Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By zzephod

Thread: Del Operator

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    USA
    Posts
    17

    Del Operator

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

    How does it follow that $\displaystyle \nabla(\theta_{0}) = - \sum_{i=1}^{k}\nabla(\theta_{i})$ ?
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member zzephod's Avatar
    Joined
    Apr 2012
    From
    Erewhon
    Posts
    261
    Thanks
    153

    Re: Del Operator

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

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

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

    .
    Last edited by zzephod; Sep 10th 2013 at 10:53 PM.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2012
    From
    USA
    Posts
    17

    Re: Del Operator

    Thanks for response, for every open cover there exists a partition of unity.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member zzephod's Avatar
    Joined
    Apr 2012
    From
    Erewhon
    Posts
    261
    Thanks
    153

    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.

    .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2012
    From
    USA
    Posts
    17

    Re: Del Operator

    Okay fair enough. I have another short question:

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

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

    Am I missing something? Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. sum operator
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Jun 30th 2012, 02:27 AM
  2. Zero Operator
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Apr 4th 2011, 12:23 AM
  3. Operator in R^n
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Jan 21st 2010, 08:22 AM
  4. D Operator
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Sep 15th 2009, 11:54 AM
  5. The use of an operator
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Mar 20th 2008, 01:47 PM

Search Tags


/mathhelpforum @mathhelpforum