Dtyjlui

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.

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.

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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 $U$.

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)