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].

By [L1] it is smooth, equals $1$ on the inner closed ball, and has support in ${\bar{B}}_s(0)\subset B_R(0)$. It is radial because the formula depends only on $\mathrm{\parallel }x\mathrm{\parallel }$.

This is the required Euclidean example.

Write a point of $S^1$ as $(x,y)$ and set $b(t)=h(t+1\mathrm{/}2)$ using [F1]. Then $b$ is nonnegative and smooth, is zero for $t\le -1\mathrm{/}2$, and is positive for $t>-1\mathrm{/}2$.

Put $g_1(x,y)=b(x)$ and $g_2(x,y)=b(-x)$. At least one of $x,-x$ is nonnegative, so $g_1+g_2>0$ everywhere. Define ${\varphi }_i=g_i\mathrm{/}(g_1+g_2)$. These functions are smooth, take values in $[0,1]$, and sum to one.

The support of ${\varphi }_1$ is contained in the closed arc $\{x\ge -1\mathrm{/}2\}$, which misses $(-1,0)$, and the support of ${\varphi }_2$ is contained in $\{x\le 1\mathrm{/}2\}$, which misses $(1,0)$. Thus $\mathrm{supp}({\varphi }_i)\subseteq U_i$ as required, with no use of the general partition theorem.

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.

For $c<1$ the sublevel set $\{h\le c\}$ is empty. For $c\ge 1$, [A1] makes it a closed ball of radius $\sqrt{1-1\mathrm{/}c}<1$, hence compact in $B_1(0)$.

If $C\subseteq \mathbb{R}$ is compact, it is closed and bounded above by some $c$. Continuity of $h$ makes $h^{-1}(C)$ closed in $B_1(0)$, and it lies in the compact set $\{h\le c\}$ from step 2.1. Thus $h^{-1}(C)$ is compact, so $h$ is proper and smooth.

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. This explicit construction uses no choice principle.

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 constant function $f=1$ is smooth on the open set $(0,\mathrm{\infty })$ by [L2]. Its prescribed zero extension is $F(x):=1$ for $x>0$ and $F(x):=0$ for $x\le 0$.

For $\epsilon =1\mathrm{/}2$ and every $\delta >0$, put $x=\delta \mathrm{/}2$. Then $\mathrm{\mid }x\mathrm{\mid }<\delta$ but $\mathrm{\mid }F(x)-F(0)\mathrm{\mid }=1$. Thus $F$ is not continuous at zero by [L2], and cannot be smooth by [L1].

This refutes the universal zero-extension claim. It does not assert that every smooth zero extension requires a cutoff or support separated from the boundary; the counterexample proves only that the recipe fails in general. No choice principle or existence-of-cutoff theorem is used.