# What is the strongest known form of Cauchy's theorem?

I am confused by the different statements of Cauchy's theorem. In another question, some users pointed out that the homology version is stronger than the homotopy version:

1. (The homotopy version.) Let $$\gamma$$ be a piecewise smooth closed path in a domain $$\Omega$$. If $$\gamma$$ is homotopic to zero, then $$\int_\gamma f(z)dz=0$$ for every $$f$$ holomorphic on $$\Omega$$.
2. (The homology version.) Let $$\gamma$$ be a piecewise smooth closed path in a domain $$\Omega$$. If the index of $$\gamma$$ about every point $$z\not\in\Omega$$ is zero, then $$\int_\gamma f(z)dz=0$$ for every $$f$$ holomorphic on $$\Omega$$.

However, I have also read about a "strong form" of Cauchy's theorem, in which we don't need $$f$$ to be holomorphic on an open set containing $$\gamma$$:

• If $$f$$ is holomorphic inside the Jordan domain enclosed by a Jordan curve $$\gamma$$ and it is also continuous on $$\gamma$$, then $$\int_\gamma f(z)dz=0$$.

(See Lynn H. Loomis, An elementary proof of the strong form of the Cauchy theorem, Bull. Amer. Math. Soc., Volume 50, Number 12 (1944), 831-833.)

Where does this strong form stand? Is the homology version still stronger than this strong form? If not, is there a strongest known form of Cauchy's theorem and is it proved in any introductory text?

• A "strongest form" may be hard to determine. "If $A'$, then $B'$" is stronger than "If $A$ then $B$" if the latter can be shown from the former, as in $A$ implies $A'$ and $B'$ implies $B$. So in a way, the strongest you can get by playing with $A'$ is "If $\int_\gamma f(z)\,\mathrm dz=0$, then $\int_\gamma f(z)\,\mathrm dz=0$" -- which is totally useless. Likewise, You can always make the claim stronger by adding "... and there are infinitely many primes" to the conclusion. – Hagen von Eitzen Sep 15 '19 at 10:15
• In term of proof the main difference is between analytic functions, holomorphic functions with continuous derivative, holomorphic functions, holomorphic distributions. The versions in term of homotopy, extension to continuous boundary and other residue theorem stuffs are mostly easy corollaries. – reuns Sep 15 '19 at 12:18
• I wonder if there is a more general homology version of the theorem that implies the "strong form" as described in the question. – William McGonagall Sep 15 '19 at 13:03