Proof of Existence of a Smooth Plateau Function on Euclidean Space

Each result cited is universally quantified over the data in its own statement. Continuity of a real-valued function on $\mathbb{R}^{q}$ is that of Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation §extrema: continuity at every point relative to $\mathbb{R}^{q}$, as a map from $(\mathbb{R}^{q},d_{E})$ into $(\mathbb{R},d_{\mathbb{R}})$ with $d_{\mathbb{R}}$ the metric of The Absolute Value Metric on the Real Line; continuity of a function $\mathbb{R}\to\mathbb{R}$ means continuity at every point relative to $\mathbb{R}$ as a map from $(\mathbb{R},d_{\mathbb{R}})$ to itself. By The Euclidean Distance on the Real Line is the Absolute Value Metric, $d_{\mathbb{R}}$ is the Euclidean distance on $\mathbb{R}=\mathbb{R}^{1}$, which is the metric on the target used in Convolution of a Continuous Function with a Compactly Supported Continuous Kernel. We write $x-y$ for $x+(-y)$ in $\mathbb{R}$ and use freely the field axioms of Field; $2$ denotes $1+1$, which is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field, so that $2^{-1}$ exists, and $\tfrac{u}{2}$ denotes $u\cdot2^{-1}$. For every real $u$, distributivity gives $u+u=u\cdot2$, hence $\tfrac{u+u}{2}=u\cdot2\cdot2^{-1}=u$; and $\tfrac{2}{2}=1$. Non-strict compatibility of the order with addition is clause 1 of Ordered Field, and $|-1|=|1|=1$ by Absolute Value in an Ordered Field and claim 2 of Properties of the Absolute Value in an Ordered Field. For points $x,y$ of $\mathbb{R}^{q}$, $x-y=x+(-y)$ with $-y=(-1)y$ (claims 2 and 3 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space), so $\lVert -y\rVert=|-1|\,\lVert y\rVert=\lVert y\rVert$ and $\lVert y-x\rVert=\lVert(-1)(x-y)\rVert=\lVert x-y\rVert$ by claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, and $d_{E}(x,y)=\lVert x-y\rVert$ by claim 2 there.

Step 1: a continuous plateau. Since $r<s$, claim 1 of Elementary Order Arithmetic in an Ordered Field gives $0<s-r$; by claim 8 there, $\eta=\tfrac{s-r}{2}$ and $\delta=\tfrac{\eta}{2}$ are positive with $\eta+\eta=s-r$ and $\delta+\delta=\eta$. Write $2\delta$ for $\delta+\delta$; then $2\delta=\eta>0$, $s-r=2\delta+2\delta$, and $(2\delta)^{-1}>0$ by claim 7 there. Define $a,b,m:\mathbb{R}\to\mathbb{R}$ and $f:\mathbb{R}^{q}\to\mathbb{R}$ by

By Absolute Value in an Ordered Field, $|a(t)|=a(t)$ if $0\le a(t)$ and $|a(t)|=-a(t)$ otherwise; so $b(t)=\tfrac{a(t)+a(t)}{2}=a(t)$ if $0\le a(t)$, and $b(t)=\tfrac{a(t)-a(t)}{2}=\tfrac{0}{2}=0$ if $a(t)<0$. In every case $0\le b(t)$. Likewise, if $b(t)\le1$ then $0\le1-b(t)$ (clause 1 of Ordered Field), so $|1-b(t)|=1-b(t)$ and $m(t)=\tfrac{b(t)+b(t)}{2}=b(t)$; and if $1<b(t)$ then $1-b(t)<0$ (claim 1 of Elementary Order Arithmetic in an Ordered Field, adding $-b(t)$), so $|1-b(t)|=-(1-b(t))=b(t)-1$ (claim 6 of Additive Cancellation and Elementary Additive Identities in a Field) and $m(t)=\tfrac{1+b(t)-(b(t)-1)}{2}=\tfrac{2}{2}=1$. Hence $0\le m(t)\le1$ for every $t$, and $0\le f(x)\le1$ for every $x$. Two further values are needed.

(a) If $\lVert x-x_{0}\rVert\le r+\delta$ then $f(x)=1$. Indeed, by claim 4 of Elementary Order Arithmetic in an Ordered Field and clause 1 of Ordered Field, $(s-\delta)-\lVert x-x_{0}\rVert\ge(s-\delta)-(r+\delta)=s-r-2\delta=2\delta$, so $a(\lVert x-x_{0}\rVert)\ge2\delta\,(2\delta)^{-1}=1$ by claim 5 of Elementary Arithmetic in an Ordered Field; thus $b(\lVert x-x_{0}\rVert)=a(\lVert x-x_{0}\rVert)\ge1$, and $m(\lVert x-x_{0}\rVert)=1$ in both cases $b=1$ and $b>1$.

(b) If $\lVert x-x_{0}\rVert\ge s-\delta$ then $f(x)=0$. Indeed, clause 1 of Ordered Field gives $(s-\delta)-\lVert x-x_{0}\rVert\le0$, so $a(\lVert x-x_{0}\rVert)\le0\cdot(2\delta)^{-1}=0$ by claim 5 of Elementary Arithmetic in an Ordered Field and claim 1 of Zero Products and Elementary Identities in a Field; then $b(\lVert x-x_{0}\rVert)=0$ (both cases $a<0$ and $a=0$ give $0$), so $b(\lVert x-x_{0}\rVert)\le1$ and $m(\lVert x-x_{0}\rVert)=b(\lVert x-x_{0}\rVert)=0$.

(c) $f$ is continuous on $\mathbb{R}^{q}$. The function $N(x)=\lVert x-x_{0}\rVert$ is continuous on $\mathbb{R}^{q}$: for $x,y\in\mathbb{R}^{q}$, claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ applied to $x-x_{0}=(y-x_{0})+(x-y)$ and to $y-x_{0}=(x-x_{0})+(y-x)$ gives $N(x)-N(y)\le\lVert x-y\rVert$ and $N(y)-N(x)\le\lVert y-x\rVert=\lVert x-y\rVert$ (clause 1 of Ordered Field), the second of which is $-\lVert x-y\rVert\le N(x)-N(y)$ by claim 4 of Elementary Order Arithmetic in an Ordered Field; so $|N(x)-N(y)|\le\lVert x-y\rVert=d_{E}(x,y)$ by claim 6 of Properties of the Absolute Value in an Ordered Field, and given $\varepsilon>0$, $\delta_{0}=\varepsilon$ serves in Continuous Map Between Metric Spaces. The absolute value $t\mapsto|t|$ is continuous on $\mathbb{R}$ by the reverse triangle inequality (claim 7 of Properties of the Absolute Value in an Ordered Field), again with $\delta_{0}=\varepsilon$, and the identity map of $\mathbb{R}$ is continuous with $\delta_{0}=\varepsilon$. By claims 1 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space (constants, sums, scalar multiples), $a$ is continuous on $\mathbb{R}$; by claim 3 of Semicontinuity and Continuity Under Composition with a Continuous Map, $|a|$ is continuous on $\mathbb{R}$, hence so is $b$ by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space; in the same way $1-b$, $|1-b|$ and then $m$ are continuous on $\mathbb{R}$; and finally $f=m\circ N$ is continuous on $\mathbb{R}^{q}$ by claim 3 of Semicontinuity and Continuity Under Composition with a Continuous Map.

Step 2: mollification. By claim 2 of Existence of Mollifier Kernels of Every Radius there is a mollifier kernel $\rho:\mathbb{R}^{q}\to\mathbb{R}$ of radius $\delta$ on $\mathbb{R}^{q}$. By clause 1 of that definition $\rho$ is smooth, hence continuous on $\mathbb{R}^{q}$ by claim 3 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous; by clause 3 there, $\rho(y)=0$ whenever $\lVert y\rVert>\delta$; by clause 2, $0\le\rho$; and by clause 4, $\rho$ is integrable with respect to Lebesgue measure $\lambda_{q}$ with $\int_{\mathbb{R}^{q}}\rho\,d\lambda_{q}=1$.

Apply Convolution of a Continuous Function with a Compactly Supported Continuous Kernel with $n=q$, $\Omega=\mathbb{R}^{q}$ (open in $(\mathbb{R}^{q},d_{E})$ by claim 1 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous and Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$), the function $f$ of Step 1 (continuous into $(\mathbb{R},d_{\mathbb{R}})$, which is the required target metric as recalled above), the radius $\delta$ and the kernel $\rho$. Since every closed ball is a subset of $\mathbb{R}^{q}$, the set $\Omega^{\delta}$ there is $\mathbb{R}^{q}$, and for every $x\in\mathbb{R}^{q}$ the integrand $h_{x}$ there is $h_{x}(y)=f(x-y)\rho(y)$ for every $y\in\mathbb{R}^{q}$; it is integrable by claim 1 of The Convolution Integrand is Continuous, Compactly Supported and Integrable. Put $\chi=f*\rho:\mathbb{R}^{q}\to\mathbb{R}$, so that $\chi(x)=\int_{\mathbb{R}^{q}}h_{x}\,d\lambda_{q}$ for every $x$. By claim 2 of Convolution with a $C^k$ Kernel is of Class $C^k$, $\chi$ is smooth on $\mathbb{R}^{q}$. We also record that the zero function on $\mathbb{R}^{q}$ equals $0\cdot\rho$, so it is integrable with $\int_{\mathbb{R}^{q}}0\,d\lambda_{q}=0\cdot\int_{\mathbb{R}^{q}}\rho\,d\lambda_{q}=0$ by the linearity of the integral for integrable functions (claim 2 of Linearity and Monotonicity of the Lebesgue Integral).

Step 3: the values of $\chi$. Fix $x\in\mathbb{R}^{q}$. For every $y$ with $\lVert y\rVert>\delta$ we have $\rho(y)=0$, hence $h_{x}(y)=0$.

Suppose $\lVert x-x_{0}\rVert\le r$. If $\lVert y\rVert\le\delta$, then $(x-y)-x_{0}=(x-x_{0})+(-y)$, so $\lVert(x-y)-x_{0}\rVert\le\lVert x-x_{0}\rVert+\lVert -y\rVert\le r+\delta$ by claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and clause 1 of Ordered Field (applied twice), so $f(x-y)=1$ by Step 1(a) and $h_{x}(y)=\rho(y)$; if $\lVert y\rVert>\delta$, then $h_{x}(y)=0=\rho(y)$. Hence $h_{x}=\rho$ and $\chi(x)=\int_{\mathbb{R}^{q}}\rho\,d\lambda_{q}=1$.

Suppose $\lVert x-x_{0}\rVert\ge s$. If $\lVert y\rVert\le\delta$, then $x-x_{0}=((x-y)-x_{0})+y$, so $s\le\lVert x-x_{0}\rVert\le\lVert(x-y)-x_{0}\rVert+\lVert y\rVert\le\lVert(x-y)-x_{0}\rVert+\delta$ by claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and clause 1 of Ordered Field, whence $s-\delta\le\lVert(x-y)-x_{0}\rVert$ by the same clause, $f(x-y)=0$ by Step 1(b), and $h_{x}(y)=0$; if $\lVert y\rVert>\delta$, then $h_{x}(y)=0$. Hence $h_{x}$ is the zero function and $\chi(x)=0$ by the record at the end of Step 2.

Finally, for every $x$ and $y$, $0\le f(x-y)\le1$ and $0\le\rho(y)$ give $0\le h_{x}(y)\le\rho(y)$ by claim 5 of Elementary Arithmetic in an Ordered Field. The zero function, $h_{x}$ and $\rho$ are integrable, so the monotonicity of the integral for integrable functions (claim 2 of Linearity and Monotonicity of the Lebesgue Integral) gives $0=\int_{\mathbb{R}^{q}}0\,d\lambda_{q}\le\chi(x)\le\int_{\mathbb{R}^{q}}\rho\,d\lambda_{q}=1$. Thus $\chi$ has all the asserted properties.