Show that \[ 4\frac{\partial}{\partial z}\frac{\partial}{\partial\overline z} = 4\frac{\partial}{\partial\overline z}\frac{\partial}{\partial z} = \Delta, \] where \(\Delta\) is the Laplacian \[ \Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}. \]
Solution
1. On a \(C^2\) function \(f\), the Wirtinger derivatives are\[ \frac{\partial}{\partial z} =\frac12\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right), \qquad \frac{\partial}{\partial\overline z} =\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right). \]
These are the definitions of the Wirtinger differential operators for \(z=x+iy\).
2. One has\[ 4\frac{\partial}{\partial z}\frac{\partial}{\partial\overline z}f = \frac{\partial^2f}{\partial x^2} +i\frac{\partial^2f}{\partial x\partial y} -i\frac{\partial^2f}{\partial y\partial x} +\frac{\partial^2f}{\partial y^2}. \]
Expand the product of the two first-order operators in 1 and apply it to \(f\).
3. Hence\[ 4\frac{\partial}{\partial z}\frac{\partial}{\partial\overline z}f=\Delta f. \]
Since \(f\) is \(C^2\), Clairaut’s theorem gives \[ \frac{\partial^2f}{\partial x\partial y} = \frac{\partial^2f}{\partial y\partial x}. \] Thus the mixed terms in 2 cancel, leaving \[ \frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=\Delta f. \]
4. Likewise,\[ 4\frac{\partial}{\partial\overline z}\frac{\partial}{\partial z}f=\Delta f. \]
Expanding in the reverse order gives \[ \frac{\partial^2f}{\partial x^2} -i\frac{\partial^2f}{\partial x\partial y} +i\frac{\partial^2f}{\partial y\partial x} +\frac{\partial^2f}{\partial y^2}, \] and the same equality of mixed partials cancels the middle terms.
5. Therefore, as second-order differential operators on \(C^2\) functions,\[ 4\frac{\partial}{\partial z}\frac{\partial}{\partial\overline z} = 4\frac{\partial}{\partial\overline z}\frac{\partial}{\partial z} = \Delta. \]
Both operator compositions agree with \(\Delta\) on every \(C^2\) function by 3 and 4.