Prove that the distributional derivative of \(f(z)=1/z\) satisfies \[ \frac{\partial f}{\partial\overline z}=C\delta_0, \] and evaluate the constant \(C\). Equivalently, show that \[ -\int_{\mathbb C} f(z)\frac{\partial\phi}{\partial\overline z}\,|dz|^2=C\phi(0) \] for every \(\phi\in C_0^\infty(\mathbb C)\).
Solution
The constant is \[ \boxed{C=\pi}. \] Equivalently, \[ \frac{\partial}{\partial\overline z}\left(\frac1z\right) =\pi\delta_0 \] in the sense of distributions.
1. Away from the origin, the function \(1/z\) has vanishing\(\overline z\)-derivative.
On \(\mathbb C\setminus\{0\}\) the function \(z\mapsto1/z\) is holomorphic. Hence \[ \frac{\partial}{\partial\overline z}\left(\frac1z\right)=0 \] pointwise there. Thus any distributional contribution is supported at the origin.
2. Let \(\phi\in C_0^\infty(\mathbb C)\). For sufficiently small\(\varepsilon>0\),\[ 2i\int_{\mathbb C\setminus D(0,\varepsilon)} \frac1z\frac{\partial\phi}{\partial\overline z}\,|dz|^2 =-\int_{|z|=\varepsilon}\frac{\phi(z)}z\,dz, \]where the circle on the right is oriented counterclockwise.
Choose \(R>0\) so large that \(\phi\) vanishes near \(|z|=R\), and set \[ A_{\varepsilon,R}=\{z:\varepsilon<|z|<R\}. \] On this annulus, \[ d\left(\frac{\phi(z)}z\,dz\right) =\frac1z\frac{\partial\phi}{\partial\overline z} \,d\overline z\wedge dz. \] Since \[ d\overline z\wedge dz=2i\,dx\wedge dy =2i\,|dz|^2, \] Stokes’ theorem gives \[ 2i\int_{A_{\varepsilon,R}} \frac1z\frac{\partial\phi}{\partial\overline z}\,|dz|^2 =\int_{\partial A_{\varepsilon,R}}\frac{\phi(z)}z\,dz. \] The outer boundary integral is zero because \(\phi\) vanishes there. The inner boundary of the annulus has clockwise orientation, hence its contribution is the negative of the counterclockwise circle integral. This proves the stated identity.
3. As \(\varepsilon\to0\),\[ \int_{|z|=\varepsilon}\frac{\phi(z)}z\,dz \longrightarrow2\pi i\,\phi(0). \]
Parametrize the circle by \(z=\varepsilon e^{it}\), \(0\le t\le2\pi\). Then \[ \frac{dz}{z}=i\,dt, \] so \[ \int_{|z|=\varepsilon}\frac{\phi(z)}z\,dz =i\int_0^{2\pi}\phi(\varepsilon e^{it})\,dt. \] Smoothness of \(\phi\) implies uniform convergence \[ \phi(\varepsilon e^{it})\longrightarrow\phi(0) \] as \(\varepsilon\to0\), and therefore the integral tends to \(2\pi i\phi(0)\).
4. The punctured-disk integral converges to the full distributionalpairing:\[ \int_{\mathbb C\setminus D(0,\varepsilon)} \frac1z\frac{\partial\phi}{\partial\overline z}\,|dz|^2 \longrightarrow \int_{\mathbb C} \frac1z\frac{\partial\phi}{\partial\overline z}\,|dz|^2. \]
The function \(1/z\) is locally integrable on \(\mathbb C\): in polar coordinates, \[ \int_{|z|<\varepsilon}\frac1{|z|}\,|dz|^2 =\int_0^{2\pi}\int_0^\varepsilon\frac1r\,r\,dr\,d\theta =2\pi\varepsilon. \] Since \(\partial\phi/\partial\overline z\) is bounded, the contribution from \(|z|<\varepsilon\) tends to zero.
5. Hence\[ -\int_{\mathbb C}\frac1z \frac{\partial\phi}{\partial\overline z}\,|dz|^2 =\pi\phi(0). \]
Combine <1>2–<1>4 and let \(\varepsilon\to0\). From 2 and 3, \[ 2i\int_{\mathbb C}\frac1z \frac{\partial\phi}{\partial\overline z}\,|dz|^2 =-2\pi i\phi(0). \] Dividing by \(2i\) yields \[ \int_{\mathbb C}\frac1z \frac{\partial\phi}{\partial\overline z}\,|dz|^2 =-\pi\phi(0), \] which is exactly \[ \left\langle \frac{\partial}{\partial\overline z}\left(\frac1z\right),\phi \right\rangle =\pi\phi(0). \] Therefore \(C=\pi\).