Smooth partitions of unity

Question

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

Answer 1

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.

Answer 2

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!