A compact-support Cauchy–Pompeiu calculation

The zero extension is $C_c^1(\mathbb{C})$, and its interior $\bar{\mathrm{\partial }}$ derivative is explicit. [F1, given, algebra] The function $h(t)=(\max \{t,0\}{)}^2$ is $C^1$ on $\mathbb{R}$, so $f(z)=h(1-\mathrm{\mid }z{\mathrm{\mid }}^2)$ is $C^1$. It is zero for $\mathrm{\mid }z\mathrm{\mid }\ge 1$, hence has support in the compact closed unit disc. On $D$, the Wirtinger formula [F1] gives ${\mathrm{\partial }}_{\bar{\zeta }}f=-2\zeta (1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2)$. In particular $f=0$ on $\mathrm{\partial }D$ and $f(0)=1$.

The sphere measure in [F4] has total mass $2\pi$. [F3, F4, F5, F6, F7, F8, F9, algebra] For the closed unit disc $K\subset {\mathbb{R}}^2$, [F5] and [F6] give $\mathrm{cont}(K)=V_2(1)=\pi$, and [F7] makes $K$ Jordan measurable. By [F3], full AC supplies the countable-choice premise of [F8], so ${\lambda }_2(K)=\pi$. Also [F9] gives ${\lambda }_2(\{0\})=0$. The set in [F4] for $E=S^1$ is $K\setminus \{0\}$, so its measure is $\pi$ and [F4] yields $\sigma (S^1)=2\pi$.

Applying Cauchy–Pompeiu at $z=0$ gives the asserted value of the area term. [F2, F3, F10, step 1.1, step 1.2, given, algebra] The unit disc is a bounded $C^1$ domain, and step 1.1 proves the needed $C^1$ hypothesis for $f$. Full AC [F3] supplies the premise of [F2]. Since $f=0$ on $\mathrm{\partial }D$, its boundary term vanishes. For $\zeta \mathrm{\ne }0$, step 1.1 gives $({\mathrm{\partial }}_{\bar{\zeta }}f)\mathrm{/}\zeta =-2(1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2)$, which extends continuously to $-2$ at $0$. Put $q(x)=\max \{1-\mathrm{\mid }x{\mathrm{\mid }}^2,0\}$, a nonnegative Borel function on ${\mathbb{R}}^2$. By [F10] and the sphere mass from step 1.2, ${\int }_D(1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2)dA(\zeta )={\int }_{{\mathbb{R}}^2}qd{\lambda }_2=2\pi {\int }_0^1(1-r^2)rdr=\frac{\pi }{2}\mathrm{.}$ Thus the area term equals $-\frac{1}{\pi }{\int }_D\frac{{\mathrm{\partial }}_{\bar{\zeta }}f}{\zeta }dA=\frac{2}{\pi }{\int }_D(1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2)dA=1=f(0)$, as claimed. ∎