Proof of Restriction of a $C^k$ Map to an Open Subset

For $h\in\mathbb{R}$ write $x^{h}=(x_{1},\dots,x_{i-1},x_{i}+h,x_{i+1},\dots,x_{n})$, the point appearing in Partial Derivative on a Euclidean Open Set. All but the $i$th summand of $\sum_{l=1}^{n}(x^{h}_{l}-x_{l})^{2}$ vanish, so claim 7 of Properties of Finite Sums gives

$$\sum_{l=1}^{n}(x^{h}_{l}-x_{l})^{2}=h^{2},$$

the finite sum being that of the real numbers. Writing $|h|$ for the absolute value of $h$, claim 1 of Properties of the Absolute Value in an Ordered Field gives $|h|=h$ or $|h|=-h$, and in either case $|h|^{2}=h^{2}$.

Since $V$ is open, Open Subset of Euclidean Space provides a real $\sigma>0$ such that every $w=(w_{1},\dots,w_{n})\in\mathbb{R}^{n}$ with $\sum_{l=1}^{n}(w_{l}-x_{l})^{2}<\sigma^{2}$ belongs to $V$. Consequently, if $h\in\mathbb{R}$ satisfies $0<|h|<\sigma$, then $\sum_{l=1}^{n}(x^{h}_{l}-x_{l})^{2}=|h|^{2}<\sigma^{2}$ by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, both $|h|$ and $\sigma$ being nonnegative; hence $x^{h}\in V$, and therefore also $x^{h}\in U$.

Claim 1. Suppose first that the partial derivative of $g$ with respect to the $i$th variable exists at $x$ with value $L$, and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Let $\delta_{1}>0$ be as in Partial Derivative on a Euclidean Open Set for $g$, and let $\delta$ be the lesser of $\delta_{1}$ and $\sigma$, which exists by claim 9 of Elementary Order Arithmetic in an Ordered Field and is positive, being equal to one of the two. Let $h\in\mathbb{R}$ with $0<|h|<\delta$. Then $|h|<\sigma$, so $x^{h}\in V$ by the previous paragraph, and $|h|<\delta_{1}$, so

$$\left|\frac{g(x^{h})-g(x)}{h}-L\right|<\varepsilon .$$

Since $x$ and $x^{h}$ lie in $V$ and $g|_{V}$ agrees with $g$ there, the same estimate holds with $g|_{V}$ in place of $g$. As $\varepsilon$ was arbitrary, the partial derivative of $g|_{V}$ with respect to the $i$th variable exists at $x$ with value $L$.

Conversely, suppose the partial derivative of $g|_{V}$ with respect to the $i$th variable exists at $x$ with value $L$, and let $\varepsilon>0$. Let $\delta>0$ be as in Partial Derivative on a Euclidean Open Set for $g|_{V}$, and let $h\in\mathbb{R}$ with $0<|h|<\delta$. Then $x^{h}\in V$ and

$$\left|\frac{g|_{V}(x^{h})-g|_{V}(x)}{h}-L\right|<\varepsilon ,$$

and $x^{h}\in V\subseteq U$ while $g|_{V}$ agrees with $g$ at $x$ and $x^{h}$, so the corresponding statement for $g$ holds. Hence the partial derivative of $g$ with respect to the $i$th variable exists at $x$ with value $L$. This proves the equivalence; the two partial derivatives therefore have exactly the same admissible values at $x$, which justifies the notation $\partial_{i}(g|_{V})(x)=\partial_{i}g(x)$.

Claim 2. Let $\varepsilon>0$ and let $\delta>0$ be as in Continuity at a Point for Maps Between Euclidean Spaces for $f$ at $x$, the domain there being $U$. Every $w\in V$ with $\sum_{l=1}^{n}(w_{l}-x_{l})^{2}<\delta^{2}$ lies in $U$, hence satisfies $\sum_{j=1}^{m}\bigl(f_{j}(w)-f_{j}(x)\bigr)^{2}<\varepsilon^{2}$; and $f|_{V}$ agrees with $f$ on $V$. So the same $\delta$ witnesses continuity of $f|_{V}$ at $x$, its domain being $V$. The argument for $g$ is the same, with $m=1$.

Claim 3. Claims 1 and 2 were proved for arbitrary data satisfying the hypotheses of this lemma. Let $A$ be the set of natural numbers $k$ with the following property: for all natural numbers $n',m'$, every open subset $U'$ of $\mathbb{R}^{n'}$, every open $V'\subseteq U'$ and every map $F:U'\to\mathbb{R}^{m'}$ of class $C^{k}$ on $U'$, the restriction $F|_{V'}$ is of class $C^{k}$ on $V'$. We show $A=\mathbb{N}$ using Principle of Induction for the Natural Numbers.

Base case. Let $F=(F_{1},\dots,F_{m'}):U'\to\mathbb{R}^{m'}$ be of class $C^{1}$ on $U'$, and let $V'\subseteq U'$ be open. By clause 1 of C^k Maps on a Euclidean Open Set, each $F_{j}$ is continuous at every point of $U'$, each partial derivative $\partial_{i}F_{j}$ exists at every point of $U'$, and each function $\partial_{i}F_{j}:U'\to\mathbb{R}$ is continuous at every point of $U'$. By claim 2, each $F_{j}|_{V'}$ is continuous at every point of $V'$. By claim 1, applied at each point of $V'$ to the function $F_{j}$, the partial derivative $\partial_{i}(F_{j}|_{V'})$ exists at every point of $V'$ and takes there the same values as $\partial_{i}F_{j}$; that is, $\partial_{i}(F_{j}|_{V'})=(\partial_{i}F_{j})|_{V'}$ as functions on $V'$. By claim 2 again, this function is continuous at every point of $V'$. Hence $F|_{V'}$ is of class $C^{1}$ on $V'$, and $1\in A$.

Inductive step. Let $k\in A$ and let $F:U'\to\mathbb{R}^{m'}$ be of class $C^{S(k)}$ on $U'$, where $S$ is the successor map and $S(k)=k+1$ by claim 1 of Arithmetic of Addition on the Natural Numbers. By clause 2 of C^k Maps on a Euclidean Open Set, $F$ is of class $C^{1}$ on $U'$ and each $\partial_{i}F_{j}:U'\to\mathbb{R}$ is of class $C^{k}$ on $U'$. The base case gives that $F|_{V'}$ is of class $C^{1}$ on $V'$, and there $\partial_{i}(F_{j}|_{V'})=(\partial_{i}F_{j})|_{V'}$. Since $k\in A$, applied to the map $\partial_{i}F_{j}$ from $U'$ into $\mathbb{R}^{1}$ in the sense of clause 3 of C^k Maps on a Euclidean Open Set, the restriction $(\partial_{i}F_{j})|_{V'}$ is of class $C^{k}$ on $V'$. By clause 2 of that definition, $F|_{V'}$ is of class $C^{S(k)}$ on $V'$; hence $S(k)\in A$.

By the principle of induction $A=\mathbb{N}$, which is claim 3; the statement for $g$ is the case $m'=1$ under the scalar convention.

Claim 4. If $f$ is smooth on $U$, then by Smooth Map on a Euclidean Open Set it is of class $C^{k}$ on $U$ for every natural number $k$, so by claim 3 the restriction $f|_{V}$ is of class $C^{k}$ on $V$ for every natural number $k$, that is, $f|_{V}$ is smooth on $V$. The same argument applies to $g$.