Partition of Unity in Spivak's Calculus on Manifolds
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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.
real-analysis multivariable-calculus
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add a comment |
$begingroup$
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.
real-analysis multivariable-calculus
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Should I include pictures of the two pages with the theorem and proof?
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– Pratyush Sarkar
Aug 24 '13 at 13:24
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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).
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– Anthony Carapetis
Aug 24 '13 at 14:00
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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.
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– A. Chu
Jul 10 '15 at 14:16
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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.
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– vaoy
Apr 18 '18 at 10:51
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@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
add a comment |
$begingroup$
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.
real-analysis multivariable-calculus
$endgroup$
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.
real-analysis multivariable-calculus
real-analysis multivariable-calculus
edited Apr 24 '18 at 10:46
Arkya Chatterjee
541315
541315
asked Aug 24 '13 at 13:23
Pratyush SarkarPratyush Sarkar
2,7801127
2,7801127
$begingroup$
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.
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– A. Chu
Jul 10 '15 at 14:16
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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.
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– vaoy
Apr 18 '18 at 10:51
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@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
add a comment |
$begingroup$
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
$begingroup$
Should I include pictures of the two pages with the theorem and proof?
$endgroup$
– Pratyush Sarkar
Aug 24 '13 at 13:24
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
add a comment |
3 Answers
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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 $sum_{i = 1}^npsi_i > 0$. 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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I think you mean $varphi_{i}$ (not $psi_{i}$) does not usually go to zero at the boundary of $text{supp}(psi_{i})$.
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– fourierwho
May 31 '16 at 16:37
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@fourierwho Yes. Thanks for the correction.
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– Pratyush Sarkar
Jul 25 '16 at 23:14
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@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 $f$.
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– George
Jan 29 '17 at 18:28
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@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 $1$.
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– Pratyush Sarkar
Feb 17 '17 at 5:56
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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 $b$.
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– Pratyush Sarkar
Feb 17 '17 at 6:03
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show 2 more comments
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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 $f=sum_{i=1}^{n}fcdotvarphi_{i}$, 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:
An application of partitions of unity: integrating over open sets and here
Do we need additional assumptions for problem 3-37 (b) in Spivak´s calculus on Manifolds?
Problem 3-38 in Spivak´s Calculus on Manifolds
Extended integral in Spivak’s Calculus on Manifolds
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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?
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– Pratyush Sarkar
Aug 24 '13 at 15:16
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@Pratyush: I have looked at this more carefully and updated my answer.
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– John
Aug 24 '13 at 21:10
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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.
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– Pratyush Sarkar
Aug 24 '13 at 22:51
add a comment |
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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 $fcdotvarphi_{i}$ 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}$$- $to the whole of $mathbf {R}^{n}$ smoothly and then guarantee such those extended functions $fcdot varphi_{i}$ satisfying all of four conditions.
The following is details:
Def. $;phi:X rightarrow mathbf {R}$ is a continuous real-valued function whose domain is an arbitrary set $X$ in $mathbf {R}^{n}$, the support of $phi$ is defined as the closure of the subset of $X$ where $phi$ is non-zero $i.e.,:$$supp(phi):=$the closure of the set $left{mathbf{x} in X: phi(mathbf{x})ne0 right }.$
Def. $;phiin C^{infty}(A,B)$ denotes $phi:(mathbf {R}^{n}supset )Arightarrow B(subsetmathbf {R}^{m})$ 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 $fcdot varphi_{i}$ briefly,we can choose a specific open subset$- U:=bigcup_{i=1}^{n}int D_{i}$$- $to set forth.If in the simplest case we need $varphi_{i}$ multiplied by $f$,not to mention in most cases.
$textbf{1.}$
$psi_iin C^{infty}(U_i,mathbf{R}),$ which is positive on $D_i$ and $0$ outside of some closed set contained in $U_i$ .(problem 2-26$^*$(d))
Define $$widetilde psi_{i}:=left{begin{matrix}
psi_{i}& xin U_i\
0& xin {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 : xin U,: sum_{k=1}^{n}widetildepsi_{k}ne 0 ;;widetildepsi_{k}in C^{infty}(mathbf{R}^{n},mathbf{R}).$$ $$Longrightarrow varphi_{i}=frac{psi_{i}}{sum_{k=1}^{n}psi_{k}}=frac{widetildepsi_{i}}{sum_{k=1}^{n}widetildepsi_{k}}in C^{infty}(U,mathbf{R}) quad (i=1,2,cdots,n).$$
$textbf{2.}$
$fin C^{infty}(U,mathbf{R}),$ which value is $1$ on $A$ and $0$ outside of some closed set contained in $U$.(problem 2-26$^*$(d))
$\$Obviously,we have $fcdot varphi_{i}in C^{infty}(U,mathbf{R});(i=1,2,cdots,n).$ Note that each $fcdot varphi_{i}in C^{infty}(U,mathbf{R})$ is only constructed on $U:!$
$textbf{3.}$
Finally,we can extend $fcdot varphi_{i}$ from $U$ to the whole of $mathbf {R}^{n}$ smoothly.
In fact,since $supp(fcdotvarphi_{i})subset U$,
define $$widetilde{fcdot varphi_{i}}:=left{begin{matrix}
fcdot varphi_{i}& xin supp(fcdotvarphi_{i})\
0 & xnotin supp(fcdotvarphi_{i})
end{matrix}right.quad (i=1,2,cdots,n),$$ then we have $widetilde {fcdot varphi_{i}}in C^{infty}(mathbf{R}^{n},mathbf{R})$,$;widetilde {fcdot varphi_{i}}bigg|_{U}=fcdotvarphi_{i}$ and $fcdot varphi_{i}$ (each domain is $U$) can be smoothly extended to $widetilde {fcdot varphi_{i}}$ (each domian is $mathbf{R}^{n}).qquadqquadqquadqquadqquadqquadqquadqquadblacksquare$
(A schematic diagram concretising above elaboration)
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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{fcdot 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$.
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– Pratyush Sarkar
Dec 26 '18 at 23:06
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3 Answers
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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}^npsi_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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I think you mean $varphi_{i}$ (not $psi_{i}$) does not usually go to zero at the boundary of $text{supp}(psi_{i})$.
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– fourierwho
May 31 '16 at 16:37
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@fourierwho Yes. Thanks for the correction.
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– Pratyush Sarkar
Jul 25 '16 at 23:14
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@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$.
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– George
Jan 29 '17 at 18:28
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@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
|
show 2 more comments
$begingroup$
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}^npsi_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$.
$endgroup$
$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
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@fourierwho Yes. Thanks for the correction.
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– Pratyush Sarkar
Jul 25 '16 at 23:14
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@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
|
show 2 more comments
$begingroup$
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}^npsi_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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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}^npsi_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$.
edited Feb 17 '17 at 5:30
answered Jan 27 '16 at 5:20
Pratyush SarkarPratyush Sarkar
2,7801127
2,7801127
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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})$.
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– fourierwho
May 31 '16 at 16:37
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@fourierwho Yes. Thanks for the correction.
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– Pratyush Sarkar
Jul 25 '16 at 23:14
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@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$.
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– George
Jan 29 '17 at 18:28
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@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$.
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– Pratyush Sarkar
Feb 17 '17 at 5:56
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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$.
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– Pratyush Sarkar
Feb 17 '17 at 6:03
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show 2 more comments
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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})$.
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– fourierwho
May 31 '16 at 16:37
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@fourierwho Yes. Thanks for the correction.
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– Pratyush Sarkar
Jul 25 '16 at 23:14
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@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$.
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– George
Jan 29 '17 at 18:28
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@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$.
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– Pratyush Sarkar
Feb 17 '17 at 5:56
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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$.
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– Pratyush Sarkar
Feb 17 '17 at 6:03
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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})$.
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– fourierwho
May 31 '16 at 16:37
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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})$.
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– fourierwho
May 31 '16 at 16:37
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@fourierwho Yes. Thanks for the correction.
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– Pratyush Sarkar
Jul 25 '16 at 23:14
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@fourierwho Yes. Thanks for the correction.
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– Pratyush Sarkar
Jul 25 '16 at 23:14
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@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$.
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– George
Jan 29 '17 at 18:28
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@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$.
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– George
Jan 29 '17 at 18:28
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@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$
@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$.
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– Pratyush Sarkar
Feb 17 '17 at 6:03
$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$.
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– Pratyush Sarkar
Feb 17 '17 at 6:03
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show 2 more comments
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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}fcdotvarphi_{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:
An application of partitions of unity: integrating over open sets and here
Do we need additional assumptions for problem 3-37 (b) in Spivak´s calculus on Manifolds?
Problem 3-38 in Spivak´s Calculus on Manifolds
Extended integral in Spivak’s Calculus on Manifolds
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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?
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– Pratyush Sarkar
Aug 24 '13 at 15:16
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@Pratyush: I have looked at this more carefully and updated my answer.
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– John
Aug 24 '13 at 21:10
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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.
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– Pratyush Sarkar
Aug 24 '13 at 22:51
add a comment |
$begingroup$
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}fcdotvarphi_{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:
An application of partitions of unity: integrating over open sets and here
Do we need additional assumptions for problem 3-37 (b) in Spivak´s calculus on Manifolds?
Problem 3-38 in Spivak´s Calculus on Manifolds
Extended integral in Spivak’s Calculus on Manifolds
$endgroup$
$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?
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– Pratyush Sarkar
Aug 24 '13 at 15:16
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@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
add a comment |
$begingroup$
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}fcdotvarphi_{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:
An application of partitions of unity: integrating over open sets and here
Do we need additional assumptions for problem 3-37 (b) in Spivak´s calculus on Manifolds?
Problem 3-38 in Spivak´s Calculus on Manifolds
Extended integral in Spivak’s Calculus on Manifolds
$endgroup$
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}fcdotvarphi_{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:
An application of partitions of unity: integrating over open sets and here
Do we need additional assumptions for problem 3-37 (b) in Spivak´s calculus on Manifolds?
Problem 3-38 in Spivak´s Calculus on Manifolds
Extended integral in Spivak’s Calculus on Manifolds
edited Apr 13 '17 at 12:58
Community♦
1
1
answered Aug 24 '13 at 14:28
JohnJohn
1,533924
1,533924
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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?
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– Pratyush Sarkar
Aug 24 '13 at 15:16
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@Pratyush: I have looked at this more carefully and updated my answer.
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– John
Aug 24 '13 at 21:10
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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
add a comment |
$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
$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$
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
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@Pratyush: I have looked at this more carefully and updated my answer.
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– John
Aug 24 '13 at 21:10
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@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
$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
add a comment |
$begingroup$
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 $fcdotvarphi_{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 $fcdot varphi_{i}$ 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.,:$$supp(phi):=$the closure of the set $left{mathbf{x} in X: phi(mathbf{x})ne0 right }.$
Def. $;phiin C^{infty}(A,B)$ denotes $phi:(mathbf {R}^{n}supset )Arightarrow B(subsetmathbf {R}^{m})$ is a $C^{infty}$ function.
From the case 1,the compact sets $D_{i}(i=1,cdots,n)$ whose interiors cover $A$.In order to explain why the author modified $varphi_{i}$ with $fcdot varphi_{i}$ briefly,we can choose a specific open subset$- U:=bigcup_{i=1}^{n}int D_{i}$$- $to set forth.If in the simplest case we need $varphi_{i}$ multiplied by $f$,not to mention in most cases.
$textbf{1.}$
$psi_iin C^{infty}(U_i,mathbf{R}),$ which is positive on $D_i$ and $0$ outside of some closed set contained in $U_i$ .(problem 2-26$^*$(d))
Define $$widetilde psi_{i}:=left{begin{matrix}
psi_{i}& xin U_i\
0& xin {U_i}^{c}
end{matrix}right.quad (i=1,2,cdots,n),$$ 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 $U$. $$ forall : xin U,: sum_{k=1}^{n}widetildepsi_{k}ne 0 ;;widetildepsi_{k}in C^{infty}(mathbf{R}^{n},mathbf{R}).$$ $$Longrightarrow varphi_{i}=frac{psi_{i}}{sum_{k=1}^{n}psi_{k}}=frac{widetildepsi_{i}}{sum_{k=1}^{n}widetildepsi_{k}}in C^{infty}(U,mathbf{R}) quad (i=1,2,cdots,n).$$
$textbf{2.}$
$fin C^{infty}(U,mathbf{R}),$ which value is $1$ on $A$ and $0$ outside of some closed set contained in $U$.(problem 2-26$^*$(d))
$\$Obviously,we have $fcdot varphi_{i}in C^{infty}(U,mathbf{R});(i=1,2,cdots,n).$ Note that each $fcdot varphi_{i}in C^{infty}(U,mathbf{R})$ is only constructed on $U:!$
$textbf{3.}$
Finally,we can extend $fcdot varphi_{i}$ from $U$ to the whole of $mathbf {R}^{n}$ smoothly.
In fact,since $supp(fcdotvarphi_{i})subset U$,
define $$widetilde{fcdot varphi_{i}}:=left{begin{matrix}
fcdot varphi_{i}& xin supp(fcdotvarphi_{i})\
0 & xnotin supp(fcdotvarphi_{i})
end{matrix}right.quad (i=1,2,cdots,n),$$ then we have $widetilde {fcdot varphi_{i}}in C^{infty}(mathbf{R}^{n},mathbf{R})$,$;widetilde {fcdot varphi_{i}}bigg|_{U}=fcdotvarphi_{i}$ and $fcdot varphi_{i}$ (each domain is $U$) can be smoothly extended to $widetilde {fcdot varphi_{i}}$ (each domian is $mathbf{R}^{n}).qquadqquadqquadqquadqquadqquadqquadqquadblacksquare$
(A schematic diagram concretising above elaboration)
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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 $widetilde{fcdot varphi_{i}}$ 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 $U_i$.
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– Pratyush Sarkar
Dec 26 '18 at 23:06
add a comment |
$begingroup$
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 $fcdotvarphi_{i}$ 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}$$- $to the whole of $mathbf {R}^{n}$ smoothly and then guarantee such those extended functions $fcdot varphi_{i}$ satisfying all of four conditions.
The following is details:
Def. $;phi:X rightarrow mathbf {R}$ is a continuous real-valued function whose domain is an arbitrary set $X$ in $mathbf {R}^{n}$, the support of $phi$ is defined as the closure of the subset of $X$ where $phi$ is non-zero $i.e.,:$$supp(phi):=$the closure of the set $left{mathbf{x} in X: phi(mathbf{x})ne0 right }.$
Def. $;phiin C^{infty}(A,B)$ denotes $phi:(mathbf {R}^{n}supset )Arightarrow B(subsetmathbf {R}^{m})$ 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 $fcdot varphi_{i}$ briefly,we can choose a specific open subset$- U:=bigcup_{i=1}^{n}int D_{i}$$- $to set forth.If in the simplest case we need $varphi_{i}$ multiplied by $f$,not to mention in most cases.
$textbf{1.}$
$psi_iin C^{infty}(U_i,mathbf{R}),$ which is positive on $D_i$ and $0$ outside of some closed set contained in $U_i$ .(problem 2-26$^*$(d))
Define $$widetilde psi_{i}:=left{begin{matrix}
psi_{i}& xin U_i\
0& xin {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 : xin U,: sum_{k=1}^{n}widetildepsi_{k}ne 0 ;;widetildepsi_{k}in C^{infty}(mathbf{R}^{n},mathbf{R}).$$ $$Longrightarrow varphi_{i}=frac{psi_{i}}{sum_{k=1}^{n}psi_{k}}=frac{widetildepsi_{i}}{sum_{k=1}^{n}widetildepsi_{k}}in C^{infty}(U,mathbf{R}) quad (i=1,2,cdots,n).$$
$textbf{2.}$
$fin C^{infty}(U,mathbf{R}),$ which value is $1$ on $A$ and $0$ outside of some closed set contained in $U$.(problem 2-26$^*$(d))
$\$Obviously,we have $fcdot varphi_{i}in C^{infty}(U,mathbf{R});(i=1,2,cdots,n).$ Note that each $fcdot varphi_{i}in C^{infty}(U,mathbf{R})$ is only constructed on $U:!$
$textbf{3.}$
Finally,we can extend $fcdot varphi_{i}$ from $U$ to the whole of $mathbf {R}^{n}$ smoothly.
In fact,since $supp(fcdotvarphi_{i})subset U$,
define $$widetilde{fcdot varphi_{i}}:=left{begin{matrix}
fcdot varphi_{i}& xin supp(fcdotvarphi_{i})\
0 & xnotin supp(fcdotvarphi_{i})
end{matrix}right.quad (i=1,2,cdots,n),$$ then we have $widetilde {fcdot varphi_{i}}in C^{infty}(mathbf{R}^{n},mathbf{R})$,$;widetilde {fcdot varphi_{i}}bigg|_{U}=fcdotvarphi_{i}$ and $fcdot varphi_{i}$ (each domain is $U$) can be smoothly extended to $widetilde {fcdot varphi_{i}}$ (each domian is $mathbf{R}^{n}).qquadqquadqquadqquadqquadqquadqquadqquadblacksquare$
(A schematic diagram concretising above elaboration)
$endgroup$
$begingroup$
Thanks for elaborating, this makes sense. A few comments: 1. Actually the functions $varphi_i$ do not satisfy property 4 of the theorem which is why we need $f$ to kill it off near some parts of the boundary of $U_i$. 2. In your method, $U$ is defined slightly differently but the result is the same, support of $widetilde{fcdot varphi_{i}}$ is contained inside $U_i$ as required by property 4. I want to point out that this can't be made stronger to support contained in $D_i$. So I think I prefer the $U$ as defined by Spivak because it is easier to see the containment in $U_i$.
$endgroup$
– Pratyush Sarkar
Dec 26 '18 at 23:06
add a comment |
$begingroup$
Merry Christmas ! I am sorry to bother you.I revised my answer.Looking foward to your comments.
First of all, as Mr.John mentioned in his answer,the functions $varphi_{i}$ have already satisfied all of the conditions of Th 3-11 but except for $textit{(4)}$. A crucial question in Spivak's proof is: why the author "redundantly" required $fcdotvarphi_{i}$ rather than $varphi_{i}$ ?
In my way of thinking , in accord with problem 2-26$^*$(d),the author wanted to extend $varphi_{i}$ from its domain$- $an open subset in $mathbf {R}^{n}$$- $to the whole of $mathbf {R}^{n}$ smoothly and then guarantee such those extended functions $fcdot varphi_{i}$ 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.,:$$supp(phi):=$the closure of the set $left{mathbf{x} in X: phi(mathbf{x})ne0 right }.$
Def. $;phiin C^{infty}(A,B)$ denotes $phi:(mathbf {R}^{n}supset )Arightarrow B(subsetmathbf {R}^{m})$ is a $C^{infty}$ function.
From the case 1,the compact sets $D_{i}(i=1,cdots,n)$ whose interiors cover $A$.In order to explain why the author modified $varphi_{i}$ with $fcdot varphi_{i}$ briefly,we can choose a specific open subset$- U:=bigcup_{i=1}^{n}int D_{i}$$- $to set forth.If in the simplest case we need $varphi_{i}$ multiplied by $f$,not to mention in most cases.
$textbf{1.}$
$psi_iin C^{infty}(U_i,mathbf{R}),$ which is positive on $D_i$ and $0$ outside of some closed set contained in $U_i$ .(problem 2-26$^*$(d))
Define $$widetilde psi_{i}:=left{begin{matrix}
psi_{i}& xin U_i\
0& xin {U_i}^{c}
end{matrix}right.quad (i=1,2,cdots,n),$$ 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 $U$. $$ forall : xin U,: sum_{k=1}^{n}widetildepsi_{k}ne 0 ;;widetildepsi_{k}in C^{infty}(mathbf{R}^{n},mathbf{R}).$$ $$Longrightarrow varphi_{i}=frac{psi_{i}}{sum_{k=1}^{n}psi_{k}}=frac{widetildepsi_{i}}{sum_{k=1}^{n}widetildepsi_{k}}in C^{infty}(U,mathbf{R}) quad (i=1,2,cdots,n).$$
$textbf{2.}$
$fin C^{infty}(U,mathbf{R}),$ which value is $1$ on $A$ and $0$ outside of some closed set contained in $U$.(problem 2-26$^*$(d))
$\$Obviously,we have $fcdot varphi_{i}in C^{infty}(U,mathbf{R});(i=1,2,cdots,n).$ Note that each $fcdot varphi_{i}in C^{infty}(U,mathbf{R})$ is only constructed on $U:!$
$textbf{3.}$
Finally,we can extend $fcdot varphi_{i}$ from $U$ to the whole of $mathbf {R}^{n}$ smoothly.
In fact,since $supp(fcdotvarphi_{i})subset U$,
define $$widetilde{fcdot varphi_{i}}:=left{begin{matrix}
fcdot varphi_{i}& xin supp(fcdotvarphi_{i})\
0 & xnotin supp(fcdotvarphi_{i})
end{matrix}right.quad (i=1,2,cdots,n),$$ then we have $widetilde {fcdot varphi_{i}}in C^{infty}(mathbf{R}^{n},mathbf{R})$,$;widetilde {fcdot varphi_{i}}bigg|_{U}=fcdotvarphi_{i}$ and $fcdot varphi_{i}$ (each domain is $U$) can be smoothly extended to $widetilde {fcdot varphi_{i}}$ (each domian is $mathbf{R}^{n}).qquadqquadqquadqquadqquadqquadqquadqquadblacksquare$
(A schematic diagram concretising above elaboration)
$endgroup$
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 $fcdotvarphi_{i}$ 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}$$- $to the whole of $mathbf {R}^{n}$ smoothly and then guarantee such those extended functions $fcdot varphi_{i}$ satisfying all of four conditions.
The following is details:
Def. $;phi:X rightarrow mathbf {R}$ is a continuous real-valued function whose domain is an arbitrary set $X$ in $mathbf {R}^{n}$, the support of $phi$ is defined as the closure of the subset of $X$ where $phi$ is non-zero $i.e.,:$$supp(phi):=$the closure of the set $left{mathbf{x} in X: phi(mathbf{x})ne0 right }.$
Def. $;phiin C^{infty}(A,B)$ denotes $phi:(mathbf {R}^{n}supset )Arightarrow B(subsetmathbf {R}^{m})$ 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 $fcdot varphi_{i}$ briefly,we can choose a specific open subset$- U:=bigcup_{i=1}^{n}int D_{i}$$- $to set forth.If in the simplest case we need $varphi_{i}$ multiplied by $f$,not to mention in most cases.
$textbf{1.}$
$psi_iin C^{infty}(U_i,mathbf{R}),$ which is positive on $D_i$ and $0$ outside of some closed set contained in $U_i$ .(problem 2-26$^*$(d))
Define $$widetilde psi_{i}:=left{begin{matrix}
psi_{i}& xin U_i\
0& xin {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 : xin U,: sum_{k=1}^{n}widetildepsi_{k}ne 0 ;;widetildepsi_{k}in C^{infty}(mathbf{R}^{n},mathbf{R}).$$ $$Longrightarrow varphi_{i}=frac{psi_{i}}{sum_{k=1}^{n}psi_{k}}=frac{widetildepsi_{i}}{sum_{k=1}^{n}widetildepsi_{k}}in C^{infty}(U,mathbf{R}) quad (i=1,2,cdots,n).$$
$textbf{2.}$
$fin C^{infty}(U,mathbf{R}),$ which value is $1$ on $A$ and $0$ outside of some closed set contained in $U$.(problem 2-26$^*$(d))
$\$Obviously,we have $fcdot varphi_{i}in C^{infty}(U,mathbf{R});(i=1,2,cdots,n).$ Note that each $fcdot varphi_{i}in C^{infty}(U,mathbf{R})$ is only constructed on $U:!$
$textbf{3.}$
Finally,we can extend $fcdot varphi_{i}$ from $U$ to the whole of $mathbf {R}^{n}$ smoothly.
In fact,since $supp(fcdotvarphi_{i})subset U$,
define $$widetilde{fcdot varphi_{i}}:=left{begin{matrix}
fcdot varphi_{i}& xin supp(fcdotvarphi_{i})\
0 & xnotin supp(fcdotvarphi_{i})
end{matrix}right.quad (i=1,2,cdots,n),$$ then we have $widetilde {fcdot varphi_{i}}in C^{infty}(mathbf{R}^{n},mathbf{R})$,$;widetilde {fcdot varphi_{i}}bigg|_{U}=fcdotvarphi_{i}$ and $fcdot varphi_{i}$ (each domain is $U$) can be smoothly extended to $widetilde {fcdot varphi_{i}}$ (each domian is $mathbf{R}^{n}).qquadqquadqquadqquadqquadqquadqquadqquadblacksquare$
(A schematic diagram concretising above elaboration)
edited Dec 27 '18 at 7:03
answered Dec 25 '18 at 5:28
user553010user553010
688
688
$begingroup$
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 $widetilde{fcdot varphi_{i}}$ 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
add a comment |
$begingroup$
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 $widetilde{fcdot varphi_{i}}$ 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
$begingroup$
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 $widetilde{fcdot varphi_{i}}$ 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
$begingroup$
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 $widetilde{fcdot varphi_{i}}$ 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
add a comment |
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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