Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question about partitions of unity specifically in the book Calculus on Manifolds by Spivak. In case 1 for the proof of existence of partition of unity, why is there a need for the function $f$? The set $\Phi = \{\varphi_1, \dotsc, \varphi_n\}$ looks like is already the desired partition of unity. Following is the theorem and proof. Only Case 1 in the proof is relevant.

enter image description here enter image description here

share|cite|improve this question
Should I include pictures of the two pages with the theorem and proof? – Pratyush Sarkar Aug 24 '13 at 13:24
That would certainly help; or at least type in the relevant details (what are the $\phi_i, f$ and what is the proposed partition of unity). – Anthony Carapetis Aug 24 '13 at 14:00
I am reading calculus on manifolds too. And as I came across this proof today, I have exactly the same doubt. Glad that I find this post. – ᴊ ᴀ s ᴏ ɴ Jul 10 '15 at 14:16
up vote 5 down vote accepted

I belive that your assertion is correct. The functions $\varphi_{i}$ satisfy all of the conditions of Theorem 3-11. I don't see why Spivak used such an $f$. Particularly since the support of $f$ contains $A$. If at least the support of $f$ lied in $A$ then $f=\sum_{i=1}^{n}f\cdot\varphi_{i}$, thus giving a representation of $f$ as a sum of functions with small supports.

Since $A$ is compact we may assume WLOG that the $U_{i}$ are bounded. Therefore, by construction, the supports of the $\psi_{i}$ are compact. Hence, the word "closed" in item ($4$) of Theorem 3-11 can be changed to "compact". The proof remains unchanged. This helps clarify the first statement of the proof of Theorem 3-12.

Also, note as well that the functions $\varphi_{i}$ are $C^{\infty}$. This basically follows from Problem 2-26.

Posts related to the section:

  1. An application of partitions of unity: integrating over open sets and here

  2. Do we need additional assumptions for problem 3-37 (b) in Spivak´s calculus on Manifolds?

  3. Problem 3-38 in Spivak´s Calculus on Manifolds

  4. Extended integral in Spivak’s Calculus on Manifolds

share|cite|improve this answer
Firstly, in property 4, isn't the closed set bounded? Which would mean it's compact? Secondly, in the proof it says "let $\psi_i$ be non-negative $C^\infty$ function positive on $D_i$ and $0$ outside some closed set contained in $U_i$". Denote the closed set as $S_i$. Then $\varphi_i$ has the same property, namely there is the set $U_i$ such that $\varphi_i = 0$ outside $S_i$ contained in $U_i$. So isn't property 4 already fulfilled? – Pratyush Sarkar Aug 24 '13 at 15:16
@Pratyush: I have looked at this more carefully and updated my answer. – John Aug 24 '13 at 21:10
Thanks for your help. I read the theorem and proof multiple times before and I couldn't figure out the purpose of the function $f$. I thought I was missing something really obvious. – Pratyush Sarkar Aug 24 '13 at 22:51

I was looking back at my question today for some reason and immediately saw why the function $f$ is required. Although $\psi_i$ is smooth with compact support in $U_i$, the functions $\varphi_i$ can only be defined on $U$ where $\sum_{i = 1}^n\psi_i > 0$. The problem is that $\varphi_i$ usually does not go to zero at the boundary of $\operatorname{supp}(\psi_i)$ (much less smoothly extend to zero outside the boundary). You can see this at the boundary of a $D_i$ which is away from all other $D_j$. Near this boundary $\varphi_i(x) = \frac{\psi_i(x)}{\psi_i(x)} = 1$. The solution is to use a cutoff function $f$ which forces everything to smoothly go to $0$ near the boundary of $U$.

share|cite|improve this answer
I think you mean $\varphi_{i}$ (not $\psi_{i}$) does not usually go to zero at the boundary of $\text{supp}(\psi_{i})$. – fourierwho May 31 at 16:37
@fourierwho Yes. Thanks for the correction. – Pratyush Sarkar Jul 25 at 23:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.