Smooth Partitions of Unity and Exhaustions

For $t>0$, put $x:=1\mathrm{/}t$; then $x\to +\mathrm{\infty }$ as $t\to 0^{+}$ and $\beta (t)\mathrm{/}{t}^m=x^m\mathrm{/}\exp (x)$ by [F1].

The expression $x^m\mathrm{/}\exp (x)$ tends to $0$ by [L1] with $a=1$.

Therefore $\beta (t)\mathrm{/}{t}^m\to 0$ as $t\to 0^{+}$.

Repeatedly applying [L2] on $(0,\mathrm{\infty })$ proves [A1] by a routine induction on $n$.

For each $n$, both ${\beta }^{(n)}(t)$ and ${\beta }^{(n)}(t)\mathrm{/}t$ are finite linear combinations of terms $\beta (t)\mathrm{/}{t}^m$, so both tend to $0$ as $t\to 0^{+}$ by [L1] and step 1.1.

We prove recursively that $\beta$ is $C^n$, that ${\beta }^{(n)}$ vanishes on $(-\mathrm{\infty },0]$, and that ${\beta }^{(n)}(0)=0$. The case $n=0$ is [F1]. If the claim holds for $n$, then the left derivative of ${\beta }^{(n)}$ at $0$ is $0$ because ${\beta }^{(n)}$ is zero on $(-\mathrm{\infty },0]$, and the right derivative is ${\lim }_{t\to 0^{+}}{\beta }^{(n)}(t)\mathrm{/}t=0$ by step 2.1. Thus ${\beta }^{(n+1)}(0)$ exists and equals $0$, and step 2.1 also gives continuity at $0$.

Hence $\beta$ is smooth on $\mathbb{R}$ and all of its derivatives vanish at $0$.

Put $s=(r+R)\mathrm{/}2$ and $d=s^2-r^2>0$, and define $u(x)=(s^2-{\sum }_{j=1}^n{x}_j^2)\mathrm{/}d$ and $\rho (x)=\sigma (u(x))$. The coordinate derivatives of $u$ are ${\mathrm{\partial }}_{i}u=-2x_i\mathrm{/}d$, ${\mathrm{\partial }}_j{\mathrm{\partial }}_{i}u=-2{\delta }_{ij}\mathrm{/}d$, and zero at every higher order, so $u$ is smooth. Here ${\delta }_{ij}$ is $1$ when $i=j$ and $0$ otherwise.

Every iterated coordinate derivative of $\rho$ is a finite sum of products of derivatives of $u$ and factors ${\sigma }^{(k)}(u)$: induction follows by applying the chain and product rules in [L1] to each factor. These factors are continuous by [F1] and step 1.1, so $\rho$ is smooth.

If $\mathrm{\parallel }x\mathrm{\parallel }\le r$, then $u(x)\ge 1$, so $\rho (x)=1$; if $\mathrm{\parallel }x\mathrm{\parallel }\ge s$, then $u(x)\le 0$, so $\rho (x)=0$. Also $0\le \rho \le 1$ by [F1].

The nonzero locus lies in $B_s(0)$, whose closure is contained in the closed ball ${\bar{B}}_s(0)$. Thus [F2] gives $\mathrm{supp}\rho \subseteq {\bar{B}}_s(0)\subseteq B_R(0)$, since $s<R$. Together with steps 2.1 and 2.2 this proves the claim.

If $K=\mathrm{\varnothing }$, the zero function has empty support and satisfies the claim. Suppose henceforth that $K\mathrm{\ne }\mathrm{\varnothing }$. Consider all triples $(p,r,R)$ with $p\in K$, $0<r<R$, and $B_R(p)\subseteq U$. Their inner balls $B_r(p)$ form an indexed open cover of $K$: openness gives a ball $B_R(p)\subseteq U$ at each $p$, and $r=R\mathrm{/}2$ is admissible. This defines the whole family without making a simultaneous choice at every point.

By [L1], finitely many admissible triples $(p_i,r_i,R_i)$, $1\le i\le m$, have inner balls covering $K$. For each of these finitely many triples use the explicit formula in [L2], with $x$ replaced by $x-p_i$, to define ${\rho }_i$. Translation preserves the coordinate-derivative formulas, so each ${\rho }_i$ is smooth, lies in $[0,1]$, equals $1$ on ${\bar{B}}_{r_i}(p_i)$, and has support contained in $B_{R_i}(p_i)\subseteq U$.

Define $\rho =1-{\prod }_{i=1}^m(1-{\rho }_i)$. Repeated use of [L3] shows that each iterated coordinate derivative is a finite sum of products of continuous derivatives of the ${\rho }_i$, so $\rho$ is smooth. Since every factor lies in $[0,1]$, so does $\rho$. On each inner ball one factor is zero; hence $\rho =1$ on their union, an open neighbourhood of $K$.

Let $F={\bigcup }_{i=1}^m\mathrm{supp}{\rho }_i$. Each support is closed by [F1], so the finite union $F$ is closed, and step 2.1 gives $F\subseteq U$. Off $F$, every ${\rho }_i$ vanishes, so the formula in step 3.1 gives $\rho =0$. Therefore the closure of the nonzero locus of $\rho$ lies in $F$: $\mathrm{supp}\rho \subseteq F\subseteq U$.

Steps 3.1 and 4.1 give all three required properties; step 1.1 handled the empty case.

If $M$ has dimension zero, each singleton is open by the chart condition and is closed since $M$ is Hausdorff. Define $\rho (p)=1$ and $\rho =0$ elsewhere. This is locally constant, hence smooth in charts, and its support is $\{p\}\subseteq W$. Henceforth assume the dimension $n$ is positive.

Choose a smooth chart $(U,\phi )$ with $p\in U$ and put $a=\phi (p)$. Since $\phi (W\cap U)$ is open, choose $\epsilon >0$ with $B_{\epsilon }(a)\subseteq \phi (W\cap U)$, and put $R=\epsilon \mathrm{/}2$, $r=R\mathrm{/}2$. Then ${\bar{B}}_R(a)\subseteq \phi (W\cap U)$. By [L3], ${\bar{B}}_r(a)$ is compact. Apply [L1] inside $B_R(a)$ to obtain a smooth $\tilde{\rho }:{\mathbb{R}}^n\to [0,1]$ equal to one on ${\bar{B}}_r(a)$ with support in $B_R(a)$.

The closed ball ${\bar{B}}_R(a)$ is compact by [L3]. Its image $K={\phi }^{-1}({\bar{B}}_R(a))$ under the continuous inverse chart, viewed as a map into $M$, is compact by [L4]. Thus $K$ is closed in $M$ and $K\subseteq W\cap U$.

On the open cover $U\cup (M\setminus K)$, use $\rho =\tilde{\rho }\circ \phi$ on $U$ and $\rho =0$ on $M\setminus K$. The functions agree on $U\setminus K$, since its coordinate image misses ${\bar{B}}_R(a)$ and hence the support of $\tilde{\rho }$. Composition and pasting in [L2] give a smooth global real-valued function whose values lie in $[0,1]$.

The function equals one at $p$ by step 2.1. Its nonzero locus is contained in the closed set $K$ by step 4.1, so [F2] gives $\mathrm{supp}\rho \subseteq K\subseteq W$. This proves the positive-dimensional case, and step 1.1 proves the zero-dimensional case.

If $K=\mathrm{\varnothing }$, take $\rho =0$ and the empty open neighbourhood. Otherwise let $\mathcal{B}$ be the set of all smooth $b:M\to [0,1]$ whose support lies in $W$. For each $b\in \mathcal{B}$, put $V_b=\{x:b(x)>1\mathrm{/}2\}$. These sets are open by [L3] and cover $K$ by [L1]. The family uses all eligible functions, not a simultaneous choice of one function at each point.

By [L2], there are finitely many $b_1,\dots ,b_m\in \mathcal{B}$ such that $V={\bigcup }_{i=1}^m{V}_{b_i}$ contains $K$. Define $s={\sum }_{i=1}^m{b}_i$. In every chart each iterated coordinate derivative of $s$ is the sum of the corresponding continuous derivatives of the $b_i$, by repeated use of [L4]. Hence $s$ and $2s$ are smooth. Since all summands are nonnegative, $2s>1$ throughout the open set $V$.

Define $\rho =\sigma \circ (2s)$. It is smooth by [F1] and [L3], takes values in $[0,1]$, and equals one on $V$ by step 2.1.

The finite union $F={\bigcup }_{i=1}^m\mathrm{supp}{b}_i$ is closed by [F2], and lies in $W$. Off $F$ every $b_i$ is zero, so $s=0$ and $\rho =\sigma (0)=0$. Taking the closure of the nonzero locus gives $\mathrm{supp}\rho \subseteq F\subseteq W$. Together with step 3.1 and the empty case in step 1.1, this proves the claim.

Fix $x\in X$; by [F1], there is a neighbourhood $N$ of $x$ meeting only finitely many supports $\mathrm{supp}(f_i)$.

If $N$ meets the cozero set of $f_i$, then it meets $\mathrm{supp}(f_i)$ by [A1], so only finitely many cozero sets can meet $N$.

Since $x$ was arbitrary, the cozero family is locally finite by [F1].

Fix $p\in M$; by [L1], there is an open neighbourhood $U$ of $p$ meeting only finitely many cozero sets, say those with indices $i_1,\dots ,i_m$.

On $U$, the pointwise sum equals the finite sum $f_{i_1}+\cdots +f_{i_m}$, so it is well defined and smooth by [A1].

Because every point has such a neighbourhood, [L2] implies that $f$ is smooth on all of $M$.

By [L1], the sum $G:={\sum }_i{g}_i$ is smooth; because the $g_i$ are nonnegative and some $g_i(p)$ is positive at each point, one has $G(p)>0$ for all $p\in M$.

Fix a chart near $p$. Since $G(p)>0$, continuity gives a smaller chart neighbourhood on which $G>0$. The real function $r(t)=1\mathrm{/}t$ is smooth on $(0,+\mathrm{\infty })$, with $r^{(k)}(t)=(-1{)}^{k}k!t^{-k-1}$; the chartwise chain rule therefore makes $1\mathrm{/}G=r\circ G$ smooth there. As $p$ was arbitrary, $1\mathrm{/}G$ is smooth on $M$, and the product rule makes every ${\varphi }_i=g_i\mathrm{/}G$ smooth and nonnegative.

Since $G>0$, ${\varphi }_i(p)>0$ exactly when $g_i(p)>0$. Their cozero sets, and hence their supports, are equal. Thus the supports of $({\varphi }_i)$ are locally finite and satisfy $\mathrm{supp}({\varphi }_i)=\mathrm{supp}(g_i)\subseteq U_i$; their cozero sets are locally finite by [L2]. Finally ${\sum }_i{\varphi }_i=(1\mathrm{/}G){\sum }_i{g}_i=1$ pointwise. These are exactly the conditions of [L3].

Form the set $\mathcal{B}$ of all coordinate balls $B$ for which $\bar{B}$ is compact and $\bar{B}\subseteq U$ for at least one $U\in \mathcal{U}$. No member is chosen for each point. This family covers $M$: for a given $p$, one cover member $U$ contains it, and [L1] supplies one such ball within $U$.

The family $\mathcal{B}$ is an open cover, so [A1] and [L2] give a finite or countable subcover $(B_n)$. By its definition, each $B_n$ has a nonempty set of witness members $U\in \mathcal{U}$ with $\overline{B_n}\subseteq U$. Use ${\mathrm{A}\mathrm{C}}_{\omega }$ once to choose a witness $U_n$ for each listed ball (finite choice suffices in the finite case). Thus $\overline{B_n}$ is compact and contained in $U_n$.

Thus the original cover has a countable subordinate cover by relatively compact coordinate balls.

Put $H_r:={\bigcup }_{i=1}^r\overline{U_i}$ for $r\ge 1$. Each $H_r$ is compact, and the interiors of the $H_r$ cover $M$ because $U_r\subseteq \mathrm{int}(H_r)$.

Set $r_1=1$. Recursively, compactness of $H_{r_m}$ and the open cover $(\mathrm{int}(H_r){)}_{r\ge 1}$ give an integer $r_{m+1}>r_m$ such that $H_{r_m}\subseteq \mathrm{int}(H_{r_{m+1}})$. Put $K_m:=H_{r_m}$ and $K_{-1}=K_0=\mathrm{\varnothing }$. Then $K_m\subseteq \mathrm{int}(K_{m+1})$ and the interiors of the $K_m$ cover $M$.

Put $A_1:=K_1$ and $A_m:=K_m\setminus \mathrm{int}(K_{m-1})$ for $m\ge 2$; each $A_m$ is compact by [L3]. For each $m$, consider all tuples $(n,W,V)$ with $V$ a coordinate ball, $W$ open, and $\bar{W}\subseteq V\subseteq \bar{V}\subseteq U_n\cap \mathrm{int}(K_{m+1})\setminus K_{m-2}\mathrm{.}$ Their $W$-sets cover $A_m$: for an individual $x\in A_m$, take any $U_n$ containing $x$ and successively use [L2] and [L1] inside the displayed open neighbourhood to construct such $V$ and $W$ around $x$. This proves existence at each point without a simultaneous point-indexed choice. By compactness, the set of finite ordered lists of these tuples whose $W$-sets cover $A_m$ is nonempty (and contains the empty list when $A_m=\mathrm{\varnothing }$).

Use [A2] once on the countable family of nonempty finite-list sets from step 3.1. Concatenate the chosen lists into an at-most-countable indexed family $(W_k,V_k{)}_{k\in I}$; take $I=\mathrm{\varnothing }$ when every list is empty. They cover $M$ because the annuli $A_m$ cover $M$. Given $y\in M$, choose $m$ with $y\in \mathrm{int}(K_m)$. If $V_k$ came from annulus $A_j$ with $j\ge m+3$, then $V_k\cap \mathrm{int}(K_m)=\mathrm{\varnothing }$ because $K_{j-2}\supseteq K_m$. Only finitely many tuples came from the remaining finitely many annuli, so $(V_k)$ is locally finite.

Hence $(W_k)$ and $(V_k)$ give the required countable locally finite shrinking.

Apply [L1] and then [L2] to obtain an at-most-countable index set $I$, open sets $W_k$ and coordinate balls $V_k$ for $k\in I$ such that $M={\bigcup }_{k\in I}{W}_k$, the family $(V_k)$ is locally finite, and each $V_k$ lies in a selected cover member $U_{j(k)}$. The set $I$ is allowed to be finite or empty.

For each $k\in I$, [L3] gives a nonempty set of smooth functions $g_k:M\to [0,1]$ equal to $1$ near $\overline{W_k}$ and supported in $V_k$. Use [A1] to select one for each $k$. The family $(g_k{)}_{k\in I}$ is locally finite and pointwise positive because every point lies in some $W_k$. When $M$ is empty, $I$ may be empty and the partition claim is vacuous.

Normalize $(g_k)$ by [L4]; the resulting nonnegative smooth family $({\psi }_k{)}_{k\in I}$ has locally finite supports, sums to one, and satisfies $\mathrm{supp}({\psi }_k)\subseteq V_k\subseteq U_{j(k)}$ for every $k$.

To index the partition by the original cover as required by Smooth partitions of unity subordinate to an open cover, put ${\varphi }_j={\sum }_{k\in I:j(k)=j}{\psi }_k$ for each $j\in J$. Every sum is locally finite and smooth; the family $(\mathrm{supp}{\varphi }_j{)}_{j\in J}$ is locally finite, since near any point only finitely many ${\psi }_k$ have support. Nonnegativity and local finiteness give $\mathrm{supp}{\varphi }_j={\bigcup }_{k:j(k)=j}\mathrm{supp}{\psi }_k\subseteq U_j$, and ${\sum }_{j\in J}{\varphi }_j={\sum }_{k\in I}{\psi }_k=1$. Thus $({\varphi }_j{)}_{j\in J}$ is the required subordinate partition.

A countable coordinate-ball cover is an open cover.

Apply [L1] to that open cover.

The resulting partition is subordinate to the given countable coordinate-ball cover.

If $M=\mathrm{\varnothing }$ or $A=\mathrm{\varnothing }$, take $f=0$, using the empty open neighbourhood of $A$. Otherwise put $O_0=U$ and $O_1=M\setminus A$. Form the set of all tuples consisting of a smooth coordinate ball $B$, its chart data, and a label $a\in \{0,1\}$ such that $\bar{B}$ is compact and lies in $O_a$. These balls cover $M$ by [F1]. Fix a countable basis. For each nonempty basis member contained in one eligible ball, [A1] chooses one such tuple. The selected balls cover $M$: every point lies in a basis member inside an eligible ball. Enumerate the selected tuples as $(B_n,a_n)$, repeating one if the nonempty family is finite. This avoids choosing a ball at every point.

Put $H_r={\bigcup }_{n\le r}\overline{B_n}$. These sets are compact and closed by [L1], and their interiors cover $M$. Put $r_1=1$ and let $r_{m+1}$ be the least integer greater than $r_m$ for which $H_{r_m}\subseteq \mathrm{int}{H}_{r_{m+1}}$. Such an integer exists by compactness and the increasing open cover $(\mathrm{int}{H}_r)$. Set $K_m=H_{r_m}$ and $K_0=K_{-1}=\mathrm{\varnothing }$. Then $K_m\subseteq \mathrm{int}{K}_{m+1}$ and their interiors cover $M$. This recursion uses least integers, not dependent choice.

The sets $C_m=K_m\setminus \mathrm{int}{K}_{m-1}$ are compact and cover $M$. For each $m$, form all tuples $(W,V,n)$ of nested smooth coordinate balls with compact closures satisfying $\bar{W}\subseteq V,\bar{V}\subseteq B_n\cap \mathrm{int}{K}_{m+1}\setminus K_{m-2}\mathrm{.}$ Their $W$-sets cover $C_m$: for $x\in C_m$, choose one $B_n$ containing $x$ and apply the small-ball construction twice inside the displayed open neighbourhood of $x$. This is an existence argument at one point, not a simultaneous point-indexed selection. Compactness provides a finite ordered list of such tuples covering $C_m$.

Apply [A1] to the countable family of nonempty sets of finite ordered covering lists, taking the empty list when $C_m$ is empty. Label a tuple from annulus $m$ by $(m,j)$. The resulting family $(W_{m,j})$ covers $M$, and $(V_{m,j})$ is locally finite: if $x\in \mathrm{int}{K}_t$, no tuple from $m\ge t+2$ meets $\mathrm{int}{K}_t$, since its $V$ misses $K_{m-2}\supseteq K_t$. Only finitely many tuples occur in the remaining finitely many annuli. The pairs $(m,j)$ form a subset of ${\mathbb{N}}^2$, hence an at most countable index set $I$.