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I was reading about Cauchy-Goursat Theorem proof (the version that works with triangles or rectangles) and I couldn't determine when it fails when we want to replicate it in the real world. In this proof, we state:

$$ f(z) = f(z_0) + f'(z_0)(z - z_0) + (z - z_0)\left[ \frac{f(z) - f(z_0)}{z - z_0} - f'(z_0) \right] $$

Then we divide the integral into those three terms. The first and second ones vanish because they have a primitive:

$$ f(z_0) \quad \text{has as its primitive} \quad z f(z_0) $$

$$ f'(z_0)(z - z_0) \quad \text{has as its primitive} \quad \frac{f'(z_0)}{2}(z - z_0)^2 $$

$$ \text{Then the last term, if we define} \text{ } h(z_0) \text{ }as: $$

$$ h(z) = \frac{f(z) - f(z_0)}{z - z_0} - f'(z_0), $$

we can prove that it's bounded and it converges to zero.

I can't find where it fails in the real case. I have read some explanations, and the fact that when we work in higher dimensions of ℝ, the differentiable matrix could be anything were the more convincing ones, but I haven't read a rigorous proof. Can you guys help me with that?

Frunobulax
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Johhny Mc
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1 Answers1

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If $U$ is an open subset of $\Bbb C$ and $f\colon U\longrightarrow\Bbb C$ is a differentiable function, then $f'$ is a function from $U$ into $\Bbb C$. So, it makes sense to ask whether or not a function $g\colon U\longrightarrow\Bbb C$ has a primitive; that would be a function $G\colon U\longrightarrow\Bbb C$ such that $G'=g$.

But that doesn't apply to functions from an open subset $U$ of $\Bbb R^2$ into $\Bbb R^2$. If $f$ is such a function and if $f$ is differentiable, then, for each $p\in U$, $f'(p)$ is not an element of $\Bbb R^2$; it's a linear map from $\Bbb R^2$ into itself. So, there are no primitives here. And therefore, the argument that $f'(z_0)(z-z_0)$ has a primitive fails in this context.

José Carlos Santos
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