Lecture 12: Pompeiu’s formula

1.  The operators $\frac{\partial }{\partial z}$ and $\frac{\partial }{\partial \overline{z}}$

When working with the Cauchy–Riemann equations, we previously needed to separate functions into their real and imaginary parts $u$ and $v$, and explicitly compute the $x$ and $y$ partial derivatives. It is often much more convenient to have operators that feel more natively complex.

To achieve this, we define the following operators:

${\displaystyle \frac{\partial }{\partial z}}$ $={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{\partial }{\partial x}}+{\displaystyle \frac{\partial }{\partial (iy)}}\right)={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{\partial }{\partial x}}-i{\displaystyle \frac{\partial }{\partial y}}\right),$

${\displaystyle \frac{\partial }{\partial \overline{z}}}$ $={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{\partial }{\partial x}}-{\displaystyle \frac{\partial }{\partial (iy)}}\right)={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{\partial }{\partial x}}+i{\displaystyle \frac{\partial }{\partial y}}\right).$

If a function $f$ is analytic, we can use the Cauchy–Riemann equations to write its derivative in multiple ways:

$$f^{\prime }(z)=\frac{\partial f}{\partial x}=\frac{\partial f}{\partial (iy)}=-i\frac{\partial f}{\partial y}.$$

By averaging the $x$ and $y$ expressions, we see that the $\frac{\partial }{\partial z}$ operator recovers the standard complex derivative:

$$f^{\prime }(z)=\frac{1}{2}(f^{\prime }(z)+f^{\prime }(z))=\frac{1}{2}\left(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right)=\frac{\partial f}{\partial z}.$$

On the other hand, if we take the difference of these equal expressions, we find a condition for the $\frac{\partial }{\partial \overline{z}}$ operator:

$$0=\frac{1}{2}(f^{\prime }(z)-f^{\prime }(z))=\frac{1}{2}\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)=\frac{\partial f}{\partial \overline{z}}.$$

This shows that if $f$ is analytic, it must satisfy $\frac{\partial f}{\partial \overline{z}}=0$.

More generally, if we express an arbitrary complex function as $f=u+iv$, applying the $\frac{\partial }{\partial \overline{z}}$ operator yields

$$\frac{\partial f}{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}+i\frac{\partial u}{\partial y}-\frac{\partial v}{\partial y}\right).$$

If we equate the real and imaginary parts of this expression to zero, we see that $\frac{\partial f}{\partial \overline{z}}=0$ holds if and only if

$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y},$$

which are precisely the standard Cauchy–Riemann equations. Consequently, the single equation

$$\frac{\partial f}{\partial \overline{z}}=0$$

is completely equivalent, and is sometimes called the complex form of the Cauchy–Riemann equations.

These new operators behave like standard partial derivative operators in many ways. They are linear combinations of partial derivatives, so they satisfy linearity:

$$\frac{\partial }{\partial z}(af+bg)=a\frac{\partial f}{\partial z}+b\frac{\partial g}{\partial z},\frac{\partial }{\partial \overline{z}}(af+bg)=a\frac{\partial f}{\partial \overline{z}}+b\frac{\partial g}{\partial \overline{z}}$$

for any constants $a,b$. They also satisfy the product rule:

$$\frac{\partial }{\partial z}(fg)=f\frac{\partial g}{\partial z}+\frac{\partial f}{\partial z}g,\frac{\partial }{\partial \overline{z}}(fg)=f\frac{\partial g}{\partial \overline{z}}+\frac{\partial f}{\partial \overline{z}}g.$$

Finally, they have useful symmetry properties with respect to complex conjugation:

$$\overline{\frac{\partial f}{\partial z}}=\frac{\partial \overline{f}}{\partial \overline{z}},\overline{\frac{\partial \overline{f}}{\partial z}}=\frac{\partial f}{\partial \overline{z}}.$$

As a quick application, suppose that both $f$ and its conjugate $\overline{f}$ are analytic. Because $\overline{f}$ is analytic, the complex Cauchy–Riemann equation gives $\frac{\partial \overline{f}}{\partial \overline{z}}=0$. But our property implies that $\overline{\frac{\partial f}{\partial z}}=0$, meaning $\frac{\partial f}{\partial z}=0$. Since $\frac{\partial f}{\partial z}$ is just the derivative $f^{\prime }(z)$ for an analytic function, $f^{\prime }(z)=0$, and so $f$ must be a constant.

Here is a slightly more serious application. If $f(z)$ is smooth (that is, smooth as a function of $x$ and $y$, i.e. on ${ℝ}^2$; equivalently, all partial derivatives with respect to $x$ and $y$ exist) then, since all partial derivatives commute with each other, we have

$$\frac{\partial }{\partial \overline{z}}\frac{\partial }{\partial z}=\frac{\partial }{\partial z}\frac{\partial }{\partial \overline{z}}.$$

In particular, if $f$ is analytic, so $\frac{\partial f}{\partial \overline{z}}=0$, then $0=\frac{\partial }{\partial z}\frac{\partial f}{\partial \overline{z}}=\frac{\partial }{\partial \overline{z}}\frac{\partial f}{\partial z}=\frac{\partial }{\partial \overline{z}}{f}^{\prime }(z)$, so $f^{\prime }(z)$ is also analytic, so $f^{\prime \prime }(z)$ exists. By induction, we find that each higher derivative $f^{(n)}(z)$ exists and is analytic, recovering a consequence of Cauchy’s formula with much less work. However, here we needed to assume that $f$ was smooth, which is not obvious from the definition of analytic functions and was not necessary for the argument via Cauchy’s formula, so that result is still important.

2.  Pompeiu’s formula

When we first introduced Cauchy’s formula, we described it as the complex version of half of Green’s theorem. In our new language, we can write it as follows: if $\frac{\partial f}{\partial \overline{z}}=0$, then

$${\int }_{\partial D}f(z)𝑑z=0.$$

We can now come back to the “full” version of Green’s theorem: recall that this says

$${\int }_{\partial D}P𝑑x+Qdy={\iint }_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)𝑑x𝑑y.$$

We have $f(z)dz=f(z)dx+if(z)dy$, so $P=f$ and $Q=if$; therefore the right-hand side is

$${\iint }_D\left(i\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)𝑑x𝑑y=i{\iint }_D\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)𝑑y=2i{\iint }_D\frac{\partial f}{\partial \overline{z}}𝑑x𝑑y.$$

That is:

Theorem (Cauchy–Green theorem).

If $D$ is a bounded domain with piecewise smooth boundary and $f$ is a smooth function on $D\cup \partial D$, then

$${\int }_{\partial D}f(z)𝑑z=2i{\iint }_D\frac{\partial f}{\partial \overline{z}}𝑑x𝑑y.$$

If $f$ is analytic, we recover Cauchy’s theorem.

Proceeding exactly how we derived Cauchy’s formula from Cauchy’s theorem using the more general formula above to incorporate a correction term, we can get the following generalization:

Theorem (Pompeiu’s formula).

If $f$ is a smooth complex-valued function on $D\cup \partial D$ as above, then for any $z\in D$ we have

$$f(z_0)=\frac{1}{2\pi i}{\int }_{\partial D}\frac{f(z)}{z-z_0}𝑑z-\frac{1}{\pi }{\iint }_D\frac{\partial f}{\partial \overline{z}}\cdot \frac{1}{z-z_0}𝑑x𝑑y.$$

When $f$ is analytic, the correction term is zero and we recover Cauchy’s formula.

Proof sketch.

Let $D_0$ be the disk of radius $r$ centered at $z_0$. In the proof of Cauchy’s formula, we used Cauchy’s theorem to relate the integral over $\partial D$ to the integral over $\partial D_0$, so that we could parametrize. Now, we use the Cauchy–Green theorem instead:

$${\int }_{\partial (D\setminus D_0)}\frac{f(z)}{z-z_0}𝑑z=2i{\iint }_{D\setminus D_0}\frac{\partial }{\partial \overline{z}}\frac{f(z)}{z-z_0}𝑑x𝑑y=2i{\iint }_{D\setminus D_0}\frac{\partial f}{\partial \overline{z}}\cdot \frac{1}{z-z_0}𝑑x𝑑y.$$

The left-hand side is

$${\int }_{\partial D}\frac{f(z)}{z-z_0}𝑑z-{\int }_{\partial D_0}\frac{f(z)}{z-z_0}𝑑z.$$

If we parametrize the circle $\partial D_0$ of radius $r$ as in the proof of Cauchy’s formula, we find that

$${\int }_{\partial D}\frac{f(z)}{z-z_0}𝑑z=2\pi iA(r);$$

but we no longer have $f$ analytic (hence harmonic), so we can’t assume $A(r)=f(z_0)$. That said, our equation now reads

$$-2\pi iA(r)+{\int }_{\partial D}\frac{f(z)}{z-z_0}=2i{\iint }_{D\setminus D_0}\frac{\partial f}{\partial \overline{z}}\cdot \frac{1}{z-z_0}𝑑x𝑑y.$$

Taking the limit as $r\to 0$, the first integral on the left-hand side approaches $-2\pi if(z_0)$, the second is independent of $r$, and the only change on the right-hand side is that the size of the omitted disk $D_0$ goes to zero; so in the limit the equation reads

$$-2\pi if(z_0)+{\int }_{\partial D}\frac{f(z)}{z-z_0}=2i{\iint }_D\frac{\partial f}{\partial \overline{z}}\cdot \frac{1}{z-z_0}𝑑x𝑑y,$$

which rearranges to the claimed formula. ∎

Note: for Cauchy’s formula, we could differentiate under the integral sign and deduce that $f$ was therefore infinitely (complex) differentiable. Can we do so here? No: if so, we’d find that every smooth function (in the real sense) is analytic, which is not true! The issue is that the integrand in the second term fails to be differentiable at the point $z=z_0$. This didn’t come up in Cauchy’s formula, where we just have the first term, since there $z\in \partial D$ rather than in $D$, so $z=z_0$ is impossible, but in the second term it does occur, so while the integral will still converge, we can’t differentiate under the integral sign. So, for example, there can be no “Pompeiu’s formula for derivatives.”

Let’s work out an example of evaluating an integral via Pompeiu’s formula, as for Cauchy’s formula. Consider $f(z)=\overline{z}$ and $D$ the open disk of radius $1$ centered at the origin. For the first term, notice that for $|z|=1$ we have $\overline{z}=\overline{e^{i\theta }}=e^{-i\theta }=\frac{1}{z}$, so this integral is the same as

$$\frac{1}{2\pi i}{\int }_{\partial D}\frac{1}{z(z-z_0)}𝑑z.$$

There are a few ways we could compute this, e.g. via explicit parametrization. Let’s instead use Cauchy’s formula: just like last week, we can reduce this to the integral around two small disks centered at $0$ and $z_0$, giving

$${\frac{1}{z}|}_{z=z_0}+{\frac{1}{z-z_0}|}_{z=0}=\frac{1}{z_0}-\frac{1}{z_0}=0.$$

Since $f(z_0)$ is nonzero in general, Cauchy’s formula alone fails! We need the correction term.

Here, observe

$$\frac{\partial f}{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial \overline{z}}{\partial x}+i\frac{\partial \overline{z}}{\partial y}\right)=\frac{1}{2}\left(1-i^2\right)=1$$

(not unsurprisingly!) so the second term is

$$-\frac{1}{\pi }{\iint }_D\frac{1}{z-z_0}𝑑x𝑑y.$$

One can, with some care, evaluate this integral directly; Pompeiu’s formula tells us much more directly that the whole thing must be equal to ${\overline{z}}_0$, or in other words

which is perhaps not a formula one would otherwise guess.