Existence of smooth partitions of unity without open condition

partial differential equations

This is a problem from Evans Chapter 5.

We assume a bounded $U$ and $U \subset \subset \cup_{i=1}^N V_i$ and want to prove the existence of $C^{\infty}$ functions $\zeta_i (i=1 \;to \; N)$ such that
\begin{align*}
\begin{cases} 0 \leq \zeta_i \leq 1, \; \text{support of } \zeta_i \subset V_i \; (i=1,…,N) \\
\sum_{i=1}^N \zeta_i = 1 \; \; \text{ on }\; U.
\end{cases}
\end{align*}

I think I should use the following preceding exercise, which gives the existence of a smooth $\zeta$ uniformly 1 on a smaller open set compactly contained in a bigger one, where $\zeta$ disappears in some neighbourhood of $\partial U.$

Let $U$, $V$ be open sets, with $V\subset\subset U$. Show there exists a smooth function $\zeta$ such that $\zeta\equiv1$ on $V$, $\zeta=0$ near $\partial U$. (Hint: Take $V\subset\subset W\subset\subset U$ and mollify $\chi_W$.)

For the problem in the image, I have a nice space to work with mollifiers and the smooth partition is there since $U, V$ were open. But now I don't have any openness involved in this problem, which is my concern. I can't even imagine how the domain of each partition should look like. How do we restrict a support of a smooth function to $V_i$ and at the same time guarantee that the partition doesn't disappear on any $U \cap \partial V_i$?

I looked through many other problems concerning this existence of smooth partition of unity, but they all needed open domains.

Best Answer

It seems to me that you can w.l.o.g. assume that $U$ is open and all the $V_i$ are open. In other words (making this wlog a bit more formal) you can just take the interiors $\mathring{U}$ and $\mathring{V}_i$ and solve the problem with the interiors.

Once you have solved the problem with the $\mathring{U}$ and the $\mathring{V}_i$ instead of $U$ and $V$ you have found functions $\zeta_i \in C^{\infty}$ such that $$ \begin{cases} 0 \leq \zeta_i \leq 1\\ \text{supp}(\zeta_i) \subset \mathring{V}_i\\ \sum_{i} \zeta_i = 1 \text{ on } \mathring{U}. \end{cases} $$ Now you can use continuity of the $\zeta_i$ and $\sum_{i} \zeta_i$ to conclude that you actually solved the problem for the original $U$ and $V_i$.

Solving the problem for the interiors can be done in a few steps (I think!! but it has been some time since I did this proof, but it can be found online probably, but some hints might be helpful if you want figure it out without full solutions first):

  • First conclude that $U \subset \subset \cup_i V_i \quad$ also holds when you substitute $U$ and $V_i$ with the interiors.
  • Use problem 5 in your image to first find $\xi_i$ for every $V_i$ (actually the interior), this can be done because you are solving the problem with the interiors (which are of course open!)
  • Some tricks to define $\zeta_i$ using a smart division of the form $\zeta_i := \frac{\xi_{i}}{\sum_{j} \xi_j}$.

Edit: An update after your comment. Indeed, we have $\cup_i(\text{int}(V_i)) \subset \text{int}(\cup_i(V_i))$, but in general not the other way around, so your comment is valid and I did not think of this beforehand.

Luckily however, things work out in the following way: Suppose we have a point $x \in \text{int}(\cup_i(V_i)) - \cup_i(\text{int}(V_i))$ and $x \in \overline{U}$. This means that $x \not \in \text{int}V_i$ for all $i \in \{1,\dots,n\}$ and $x \in \partial V_i$ for at least one $i$.

However, in this case a partition of unity subordinate to $\{V_i\}_i$ cannot exist: Indeed for arbitrary $i \in \{1,\dots,n\}$ we have $\zeta_i$ with support in $V_i$ which means that $\zeta_i(x) = 0$, since either $x \in \partial V_i$ or $x \not \in V_i$. This implies we have $\sum_i \zeta_i(x) = 0.$

However, since $x \in \overline{U}$ we get by continuity $\sum_i \zeta_i(x) = 1$, a contradiction. So we are not interested in this case, since no interesting partition of unity exists, so we can wlog assume that $x \in \overline{U} \implies x \in int(V_i)$ for some $i$, and I believe this solves the problem you correctly stated in your comment.

Related Question