## Morera’s Theorem.

### October 29, 2010

Morera’s theorem, named after the mathematician Giacinto Morera whose name is pretty sweet but is second only to his ultra-fly mustache

is an extremely important result in complex analysis: it states that if $f$ is a continuous function defined on an open set $D$ in the complex plane, and, in addition, we have that the integral around every closed curve is zero for every closed curve $C$ in $X$, then, in fact, $f$ must be complex differentiable everywhere in $D$.  This is a common tool to use in the proofs of other theorems, as well as in its own right showing that a continuous function is actually much nicer than “just” continuous.

One thing I want to mention right now is a bit of notation.  The last part of the statement of the theorem above (the “integration around every closed curve is zero” part) can be written more “mathematically” as:

$\displaystyle \oint_{C} f(z)\, dz = 0$

and perhaps a few people reading this are not comfortable with that little circle on the integral sign: don’t worry too much about it, it simply means, “take a curve which is a closed curve (the start and end points are the same), then take the intergral around that curve.”  Usually we will parameterize the curve, and usually it will be a nice circle.  But it doesn’t matter right this second, because we will be doing almost none of that for the proof.

The proof of this is surprisingly simple.  The point of the proof is as follows: we will construct an anti-derivative (?!) and since the fundamental theorem of calculus states that this anti-derivative is differentiable, and hence infinitely differentiable, and so its derivative is also infinitely differentiable.  The word for infinitely differentiable in complex analysis (which is actually equivalent to once-differentiable, but this is another topic…) is holomorphic.

Before we get on with the proof, let’s just note something. Suppose we have two paths from the point $a$ to the point $b$, let’s call them A and B as in the picture below.

Then if we have that the integral around any closed curve is 0, then the integral from $a$ to $b$ for any path we choose will be the same.  Why?

$\displaystyle \int_{A}f(z)\, dz + \int_{B}f(z)\, dz = \int_{A+B}f(z)\, dz = \int_{C}f(z)\, dz = 0$

Where $C$ is the complete closed curve.  This means that the integral from $a$ to $b$ along $A$ is the NEGATIVE of the integral from $b$ to $a$ along $B$.  But when we switch the limits in the second integral (along $B$), they are equal.  Since $A$ and $B$ were arbitrary, we have the result that if the integral around any closed curve is 0 in a domain $D$, then the value of the integral from a point $a$ to a point $b$ in our domain $D$ is the same no matter what path we take.

This is kind of a neat result in itself, but it also makes proving Morera’s theorem much easier.  Let’s prove this now.

Theorem (Morera):  Given $f$ continuous in a domain (an open set) $D$ which is a subset of the complex plane, and $f$ satisfies

$\displaystyle \oint_{C}f(z)\, dz = 0$

for every closed curve $C$ in $D$, then we have that $f$ is holomorphic in $D$.

Proof. Let’s explicitly construct an anti-derivative.  Pick an arbitrary $a$ in $D$.  Define the integral

$F(z) = \displaystyle \int_{a}^{z}f(z)\, dz$

which is well defined for every $z\in D$, since every path from $a$ to $z$ has the same value.

One can now show, using a variety of methods (for example, applying the definition of the derivative to this integral and looking at how similar that looks to Cauchy’s General Integral Formula), that

$F'(z) = f(z)$

Since $F$ is once, and therefore infinitely, differentiable, we have that it is holomorphic.  Since $f(z)$ is $F$‘s derivative, it is also infinitely complex differentiable.  Hence $f(z)$ is holomorphic on $D$, as required.

[EDIT:  In the original proof I wrote here a few years ago I used the Fundamental Theorem of Calculus to show that the derivative of $F$ was $f$.  I cannot do this, since FTC does not apply to complex functions.  It is an exercise to the reader to find a complex function which “looks” like the FTC could be applied to it but which yields the incorrect derivative (or, prove no such thing could happen).]

Last, let me just note something on terms.  There seems to be a bit of a dispute (at least in the few books I have) as to what to call a complex function which is differentiable.  I have used holomorphic here, because this is how I learned it.  Others use the term “analytic” which I use to mean “has a power-series approximation in a neighborhood of a point.”  Some authors apparently switch these terms; though, you’ll be pleased to know this causes nearly no confusion because it is an important theorem of complex analysis that, regardless of which one you call which, they are actually equivalent to each other.  Hence, above, we could have also said that $f$ was analytic on $D$ and we would have been correct.

### 3 Responses to “Morera’s Theorem.”

1. Anonymous said

nice job, good explain

2. Anonymous said

You can’t use the Fundamental Theorem of Calculus for complex functions

• James said

You’re correct; I’m not sure what I was thinking here. The integral we consider is correct, but showing that it is differentiable can be done using the general version of Cauchy’s formula, or differentiating directly. I’ll fix this soon, but I’ll get rid of the FTC part now.