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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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  • $\begingroup$ Should I include pictures of the two pages with the theorem and proof? $\endgroup$ – Pratyush Sarkar Aug 24 '13 at 13:24
  • $\begingroup$ 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). $\endgroup$ – Anthony Carapetis Aug 24 '13 at 14:00
  • $\begingroup$ 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. $\endgroup$ – A. Chu Jul 10 '15 at 14:16
  • $\begingroup$ On what set is $\sigma$ define din the second case? Because I do not see how how defining $\phi / \sigma$ gives a partition of unity, or even how the division makes sense. $\endgroup$ – vaoy Apr 18 '18 at 10:51
  • $\begingroup$ @vaoy $\Phi_i$ is a partition of unity subordinate to $\mathcal O_i$, so we can assume $\varphi \in \Phi_i$ is defined on $A$ and zero outside $\operatorname{int}(A_{i + 1}) - A_{i - 2}$. And $\sigma$ is also defined on $A$ and as explained, the sum in the definition is at most finite for all $x \in A$ and necessarily positive, so division is possible. Note that $\sigma$ is just a scaling factor to make sure the new $\varphi'$ sum up to $1$ at each $x \in A$. This scaling is required because although $\Phi_i$ is a partition of unity, $\Phi_j$ for $j = i + 1, i - 1$ can interfere with $\Phi_i$. $\endgroup$ – Pratyush Sarkar Apr 24 '18 at 18:00
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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

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  • $\begingroup$ 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? $\endgroup$ – Pratyush Sarkar Aug 24 '13 at 15:16
  • $\begingroup$ @Pratyush: I have looked at this more carefully and updated my answer. $\endgroup$ – John Aug 24 '13 at 21:10
  • $\begingroup$ 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. $\endgroup$ – Pratyush Sarkar Aug 24 '13 at 22:51
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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 near the boundary of a $\operatorname{supp}(\psi_i)$ which is away from all other $\operatorname{supp}(\psi_j)$. 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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  • $\begingroup$ I think you mean $\varphi_{i}$ (not $\psi_{i}$) does not usually go to zero at the boundary of $\text{supp}(\psi_{i})$. $\endgroup$ – fourierwho May 31 '16 at 16:37
  • $\begingroup$ @fourierwho Yes. Thanks for the correction. $\endgroup$ – Pratyush Sarkar Jul 25 '16 at 23:14
  • $\begingroup$ @PratyushSarkar, the fact of $\phi_i$ doesn't smoothly extend to zero outside the boundary of $\operatorname{supp} (\psi_i)$ doesn't contradicts the fact of $\psi_i$ are $C^{\infty}$ by the way was built? If your answer is not, so I think that I didn't understand the purpose of $f$. $\endgroup$ – George Jan 29 '17 at 18:28
  • $\begingroup$ @George No there is no contradiction. What I explained is, although $\psi_i$ obviously extends to $0$ smoothly outside $\operatorname{supp} (\psi_i)$, the biggest set where we can define $\varphi_i$ is only on $U = \bigcup_{i = 1}^n \operatorname{supp} (\psi_i)$ and it is possible that near the boundary of $U$ which, say, coincides with $\operatorname{supp} (\psi_i)$ and is away from all other $\operatorname{supp} (\psi_j)$, the value of $\varphi_i(x)$ tends to $1$. $\endgroup$ – Pratyush Sarkar Feb 17 '17 at 5:56
  • $\begingroup$ More explicitly, we can easily think of (and draw) an example where $B_i = \partial U \setminus \left(\bigcup_{j \neq i} \operatorname{supp} (\psi_j)\right)$ is nonempty and in that case we have $\lim_{x \to b} \varphi_i(x) = 1$ for any $b \in B_i$ since $\varphi_i(x) = \frac{\psi_i(x)}{\psi_i(x)} = 1$ for $x$ sufficiently close to $b$. $\endgroup$ – Pratyush Sarkar Feb 17 '17 at 6:03
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Merry Christmas ! I am sorry to bother you.I revised my answer.Looking foward to your comments.

First of all, as Mr.John mentioned in his answer,the functions $\varphi_{i}$ have already satisfied all of the conditions of Th 3-11 but except for $\textit{(4)}$. A crucial question in Spivak's proof is: why the author "redundantly" required $f\cdot\varphi_{i}$ rather than $\varphi_{i}$ ? In my way of thinking , in accord with problem 2-26$^*$(d),the author wanted to extend $\varphi_{i}$ from its domain$-\ $an open subset in $\mathbf {R}^{n}$$-\ $to the whole of $\mathbf {R}^{n}$ smoothly and then guarantee such those extended functions $f\cdot \varphi_{i}$ satisfying all of four conditions.

The following is details:

Def. $\;\phi:X \rightarrow \mathbf {R}$ is a continuous real-valued function whose domain is an arbitrary set $X$ in $\mathbf {R}^{n}$, the support of $\phi$ is defined as the closure of the subset of $X$ where $\phi$ is non-zero $i.e.,\:$$supp(\phi):=$the closure of the set $\left\{\mathbf{x} \in X: \phi(\mathbf{x})\ne0 \right \}.$

Def. $\;\phi\in C^{\infty}(A,B)$ denotes $\phi:(\mathbf {R}^{n}\supset )A\rightarrow B(\subset\mathbf {R}^{m})$ is a $C^{\infty}$ function.

From the case 1,the compact sets $D_{i}(i=1,\cdots,n)$ whose interiors cover $A$.In order to explain why the author modified $\varphi_{i}$ with $f\cdot \varphi_{i}$ briefly,we can choose a specific open subset$-\ U:=\bigcup_{i=1}^{n}int D_{i}$$-\ $to set forth.If in the simplest case we need $\varphi_{i}$ multiplied by $f$,not to mention in most cases.

$\textbf{1.}$

$\psi_i\in C^{\infty}(U_i,\mathbf{R}),$ which is positive on $D_i$ and $0$ outside of some closed set contained in $U_i$ .(problem 2-26$^*$(d))

Define $$\widetilde \psi_{i}:=\left\{\begin{matrix} \psi_{i}& x\in U_i\\ 0& x\in {U_i}^{c} \end{matrix}\right.\quad (i=1,2,\cdots,n),$$ then we have $\widetilde \psi_{i}\in C^{\infty}(\mathbf{R}^{n},\mathbf{R})$,$\;\widetilde \psi_{i}\bigg|_{U_{i}}=\widetilde \psi_{i}$ and $\psi_{i}$ (each domain is $U_{i}$) can be smoothly extended to $\widetilde \psi_{i}$(each domian is $\mathbf{R}^{n}$).

All functions $$\varphi_{i}=\frac{\psi_{i}}{\sum_{k=1}^{n}\psi_{k}}\;(i=1,\cdots,n)$$ are only constructed on $U$. $$ \forall \: x\in U,\: \sum_{k=1}^{n}\widetilde\psi_{k}\ne 0 ;\;\widetilde\psi_{k}\in C^{\infty}(\mathbf{R}^{n},\mathbf{R}).$$ $$\Longrightarrow \varphi_{i}=\frac{\psi_{i}}{\sum_{k=1}^{n}\psi_{k}}=\frac{\widetilde\psi_{i}}{\sum_{k=1}^{n}\widetilde\psi_{k}}\in C^{\infty}(U,\mathbf{R}) \quad (i=1,2,\cdots,n).$$

$\textbf{2.}$

$f\in C^{\infty}(U,\mathbf{R}),$ which value is $1$ on $A$ and $0$ outside of some closed set contained in $U$.(problem 2-26$^*$(d))

$\\$Obviously,we have $f\cdot \varphi_{i}\in C^{\infty}(U,\mathbf{R})\;(i=1,2,\cdots,n).$ Note that each $f\cdot \varphi_{i}\in C^{\infty}(U,\mathbf{R})$ is only constructed on $U\:!$

$\textbf{3.}$

Finally,we can extend $f\cdot \varphi_{i}$ from $U$ to the whole of $\mathbf {R}^{n}$ smoothly.

In fact,since $supp(f\cdot\varphi_{i})\subset U$, define $$\widetilde{f\cdot \varphi_{i}}:=\left\{\begin{matrix} f\cdot \varphi_{i}& x\in supp(f\cdot\varphi_{i})\\ 0 & x\notin supp(f\cdot\varphi_{i}) \end{matrix}\right.\quad (i=1,2,\cdots,n),$$ then we have $\widetilde {f\cdot \varphi_{i}}\in C^{\infty}(\mathbf{R}^{n},\mathbf{R})$,$\;\widetilde {f\cdot \varphi_{i}}\bigg|_{U}=f\cdot\varphi_{i}$ and $f\cdot \varphi_{i}$ (each domain is $U$) can be smoothly extended to $\widetilde {f\cdot \varphi_{i}}$ (each domian is $\mathbf{R}^{n}).\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare$


A schematic diagram (A schematic diagram concretising above elaboration)

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  • $\begingroup$ Thanks for elaborating, this makes sense. A few comments: 1. Actually the functions $\varphi_i$ do not satisfy property 4 of the theorem which is why we need $f$ to kill it off near some parts of the boundary of $U_i$. 2. In your method, $U$ is defined slightly differently but the result is the same, support of $\widetilde{f\cdot \varphi_{i}}$ is contained inside $U_i$ as required by property 4. I want to point out that this can't be made stronger to support contained in $D_i$. So I think I prefer the $U$ as defined by Spivak because it is easier to see the containment in $U_i$. $\endgroup$ – Pratyush Sarkar Dec 26 '18 at 23:06

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