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I was reading John Lee's Introduction to Smooth manifolds, and I came across this question:

  • Let $M$ be a smooth manifold, and let $\delta : M \rightarrow \mathbb{R}$ be a positive continuous function. Using a partition of unity, show that there is a smooth function $\tilde{\delta} : M \rightarrow \mathbb{R}$ such that $0 < \tilde{\delta}(x) < \delta(x)$ for all $x \in M$.

I thought about it for a while, and I'm pretty stuck on it. Does anyone have any ideas?

(Edit: It has been pointed out that one can basically assume that $M = \mathbb{R}$, because the proof should be the same in both cases. If you do not know anything about smooth manifolds, feel free to do this!)

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    The main advantage of $\mathbb R$ is you have explicit partitions of unity given by evenly-spaced bump functions. So consider $\min\{\delta(x) : n \leq x \leq n+1\}$ for various integers $n$.2010-11-06

1 Answers 1

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Have to do it with 2 coverings.

Covering 1: For each $m\in M$ chose open $V_m \subset M$ and \epsilon_m >0 with $\epsilon_m < inf \delta(V_m)$.

Covering 2: Take locally finite covering $(U_i)_{i\in I}$ finer than preceding covering. For each $i\in I$ chose $m\in M$ with $U_i\subset V_m$ and put $\delta_i =\epsilon_m$ . Remark that on $U_i$ we have $\delta_i < inf \delta (U_i).$

Then associate partition of unity $(\phi_i)_{i\in I}$ with covering $(U_i)_{i\in I}$ and wanted function is

$\tilde \delta=\sum \delta_i \phi_i $

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    Question 2: Such $U_i$ exist by definition of paracompact. Manifolds are paracompact by definition or by equivalent property . In Lee book supposed second countable and then paracompact proved page 53.2010-11-06