Lecture 11: applications of Cauchy’s formula to analyticity

1.  Morera’s theorem

Recall Cauchy’s theorem: if $f(z)$ is analytic on a domain $D$, then the differential form $f(z)dz$ is closed on $D$. Consequently, the integral of $f$ over the boundary of the domain vanishes, so by Green’s theorem ${\int }_{\partial D}f(z)𝑑z={\int }_{D}0𝑑z=0$.

Loosely speaking, Morera’s theorem establishes that the converse holds. More precisely:

Theorem (Morera’s theorem, version 1).

Let $f$ be a continuous function on a domain $D$. If for any sub-domain $D^{\prime }\subset D$ we have

$${\int }_{\partial D^{\prime }}f(z)𝑑z=0,$$

then $f$ is analytic on $D$.

It is noteworthy that in order to apply Morera’s theorem we only need $f$ to be continuous, and only check a condition on its integrals; but the conclusion is that $f$ is analytic, which we’ve seen implies that it is actually infinitely differentiable. This is analogous to the fact that not only do harmonic functions satisfy the mean value property, but in fact any continuous function satisfying the mean value property is harmonic. Characteristically, the version for analytic functions in place of harmonic functions is more flexible and more powerful, but the idea is fundamentally similar.

Actually, we do not need to check every possible sub-domain. It is sufficient to check the condition for a specific class of rectangles, giving us a more practically useful version of the theorem:

Theorem (Morera’s theorem, version 2).

Let $f$ be a continuous function on a domain $D$. If for any rectangle $R\subset D$ with sides parallel to the coordinate axes we have

$${\int }_{\partial R}f(z)𝑑z=0,$$

then $f$ is analytic on $D$.

To prove Morera’s theorem, it suffices to prove version 2. We can further simplify the problem by assuming $D$ is a disk. This is valid because we can pick a small disk around any point in $D$ and use Morera’s theorem on that disk to show $f$ is analytic there, which implies $f$ is analytic on all of $D$. Assuming $D$ is a disk centered at a point $z_0$, every closed form is exact, so we expect to find a primitive $F$ such that $f=F^{\prime }$. We define this primitive by integrating along grid lines:

$$F(z)={\int }_{z_0}^{z}f(w)𝑑w$$

where the path of integration travels horizontally and then vertically.

Fix a small complex number $h$ with magnitude . The difference in our function is given by

$$F(z+h)-F(z)={\int }_z^{z+h}f(w)𝑑w$$

where the path is defined by tracing from $z$ backwards to $z_0$ (the reverse of the previous path), and then from $z_0$ to $z+h$ as defined above. Because the integral over the boundary of any rectangle is $0$ by assumption, we can equivalently just move the line of integration to trace the edges of a small rectangle directly from $z$ to $z+h$:

Now we can analyze the local difference algebraically:

${\displaystyle {\int }_z^{z+h}}f(w)𝑑w$ $={\displaystyle {\int }_z^{z+h}}\left(f(z)+(f(w)-f(z))\right)𝑑w$

$=hf(z)+{\displaystyle {\int }_z^{z+h}}(f(w)-f(z))𝑑w.$

Since $f$ is continuous, for any chosen $\delta >0$ we can ensure that by taking $ϵ$ small enough. Furthermore, the length of the path from $z$ to $z+h$ along the rectangle edges is at most $2|h|$. By the ML bound, we then have

Dividing both sides by $h$ yields

Because we can take $\delta$ to be as small as we like as $h\to 0$, this absolute difference must go to $0$. This implies that $F$ is complex differentiable, i.e. analytic. Finally, by Cauchy’s formula for derivatives, the derivative of an analytic function is itself analytic, which tells us that $F^{\prime }=f$ is analytic, completing the proof.

2.  Goursat’s theorem

By definition, for an analytic function $f$ on a domain $D$,

• it must be complex differentiable on $D$;

• and its derivative $f^{\prime }$ must be continuous on $D$.

We previously claimed that this second condition is redundant. We are now in a position to rigorously justify this.

Theorem (Goursat’s theorem).

If $f:D\to ℂ$ is a function such that

$$f^{\prime }(z_0)=\underset{z\to z_0}{lim}\frac{f(z)-f(z_0)}{z-z_0}$$

exists for every $z_0\in D$, then $f$ is analytic (i.e. $f^{\prime }$ is automatically continuous).

To prove this, we first note that if $f$ is differentiable, it is automatically continuous, otherwise the limit defining the derivative would not exist. This continuity means we can hope to apply Morera’s theorem. To do so, let $R\subset D$ be an arbitrary rectangle. Our goal is to show that the integral over its boundary is zero:

$${\int }_{\partial R}f(z)𝑑z=0.$$

We proceed by repeatedly dividing the rectangle. First, divide $R$ into four equal, smaller rectangles $R_1,R_2,R_3,R_4$:

Because the inner edges are traversed twice in opposite directions, their contributions cancel, so the integral over the boundary of $R$ splits nicely into the sum over the sub-rectangles:

$${\int }_{\partial R}f(z)𝑑z={\int }_{\partial R_1}f(z)𝑑z+{\int }_{\partial R_2}f(z)𝑑z+{\int }_{\partial R_3}f(z)𝑑z+{\int }_{\partial R_4}f(z)𝑑z$$

By the triangle inequality, at least one of these smaller rectangles, which we will call $R_i$, must carry at least a quarter of the total value:

$$\left|{\int }_{\partial R_i}f(z)𝑑z\right|\ge \frac{1}{4}\left|{\int }_{\partial R}f(z)𝑑z\right|.$$

We relabel this chosen rectangle as $R^1$ and repeat the same subdivision process on it:

Again, one of the inner rectangles (call it $R^2$) must satisfy the proportional bound:

$$\left|{\int }_{\partial R^2}f(z)𝑑z\right|\ge \frac{1}{4}\left|{\int }_{\partial R^1}f(z)𝑑z\right|\ge \frac{1}{4^2}\left|{\int }_{\partial R}f(z)𝑑z\right|.$$

We continue this subdivision indefinitely:

This produces a nested sequence of rectangles $R\supset R^1\supset R^2\supset \mathrm{\cdots }$, where the $n$-th rectangle bounds the integral of the original rectangle by

$$\left|{\int }_{\partial R^n}f(z)𝑑z\right|\ge \frac{1}{4^n}\left|{\int }_{\partial R}f(z)𝑑z\right|.$$

Because the diameters of the rectangles $R^n$ are strictly decreasing to $0$ and they are all nested within one another, they must converge towards a single point $z_0$. By our initial assumption, $f$ is differentiable at this specific point $z_0$. This means that for any $z\in R^n$, the approximation $\frac{f(z)-f(z_0)}{z-z_0}$ is close to the derivative:

for some sequence ${ϵ}_n\to 0$ as $n\to \mathrm{\infty }$.

Let $L$ denote the perimeter length of the original rectangle $\partial R$. By construction, the perimeter length of $\partial R^n$ is $L/2^n$. The differentiability bound translates to

Notice that the function given by $f(z_0)+f^{\prime }(z_0)(z-z_0)$ is a linear polynomial in $z$, so it is analytic everywhere. By Cauchy’s theorem, its integral around any closed loop is zero:

$${\int }_{\partial R^n}(f(z_0)+f^{\prime }(z_0)(z-z_0))𝑑z=0.$$

We can subtract this zero integral from our evaluation over $R^n$ and then apply the ML bound:

$\left|{\displaystyle {\int }_{\partial R^n}}f(z)𝑑z\right|$ $=\left|{\displaystyle {\int }_{\partial R^n}}(f(z)-f(z_0)-f^{\prime }(z_0)(z-z_0))𝑑z\right|$

$\le {\displaystyle \frac{L}{2^n}}\cdot {ϵ}_n\cdot {\displaystyle \frac{L}{2^n}}$

$={\displaystyle \frac{L^2}{4^n}}\cdot {ϵ}_n.$

Combining this with our earlier inequality relating $R^n$ and $R$, we obtain

$$\left|{\int }_{\partial R}f(z)𝑑z\right|\le 4^n\left|{\int }_{\partial R^n}f(z)𝑑z\right|\le L^2{ϵ}_n$$

for all integers $n$. Since ${ϵ}_n\to 0$ as $n$ goes to infinity, the integral over the original rectangle $R$ must vanish. Because $R$ was arbitrary, Morera’s theorem implies that $f$ is indeed analytic.