# Partition of Unity in Spivak's Calculus on Manifolds

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.

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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

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.

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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 $U$.

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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