Existence of Local Primitive Theorem for Complex Analysis

**tttcomrader** · May 26th 2009, 06:36 AM

Theorem: Let f be a holomorphic function on a disc U, then there exist a function g such that g is the primitive of f and $\int _ {\gamma} f = 0$ for every closed path gamma.

Proof.

Define $g(z_1)= \int ^{z} _{z_0}f \ \ \ \ \ \forall z \in U$

Then define the path from $z_0$ to $z$ by t and from z to $z_0$ by $k^{-1}$ , so that t and k forms a circle.

Then the integral over t and k are equal and the integration is independent of path.

Now, I don't really understand why they have to be equal, and what does "independent of path" means, would someone please explain?

Thank you!

**Jose27** · May 26th 2009, 08:28 AM

Well, when you define functions like g, you have to make sure that they do not depend on the path you chose (think of it like testing for a normal function to be well-defined, ie. that two paths that join the same points do not give two different values).

the integrals over the two paths are the same because the sum of them is a closed curve, and since you have f holomorphic in a simply connected domain, Cauchy's theorem gives that the integral over the sum of paths is zero, hence they only differ in sign, but this is to be expected because you're going in opposite directions.

**ThePerfectHacker** · May 26th 2009, 10:34 AM

Originally Posted by tttcomrader

Theorem: Let f be a holomorphic function on a disc U, then there exist a function g such that g is the primitive of f and $\int _ {\gamma} f = 0$ for every closed path gamma.

Proof.

Define $g(z_1)= \int ^{z} _{z_0}f \ \ \ \ \ \forall z \in U$

Let $z_0$ be the center of the disk. Define $g(z) = \int_{[z_0,z]} f$ i.e. the integral along the line segment from $z_0$ to $z$ (we also set $g(z_0) = 0$ ).

Let $z\in U$ be some point in the disk different from $z_0$ . We know that by Cauchy's theorem (since the curve is a closed curve):
$\int_{[z_0,z]}f + \int_{[z,z+h]}f - \int_{[z_0,z+h]} f = 0 \implies \int_{[z,z+h]} f = g(z+h) - g(z)$

Notice that $hf(z) = \int_{[z,z+h]} f(z)$ .
Thus, $g(z+h) - g(z) - hf(z) = \int_{[z,z+h]} (f - f(z))$

The function $f - f(z)$ is continous on the disk centered at $z$ with radius $h$ so $|f-f(z)| \leq e$ as we shrink $h$ does to zero.
This implies, $\left| \int_{[z,z+h]} (f-f(z)) \right| \leq \int_{[z,z+h]} |f-f(z)| \leq \epsilon|h|$ .

Finally, $\left| \frac{g(z+h) - g(z)}{h} - f(z) \right| \leq \frac{1}{|h|}\cdot |h|\cdot \epsilon = \epsilon$
However we can make $\epsilon$ as small as we like by making $|h|$ sufficiently small.

Henceforth, $\lim_{h\to 0} \frac{g(z+h)-g(z)}{h} = f(z)\implies g'(z) = f(z)$ .

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