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 $0$.

Define $u(x):=(R^2-\mathrm{\parallel }x{\mathrm{\parallel }}^2)\mathrm{/}(R^2-r^2)$ and $\rho (x):=\sigma (u(x))$; then $u$ is smooth by [A1] and [L1], so $\rho$ is smooth by [F1] and [L1].

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

Thus $\rho$ maps into $[0,1]$, equals $1$ on ${\bar{B}}_r(0)$, and vanishes on ${\mathbb{R}}^n\setminus B_R(0)$, so $\mathrm{supp}(\rho )\subseteq B_R(0)$.

For each $p\in K$, choose radii as in [L1] and a bump ${\rho }_p$ as in [L2]; compactness gives finitely many points $p_1,\dots ,p_m$ such that $K\subseteq {\bigcup }_{i=1}^m{B}_{r_{p_i}}(p_i)$.

Put $s:={\rho }_{p_1}+\cdots +{\rho }_{p_m}$; then $s$ is smooth by [A1], one has $s\ge 1$ on $K$, and $\mathrm{supp}(s)\subseteq U$.

Define $\rho :=\sigma \circ s$; then $\rho$ is smooth, equals $1$ on $K$, and vanishes off $U$, so $\mathrm{supp}(\rho )\subseteq U$.

Choose a smooth chart $(U,\phi )$ with $p\in U$ and put $a:=\phi (p)$. Choose $0<r<R$ such that ${\bar{B}}_r(a)\subseteq B_R(a)\subseteq {\bar{B}}_R(a)\subseteq \phi (W\cap U)\mathrm{.}$ Applying [L1] to ${\bar{B}}_r(a)\subseteq B_R(a)$ gives a smooth $\tilde{\rho }:{\mathbb{R}}^n\to [0,1]$ equal to $1$ on ${\bar{B}}_r(a)$ and supported in $B_R(a)$. Its support is compact by [A1].

Let $K:={\phi }^{-1}(\mathrm{supp}(\tilde{\rho }))$. By step 1.1, $K$ is a compact, hence closed, subset of $W\cap U$. On the open cover $U\cup (M\setminus K)$, define $\rho =\tilde{\rho }\circ \phi$ on $U$ and $\rho =0$ on $M\setminus K$. The formulas agree on $U\setminus K$, so [L2] gives a smooth global function.

One has $\rho (p)=\tilde{\rho }(a)=1$, and $\rho$ vanishes outside $W$, so $\mathrm{supp}(\rho )\subseteq W$.

For each $p\in K$, choose a smooth bump ${\rho }_p$ from [L1] with support in $W$ and ${\rho }_p(p)=1$; compactness gives finitely many points $p_1,\dots ,p_m$ such that the open sets $V_{p_i}:={\rho }_{p_i}^{-1}((1\mathrm{/}2,1])$ cover $K$.

Put $s:={\rho }_{p_1}+\cdots +{\rho }_{p_m}$; then $s$ is smooth by [A1], one has $2s\ge 1$ on the open neighbourhood $V_{p_1}\cup \cdots \cup V_{p_m}$ of $K$, and $\mathrm{supp}(s)\subseteq W$.

Define $\rho :=\sigma \circ (2s)$; then $\rho$ is smooth, equals $1$ on an open neighbourhood of $K$, and vanishes off $W$, so $\mathrm{supp}(\rho )\subseteq W$.

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 $p\in M$.

By [A1], the reciprocal $1\mathrm{/}G$ is smooth, so each ${\varphi }_i=g_i(1\mathrm{/}G)$ is smooth and nonnegative; also $\mathrm{supp}({\varphi }_i)\subseteq \mathrm{supp}(g_i)$.

The family $({\varphi }_i)$ is locally finite by [L2], and ${\sum }_i{\varphi }_i=(1\mathrm{/}G){\sum }_i{g}_i=1$ pointwise. Therefore $({\varphi }_i)$ is a partition of unity subordinate to $(\mathrm{supp}(g_i))$.

For each $p\in M$, choose $U_p\in \mathcal{U}$ containing $p$, then choose a coordinate ball $B_p$ with $p\in B_p\subseteq \overline{B_p}\subseteq U_p$ by [L1].

The family $(B_p{)}_{p\in M}$ is an open cover of $M$, so [A1] and [L2] give a countable subcover $B_1,B_2,\dots$. Each $\overline{B_n}$ is compact and lies in some member of $\mathcal{U}$ by step 1.1.

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]. Fix $x\in A_m$, choose some $U_n$ containing $x$, and set $O_x:=U_n\cap \mathrm{int}(K_{m+1})\setminus K_{m-2}\mathrm{.}$ By step 2.1, this is an open neighbourhood of $x$. Apply [L2] to choose an open $R_x$ with $x\in R_x\subseteq \overline{R_x}\subseteq O_x$, and then [L1] to choose a coordinate ball $V_x$ with $x\in V_x\subseteq R_x$ and compact closure. Thus $\overline{V_x}\subseteq \overline{R_x}\subseteq O_x$. Applying [L2] inside $V_x$, choose an open set $W_x$ with $x\in W_x\subseteq \overline{W_x}\subseteq V_x\mathrm{.}$ Compactness of $A_m$ gives finitely many such pairs covering $A_m$.

Collect the finitely many pairs from each $A_m$ into sequences $(W_k)$ and $(V_k)$. They are countable, and they cover $M$ because the annuli $A_m$ cover $M$. Given $y\in M$, choose $m$ with $y\in \mathrm{int}(K_m)$. If $V_k$ is attached to $A_j$ with $j\ge m+3$, then step 3.1 gives $V_k\cap \mathrm{int}(K_m)=\mathrm{\varnothing }$ because $K_{j-2}\supseteq K_m$. Hence the neighbourhood $\mathrm{int}(K_m)$ meets only the finitely many families attached to $A_1,\dots ,A_{m+2}$. Thus $(V_k)$ is locally finite.

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

Apply [L1] and then [L2] to obtain countably many open sets $W_k$ and coordinate balls $V_k$ such that $M={\bigcup }_k{W}_k$, the family $(V_k)$ is locally finite, and each $V_k$ lies in some member $U_{n(k)}$ of $\mathcal{U}$.

For each $k$, apply [L3] to the compact set $\overline{W_k}\subseteq V_k$ to obtain a smooth function $g_k:M\to [0,1]$ that equals $1$ on a neighbourhood of $\overline{W_k}$ and is supported in $V_k$. The family $(g_k)$ is locally finite and pointwise positive because every point lies in some $W_k$.

Normalize $(g_k)$ by [L4]; the resulting family $({\varphi }_k)$ is a smooth partition of unity, and $\mathrm{supp}({\varphi }_k)\subseteq V_k\subseteq U_{n(k)}$ for every $k$.

Hence $({\varphi }_k)$ is subordinate to $\mathcal{U}$ in the sense of Smooth partitions of unity subordinate to an open cover.

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.

The two open sets $U$ and $M\setminus A$ cover $M$, so [L1] gives smooth functions ${\varphi }_U,{\varphi }_A:M\to [0,1]$ subordinate to this cover with ${\varphi }_U+{\varphi }_A=1$.

Because $\mathrm{supp}({\varphi }_A)\subseteq M\setminus A$, the function ${\varphi }_A$ vanishes on an open neighbourhood of $A$; hence ${\varphi }_U=1$ there, and $\mathrm{supp}({\varphi }_U)\subseteq U$ by [F1].

Taking $f:={\varphi }_U$ yields the required smooth function.

The singleton $\{p\}$ is closed and lies in the open set $M\setminus A$.

Apply [L1] to the closed set $\{p\}\subseteq M\setminus A$; the resulting function is $1$ at $p$ and vanishes on $A$ because its support lies in $M\setminus A$.

This is the required separating smooth function.

Let $\chi$ be as in [L1]; then $\chi f$ is smooth on $U$ by [A1] and equals $f$ on an open neighbourhood of $C$ because $\chi =1$ there.

Since $\mathrm{supp}(\chi )\subseteq U$, the function $\chi$ vanishes on an open neighbourhood of $M\setminus U$, so the local formula $\chi f$ on $U$ agrees with the constant zero map on an open neighbourhood of every boundary point of $U$.

Pasting these two local formulas by [L2] yields a smooth function $F:M\to \mathbb{R}$ with $F=f$ near $C$ and $\mathrm{supp}(F)\subseteq U$.

Fix $i\in I$. On the open set $U_i$ let $G_i:={\varphi }_i{f}_i$, and on the open set $M\setminus \mathrm{supp}({\varphi }_i)$ let $H_i:=0$. By [A1], the map $G_i$ is smooth on $U_i$; by [F1], the two open sets cover $M$ and on the overlap $U_i\setminus \mathrm{supp}({\varphi }_i)$ one has ${\varphi }_i=0$, so $G_i=H_i$. Therefore [L1] pastes them to a smooth global function $F_i:M\to \mathbb{R}$ with $F_i={\varphi }_i{f}_i$ on $U_i$ and $\mathrm{supp}(F_i)\subseteq \mathrm{supp}({\varphi }_i)$.

By [F1] and step 1.1, the family $(\mathrm{supp}(F_i){)}_{i\in I}$ is locally finite. Hence the sum $F:={\sum }_i{F}_i$ is well defined and smooth by [L2].

Let $p\in M$. If ${\varphi }_i(p)\mathrm{\ne }0$, then $p\in U_i$ and step 1.1 gives $F_i(p)={\varphi }_i(p)f_i(p)$. If ${\varphi }_i(p)=0$, then $p\mathrm{\notin }\mathrm{supp}({\varphi }_i)$, so step 1.1 gives $F_i(p)=0={\varphi }_i(p)f_i(p)$. Thus $F(p)={\sum }_i{\varphi }_i(p)f_i(p)$, and this pointwise formula is smooth by step 2.1.

Start from the countable cover $U_1,U_2,\dots$ of [L1]. Recursively choose integers $1\le m_1<m_2<\cdots$ such that $\overline{U_1}\subseteq {\bigcup }_{i\le m_1}{U}_i$ and, for each $k\ge 1$, the compact set ${\bigcup }_{i\le m_k}\overline{U_i}$ is contained in ${\bigcup }_{i\le m_{k+1}}{U}_i$; this is possible by [A1].

Put $K_k:={\bigcup }_{i\le m_k}\overline{U_i}$. Each $K_k$ is compact, step 1.1 gives $K_k\subseteq \mathrm{int}(K_{k+1})$, and every point of $M$ lies in some $K_k$ because the $U_i$ cover $M$.

Therefore $(K_k)$ is a compact exhaustion in the sense of Compact exhaustions of a manifold.

Let $(K_n)$ be the exhaustion from [L1], and set $K_0=K_{-1}=\mathrm{\varnothing }$. For each $n\ge 1$, apply [L2] to $A_n:=K_n\setminus \mathrm{int}(K_{n-1})$ inside the open set $\mathrm{int}(K_{n+1})\setminus K_{n-2}$ to obtain a smooth function ${\rho }_n:M\to [0,1]$ equal to $1$ on a neighbourhood of $A_n$ and supported in $\mathrm{int}(K_{n+1})\setminus K_{n-2}$.

The supports of $({\rho }_n)$ are locally finite. Indeed, if $x\in A_m:=K_m\setminus \mathrm{int}(K_{m-1})$, then $\mathrm{supp}({\rho }_n)\subseteq \mathrm{int}(K_{n+1})\setminus K_{n-2}$ from step 1.1. If $n\le m-3$, then $\mathrm{int}(K_{n+1})\subseteq \mathrm{int}(K_{m-2})\subseteq \mathrm{int}(K_{m-1})$, so $x\mathrm{\notin }\mathrm{supp}({\rho }_n)$. If $n\ge m+3$, then $K_{n-2}\supseteq K_{m+1}\supseteq K_m$, so again $x\mathrm{\notin }\mathrm{supp}({\rho }_n)$. Thus only ${\rho }_{m-2},{\rho }_{m-1},{\rho }_m,{\rho }_{m+1},{\rho }_{m+2}$ can be nonzero at $x$. Therefore $h:={\sum }_{n\ge 1}n{\rho }_n$ is a smooth nonnegative function by [L3].

If $c\ge 0$ and $N>c$, then any point $x\mathrm{\notin }{K}_N$ lies in some annulus $A_n$ with $n>N$, so ${\rho }_n(x)=1$ and hence $h(x)\ge n>N>c$. Thus $\{x:h(x)\le c\}\subseteq K_N$, and this sublevel set is compact by [A1].