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Suppose that $G\subseteq \mathbb{C}$ is a convex open set and that $f:G\to \mathbb{C}$ is a continuous function such that $\int_{\partial T}f(z)dz=0 $ for all triangles $T\subset G$. Then $\exists$ function $F:G\to\mathbb{C}$ such that $F'(z)=f(z)$, $\forall z\in G$.

What does this proposition exactly imply? Does it say that if the above conditions are satisfied then $f$ is integrable on $G$? But if $f$ is continuous, we know that it is integrable on $G$. Also, one of the conditions is an integral, so no. But it seems to me that this proposition states something apparent - that a continuous function has an antiderivative.

Please clarify this for me.

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    What exactly do you mean by "integrable"? This tells us there is an antiderivative, and that is a very strong property for functions of a complex variable, even continuous ones, as it implies the function is complex analytic. E.g. recall $f(z) = \overline{z}$ has no antiderivative. – 2017-02-09
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    I mean has an antiderivative. – 2017-02-09
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    That isn't the meaning of the word "integrable". You rightly point out it states the existence of an antiderivative. That's not the same thing. – 2017-02-09
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    @UmbertoP. Thanks for the point, I'll study the difference. But doesn't every continuous function have an antiderivative? – 2017-02-09
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    In the complex plane, no. There is a huge difference between real differentiability and complex differentiability. – 2017-02-09
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    @UmbertoP. the definition of complex differentiability appears to be similar to that of real differentiability, except that we're dealing with 2-vectors in the former case, and so, if $f$ is complex-differentiable at $z_0$, then it has an $\mathbb{R}^2$ Jacobian which is continuous at $z_0$. But you're right, of course, since continuity in $\mathbb{C}$ does not imply differentiability. – 2017-02-09
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    This is just a version of Morera's theorem, which gives an extremely useful criterion for a continuous function to be analytic. It's used in practice for a variety of purposes. See the wikipedia article for some specific examples: https://en.wikipedia.org/wiki/Morera's_theorem – 2017-02-09
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    One way to interpret the difference is that $f: \mathbb R^2 \to \mathbb R^2$ is differentiable if it has a linear approximation, but for $f : \mathbb C \to \mathbb C$ that linear approximation must be *multiplication by a complex number*. That is a restriction not present in the real case. – 2017-02-09
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    @UmbertoP. Can you please clarify what a linear approximation with multiplication by a complex number looks like? – 2017-02-10
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    @sequence: It looks like a dilation-rotation. Locally a complex differentiable function rescales by a positive factor and rotates, and thus locally it preserves angles. – 2017-02-10

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