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In Stein and Shakarchi, Complex Analysis, Princeton lectures in Analysis, Chapter 2, Problem 2 an interesting question is posed. The problem section in each chapter contains more complicated problems, with a research taste.

Morera's theorem simply states that if a function $f$ is continuous on $\Bbb{C}$ and $\int_D f(z)dz=0$ for any triangle(rectangle) $D$, then $f$ is holomorphic in $\Bbb{C}$. (the theorem is still valid if we replace $\Bbb{C}$ by a disk).

The problem presented above, states that

Morera's theorem is still valid if we replace the contours of integration from triangles/rectangles to circles, and more generally, to any contour which is a translate and dilate of a toy contour $\Gamma$.

Is there a simple proof for this problem, or maybe a reference to an article in which I can find the proofs?

I initially posted it on MO, but I didn't get an answer, and I was told that that was not the place for such questions. I received a comment with an idea of solution, but I don't seem to get it:

Convolve with a mollifier, apply Green, conclude that $\overline{\partial}$ of the convolution is $0$, recall that the uniform limit of analytic functions is analytic.

I would be glad if you could explain a bit how the answer above works for solving the initial problem.

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1 Answer 1

up vote 3 down vote accepted

Here is what the hint (probably) means:

  1. We only know that $f$ is continuous, but for what follows we need a smooth function ($C^\infty$ - although less smooth would be enough). To get it, take a smooth function $\phi(x,y)$ such that $\phi(x,y)=0$ if $x^2+y^2>1$ and $\int\phi dxdy=1$. Let $\phi_\epsilon(x,y)=\phi(x/\epsilon,y/\epsilon)/\epsilon^2$ (so that $\phi_\epsilon(x,y)=0$ if $x^2+y^2>\epsilon^2$ and $\int\phi_\epsilon dxdy=1$). Now set $f_\epsilon=f*\phi_\epsilon$, i.e. $f_\epsilon(x,y)=\int f(x-x',y-y')\phi(x',y')dx'dy'$. We have $f_\epsilon\to f$ uniformly as $\epsilon\to 0$, and $f_\epsilon$ are smooth.

  2. If $\gamma$ is a contour bounding a domain $D$ then by Stokes theorem $\int_\gamma f_\epsilon dz = \int_D df_\epsilon\wedge dz = 2i\int_D\bar\partial f_\epsilon dx dy$. As $\int_\gamma f_\epsilon dz=0$ for all translations and rescaling of some $\gamma_0$ bounding some $D_0$, we get $\bar\partial f_\epsilon=0$ (e.g. by $0=\lim_{\delta\to 0} \int_{z_0+\delta D_0} \partial f_\epsilon dx dy/\text{Area}(\delta D_0)=\partial f_\epsilon(z_0)$), i.e. $f_\epsilon$ is holomorphic.

  3. As $f_\epsilon\to f$ uniformly and $f_\epsilon$ are holomorphic, $f$ is also holomorphic.

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Thank you for the additional details. I'm sure I'll be able to understand the proof now. –  Beni Bogosel May 30 '11 at 20:27

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