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I'm currently taking a complex analysis course, and upon seeing my professor's proof of Goursat's Theorem, I noticed that it didn't inherently rely on the function we were integrating over being holomorphic.

I've been having a somewhat heated discussion for the last 5 hours with other students/my professor, and none of us know if

Lemma

would hold for a general $$L :\mathbb{R}^2 \mapsto \mathbb{R}^2$$

linear map, or at least a linear map that does not satisfy the Cauchy Riemann Equations.

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    I'm not personally sure I understand your question. You can't apply that theorem to a general map in $\mathbb{R}^2$, for the reason that the algebra involved in defining that integral is for complex numbers. – 2017-02-09
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    I mean is there an equivalent form for a linear map $$\mathbb{R}^2 \mapsto \mathbb{R}^2$$. Obviously I can't apply this theorem since it only works for complex values, but replace $\mathbb{C}$ with $\mathbb{R}^2$. Does the theorem still work? Perhaps not for *all* linear maps, but at least a linear map that doesn't satisfy CRE. – 2017-02-09
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    But in $\mathbb{R}^2$ how would you perform the product $f(\gamma(t)) \gamma'(t)$? which basically are two vectors, and product between two vector is not defined. – 2017-02-09
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    That's not the same thing... in the case of complex analysis the integral returns a complex number, which is a vector in $\mathbb{R}^2$. In the case you pointed out the integral returns in general a real number. And for those cases there are conditions that tell you when the integral is 0 (for example if the differential form is exact or not). – 2017-02-09
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    So Goursat's theorem absolutely does require that the function we're integrating over is holomorphic? – 2017-02-09
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    That fact that $F'(z) = f(z)$ implies $F(z)$ is holomorphic, then $f(z)$ is also holomorphic. This statement of the theorem might help http://math.stackexchange.com/questions/651658/is-moreras-theorem-the-inverse-theorem-of-goursats-theorem – 2017-02-09
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    The proof you posted just pushes the ideas somewhere else. Look at the Wikipedia article's proof of the theorem that appeals to Green's theorem: https://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem This really helps make clear where the function being holomorphic shows up. As a matter of fact, this shows that a complex valued $C^1$ function satisfies Cauchy-Goursat for all contractible curves if and only if it is holomorphic. – 2017-02-09
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    I wasn't looking at the "proof", but just that statement of the Goursat theorem. It says "whenever $f$ is holomorphic". – 2017-02-09

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user8469759 is correct that the question as posed doesn't make a whole lot of sense because it has an underlying arithmatic confusion. However, there is an analog to Goursat's theorem for functions from $\mathbb{R}^n\to\mathbb{R}$ which is known as the of the Gradient Theorem. The Gradient Theorem says that if $\varphi:U\subseteq\mathbb{R}^n\to\mathbb{R}$ and $\gamma$ is a curve from $u$ to $v$ then

$$\varphi(q)-\varphi(p)=\int_\gamma \nabla\varphi(r)\cdot dr $$

The converse of the gradient theorem is also true, and it says that if $F$ is a path-independent vector field, then $F$ is the gradient of some scalar-valued function. This gives a precise criterion for when a function from $\mathbb{R}^n\to\mathbb{R}$ has a path-independent integral.