Proof of The Exponential Bump Building Block is Smooth on the Real Line

Throughout, $\mathbb{R}$ is an interval all of whose points are interior points of it, by claim 1 of One-Dimensional Derivatives, Partial Derivatives, and Smoothness on the Real Line; differentiability of a function defined on an interval, at an interior point of that interval, is that of Derivative at an Interior Point. Absolute values are those of that definition, with the properties collected in Properties of the Absolute Value in an Ordered Field; the order arithmetic used below is that of Elementary Order Arithmetic in an Ordered Field, and $2$ denotes $1+1$. For $w\in\mathbb{R}$ we abbreviate $w\cdot w$ by $w^{2}$.

Step 1 (the two open half-lines). Put $P=\{s\in\mathbb{R}:0<s\}$ and $N=\{s\in\mathbb{R}:s<0\}$. We check that each is an interval all of whose points are interior points of it.

$P$ is an interval: if $x,z\in P$ and $y\in\mathbb{R}$ satisfies $x\le y\le z$, then $0<x$ and $x\le y$, so $0<y$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, i.e. $y\in P$. Let $x\in P$. By claim 8 there, $0<x\cdot 2^{-1}$ and $x\cdot 2^{-1}<x$, so $u=x\cdot 2^{-1}$ lies in $P$ and $u<x$. By claim 6 there $0<1$, so $x<x+1$ by claim 1 there, and $v=x+1\in P$ by claim 2. Thus $u<x<v$ with $u,v\in P$, so $x$ is an interior point of $P$.

$N$ is an interval: if $x,z\in N$ and $x\le y\le z$, then $y\le z$ and $z<0$, so $y<0$ by claim 2. Let $x\in N$. From $0<1$ and claim 4 we get $-1<0$, hence $x+(-1)<x$ by claim 1, and $x+(-1)<x<0$ gives $x+(-1)\in N$ by claim 2. Put $w=(-x)\cdot 2^{-1}$; since $0<-x$ by claim 4, claim 8 gives $0<w$, so $x<x+w$ by claim 1, while $x+w<x+(-x)=0$ by claim 1 and $w<-x$ (claim 8). Thus $x+w\in N$ and $x+(-1)<x<x+w$, so $x$ is an interior point of $N$.

Step 2 (localisation). We record the following elementary fact. Let $f:\mathbb{R}\to\mathbb{R}$, let $x_{0}\in\mathbb{R}$, let $J\subseteq\mathbb{R}$ be an interval containing $x_{0}$ as an interior point, and let $\theta\in\mathbb{R}$ with $0<\theta$ be such that every $z\in\mathbb{R}$ with $|z-x_{0}|<\theta$ belongs to $J$. If the restriction $f|_{J}$ is differentiable at $x_{0}$ with derivative $L$, then $f$ is differentiable at $x_{0}$ with derivative $L$.

Indeed, let $\varepsilon>0$ and let $\delta>0$ be as in Derivative at an Interior Point for $f|_{J}$ at $x_{0}$ with the value $L$ and this $\varepsilon$. Let $\delta'$ be the smaller of $\delta$ and $\theta$ (claim 9 of Elementary Order Arithmetic in an Ordered Field); then $0<\delta'$. If $h\in\mathbb{R}$ satisfies $0<|h|<\delta'$, then $|(x_{0}+h)-x_{0}|=|h|<\theta$, so $x_{0}+h\in J$; since also $0<|h|<\delta$, the choice of $\delta$ gives

because $f$ and $f|_{J}$ take the same values at $x_{0}$ and at $x_{0}+h$. As $\varepsilon>0$ was arbitrary, $f$ is differentiable at $x_{0}$ with derivative $L$.

We apply this in two situations. If $0<x_{0}$ we take $J=P$ and $\theta=x_{0}$: if $|z-x_{0}|<x_{0}$ then $-x_{0}<z-x_{0}$ by claim 9 of Properties of the Absolute Value in an Ordered Field, so $0<z$ by claim 1 of Elementary Order Arithmetic in an Ordered Field. If $x_{0}<0$ we take $J=N$ and $\theta=-x_{0}$, which is positive by claim 4 of that lemma: if $|z-x_{0}|<-x_{0}$ then $z-x_{0}<-x_{0}$ by claim 9 of the absolute value lemma, so $z<0$ by claim 1.

Step 3 (the family $D$). For a polynomial function $q$ on $\mathbb{R}$ define $\psi_{q}:\mathbb{R}\to\mathbb{R}$ by

and let $D$ be the set of all functions from $\mathbb{R}$ to $\mathbb{R}$ of the form $\psi_{q}$ with $q$ a polynomial function on $\mathbb{R}$.

Fix a polynomial function $q$ on $\mathbb{R}$, and let $q^{\ast}$ be the polynomial function supplied by claim 2 of Derivative of a Polynomial Function on the Real Line, so that for every interval $I$ and every interior point $x_{0}$ of $I$ the restriction $q|_{I}$ is differentiable at $x_{0}$ with derivative $q^{\ast}(x_{0})$. Define $q^{\sharp}:\mathbb{R}\to\mathbb{R}$ by

By claim 1 of Properties of Natural Number Powers in a Field we have $x^{1}=x$ for the natural number powers of $\mathbb{R}$, so the map $x\mapsto x$ is the map $x\mapsto x^{1}$ and is a polynomial function by claim 1 of Constants, Powers, Sums, Scalar Multiples and Products of Polynomial Functions; by claims 2 and 3 there, $q^{\sharp}$ is a polynomial function on $\mathbb{R}$, so $\psi_{q^{\sharp}}\in D$. Steps 4 to 6 show that $\psi_{q}$ is differentiable at every point of $\mathbb{R}$ and that its derivative function is $\psi_{q^{\sharp}}$.

Step 4 (points of $P$). Let $s_{0}\in P$ and put $u_{0}=s_{0}^{-1}$; then $0<u_{0}$ by claim 7 of Elementary Order Arithmetic in an Ordered Field, so $u_{0}\in P$. Let $r:P\to\mathbb{R}$ be given by $r(s)=s^{-1}$. Since $0\notin P$ and every point of $P$ is an interior point of $P$ by Step 1, claim 2 of Reciprocal Rule for One-Dimensional Derivatives shows that $r$ is differentiable at $s_{0}$ with $r'(s_{0})=-\bigl(u_{0}\bigr)^{2}$. Moreover $r(s)\in P$ for every $s\in P$, again by claim 7.

(a) The point $r(s_{0})=u_{0}$ is an interior point of $P$ at which $q|_{P}$ is differentiable with derivative $q^{\ast}(u_{0})$. Hence Chain Rule for One-Dimensional Derivatives applies to $\gamma=r$ and $g=q|_{P}$ and gives that $q\circ r:P\to\mathbb{R}$ is differentiable at $s_{0}$ with

(b) Let $\gamma:P\to\mathbb{R}$ be given by $\gamma(s)=(-1)\,r(s)$, so that $\gamma(s)=-\bigl(s^{-1}\bigr)$ by claim 2 of Zero Products and Elementary Identities in a Field. By claim 2 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives, $\gamma$ is differentiable at $s_{0}$ with $\gamma'(s_{0})=(-1)\bigl(-(u_{0})^{2}\bigr)=(u_{0})^{2}$. Every point of $\mathbb{R}$ is an interior point of $\mathbb{R}$, and $\exp$ is differentiable at every point with $\exp'=\exp$ by claim 3 of Basic Properties of the Exponential Function. Hence Chain Rule for One-Dimensional Derivatives applies to $\gamma$ and $g=\exp$ and gives that $E=\exp\circ\gamma:P\to\mathbb{R}$ is differentiable at $s_{0}$ with

(c) For $s\in P$ we have $\psi_{q}(s)=(q\circ r)(s)\,E(s)$, so the restriction $\psi_{q}|_{P}$ is the pointwise product of $q\circ r$ and $E$. By the product rule (claim 3 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives) it is differentiable at $s_{0}$ with derivative

which is $q^{\sharp}(u_{0})\exp\bigl(-\bigl(s_{0}^{-1}\bigr)\bigr)=\psi_{q^{\sharp}}(s_{0})$. By Step 2 with $J=P$ and $\theta=s_{0}$, $\psi_{q}$ is differentiable at $s_{0}$ with $\psi_{q}'(s_{0})=\psi_{q^{\sharp}}(s_{0})$.

Step 5 (points of $N$). Let $s_{0}\in N$. Every $s\in N$ satisfies $s\le 0$, so $\psi_{q}|_{N}$ is the function on $N$ with constant value $0$, which by claim 1 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives is differentiable at $s_{0}$ with derivative $0$. By Step 2 with $J=N$ and $\theta=-s_{0}$, $\psi_{q}$ is differentiable at $s_{0}$ with $\psi_{q}'(s_{0})=0$; and $\psi_{q^{\sharp}}(s_{0})=0$ because $s_{0}\le 0$.

Step 6 (the origin). Let $m:\mathbb{R}\to\mathbb{R}$ be given by $m(x)=x\cdot q(x)$; it is a polynomial function on $\mathbb{R}$ by claims 1 and 3 of Constants, Powers, Sums, Scalar Multiples and Products of Polynomial Functions, using again that $x\mapsto x$ is a polynomial function. Let $\varepsilon>0$. By claim 2 of The Exponential Function Dominates Every Polynomial Function applied to $m$ there is $\delta\in\mathbb{R}$ with $0<\delta$ such that

Since $\psi_{q}(0)=0$, the difference quotient of $\psi_{q}$ at $0$ for $h\ne 0$ equals $h^{-1}\psi_{q}(h)$. Let $h\in\mathbb{R}$ with $0<|h|<\delta$. If $h<0$ then $\psi_{q}(h)=0$, so the quotient is $0$ and its distance to $0$ is $0<\varepsilon$. If $0<h$ then $h=|h|<\delta$ and

so, since $\exp$ takes positive values (claim 2 of Basic Properties of the Exponential Function) and by claims 1 and 4 of Properties of the Absolute Value in an Ordered Field,

As $\varepsilon>0$ was arbitrary, $\psi_{q}$ is differentiable at $0$ with $\psi_{q}'(0)=0$; and $\psi_{q^{\sharp}}(0)=0$.

Step 7 (proof of claim 2). By Steps 4 to 6, every element of $D$ is differentiable at every point of $\mathbb{R}$, and the function sending $s$ to $\psi_{q}'(s)$ is $\psi_{q^{\sharp}}$, which again lies in $D$. Hence claim 3 of One-Dimensional Derivatives, Partial Derivatives, and Smoothness on the Real Line applies to $D$ and shows that every element of $D$ is a smooth map on $\mathbb{R}^{1}$. Let $q_{1}$ be the polynomial function on $\mathbb{R}$ with constant value $1$ (claim 1 of Constants, Powers, Sums, Scalar Multiples and Products of Polynomial Functions). For $0<s$ we have $\psi_{q_{1}}(s)=1\cdot\exp\bigl(-\bigl(s^{-1}\bigr)\bigr)=\varphi(s)$, and for $s\le 0$ we have $\psi_{q_{1}}(s)=0=\varphi(s)$. Therefore $\varphi=\psi_{q_{1}}\in D$ is smooth on $\mathbb{R}^{1}$.

Step 8 (proof of claim 1). If $s\le 0$ then $\varphi(s)=0$, so $0\le\varphi(s)$ and $0<\varphi(s)$ fails. If $0<s$ then $\varphi(s)=\exp\bigl(-\bigl(s^{-1}\bigr)\bigr)$, which is positive by claim 2 of Basic Properties of the Exponential Function; hence $0<\varphi(s)$, and $0\le\varphi(s)$. Both assertions of claim 1 follow. $\blacksquare$