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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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Can you do it when $M=\mathbb R$? If you can, then do the same thing in the general case; if not, then you could edit the question to ask this instead, as this is where the problem is :) –  Mariano Suárez-Alvarez Nov 6 '10 at 18:30
    
This exercise is something of a step towards showing that smooth functions are dense (in the $C^0$-topology) in the space of continuous functions. The proof is very similar to the proof that smooth manifolds admit Riemann metrics. Have you read that proof? –  Ryan Budney Nov 6 '10 at 18:53
    
If you can partition 1, you have a way to partition any constant, including those between 0 and 1. –  whuber Nov 6 '10 at 20:16
    
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 $n$. –  Ryan Budney Nov 6 '10 at 20:45
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1 Answer 1

up vote 2 down vote accepted

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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How do you know that $\inf \delta(V_m) > 0$ ?. I think you need to make some sort of assumption like "These $V_m$ are precompact" for that to be true. Also, how do you construct the $U_i$? –  Eric Haengel Nov 6 '10 at 22:34
    
Question 1: I take for $V_m$ set of points $x\in M$ where $\delta (x)>\delta (m)/2$. –  evgeniamerkulova Nov 6 '10 at 23:16
    
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. –  evgeniamerkulova Nov 6 '10 at 23:27
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