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Morera's Theorem states that

If $f$ is continuous in a region $D$ and satisfies $\oint_{\gamma} f = 0$ for any closed curve $\gamma$ in $D$, then $f$ is analytic in $D$.

I have two questions:

  1. If $f$ is continuous in $D$ and $\oint_C f = 0$ for any circle $C$ in $D$, can we deduce that $\oint_{\gamma} f = 0$ for any closed curve $\gamma$ in $D$?

  2. (more ambitiously) If $f$ is continuous and $\oint_C f = 0$ for any circle $C$ in $D$, is $f$ analytic in $D$ ?

Partial ansers for question 2 seem to be here, but I doubt their argument, specificly, the construction of the original function.

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    $\begingroup$ Number 2 doesn't seem more ambitious to me in light of Morera's theorem. $\endgroup$
    – Dan Petersen
    Commented Sep 27, 2013 at 6:12
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    $\begingroup$ Some googling reveals anhngq.wordpress.com/2009/07/20/… $\endgroup$
    – Sean Eberhard
    Commented Sep 27, 2013 at 9:06
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    $\begingroup$ @Sean: Very nice proof on that webpage. I suggest that you add your comment as an answer, because it really answers the original question. $\endgroup$
    – GH from MO
    Commented Sep 27, 2013 at 9:59
  • $\begingroup$ "If $\oint_\gamma f=0$ for any closed curve $\gamma$" is ambiguous. A reasonable reader could think it means "If there is any closed curve $\gamma$ for which $\oint_\gamma f=0$". But I don't think that's what was intended in this case. Simply changing "any" to "every" would disambiguate it. $\endgroup$
    – Michael Hardy
    Commented Sep 28, 2013 at 5:18

2 Answers 2

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The answer is yes, and a proof can be found for example on this webpage: http://anhngq.wordpress.com/2009/07/20/a-generalization-of-the-morera%E2%80%99s-theorem/

A brief summary: Suppose $f$ is continuous and $\int_C f = 0$ for every circle $C$, but $\int_\gamma f \neq 0$ for some closed curve $\gamma$. By convolving $f$ with a smooth approximation to the identity, we may assume $f$ is smooth. But then by applying Green's formula to $\int_C f = 0$ for small circles $C$, we see that $f$ must satisfy the Cauchy-Riemann equations, so $\int_\gamma f = 0$, a contradiction.

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  • $\begingroup$ Thanks. The proof seems smooth, but how do you prove you can interchange the order of integration in the 4th formula of that proof ? $\endgroup$
    – booksee
    Commented Sep 28, 2013 at 16:08
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    $\begingroup$ Fubini's theorem (en.wikipedia.org/wiki/Fubini's_theorem) states that interchanging the order of integration is nice and friendly. $\endgroup$
    – Sean Eberhard
    Commented Sep 28, 2013 at 16:37
  • $\begingroup$ Surely, I know. Is that function Lebesgue integrable over $\mathbb{C}\times C$ ? $\endgroup$
    – booksee
    Commented Sep 28, 2013 at 16:42
  • $\begingroup$ Just take $\phi$ (the approximation to the identity) to be compactly supported, and then it's obvious, as $\mathbb{C}\times C$ can be replaced in that equation throughout by $\text{supp}\,\phi\times C$. $\endgroup$
    – Sean Eberhard
    Commented Sep 29, 2013 at 8:38
  • $\begingroup$ Sorry, my question was not well-posed. The correct question is: since $z-w$ would go over circles everywhere in $\mathbb{C}$ as $w$ go over $\mathbb{C}$, so we must require the function be integrable over $\mathbb{C}\times\mathbb{C}$ instead of $\mathbb{C}\times C$. How do we prove this ? It's easy to prove it is locally integrable in $\mathbb{C}\times\mathbb{C}$. $\endgroup$
    – booksee
    Commented Sep 29, 2013 at 16:08
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First of all, your questions 1 and 2 are equivalent (by the usual Morera theorem).

Second, even stronger generalizations of Morera are available (one does not need all circles). There is an old nice survey of Zalcman, Offbeat Integral Geometry, in the Monthly. In particular it contains the following result for the case $D=C$: if the integrals over all circles of two fixed radii $r_1$ and $r_2$ are zero, then the function is analytic, unless the ratio of these radii is a zero of Bessel's function $J_1$. On some more modern research on the topic, I recommend the papers of Hansen, Nadirashvili and Tumanov MR2046196.

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  • $\begingroup$ Thanks a lot but is the deduction after formula (3) wrong in Zalcman's paper? $\int_C fds\neq \int_0^{2\pi} f(z+re^{i\theta}) d\theta$. $\endgroup$
    – booksee
    Commented Sep 28, 2013 at 18:17
  • $\begingroup$ btw, the world is so small that we're in the same building... $\endgroup$
    – booksee
    Commented Sep 28, 2013 at 18:20
  • $\begingroup$ I am glad to now this:-) $\endgroup$
    – Alexandre Eremenko
    Commented Sep 29, 2013 at 4:52

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