The gradient estimate of the partition of unity

Question

If $M$ is a compact Riemannian manifold with metric $g$, can we find a constant $C>0$, which is independent of $M$ and $g$, such that for any finite open covering $\{U_i\}$ of $M$, we can find a subordinate partition of unity $\{\varphi^2_i\}$ which satisfies the estimation $$\sum_i |\nabla\varphi_i| \le C $$

Answer 1

No. Cover the circle (length $2\pi$) by $n+1$ arcs of length $2\pi /n$ so that the midpoints of the arcs are equidistributed along the circle. Note that each midpoint $m_i$ is covered by the $i$^th arc only. Thus, any subordinate partition of unity will satisfy $\varphi_i(m_i)=1$, hence $\sup |\nabla \varphi_i|\ge n/\pi$.