The radial composite $x\mapsto u(\mid x\mid )$ on an annulus: $C^k$ regularity and $\mathrm{\nabla }(u\circ \mid \cdot \mid )(x)=u^{\mathrm{\prime }}(\mid x\mid )x\mathrm{/}\mid x\mid$

The norm is continuous on ${\mathbb{R}}^3$ by the reverse triangle inequality [L2], so $A$, the preimage of the open set $J_0$ under $N$, is open; and $N(0)=0\mathrm{\notin }{J}_0$ gives $0\mathrm{\notin }A$. This is clause 1.

Fix $x\in A$ and $h\in {\mathbb{R}}^3$ with $x+h\in A$. By bilinearity, symmetry and [L1], $\mid x+h{\mid }^2-\mid x{\mid }^2=2⟨x,h⟩+\mid h{\mid }^2$, and therefore $\mid x+h\mid -\mid x\mid =(2⟨x,h⟩+\mid h{\mid }^2)\mathrm{/}(\mid x+h\mid +\mid x\mid )$, the denominator being at least $\mid x\mid >0$.

Remainder estimate at $x$: subtracting $⟨x,h⟩\mathrm{/}\mid x\mid$ from step 1.2 and putting over the common denominator $\mid x\mid (\mid x+h\mid +\mid x\mid )$ gives $\mid \mid x+h\mid -\mid x\mid -\frac{⟨x,h⟩}{\mid x\mid }\mid \le \frac{\mid x\mid \mid h{\mid }^2+\mid ⟨x,h⟩\mid \mid \mid x\mid -\mid x+h\mid \mid }{\mid x\mid (\mid x+h\mid +\mid x\mid )}\le \frac{2\mid h{\mid }^2}{\mid x\mid },$ where [L3] bounds the first numerator term by $\mid x\mid \mid h{\mid }^2$, [L2] bounds the second by $\mid x\mid \mid h\mid \cdot \mid h\mid$, and the denominator is at least $\mid x{\mid }^2$. Hence the normalized remainder tends to $0$ as $h\to 0$: $N$ is totally differentiable at $x$ with $DN(x)h=⟨x,h⟩\mathrm{/}\mid x\mid$.

The representing vector: for every $v$, [L4] gives $D_{v}N(x)=DN(x)v=⟨x,v⟩\mathrm{/}\mid x\mid =⟨x\mathrm{/}\mid x\mid ,v⟩$; comparing with $D_{v}N(x)=⟨\mathrm{\nabla }N(x),v⟩$ and taking $v$ over the standard basis shows $\mathrm{\nabla }N(x)=x\mathrm{/}\mid x\mid$. In particular every first partial derivative of $N$ agrees on $A$ with a component of $x\mathrm{/}\mid x\mid$ and is continuous, so $N$ is $C^1$ by $C^k$ Euclidean maps and diffeomorphisms.

Chain rule for the composite $U=u\circ N$: $u$ is $C^k$ hence totally differentiable, $N$ is totally differentiable at every point of $A$ by step 2.1, and the domains compose, so [L4] gives $DU(x)=Du(N(x))\circ DN(x)$; and $DU(x)h=u^{\mathrm{\prime }}(\mid x\mid )\cdot DN(x)h=u^{\mathrm{\prime }}(\mid x\mid )⟨x,h⟩\mathrm{/}\mid x\mid$ because the total derivative of the real function $u$ at $t$ is multiplication by $u^{\mathrm{\prime }}(t)$ by [L4]. This is the displayed formula for $DU$.

The induction on smoothness: the base case $m=1$ is step 3.1. Assume $m\ge 1$ and $N\in C^m(A)$. The map $t\mapsto t^{-1}$ is $C^{\mathrm{\infty }}$ on $(0,\mathrm{\infty })$ by [L6] with $\alpha =-1$ iterated, so $x\mapsto 1\mathrm{/}\mid x\mid$ is $C^m$ on $A$ as the composite $t\mapsto t^{-1}$ after $N$. The coordinate maps $x\mapsto x_j$ are the components of the identity, whose partial derivatives are the constants ${\delta }_{jk}$; they are therefore $C^{\mathrm{\infty }}$, hence $C^m$, by $C^k$ Euclidean maps and diffeomorphisms and Directional derivatives and partial derivatives of a map $U\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^n$. So $\mathrm{\nabla }N(x)=x\mathrm{/}\mid x\mid$, a finite product of $C^m$ maps, is $C^m$; since a map is $C^{m+1}$ exactly when its first partial derivatives are $C^m$ ($C^k$ maps and multi-index derivative notation in Euclidean space), $N$ is $C^{m+1}$. By induction [L7], $N\in C^m(A)$ for every $m$, that is $N\in C^{\mathrm{\infty }}(A)$. This completes clause 2.

The gradient and the regularity: the display of step 3.2 writes $DU(x)$ as the linear map $h\mapsto ⟨u^{\mathrm{\prime }}(\mid x\mid )x\mathrm{/}\mid x\mid ,h⟩$, so $\mathrm{\nabla }U(x)=u^{\mathrm{\prime }}(\mid x\mid )x\mathrm{/}\mid x\mid$ by the same uniqueness argument as step 3.1; and $U$, a composite of the $C^k$ maps $u$ and $N$ ($C^k$ Euclidean maps and diffeomorphisms), is $C^k$ on $A$ by [L5]. This is clause 3 and completes the proof.