TITLE: Partitions of Unity
QUESTION [5 upvotes]: Fix a metric $g$ on a smooth, closed manifold $\mathcal{M}$.  Take a finite subcover of the manifold from its atlas.  Is it true that any smooth partition of unity subordinate to this cover has uniformly bounded derivatives in $L^p (\mathcal{M})$ for each $p \geq 1$ ?

REPLY [4 votes]: A counterexample: Let $M$ be the unit circle. Let the two charts be the arcs $A = (0,2\pi)$ and $B = (\pi/2, 5\pi/2)$. For each $n$, consider the partition of unity subordinate to $\{A,B\}$ given by 
$$ \psi_{A,n} = \sin^2 ( (2n+1) \theta ) $$
and
$$ \psi_{B,n} = \cos^2 ( (2n+1) \theta ) $$
Check that $\psi_{A,n} (0) = 0 = \psi_{B,n} (\pi/2)$, and clearly $\psi_{A,n} + \psi_{B,n} = 1$. 
By the scaling property it is easy to check that 
$$ \| \partial^k\psi_{A,n} \| = (2n+1)^k \| \partial^k\psi_{A,1} \| $$
And using that 
$$ \sin^2(\theta) = \frac12 (1-\cos 2\theta) $$
you see immediately that your desired uniform bound is impossible.