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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!
$\displaystyle \theta_0=1-\sum_{i=1}^k \theta_i$
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.
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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.Quote:
Originally Posted by Johnyboy
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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.
