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I have a question about how well one can choose a partition of unity.

So suppose $(M,g)$ is an open Riemannian manifold. I would like to have the following statement to be true:

"There exists an open cover $\{U_n\}$ of $M$ which is uniformly locally finite (i.e. each point of $p \in M$ lies in at most $q$ sets) and uniformly bounded (i.e. $\mathcal{U} := \bigcup U_n \times U_n$ is a controlled subset of $M \times M$, i.e. $\sup_{p \in \mathcal{U}} d(\pi_1(p), \pi_2(p)) < \infty$, where the $\pi_i$ are the projections on the first and second coordinate).

Furthermore, there exists an subordinate partition of unity $\{g_n\}$ with the properties that the functions $\{g_n^{1/2}\}$ are also smooth and for every $l \in \mathbb{N}$ the $i$-th derivatives $(g_n^{1/2})^{(i)}$ are uniformly bounded for all $i \le l$, i.e. $\|(g_n^{1/2})^{(i)}\|_\infty := \sup_{p \in M} |\nabla_{v_1, \ldots, v_i} g_n^{1/2}(p)| \le G_l$, where the $v_i$ are unit vectors at the point $p \in M$."

I think the most critical point is the uniform boundedness of the derivatives. Maybe someone knows a reference where it is proven that such a partition of unity always exists (or maybe just one, which has only the property of uniform boundedness of the derivatives)? Is this statement even true?

Thanks, Alex

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Asked it at MathOverflow: mathoverflow.net/questions/59182 –  AlexE Mar 22 '11 at 12:59
This question can be closed since it got answered at MathOverflow. –  AlexE Mar 23 '11 at 8:17
Actually IIRC the agreed-upon protocol is to repost the accepted answer as community wiki (so you don't get points for it) and then accept it. That way the site has fewer closed or unanswered questions. –  Aaron Mazel-Gee Mar 24 '11 at 8:03

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