}) is a vector field, does the notion of compact support extend to vector fields?
    If so how?

    Secondly it is stated that \nabla(\theta_{0}) \in L^{\infty}(\mathbb{R}^n) , but \nabla(\theta_{0}) too is a vector field.
    My understanding is that f  \in L^{\infty}(\mathbb{R}^n) implies f: \mathbb{R}^n \rightarrow \mathbb{R} where f is measurable and f is bounded a.e..

    Am I missing something? Thanks.
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