Proof of A Continuously Differentiable Map Whose Jacobian Matrix is Close to the Identity Maps a Cube into a Slightly Larger Cube

Each result cited below is universally quantified over the data in its own statement.

Let $c$, $r$ and $\eta$ be as in A Continuously Differentiable Map Whose Jacobian Matrix is Close to the Identity Maps a Cube into a Slightly Larger Cube §cube, fix $x\in Q(c,r)$, and put $h=x-c$, so that $h_{k}=x_{k}-c_{k}$ and $|h_{k}|\le r$ for every $k\in[d]$. Fix $i\in[d]$. The $i$th coordinate of $G(c)$ is $G_{i}(c)$, so it suffices to show $|G_{i}(x)-G_{i}(c)|\le(1+\eta)r$; note that $(1+\eta)r$ is nonnegative.

Step 1 (Derivative along the segment). The function $G_{i}$ is of class $C^{1}$ on the open set $\mathbb{R}^{d}$ (read through clause 3 of C^k Maps on a Euclidean Open Set), so by A Real-Valued C^1 Function is Differentiable at Every Point it is differentiable at every point of $\mathbb{R}^{d}$. Let $J=(-1,2)=\{\tau\in\mathbb{R}:-1<\tau<2\}$. It is an interval: if $u,w\in J$ and $u\le v\le w$, then $-1<v<2$ by claim 2 of Elementary Order Arithmetic in an Ordered Field. Every $\tau\in J$ is an interior point of $J$, the midpoints $\frac{\tau-1}{2}$ and $\frac{\tau+2}{2}$ lying in $J$ and satisfying $\frac{\tau-1}{2}<\tau<\frac{\tau+2}{2}$. Define $F:J\to\mathbb{R}$ by $F(\tau)=G_{i}(c+\tau h)$. For every $\tau\in J$, Chain Rule Along an Affine Path (with $U=\mathbb{R}^{d}$, open by claim 1 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous, base point $c$, direction $h$ and $\tau_{0}=\tau$) shows that $F$ is differentiable at $\tau$ with

Step 2 (Mean value). By Mean Value Theorem on an Open Interval, applied to $F$ on the open interval $(-1,2)$ with the points $0<1$ of that interval, there is $\xi\in(0,1)$ with $F(1)-F(0)=F'(\xi)(1-0)$. Since $F(1)=G_{i}(x)$ and $F(0)=G_{i}(c)$, with $z=c+\xi h$ we get

Step 3 (The intermediate point lies in the cube). For $k\in[d]$ we have $z_{k}-c_{k}=\xi h_{k}$, and $|\xi h_{k}|=|\xi|\,|h_{k}|=\xi|h_{k}|$ by claim 4 of Properties of the Absolute Value in an Ordered Field and Absolute Value in an Ordered Field, as $0<\xi$. Since $\xi\le1$ and $0\le|h_{k}|$ (claim 1 of Properties of the Absolute Value in an Ordered Field), claim 5 of Elementary Arithmetic in an Ordered Field gives $\xi|h_{k}|\le|h_{k}|\le r$. Hence $z\in Q(c,r)$, and the hypothesis gives $|\partial_{j}G_{i}(z)-\delta_{ij}|\le\frac{\eta}{d}$ for every $j\in[d]$.

Step 4 (Estimate). By claim 7 of Properties of Finite Sums, $\sum_{j=1}^{d}\delta_{ij}h_{j}=h_{i}$, so by Step 2 and claims 2 and 3 of Properties of Finite Sums,

By claim 4 of Properties of the Absolute Value in an Ordered Field, Step 3, $|h_{j}|\le r$ and claim 5 of Elementary Arithmetic in an Ordered Field (all factors being nonnegative), $|a_{j}|\le\frac{\eta}{d}\,r$, hence $-\frac{\eta r}{d}\le a_{j}\le\frac{\eta r}{d}$ by claim 6 of Properties of the Absolute Value in an Ordered Field. Applying claim 5 of Properties of Finite Sums to the nonnegative summands $\frac{\eta r}{d}-a_{j}$ and $\frac{\eta r}{d}+a_{j}$, and using claims 2 and 3 there together with $\sum_{j=1}^{d}\frac{\eta r}{d}=d\cdot\frac{\eta r}{d}=\eta r$, we get $-\eta r\le\sum_{j}a_{j}\le\eta r$, so

by claim 6 of Properties of the Absolute Value in an Ordered Field. Finally, by claim 5 of Properties of the Absolute Value in an Ordered Field,

As $i\in[d]$ was arbitrary, $G(x)\in Q\bigl(G(c),(1+\eta)r\bigr)$.