Del Operator

Printable View

September 10th 2013, 11:27 AM
Johnyboy

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!
September 10th 2013, 11:47 PM
zzephod

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.

.
September 11th 2013, 04:48 AM
Johnyboy

Re: Del Operator

Thanks for response, for every open cover there exists a partition of unity.
September 11th 2013, 08:57 AM
zzephod

Re: Del Operator

Quote:

Originally Posted by Johnyboy

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.

.
September 11th 2013, 02:19 PM
Johnyboy

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.