Proof of Existence of Smooth Bump Functions on Euclidean Space

Throughout, continuity of real-valued functions is continuity at every point in the Euclidean sense. For $m\in\{1,n\}$ regard $\mathbb{R}^m$ as a metric space through the Euclidean distance $d_E$, a metric by Euclidean Distance is a Metric on $\mathbb{R}^n$, and regard the codomain $\mathbb{R}$ as a metric space through the metric $d_{\mathbb{R}}$ of The Absolute Value Metric on the Real Line. By claim 1 of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions, a real-valued function on a subset $E\subseteq\mathbb{R}^m$ is continuous at a point of $E$ in the Euclidean sense if and only if it is continuous there relative to $E$ as a map into $(\mathbb{R},d_{\mathbb{R}})$. Combining this equivalence with claims 2 and 3 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, and inducting on the number of terms and of factors, gives the principle used repeatedly below:

(SP) a finite sum of finite products of real-valued functions that are continuous at every point of a subset $E\subseteq\mathbb{R}^m$ in the Euclidean sense is itself continuous at every point of $E$ in the Euclidean sense.

A polynomial function on $\mathbb{R}$ or on $\mathbb{R}^n$ means a finite sum of finite products of constant functions and coordinate functions. Constant functions and coordinate functions are continuous directly from the definition of Euclidean continuity (for a coordinate function take $\delta=\varepsilon$; for a constant any $\delta$), so polynomial functions are continuous by (SP).

Step 0 (the exponential function). Define $\exp:\mathbb{R}\to\mathbb{R}$ by $\exp(u)=\sum_{k=0}^{\infty}u^{k}/k!$, with factorials; the partial sums form a Cauchy sequence, since for $k>2|u|$ the terms are dominated in absolute value by a geometric sequence with ratio $1/2$, and hence converge by Every Cauchy Sequence of Real Numbers Converges; the same domination shows the series converges absolutely. We use three properties. (i) For $u\ge 0$ all terms are nonnegative, so $\exp(u)>0$ and $\exp(u)\ge u^{M}/M!$ for every exponent $M$. (ii) $\exp(u+s)=\exp(u)\exp(s)$, by the Cauchy product of the two series together with the binomial expansion of $(u+s)^k$; the rearrangement of the doubly indexed series is justified by absolute convergence, comparing all partial sums with the product of the two absolute-value series. (iii) $\exp$ is differentiable with $\exp'=\exp$, where the derivative is the one-dimensional one: by (ii), $\bigl(\exp(u+s)-\exp(u)\bigr)/s=\exp(u)\bigl(\exp(s)-1\bigr)/s$, and the series gives $|\exp(s)-1-s|\le|s|^{2}\exp(|s|)$, so $\bigl(\exp(s)-1\bigr)/s\to 1$ as $s\to 0$. Iterating (iii), $\exp$ has derivatives of every order, all equal to $\exp$, and in particular $\exp$ is continuous.

Step 1 ($h$ is a smooth map). Define $h:\mathbb{R}\to\mathbb{R}$ by $h(t)=\exp(-1/t)$ for $t>0$ and $h(t)=0$ for $t\le 0$. On $\{t<0\}$, $h$ vanishes identically, so derivatives of all orders exist and are $0$. On $\{t>0\}$, an induction using claims 2 and 3 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives and the one-dimensional chain rule, together with property (iii) of Step 0 and the derivative $-1/t^{2}$ of $t\mapsto 1/t$, shows that for every order $m\ge 1$ the $m$th derivative exists and has the form $h^{(m)}(t)=p_m(1/t)\exp(-1/t)$, where the polynomial functions $p_m$ are determined recursively by

$$p_1(u)=u^2,\qquad p_{m+1}(u)=u^{2}\bigl(p_m(u)-p_m'(u)\bigr);$$

by induction, $p_m$ has degree $2m$. Decay estimate. Fix $m$, write $p_m(u)=\sum_{l=0}^{2m}c_l u^{l}$, and set $C_m=(2m+2)!\sum_{l=0}^{2m}|c_l|$. Note that the exponent in property (i) may be chosen independently of $m$, and that $M=m+1$ would not suffice here since $p_m$ has degree $2m$; we take $M=2m+2$. For $0<t\le 1$, property (i) with $u=1/t$ and $M=2m+2$ gives $0\le\exp(-1/t)\le(2m+2)!\,t^{2m+2}$, while $|p_m(1/t)|\le\bigl(\sum_l|c_l|\bigr)t^{-2m}$; hence

$$\bigl|h^{(m)}(t)\bigr|=\bigl|p_m(1/t)\bigr|\exp(-1/t)\le C_m\,t^{2}\longrightarrow 0\quad(t\to 0^{+}),$$

and likewise the right difference quotient of $h^{(m)}$ at $0$ satisfies

$$\Bigl|\frac{h^{(m)}(t)-0}{t}\Bigr|=\frac{1}{t}\bigl|p_m(1/t)\bigr|\exp(-1/t)\le C_m\,t\longrightarrow 0\quad(t\to 0^{+}).$$

By induction on $m$ (using the left-sided quotients being identically $0$), each $h^{(m)}(0)$ exists and equals $0$, and each $h^{(m)}$ is continuous at $0$ by the first display. Hence $h$ is a smooth map on $\mathbb{R}$.

Step 2 (construction of $\chi$). Let $q:\mathbb{R}^n\to\mathbb{R}$ be given by

$$q(x)=d(x,x_0)^2=\sum_{i=1}^{n}\bigl(x_i-(x_0)_i\bigr)^2,$$

using the Euclidean distance. Then $q$ is a polynomial function; its first partial derivatives are $2\bigl(x_i-(x_0)_i\bigr)$, its partial derivatives of order two are constant and all its partial derivatives of order at least three vanish, all orders being taken in the order-$\alpha$ sense used by the definition of a smooth map; all of these are polynomial functions, hence continuous by the opening paragraph, so $q$ is a smooth map. Define $g:\mathbb{R}\to\mathbb{R}$ by

$$g(t)=\frac{h(s^2-t)}{h(s^2-t)+h(t-r^2)}.$$

The denominator is everywhere positive: since $0<r<s$ we have $r^2<s^2$, so for each $t$ at least one of $s^2-t>0$, $t-r^2>0$ holds, and $h>0$ on positive arguments (Step 0 (i)) while $h\ge 0$ everywhere. The functions $t\mapsto h(s^2-t)$ and $t\mapsto h(t-r^2)$ are smooth: by Chain Rule for One-Dimensional Derivatives, applied with the affine map as the inner function and with the derivative of an affine map computed from claims 1, 2 and 3 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives, their derivatives of all orders are $\pm$ the corresponding derivatives of $h$ evaluated along that affine map. Their sum is smooth by claim 2 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives, applied at each order; so $g$ is smooth by Products and Quotients of C^k Real-Valued Maps on Euclidean Open Sets Are C^k applied to the quotient with nonvanishing denominator.

Set $\chi=g\circ q$. Fix $a\in\mathbb{R}^n$ and an index $i$. The slice of $\chi$ at $a$ in the $i$th variable is the composite of $g$ with the slice of $q$ at $a$ in the $i$th variable, both slices being defined on the interval supplied by claim 1 of Slice Function and the Partial Derivative applied on the open set $\mathbb{R}^n$. Since $q$ is a polynomial function its first partial derivatives exist, so by claim 2 of that lemma the slice of $q$ is differentiable at $a_i$ with derivative $\partial_i q(a)$; and $g$ is differentiable at $q(a)$, which is an interior point of $\mathbb{R}$, by the previous paragraph. Hence Chain Rule for One-Dimensional Derivatives gives that the slice of $\chi$ is differentiable at $a_i$ with derivative $g'(q(a))\,\partial_i q(a)$, and claim 2 of the slice lemma converts this back into

$$\partial_i\chi=(g'\circ q)\,\partial_i q\qquad\text{on }\mathbb{R}^n .$$

The derivative $g'$ is again smooth on $\mathbb{R}$ and $\partial_i q$ is again a polynomial function, so the same argument applies with $\partial_i\chi$ in place of $\chi$. By induction on the order, every partial derivative of $\chi$ of every order $\alpha$ exists on $\mathbb{R}^n$ and is a finite sum of finite products of polynomial functions with functions of the form $g^{(k)}\circ q$. Each $g^{(k)}$ is differentiable on $\mathbb{R}$, hence continuous there by Differentiability at an Interior Point Implies Continuity There and claim 1 of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions, so each $g^{(k)}\circ q$ is continuous by Composition of Continuous Euclidean Maps. Therefore every such partial derivative is continuous by (SP), and $\chi$ is smooth.

Step 3 (the three properties). Since $h\ge 0$, we have $0\le g(t)\le 1$ for all $t$, so property 1 holds. If $d(x,x_0)\le r$ then $q(x)\le r^2$, so $h(q(x)-r^2)=0$ and $\chi(x)=g(q(x))=1$; this is property 2. If $d(x,x_0)\ge s$ then $q(x)\ge s^2$, so $h(s^2-q(x))=0$ and $\chi(x)=0$; this is property 3. $\blacksquare$