Smooth Partitions of Unity and Exhaustions — Examples

On $(-1,1)$ the function is the composite from [A1], and on $\mathrm{\mid }t\mathrm{\mid }\ge 1$ it is identically zero.

At $t=\pm 1$, the inner variable $1-t^2$ tends to $0^{+}$, so [L1] shows that all derivatives from the inside tend to $0$ and match the outer zero branch.

Therefore $\rho$ is smooth, positive on $(-1,1)$, and supported on $[-1,1]$.

The displayed function is the explicit construction used in [L1].

Therefore it is smooth, radial, equal to $1$ on the inner closed ball, and supported in the outer open ball.

This is the required Euclidean example.

The sets $U_1$ and $U_2$ are open and cover $S^1$.

Apply [L1] to this cover to obtain the required functions ${\varphi }_1,{\varphi }_2$.

Thus the circle carries a two-function smooth partition subordinate to the chosen arcs.

Because $x\mapsto (x_1+1)\mathrm{/}2$ is smooth, so are ${\varphi }_{+}$ and ${\varphi }_{-}=1-{\varphi }_{+}$.

The functions are nonnegative and sum to $1$; moreover ${\varphi }_{+}=0$ when $x_1\le -1$, so $\mathrm{supp}({\varphi }_{+})\subseteq \{x:x_1\ge -1\}\subseteq U_{+}$, and ${\varphi }_{-}=0$ when $x_1\ge 1$, so $\mathrm{supp}({\varphi }_{-})\subseteq \{x:x_1\le 1\}\subseteq U_{-}$.

Hence $({\varphi }_{-},{\varphi }_{+})$ is the required smooth partition.

The function $h$ is polynomial in the coordinates, hence smooth.

By [A1], every sublevel set is compact, so $h$ is proper.

This is an explicit example of [L1].

The denominator is positive on $B_1(0)$, so $h$ is smooth there.

By [A1], every sublevel set is a closed ball of radius strictly less than $1$, hence compact in $B_1(0)$.

Therefore $h$ is a proper smooth function on the open ball, as predicted by [L1].

The function $g$ is smooth and nonnegative.

By [A1], the equation $g(x)=0$ holds exactly when $x\in \mathbb{Z}$.

Thus $g$ realizes the closed set $\mathbb{Z}$ as a smooth zero set, as promised abstractly by [L1].

For every fixed $x$, only finitely many $f_n(x)$ are nonzero, so the family $(f_n{)}_{n\ge 1}$ is pointwise finite; however $f_n(0)=0$ and $f_n(1\mathrm{/}n)=1$ for every $n\ge 1$.

The sum $F(x):={\sum }_n{f}_n(x)$ therefore satisfies $F(0)=0$ and $F(1\mathrm{/}n)\ge 1$ for every $n$, so $F$ is not continuous at $0$.

This refutes the statement and exhibits why [L1] needs local finiteness.

The naive zero extension is the step function $F(x):=1$ for $x>0$ and $F(x):=0$ for $x\le 0$.

The function $F$ is not continuous at $0$, so it is not smooth.

Hence the hypothesis singled out in [L1] is essential.