Proof of A Bounded-Derivative Truncation of the Cube Map on the Real Line

Each result cited below is universally quantified over the data appearing in its own statement, and is applied here to the data named in the step in question. Throughout, $D$ denotes the function from $\mathbb{R}$ to $\mathbb{R}$ with $D(s)=1+\varepsilon s^{2}$, and powers are those of Natural Number Power of an Element of a Field, so that $s^{1}=s$ and $s^{S(m)}=s^{m}\,s$ for every natural number $m$ with successor $S(m)$, by claim 1 of Properties of Natural Number Powers in a Field; in particular $s^{2}=s\,s$, $s^{3}=s^{2}s$ and $s^{4}=s^{3}s$. By claim 1 of One-Dimensional Derivatives, Partial Derivatives, and Smoothness on the Real Line, $\mathbb{R}$ is an interval every point of which is an interior point of it; all applications below of the one-dimensional differentiation rules take the interval to be $\mathbb{R}$ and the interior point to be the point under discussion.

A preliminary remark. For a positive $y\in\mathbb{R}$ one has $|y|=y$. Indeed, by claim 1 of Properties of the Absolute Value in an Ordered Field either $|y|=y$ or $|y|=-y$; in the second case $0\le|y|=-y$ gives $y\le0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field, contradicting $0<y$.

Claim 1. Let $s\in\mathbb{R}$. By claim 1 of Properties of the Absolute Value in an Ordered Field either $|s|=s$ or $|s|=-s$, and in both cases $|s|\,|s|=s\,s=s^{2}$, by the sign rules of the field $\mathbb{R}$. Since $0\le|s|$ by that same claim, claim 5 of Elementary Arithmetic in an Ordered Field, applied with $0\le|s|$ and the nonnegative multiplier $|s|$, gives $0\,|s|\le|s|\,|s|$, and $0\,|s|=0$ by claim 1 of Zero Products and Elementary Identities in a Field; hence $0\le s^{2}$. Applying claim 5 of Elementary Arithmetic in an Ordered Field again, with $0\le s^{2}$ and the nonnegative multiplier $\varepsilon$, gives $0\le\varepsilon s^{2}$, so $1\le1+\varepsilon s^{2}=D(s)$ by claim 3 of Elementary Arithmetic in an Ordered Field. Since $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, claim 2 of that lemma gives $0<D(s)$; in particular $D(s)\ne0$, so its multiplicative inverse exists, and that inverse is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field. Write $w(s)=D(s)^{-1}$, so that $D(s)\,w(s)=1$ and $0<w(s)$. Finally, from $1\le D(s)$ and the nonnegative multiplier $w(s)$, claim 5 of Elementary Arithmetic in an Ordered Field gives $w(s)=1\,w(s)\le D(s)\,w(s)=1$. This proves the assertions of claim 1 and legitimises the formula defining $\phi_{\varepsilon}$.

Claim 2, the derivative. By claim 1 of Derivative of a Polynomial Function on the Real Line the map $s\mapsto s^{1}$, which is the identity map of $\mathbb{R}$, is differentiable at every point with derivative $1$. By the product rule, claim 3 of Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers, the map $s\mapsto s^{2}$ is therefore differentiable at every point with derivative $1\,s+s\,1=2s$, where $2=1+1$; and, applying that rule once more to $s^{3}=s^{2}s$, the map $s\mapsto s^{3}$ is differentiable at every point with derivative $2s\,s+s^{2}\,1=3s^{2}$. By claims 1 and 2 of Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers, $D$ is differentiable at every point with $D'(s)=\varepsilon\,2s=2\varepsilon s$. Since $D$ vanishes nowhere, claim 1 of Reciprocal Rule for One-Dimensional Derivatives applies and shows that $w$ is differentiable at every point with

$$w'(s)=-D'(s)\,\bigl(D(s)^{-1}\bigr)^{2}=-2\varepsilon s\,w(s)^{2}.$$

Since $\phi_{\varepsilon}(s)=s^{3}\,w(s)$, the product rule gives that $\phi_{\varepsilon}$ is differentiable at every point with

$$\phi_{\varepsilon}'(s)=3s^{2}\,w(s)+s^{3}\bigl(-2\varepsilon s\,w(s)^{2}\bigr)=3s^{2}\,w(s)-2\varepsilon s^{4}\,w(s)^{2}.$$

Now $D(s)\,w(s)=1$ gives $w(s)=D(s)\,w(s)^{2}=(1+\varepsilon s^{2})\,w(s)^{2}$, so $3s^{2}w(s)=(3s^{2}+3\varepsilon s^{4})\,w(s)^{2}$ and therefore

$$\phi_{\varepsilon}'(s)=(3s^{2}+3\varepsilon s^{4}-2\varepsilon s^{4})\,w(s)^{2}=(3s^{2}+\varepsilon s^{4})\,w(s)^{2},$$

which is the asserted formula.

Claim 2, continuity and the class $C^{1}$. We use the two readings of continuity of a function from $\mathbb{R}$ to $\mathbb{R}$ fixed in Euclidean, Metric and Sequential Continuity of a Real Function of a Real Variable, which agree by Euclidean, Metric and Sequential Continuity of a Real Function of a Real Variable (ii).

First, $w$ is continuous at every $a\in\mathbb{R}$. Indeed, let $x\in\mathbb{R}$ satisfy $|x-a|<1$. From $D(x)w(x)=1=D(a)w(a)$ one computes

$$w(x)-w(a)=\bigl(D(a)-D(x)\bigr)\,w(x)\,w(a),$$

and, since $0<w(x)\le1$ and $0<w(a)\le1$ by claim 1, two applications of claim 5 of Elementary Arithmetic in an Ordered Field give $w(x)\,w(a)\le1$; hence, by claims 4 and 1 of Properties of the Absolute Value in an Ordered Field and claim 5 of Elementary Arithmetic in an Ordered Field,

$$|w(x)-w(a)|=|D(a)-D(x)|\,w(x)\,w(a)\le|D(a)-D(x)| .$$

Moreover $D(a)-D(x)=\varepsilon(a^{2}-x^{2})=\varepsilon(a-x)(a+x)$, so $|D(a)-D(x)|=\varepsilon\,|x-a|\,|a+x|$ by claim 4 of Properties of the Absolute Value in an Ordered Field, using $|a-x|=|x-a|$ from claim 2 of that lemma. By the triangle inequality, claim 5 of Properties of the Absolute Value in an Ordered Field, $|a+x|\le|a|+|x|\le|a|+\bigl(|x-a|+|a|\bigr)<2|a|+1$. Hence $|w(x)-w(a)|\le L\,|x-a|$ with $L=\varepsilon\,(2|a|+1)$, which is nonnegative, for every $x$ with $|x-a|<1$. By Euclidean, Metric and Sequential Continuity of a Real Function of a Real Variable (iii), applied with $\rho=1$, the function $w$ is continuous at $a$ in both readings.

Second, for every natural number $m$ the map $s\mapsto s^{m}$ is continuous at every point of $\mathbb{R}$ in both readings, and so is $s\mapsto3s^{2}+\varepsilon s^{4}$. Indeed, by claim 2 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions the constant functions and the single coordinate function of $\mathbb{R}^{1}$, which is the identity map, are smooth on $\mathbb{R}^{1}$, and by claim 3 of that lemma sums, scalar multiples and products of smooth functions are smooth; so each $s\mapsto s^{m}$, being an iterated product of the identity map, and also $s\mapsto3s^{2}+\varepsilon s^{4}$, are smooth on $\mathbb{R}^{1}$. By claim 3 of Euclidean Space Is Open in Itself they are continuous at every point of $\mathbb{R}^{1}$ in both readings.

Third, by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, applied in the metric space $(\mathbb{R},d_{\mathbb{R}})$ with $A=\mathbb{R}$, a pointwise product of functions continuous on $\mathbb{R}$ in the metric reading is continuous on $\mathbb{R}$ in that reading. Applying this to $\phi_{\varepsilon}(s)=s^{3}w(s)$, and twice to $\phi_{\varepsilon}'(s)=(3s^{2}+\varepsilon s^{4})\,w(s)\,w(s)$ — first to the product of $w$ with itself and then to the product of the result with $s\mapsto3s^{2}+\varepsilon s^{4}$ — shows that $\phi_{\varepsilon}$ and $\phi_{\varepsilon}'$ are continuous at every point of $\mathbb{R}$ in the metric reading, hence also in the Euclidean reading by Euclidean, Metric and Sequential Continuity of a Real Function of a Real Variable (ii).

Finally, by claim 2 of One-Dimensional Derivatives, Partial Derivatives, and Smoothness on the Real Line the partial derivative of $\phi_{\varepsilon}$ with respect to the first variable exists at every point of $\mathbb{R}^{1}$ and $\partial_{1}\phi_{\varepsilon}(s)=\phi_{\varepsilon}'(s)$. Thus the single coordinate function of $\phi_{\varepsilon}$ is continuous at every point of $\mathbb{R}^{1}$ in the Euclidean sense, its single first-order partial derivative exists at every point, and that partial derivative is continuous at every point in the Euclidean sense; by clauses 1 and 3 of Functions of Class Cᵏ on an Open Subset of ℝⁿ, $\phi_{\varepsilon}$ is of class $C^{1}$ on $\mathbb{R}^{1}$.

Claim 3. Let $s\in\mathbb{R}$. By the argument of claim 1 applied to $s$ and to $s^{2}$ one has $0\le s^{2}$ and $0\le s^{4}=(s^{2})^{2}$, hence $0\le3s^{2}$ and $0\le\varepsilon s^{4}$ and so $0\le3s^{2}+\varepsilon s^{4}$; and $0\le w(s)^{2}$ likewise. By clause 2 of Ordered Field the product of two nonnegative elements is nonnegative, so $0\le\phi_{\varepsilon}'(s)$.

For the upper bound, expand

$$4\varepsilon^{-1}\,D(s)^{2}=4\varepsilon^{-1}\bigl(1+2\varepsilon s^{2}+\varepsilon^{2}s^{4}\bigr)=4\varepsilon^{-1}+8s^{2}+4\varepsilon s^{4},$$

using $\varepsilon^{-1}\varepsilon=1$. Hence

$$4\varepsilon^{-1}\,D(s)^{2}-\bigl(3s^{2}+\varepsilon s^{4}\bigr)=4\varepsilon^{-1}+5s^{2}+3\varepsilon s^{4},$$

which is nonnegative, since $\varepsilon^{-1}$ is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field and since $0\le s^{2}$ and $0\le\varepsilon s^{4}$. Multiplying by the nonnegative number $w(s)^{2}$ (clause 2 of Ordered Field) and using $D(s)^{2}w(s)^{2}=(D(s)w(s))^{2}=1$ gives

$$0\le4\varepsilon^{-1}-\bigl(3s^{2}+\varepsilon s^{4}\bigr)w(s)^{2}=4\varepsilon^{-1}-\phi_{\varepsilon}'(s),$$

so $\phi_{\varepsilon}'(s)\le4\varepsilon^{-1}$ by claim 3 of Elementary Arithmetic in an Ordered Field.

Claim 4. Let $s\in\mathbb{R}$. Then $s\,\phi_{\varepsilon}(s)=s\,s^{3}w(s)=s^{4}\,w(s)$, which is nonnegative because $0\le s^{4}$ and $0<w(s)$, by clause 2 of Ordered Field. Next, by claim 4 of Properties of the Absolute Value in an Ordered Field, applied twice, $|s^{3}|=|s|^{3}$; and $|w(s)|=w(s)$ by the preliminary remark, $w(s)$ being positive. Hence, again by claim 4 of Properties of the Absolute Value in an Ordered Field,

$$|\phi_{\varepsilon}(s)|=|s^{3}|\,|w(s)|=|s|^{3}\,w(s)\le|s|^{3},$$

the last step by claim 5 of Elementary Arithmetic in an Ordered Field, applied with $w(s)\le1$ and the nonnegative multiplier $|s|^{3}$.

Claim 5. Let $s\in\mathbb{R}$. From $D(s)w(s)=1$ we get $w(s)-1=w(s)-D(s)w(s)=\bigl(1-D(s)\bigr)w(s)=-\varepsilon s^{2}\,w(s)$, whence

$$\phi_{\varepsilon}(s)-s^{3}=s^{3}w(s)-s^{3}=s^{3}\bigl(w(s)-1\bigr)=-\varepsilon\,s^{5}\,w(s).$$

Taking absolute values and using claim 4 of Properties of the Absolute Value in an Ordered Field repeatedly, together with $|-y|=|y|$ from claim 2 of that lemma and, by the preliminary remark, $|\varepsilon|=\varepsilon$ and $|w(s)|=w(s)$,

$$\bigl|\phi_{\varepsilon}(s)-s^{3}\bigr|=\varepsilon\,|s|^{5}\,w(s)\le\varepsilon\,|s|^{5},$$

the last step by claim 5 of Elementary Arithmetic in an Ordered Field, applied with $w(s)\le1$ and the nonnegative multiplier $\varepsilon\,|s|^{5}$.