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Suppose $\{U_\alpha\}$ is an atlas of coordinate patches of a (noncompact) smooth manifold $M$ of dimension $n$, with coordinates $(x_\alpha^1,\dots,x_\alpha^n)$ on $U_\alpha$. Furthermore we assume that $\{U_\alpha\}$ is locally finite and countable whose members all have compact closure. My question is:

Q: Does there exist a uniform constant $C>0$ (which only depends on the cover $\{U_\alpha\} $ and the chosen coordinate system) and two sets of smooth partition of unity $\{\rho_\alpha\}$ and $\{\phi_\alpha\}$ such that

(1) $\mathrm{supp}\,\rho_\alpha \subset \mathrm{int}(\mathrm{supp}\phi _\alpha)\subset \mathrm{supp}\,\phi_\alpha \subset U_\alpha$ for every $\alpha$ and

(2) $|\nabla \rho_\alpha|\leqslant C\phi_\alpha$ on $U_\alpha$ for every $\alpha$, where the gradient is computed with respect to THE COORDINATE SYSTEM $(x_\alpha^1,\dots,x_\alpha^n)$?

I have found a seemingly related post Existence of a partition of unity with uniformly bounded derivatives. but there the gradient is taken with respect to some riemannian metric on $M$. I wonder whether my question, which seems more elementary and local, could have an affirmative answer.

Many thanks! Anar

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    $\begingroup$ No. For example $M=\mathbb R$, $U_n=(\log n-2^{-n},\log (n+1)+2^{-n})$ (and other $U$'s that cover the negative part of $\mathbb R$). Then since $|U_n|\to 0$, you cannot keep $\rho'_n$ uniformly bounded. $\endgroup$ Commented Feb 26 at 15:17
  • $\begingroup$ Christian, you seem to have examined the statement for a specific partition of unity, but my question asks for existence of not only the constant $C$, but also two sets of partition of unity, which, if exist, could be very different ones from the one you mentioned. $\endgroup$
    – Anar C
    Commented Feb 27 at 18:29
  • $\begingroup$ Not at all. What I'm saying is that whatever $\rho_n,\phi_n$ are for these $U_n$, it can never be true that $\rho'_n\le C\phi_n\le C$ for all $n$, for any $C$. $\endgroup$ Commented Feb 27 at 21:59
  • $\begingroup$ You are very right! Thank you very much. $\endgroup$
    – Anar C
    Commented Mar 1 at 19:51

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