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Let $ M $ be a Riemannian manifold and let $ \{U_i\} $ be a countable covering of $ M $. It is well known that there exists a countable collection of smooth function with compact support $ \{\rho_i\} $ (called smooth partition of unity subordinated to $ \{U_i\} $) such that the collection of supports is locally finite and

$$ support(\rho_i) \subset U_i $$ $$ \sum_i \rho_i = 1 \;\;\; on \; M $$ $$ 0 \leq \rho_i \leq 1 $$

My question: is it possible to find a partion of unity subordinated to $ \{U_i\} $ with the following ADDITIONAL condition: there exists a constant independent on $ x \in M $ such that

$$ \sum_i |\nabla \rho_i(x)| < C \; \; \textrm{for every} \; x \in M \; ? $$

Thanks

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    $\begingroup$ The support of the $\nabla \rho_i$ is locally finite, so that is a locally finite sum, and as such finite. $\endgroup$ – Daniel Fischer Jul 16 '13 at 10:15
  • $\begingroup$ Thanks for your answer! I've just edited the question since in its previous form it could be not clear. My question is substantially distinct from the previous one. (the current question is actually what i want to know) $\endgroup$ – user55449 Jul 16 '13 at 10:40
  • $\begingroup$ Ah, globally bounded steepness. I think you can't have that in general, you'd need nice enough covers. But I'm not sure. $\endgroup$ – Daniel Fischer Jul 16 '13 at 10:46
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The answer to your revised question is no. Here's a counterexample: Let $M$ be the interval $(0,1)$ with the Euclidean metric, and consider the open cover consisting of the intervals $U_i=(\tfrac{1}{i+2},\tfrac{1}{i})$ for positive integers $i$. Every point of $M$ is in at most two of these open sets. If $x\in U_i\cap U_{i+1}$, then $\rho_i(x)+\rho_{i+1}(x)=1$, so at least one of these two functions must have a value greater than or equal to $\tfrac12$. If $\rho_i$ reaches a value greater than or equal to $\tfrac12$, then since its support is an interval of width $\tfrac{1}{i} - \tfrac{1}{i+2} = \tfrac{2}{i(i+2)}$, the magnitude of its derivative must be at least $\tfrac{i(i+2)}{2}$ somewhere. If this is not true for $\rho_i$, then it must be true for $\rho_{i+1}$, and thus the derivatives of the $\rho_i$'s are unbounded.

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The post is community-wiki as the solution has essentially been noted by Daniel Fischer above:

If $\rho_i$ vanishes in an open set, then its gradient also vanishes in that open set by the definition of the gradient. The hypothesis that the collection of the supports of the $\rho_i$ is locally finite implies that each point of $M$ possesses a neighbourhood which intersects finitely many such supports. Therefore, on that neighbourhood, all but finitely many $\rho_i$ will vanish and so all but finitely many $\nabla \rho_i$ will vanish. So, your sum is finite at each point of $M$.

I hope this helps!

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  • $\begingroup$ Thanks for your answer! I've just edited the question since in its previous form it could be not clear. My question is substantially distinct from the previous one. (the current question is actually what i want to know) $\endgroup$ – user55449 Jul 16 '13 at 10:42

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