Zero Simple Staircase Integral Condition for Primitive

Theorem
Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.

Let $\displaystyle \oint_C f \left({z}\right) \rd z = 0$ for all simple closed staircase contours $C$ in $D$.

Then $f$ has a primitive $F: D \to \C$.

Proof
Let $C$ be a closed staircase contour in $D$, not necessarily simple.

If we show that $\displaystyle \oint_C f \left({z}\right) \rd z = 0$, then the result follows from Zero Staircase Integral Condition for Primitive.

The staircase contour $C$ is a concatenation of $C_1, \ldots, C_n$, where the image of each $C_k$ is a line segment parallel with either the real axis or the imaginary axis.

Denote the parameterization of $C$ as $\gamma: \left[{a\,.\,.\,b}\right] \to \C$, where $\left[{a \,.\,.\, b}\right]$ is a closed real interval.

Denote the parameterization of $C_k$ as $\gamma_k: \left[{a_k \,.\,.\, b_k}\right] \to \C$.

Splitting up the Contour
The lemma shows that given a staircase contour $C$, we can assume that for $k \in \left\{ {1, \ldots, n-1} \right\}$, the intersection of the images of $C_k$ and $C_{k+1}$ is equal to their common end point $\gamma_k \left({b_k}\right)$.

This means that in order to intersect itself, $C$ must be a concatenation of at least $4$ directed smooth curves.

Now, we prove the main requirement for Zero Staircase Integral Condition for Primitive, that $\displaystyle \oint_C f \left({z}\right) \ \mathrm dz = 0$.

The proof is by induction over $n \in \N$, the number of directed smooth curves that $C$ is a concatenation of.

Basis for the Induction
For $n = 1$, $C$ can only be a closed staircase contour if $\gamma$ is constant, so:

For $n = 4$, $C$ can only be a closed staircase contour if $C$ is a simple contour.

Then, $\displaystyle \oint_C f \left({z}\right) \rd z = 0$ by the original assumption of this theorem.

Induction Hypothesis
For $N \in \N$, if $C$ is a closed staircase contour that is a concatenation of $n$ directed smooth curves with $n \le N$, then:


 * $\displaystyle \oint_C f \left({z}\right) \rd z = 0$

Induction Step
Suppose that $C$ is a closed staircase contour that is a concatenation of $n+1$ directed smooth curves.

If $C$ is a simple contour, the induction hypothesis is true by the original assumption of this theorem.

Otherwise, define $t_0 = a$, and $t_3 = b$.

Define $t_1 \in \left[{a \,.\,.\, b}\right]$ as the infimum of all $t \in \left[{a\,.\,.\,b}\right]$ for which $\gamma$ intersects itself.

Then, define $t_2 \in \left({t_1 \,.\,.\, b}\right]$ as the infimum of all $t \in \left({t_1 \,.\,.\, b}\right]$ for which $\gamma \left({t}\right) = \gamma \left({t_1}\right)$.

For $k \in \left\{ {1, \ldots, 3}\right\}$, define $\tilde C_k$ as the staircase contour with parameterization $\gamma \restriction{\left[{t_{k-1} \,.\,.\, t_k}\right]}$.

Then $\tilde C_2$ is a closed staircase contour that is a concatenation of at least $4$ directed smooth curves.

Then both $\tilde C_1 \cup \tilde C_3$ and $\tilde C_2$ are a concatenation of less than $n+1$ directed smooth curves, so: