# What would be a counterexample to Cauchy's integral formula or Cauchy's theorem?

Here is the Cauchy's theorem.

Let $G$ be open in $\mathbb{C}$. (Not necessarily connected)

Let $f:G\rightarrow \mathbb{C}$ be a holomorphic function.

Let $\gamma_k$ be closed rectifiable curves in $G$ for $1\leq k \leq n$.

If $\sum Wnd(\gamma_k,z)=0$ for all $z\in \mathbb{C}\setminus G=0$, then $\sum \int_{\gamma_k} f(z) dz = 0$.

As you can see, it requires curves to lie inside $G$.

Now, let's consider this case:

Let $\gamma$ be a simple closed rectifiable curve in the plane.

Let $G$ be the interior of $\gamma$ and $f:G\cup\{\gamma\} \rightarrow \mathbb{C}$ be a function holomorphic on $G$.

Assume that $f$ is complex-differentiable on $\{\gamma\}$. (I mean complex differentiablity at points, not holomorphy)

Then, is $\int_\gamma f(z) dz=0$?

What would be a counterexample?

• What do you mean complex differentiable on $\gamma?$ That would require that for each point $a$ on $\gamma,$ there is a full disc $D(a,r)$ where $f$ is defined? – zhw. Nov 20 '15 at 21:29
• @zhw. Since $\gamma$ is a simple closed curve, every point $p$ on the trace $\{\gamma\}$ is a limit point of the interior $G$. So, $\lim_{z\to p} [f(z)-f(p)]/(z-p)$ is well-defined. I meant this. Differentiability on the neighborhood $B(p,r)$ of $p$ is not required. $f$ need not be defined on the whole $B(p,r)$. – Rubertos Nov 20 '15 at 21:33
• No, it is not well defined. That symbol requires $f$ to be defined in some $D(p,r).$ That is the meaning of it. Perhaps you mean $\lim [f(z)-f(p)]/(z-p)$ as $z\to p$ within $G$? Or is it $z\to p$ within $\overline G$? You should make this clear. – zhw. Nov 20 '15 at 21:38
• The latter one. $z\to p$ within $\overline G$. $f$ need not be defined in the exterior of $\gamma$. Since I only mentioned that $f$ is only defined on $\overline{G}=G\cup\{\gamma\}$, I think the above limit is well-defined though.. – Rubertos Nov 20 '15 at 21:40
• You really should edit your question so that people know that "complex differentiable on $\{\gamma\}$" has a technical meaning here different from the usual one. – zhw. Nov 21 '15 at 1:39

This is true, even under slightly weaker assumptions, but is fairly technical to prove.

In fact, we have

Theorem If $\gamma$ is a rectifiable Jordan curve and $f$ is holomorphic on the interior $G$ of $\gamma$ and continuous on $\bar G = G \cup \gamma$, then $$\int_\gamma f(z)\,dz = 0.$$

The proof is apparently due to Denjoy and appeared in Compt. Rend., 196, 29-33 (1933). I also managed to find a reference to a different proof by Walsh from 1933.

Partial result: Suppose $\gamma$ encloses a convex region $G.$ For convenience, assume $0\in G.$ Then for $0<r<1,$ $r\gamma$ is a closed contour in $G.$ Since $G$ is simply connected, $\int_{r\gamma} f(z)\,dz = 0$ by Cauchy. As $r\to 1^-,$ the uniform continuity of $f$ on $\overline G$ shows

$$\int_{r\gamma} f(z)\,dz \to \int_{\gamma} f(z)\,dz.$$

Therefore $\int_{\gamma} f(z)\,dz =0.$

Here's the thing: The same result would be true if we merely assumed $f$ is continuous on $\overline G$ and $f\in H(G).$ So at this point I can't tell if this "complex differentiability" at points of $\gamma$ hypothesis gives any advantage over continuity on $\overline G.$

• G is not necessarily convex but I believe we can prove that if $\gamma$ can be represented by a continuous function $g:S^1\to C$ then, for any $e>0$ there is a simple closed curve $h:S^1\to G$ with $\forall t\in S^1 (|g(t)-h(t)|<e)$. – DanielWainfleet Nov 21 '15 at 2:19
• Can't see how that helps. Any simple closed curve can be represented that way. – zhw. Nov 21 '15 at 2:26
• The same way as in your answer.The integral of f along $\gamma$ would be arbitrarily close to the integral of along some g, and the latter integral is 0. – DanielWainfleet Nov 21 '15 at 3:02
• OK, I see I mistook $h:S^1\to G$ as $h:S^1\to \overline G.$ There's a lot of work left to do in your scenario. – zhw. Nov 21 '15 at 3:05
• That's why I asked in a prior comment about analytic continuation as an alternative approach.I've forgotten a lot of that. – DanielWainfleet Nov 21 '15 at 4:15