Proof of Partial Derivatives, Continuity and $C^k$ Regularity under a Scaling Substitution

Claims 1, 2 and 3 are established below for arbitrary data $n,m,c,\lambda,\mu,U,f$ of the form fixed in the statement, and are applied in the course of the argument to other such data.

Throughout, $T(x)=c+\lambda x$ and $S(y)=\lambda^{-1}(y-c)=(-\lambda^{-1}c)+\lambda^{-1}y$; sums, differences and scalar multiples of points are those of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space and Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$. For $z\in\mathbb{R}^n$ and a real number $t$, write $z[t]$ for the point of $\mathbb{R}^n$ whose $i$th coordinate is $t$ and whose $k$th coordinate is $z_k$ for every $k\ne i$; the index $i\in\{1,\dots,n\}$ is the one fixed in each context where this notation is used.

We record two identities. First, for $x,x'\in\mathbb{R}^n$ we have $T(x)-T(x')=\lambda(x-x')$, so claims 2 and 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ give

$$d_E(T(x),T(x'))=\lVert\lambda(x-x')\rVert=|\lambda|\,d_E(x,x').\tag{$*$}$$

Second, since $T$ acts coordinatewise by $T(z)_k=c_k+\lambda z_k$, for every $z\in\mathbb{R}^n$, every index $i\in\{1,\dots,n\}$ and every real $h$,

$$T\bigl(z[z_i+h]\bigr)=T(z)\bigl[T(z)_i+\lambda h\bigr].\tag{$**$}$$

Claim 1. As $\lambda\ne0$, $S(T(x))=\lambda^{-1}(c+\lambda x-c)=x$ and $T(S(y))=c+\lambda\lambda^{-1}(y-c)=y$ for all $x,y\in\mathbb{R}^n$, so $T$ is a bijection of $\mathbb{R}^n$ with inverse $S$. By the definition of $V$, $T$ maps $V$ into $U$; and if $y\in U$ then $T(S(y))=y\in U$, so $S(y)\in V$. Hence $T$ restricts to a bijection from $V$ onto $U$ with inverse the restriction of $S$, and $U=\{y:S(y)\in V\}$. For $y\in U$ we get $g(S(y))=\mu f(T(S(y)))=\mu f(y)$, whence $f(y)=\mu^{-1}g(S(y))$; since $\mu\ne0$ and $\lambda^{-1}\ne0$, this is a substitution of the stated form.

For the openness of $V$: by Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$ the set $U$ is open in $(\mathbb{R}^n,d_E)$, so for $x\in V$ and $a=T(x)\in U$ there is a real $r>0$ such that every $y\in\mathbb{R}^n$ with $d_E(y,a)<r$ lies in $U$. Put $s=r/|\lambda|>0$. If $d_E(x',x)<s$ then $d_E(T(x'),a)=|\lambda|\,d_E(x',x)<r$ by $(*)$, so $T(x')\in U$ and $x'\in V$. Thus $V$ is open in $(\mathbb{R}^n,d_E)$, hence Euclidean open by Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$ again.

Claim 2. Fix $x\in V$, $a=T(x)$, $i$ and $j$. For every nonzero real $h$, putting $k=\lambda h$, identity $(**)$ gives $T(x[x_i+h])=a[a_i+k]$, and therefore, whenever $a[a_i+k]\in U$,

$$\frac{g_j\bigl(x[x_i+h]\bigr)-g_j(x)}{h} =\mu\,\frac{f_j\bigl(a[a_i+k]\bigr)-f_j(a)}{h} =\mu\lambda\,\frac{f_j\bigl(a[a_i+k]\bigr)-f_j(a)}{k},\tag{$\dagger$}$$

using $h=k/\lambda$.

Suppose the partial derivative of $f_j$ with respect to the $i$th variable exists at $a$, with value $L$. Let $\varepsilon>0$ be real. By Partial Derivative on a Euclidean Open Set applied with the positive real number $\varepsilon/|\mu\lambda|$ there is a real $\delta'>0$ such that every real $k$ with $0<|k|<\delta'$ satisfies $a[a_i+k]\in U$ and

$$\Bigl|\frac{f_j(a[a_i+k])-f_j(a)}{k}-L\Bigr|<\frac{\varepsilon}{|\mu\lambda|}.$$

Put $\delta=\delta'/|\lambda|>0$ and let $h$ be real with $0<|h|<\delta$. Then $k=\lambda h$ satisfies $|k|=|\lambda|\,|h|$ by claim 4 of Properties of the Absolute Value in an Ordered Field, so $0<|k|<\delta'$; hence $a[a_i+k]\in U$, and since $T(x[x_i+h])=a[a_i+k]$ this gives $x[x_i+h]\in V$. By $(\dagger)$ and claim 4 of Properties of the Absolute Value in an Ordered Field,

$$\Bigl|\frac{g_j(x[x_i+h])-g_j(x)}{h}-\mu\lambda L\Bigr| =|\mu\lambda|\,\Bigl|\frac{f_j(a[a_i+k])-f_j(a)}{k}-L\Bigr|<\varepsilon .$$

So the partial derivative of $g_j$ with respect to the $i$th variable exists at $x$ and equals $\mu\lambda L$.

Conversely, suppose that partial derivative of $g_j$ exists at $x$. By claim 1 the pair $(f,U)$ arises from $(g,V)$ by a substitution of the same form, under which the point $a\in U$ corresponds to $S(a)=x\in V$; so the implication just proved, applied to that substitution, shows that the partial derivative of $f_j$ with respect to the $i$th variable exists at $a$ and equals $\mu^{-1}\lambda^{-1}\partial_i g_j(x)$, which is the asserted identity.

Claim 3. By claim 1 of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions it is equivalent to prove the corresponding statement for continuity relative to $U$ (respectively $V$) as maps into the metric space $(\mathbb{R},d_{\mathbb{R}})$ of The Absolute Value Metric on the Real Numbers.

Suppose $f_j$ is continuous at every point of $U$, let $x\in V$ and $a=T(x)$, and let $\varepsilon>0$ be real. Choose a real $r>0$ such that every $y\in U$ with $d_E(y,a)<r$ satisfies $|f_j(y)-f_j(a)|<\varepsilon/|\mu|$, and put $s=r/|\lambda|>0$. If $x'\in V$ satisfies $d_E(x',x)<s$, then $T(x')\in U$ and $d_E(T(x'),a)<r$ by $(*)$, so by claim 4 of Properties of the Absolute Value in an Ordered Field,

$$|g_j(x')-g_j(x)|=|\mu|\,\bigl|f_j(T(x'))-f_j(a)\bigr|<\varepsilon .$$

Hence $g_j$ is continuous at every point of $V$. The converse follows by applying this implication to the substitution of claim 1.

Claim 4. We argue by induction on $k$, the statement being proved for all data $n,m,c,\lambda,\mu,U,f$ of the given form simultaneously.

Base case $k=1$. By clauses 1 and 3 of C^k Maps on a Euclidean Open Set, $f$ is of class $C^1$ on $U$ if and only if every $f_j$ is continuous at every point of $U$ and, for all $i$ and $j$, the partial derivative of $f_j$ with respect to the $i$th variable exists at every point of $U$ and the resulting function $\partial_i f_j:U\to\mathbb{R}$ is continuous at every point of $U$; and similarly for $g$ on $V$. The first conditions are equivalent by claim 3. Since $T$ restricts to a bijection from $V$ onto $U$ (claim 1), claim 2 shows that $\partial_i f_j$ exists at every point of $U$ if and only if $\partial_i g_j$ exists at every point of $V$; and in that case claim 2 also gives

$$\partial_i g_j(x)=(\mu\lambda)\,(\partial_i f_j)(c+\lambda x)\qquad(x\in V),$$

so the pair $(\partial_i g_j,V)$ arises from $(\partial_i f_j,U)$ by the substitution of the same form with $c$ and $\lambda$ unchanged and $\mu$ replaced by the nonzero real number $\mu\lambda$. Applying claim 3 to that substitution, $\partial_i f_j$ is continuous at every point of $U$ if and only if $\partial_i g_j$ is continuous at every point of $V$. Hence $f$ is of class $C^1$ on $U$ if and only if $g$ is of class $C^1$ on $V$.

Induction step. Let $k$ be a natural number and assume the assertion for $k$, for every substitution of the given form. By clauses 2 and 3 of C^k Maps on a Euclidean Open Set, $f$ is of class $C^{k+1}$ on $U$ if and only if $f$ is of class $C^1$ on $U$ and, for all $i$ and $j$, the function $\partial_i f_j$ is of class $C^k$ on $U$; and similarly for $g$ on $V$. The $C^1$ conditions are equivalent by the base case; assume them. As shown there, $(\partial_i g_j,V)$ arises from $(\partial_i f_j,U)$ by a substitution of the given form, so the induction hypothesis applied to it gives that $\partial_i f_j$ is of class $C^k$ on $U$ if and only if $\partial_i g_j$ is of class $C^k$ on $V$. Hence $f$ is of class $C^{k+1}$ on $U$ if and only if $g$ is of class $C^{k+1}$ on $V$.

Claim 5. By Smooth Map on a Euclidean Open Set, $f$ is smooth on $U$ exactly when $f$ is of class $C^k$ on $U$ for every natural number $k$, and likewise for $g$ on $V$; the equivalence is therefore immediate from claim 4. $\blacksquare$