C¹ planar fields on a closed disk extend to a neighbourhood

Write every nonzero point in a collar of the unit circle uniquely as $q=(1+s)p$ with $p\in S^1$ and $s>-1$, fix once and for all a number $0<\epsilon <1\mathrm{/}2$, and define $E(q):=3X((1-s)p)-2X((1-2s)p)$ for $0<s<\epsilon$ while $E(q):=X(q)$ for $\mathrm{\mid }q\mathrm{\mid }\le 1$; this is an explicit finite formula with no choice.

On the unit circle, where $s=0$, the outer formula gives $3X(p)-2X(p)=X(p)$, so the two definitions agree there and $E$ is a well-defined map on $\{\mathrm{\mid }q\mathrm{\mid }<1+\epsilon \}$.

The inner formula is the restriction of $X$, which is $C^1$ up to the boundary; the outer formula is a composite of smooth scalar operations with the map $(s,p)\mapsto X((1-2s)p)$, and for $0<s<\epsilon$ the points $(1-2s)p$ lie in the interior of $D$, where $X$ is $C^1$; hence [F1] shows that $E$ is $C^1$ on each of the two open regions $\mathrm{\mid }q\mathrm{\mid }<1$ and $1<\mathrm{\mid }q\mathrm{\mid }<1+\epsilon$, with derivatives computed by the chain rule.

Parametrize the circle by $p=p(\theta )$. The chain rule gives ${\mathrm{\partial }}_{\theta }E=3(1-s)D{X}_{(1-s)p}p\mathrm{\prime }(\theta )-2(1-2s)D{X}_{(1-2s)p}p\mathrm{\prime }(\theta )$ outside the disk. As $s\downarrow 0$ this tends to $(3-2)D{X}_{p}p\mathrm{\prime }(\theta )=D{X}_{p}p\mathrm{\prime }(\theta )$, the inner tangential derivative.

The outer radial derivative is ${\mathrm{\partial }}_{s}E=-3D{X}_{(1-s)p}p+4D{X}_{(1-2s)p}p$. As $s\downarrow 0$ it tends to $(-3+4)D{X}_{p}p=D{X}_{p}p$, the inner radial derivative. Both limiting derivatives depend continuously on $p$.

The first partial derivatives of $E$ are therefore continuous across the unit circle, each side being $C^1$ with matching limits by step 3.1 and step 3.2; for $q$ on the circle and a small displacement $h$, applying [F2] on the segments on either side of the circle gives $\mathrm{\mid }E(q+h)-E(q)-D{E}_{q}h\mathrm{\mid }\le {\sup }_{0\le t\le 1}\mathrm{\parallel }D{E}_{q+th}-D{E}_q\mathrm{\parallel }\mathrm{\parallel }h\mathrm{\parallel }$, and the supremum tends to $0$ because the partial derivatives are continuous at $q$, so $E$ is differentiable there with total derivative $D{E}_q$ and hence $C^1$ on $\{\mathrm{\mid }q\mathrm{\mid }<1+\epsilon \}$. Since $E=X$ on $D$ and the derivative identity just established gives $DE=DX$ along the circle from the inner side, the value and first derivative of the extension agree with $X$ on the closed disk; the construction uses only the explicit formula of step 1.1 and finitely many evaluations.