partitions of unity in the proof of the Meyers-Serrin Theorem In my previous question, proof of the Meyers-Serrin Theorem in Evans's "Partial Differential Equations", "partitions of unity" is used in the proof (see (2) in the linked post):

Let $\{V_i\}$ be an open cover for a bounded open subset $U$ of $\Bbb{R}^n$. Then there exist $\zeta_i:U\to[0,1]$ such that 
  $
\zeta_i\in C_c^\infty(V_i)
$
  and 
  $
\sum_{i=1}^\infty\zeta_i(x)=1,\quad x\in U. 
$  $\tag{**}$

Here are my questions: 


*

*If $U$ is unbounded, do we still have $\{\zeta_i\}$ with the desired properties?

*A general definition of partitions of unity (for example in Folland's Real Analysis) requires that for any $x\in U$, there is a neighborhood of $x$ where all but finitely many $\zeta_i$ are identically zero. Do we have this property for the $\zeta_i$'s here?

*What mode of convergence of the infinite sum
$$
\sum_{i=1}^\infty\zeta_i(x)=1,\quad x\in U
$$
can we expect? Of course we have the pointwise convergence here. Do we have uniform convergence or compact convergence?

 A: In short, the answer is we still have a partition of unity if $U$ is unbounded, and we only get pointwise/compact convergence (uniform convergence cannot happen).
For simplicity why not think about just $\mathbb{R}$ first? There certainly will be a function $\zeta(x)$ with


*

*$\zeta\in C^\infty_c(\mathbb{R})$ with supp$(\zeta)=[-1,1]$,

*$\zeta$ is even

*$\zeta(x)=1$ for all $x\in[-1/2,1/2]$, $\zeta(x)>0$ for all $x\in(-1,1)$ and smoothly goes from $1$ to $0$ as $|x|$ goes from $1/2$ to $1$


(basically, imagine a smoothed out bump function that is symmetric across the $y-$axis. It wouldn't be too hard to write an explicit formula for this, but would require a little thought to make it smooth...)
Now we can piece these together so that the transitions overlap in a way that the sum over these intervals is $1$: set $\zeta_k=\zeta(x-\frac{3}{2}k)$.
Here is the picture:

Now it follows that for the open cover $\{U_k\}$ where $U_k=(-1+\frac{3}{2}k,1+\frac{3}{2}k)$ $\zeta_k\in C^\infty_c(U_k)$ and $\sum_{k}\zeta_k(x)=1$ for all $x\in\mathbb{R}$. Moreover, all but at most two terms will be locally identically zero at each $x\in\mathbb{R}$.
As far as convergence, why can't uniform convergence happen? if we have a finite partial sum $S_N:=\sum_{k=0}^N\zeta_k$, then spt$(S_N)\subset \bigcup_{k=0}^N U_k$, no? But the full sum achieves the value of $1$ everywhere, so $\|S_N-\sum_k\zeta_k\|_\infty=1$ for all $N$.
The exact same argument, adjusted slightly, will prove that we do have compact convergence. Also this should generalize to $\mathbb{R}^n$, with a little more bookkeeping...
A: A good resource for this is the wikipedia pages "partition of unity" and "paracompact space", which I'm quoting in the definition/theorem below.

Definition. A partition of unity on a topological space $U$ is a tuple of continuous functions $(\zeta_\alpha)_{\alpha \in J}$ such that:
  
  
*
  
*$\zeta_\alpha : U \to [0,1]$ for all $\alpha \in J$.
  
*For each $x \in U$, there is an open neighborhood $O_x$ of $x$ for which $\left. \zeta_\alpha \right|_{O_x} = 0$ for all but finitely many $\alpha \in J$. This property is called local finiteness.
  
*$\sum\limits_{\alpha \in J} \zeta_\alpha(x) = 1$ for all $x \in U$.
  

Generally, when speaking of partitions of unity, we have an open cover in mind. This is addressed in the following theorem:

Theorem. $U$ is paracompact iff there exists a partition of unity on $U$ which may also be chosen to satisfy one of the following "secondary" conditions:
  
  
*
  
*For any open cover $\{V_i\}_{i \in I}$, the partition of unity $(\zeta_\alpha)_{\alpha \in J}$ is subordinate to $\{V_i\}_{i \in I}$, which means that $J = I$ and for each $i \in I$, the support of $\zeta_i$ is contained in $V_i$.
  
*For any open cover $\{V_i\}_{i \in I}$, each $\zeta_\alpha$ has compact support which is contained in $V_i$ for some $i \in I$.
  
  
  If $U$ is compact, then there exists a partition of unity satisfying both secondary conditions.

Two other important theorems are 


*

*Any metrizable space is paracompact. Hence your open subset $U$ of $\mathbb R^n$ is paracompact and admits a partition of unity.

*If $U$ is an open subset of $\mathbb R^n$, then we may take the partition of unity functions $\zeta_\alpha$ to be smooth.


Using the above, we answer your questions:


*

*No.  As stated, you have $\zeta_i \in C_c^\infty(V_i)$, which requires the partition of unity to be subordinate to $\{V_i\}_{i \in I}$ (with the same index set) and to have compact supports. This requires both of the "secondary" conditions. But since $U$ is unbounded, it's not compact, and we're not guaranteed the existence of a partition of unity with both secondary conditions. A good counterexample is the one you pointed out, where $U = \mathbb R$ and the open cover consists of a single element $V_1 = \mathbb R$. For our partition of unity to be subordinate to $\{V_1\}$, we're allowed just one function $\zeta_1 : \mathbb R \to [0,1]$, and by 3., it has to be the constant function $\zeta_1 \equiv 1$. But this doesn't satisfy the compact support condition, as $\operatorname{supp} \zeta_1 = \mathbb R$ is not compact.

*Yes. All partitions of unity satisfy 2. of the definition, which is what you've asked about.

*The important thing to realize here is that, because of local finiteness (i.e., 2.), the sum $\sum_{i \in \mathbb N} \zeta_i(x)$ in 3. has only finitely many terms. Although in general the sum may not converge uniformly*, it does converge locally uniformly because the sum is finite; take $N$ in the definition of locally uniform to be $\max\{ i \in \mathbb N : \left. \zeta_i \right|_{O_x} \neq 0\}$, which exists for a sufficiently small ball $O_x$ about $x$ by local finiteness. Compact convergence follows similarly; if $K$ is compact, the open cover $\bigcup_{x \in K} O_x$, where $O_x$ is the neighborhood of $x$ supplied by local finiteness, reduces to a finite subcover $O_{x_1}, \dots, O_{x_k}$. Corresponding to $O_{x_1}, \dots, O_{x_k}$ are $N_1, \dots, N_k$ as found for locally uniform convergence, and we may take take $N = \max\{N_1, \dots, N_k\}$ in the definition of compact convergence.
*A good counterexample is given by @charlestoncrabb. For the sum $\mathbb R \ni x \mapsto \sum_{i \in \mathbb N} \zeta_i(x)$ to converge uniformly to $1$ on $\mathbb R$, we would need for any $\varepsilon > 0$ some $N_0 \in \mathbb N$ so that $\sup_{x \in \mathbb R} \left( 1 - \sum_{i=1}^N \zeta_i(x) \right) < \varepsilon$ for all $N \geq N_0$. But as we can see in the picture, if we omit any of the $\zeta_i$, the difference $1 - \sum_{j \in \mathbb N \backslash \{i\}} \zeta_j(x)$ is $1$ on some interval of length $1$ (which was before covered by $\zeta_i$), and hence not less than $\varepsilon$ for $0 < \varepsilon < 1$.
