Proof of Jacobian Matrix of a Local Smooth Extension on an Admissible Domain

Claim 1. Fix $k\in\{1,\dots,m\}$ and $j\in\{1,\dots,n\}$, and let $e_j$ denote the multi-index of length $n$ whose $j$th entry is $1$ and whose remaining entries are $0$. By Smooth Map on an Open Subset of Euclidean Space, the partial derivative of $G_k$ of order $\alpha$ exists on $W$ for every multi-index $\alpha$ of length $n$; apply this with $\alpha=e_j$. Since $e_j\ne 0$, and since $j$ is the only index $i$ with $e_i\le e_j$, the recursive clause defining partial derivatives of order $\alpha$, which Smooth Map on an Open Subset of Euclidean Space invokes, requires at $\alpha=e_j$ exactly that the ordinary partial derivative of $G_k$ with respect to the $j$th variable exist at every point of $W$: that is, for every $a\in W$ there is a real number $L$ such that for every real $\varepsilon>0$ there is a real $\delta>0$ for which every real $h$ with $0<|h|<\delta$ satisfies

This is the condition of Partial Derivative on a Euclidean Open Set except for the additional requirement there that the perturbed point lie in $W$. That requirement is met after shrinking $\delta$: since $W$ is open and $a\in W$, there is a real $\rho>0$ such that every $y\in\mathbb{R}^n$ with $\sum_{i=1}^n(y_i-a_i)^2<\rho^2$ lies in $W$, and for $0<|h|<\rho$ the point $y=(a_1,\dots,a_{j-1},a_j+h,a_{j+1},\dots,a_n)$ satisfies $\sum_{i=1}^n(y_i-a_i)^2=h^2$, with $h^2<\rho^2$ because multiplying the strict inequality $|h|<\rho$ first by the positive number $|h|$ and then by the positive number $\rho$ gives $h^2<|h|\rho<\rho^2$ (claims 10 and 2 of the order arithmetic lemma, and $h^2=|h|^2$ by claims 1 and 4 of the absolute value lemma). Replacing $\delta$ by the smaller of $\delta$ and $\rho$, which is positive by claim 9 of the order arithmetic lemma, therefore gives the condition of Partial Derivative on a Euclidean Open Set. Hence the partial derivative of $G_k$ with respect to the $j$th variable exists at every point of $W$ in that sense.

As $k$ and $j$ were arbitrary, the hypothesis of Jacobian Matrix of a Map Between Euclidean Spaces is satisfied at every $a\in W$, so $DG(a)$ is defined there. This proves claim 1.

Claim 2. Throughout write $x=(x_1,\dots,x_n)$, and let $D$ be the ambient set of $\Omega$ in the sense of Continuous n-Form, Support, and Zero Extension on a Euclidean or Half-Space Domain.

Step 1: a common radius. Since $W$ is open and $x\in W$, there is a real $r_1>0$ such that every $y\in\mathbb{R}^n$ with $\sum_{i=1}^n(y_i-x_i)^2<r_1^2$ lies in $W$; similarly there is a real $r_2>0$ doing the same for $\widetilde{W}$. Next, if $D=\mathbb{R}^n$ then $\Omega$ is open in $\mathbb{R}^n$, and we let $U=\Omega$; if $D=H^n$ then, by Continuous n-Form, Support, and Zero Extension on a Euclidean or Half-Space Domain, $\Omega$ is open in the closed upper half-space $H^n$, so by that definition there is an open subset $U\subseteq\mathbb{R}^n$ with $\Omega=H^n\cap U$. In both cases $x\in U$ and $U$ is open, so there is a real $r_3>0$ such that every $y\in\mathbb{R}^n$ with $\sum_{i=1}^n(y_i-x_i)^2<r_3^2$ lies in $U$. Let $r$ be the least of $r_1,r_2,r_3$; then $r>0$ by claim 9 of the order arithmetic lemma, applied twice.

Step 2: nearby points of the positive $j$th coordinate direction lie in $W\cap\widetilde{W}\cap\Omega$. Fix $j\in\{1,\dots,n\}$ and let $t$ be a real number with $0<t<r$. Put

Then $\sum_{i=1}^n(y_i-x_i)^2=t^2$. Let $u$ be any one of $r_1,r_2,r_3$. Since $r\le u$ and $t<r$, mixed transitivity gives $t<u$, and multiplying the strict inequality $t<u$ first by the positive number $t$ and then by the positive number $u$ gives $t^2<tu<u^2$, whence $t^2<u^2$; here we used claims 10 and 2 of the order arithmetic lemma. Taking $u=r_1$, $u=r_2$ and $u=r_3$ in turn gives $y\in W$, $y\in\widetilde{W}$ and $y\in U$.

If $D=\mathbb{R}^n$ then $U=\Omega$ and so $y\in\Omega$. If $D=H^n$, then $x\in\Omega\subseteq H^n$ gives $x_n\ge 0$, and the $n$th coordinate of $y$ equals $x_n$ when $j<n$ and equals $x_n+t$ when $j=n$; in the second case $x_n+t>0$ by claim 3 of the order arithmetic lemma applied to $0\le x_n$ and $0<t$. In either case the $n$th coordinate of $y$ is nonnegative, so $y\in H^n$ and therefore $y\in H^n\cap U=\Omega$.

Thus $y\in W\cap\widetilde{W}\cap\Omega$, and the hypothesis gives $G(y)=\widetilde{G}(y)$. The point $x$ itself lies in $W\cap\widetilde{W}\cap\Omega$, so also $G(x)=\widetilde{G}(x)$.

Step 3: the partial derivatives at $x$ agree. Fix $k\in\{1,\dots,m\}$ and $j\in\{1,\dots,n\}$. By claim 1 the partial derivatives

exist in the sense of Partial Derivative on a Euclidean Open Set. Suppose, for contradiction, that $L\ne\widetilde{L}$, and set $\varepsilon=|L-\widetilde{L}|/2$, which is positive by claim 1 of the absolute value lemma together with claim 8 of the order arithmetic lemma. Choose $\delta>0$ for $L$ and $\widetilde{\delta}>0$ for $\widetilde{L}$ as in Partial Derivative on a Euclidean Open Set for this $\varepsilon$, let $\sigma$ be the least of $\delta$, $\widetilde{\delta}$ and $r$, which is positive, and choose a real $t$ with $0<t<\sigma$, for instance $t=\sigma/2$ (claim 8 of the order arithmetic lemma).

With $y$ as in Step 2 for this $t$, set

By Step 2, $G(y)=\widetilde{G}(y)$ and $G(x)=\widetilde{G}(x)$, so their $k$th coordinates agree and therefore $q$ also equals $(\widetilde{G}_k(y)-\widetilde{G}_k(x))/t$. Since $0<t<\delta$ and $0<t<\widetilde{\delta}$, the defining condition of Partial Derivative on a Euclidean Open Set gives $|q-L|<\varepsilon$ and $|q-\widetilde{L}|<\varepsilon$. By symmetry of the absolute value and the triangle inequality (claims 2 and 5 of the absolute value lemma),

which is impossible. Hence $L=\widetilde{L}$.

Conclusion. By Step 3 every entry of $DG(x)$ equals the corresponding entry of $D\widetilde{G}(x)$, since by Jacobian Matrix of a Map Between Euclidean Spaces the entry in row $k$ and column $j$ of each is the partial derivative of the respective $k$th coordinate function with respect to the $j$th variable at $x$. Two real matrices of the same shape with equal entries are equal, so $DG(x)=D\widetilde{G}(x)$. This proves claim 2.