User:Anghel/Sandbox

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

Suppose that $\displaystyle \oint_C f \left({z}\right) \ \mathrm dz = 0$ 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 $\displaystyle \oint_C f \left({z}\right) \ \mathrm dz = 0$, 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 $\gamma_k: \left[{a_k \,.\,.\, b_k}\right] \to \C$.

Avoiding Immediate Overlap
First, we show that there exists a contour $C'$ such that $\displaystyle \oint_C f \left({z}\right) \ \mathrm dz = \displaystyle \oint_{C'} f \left({z}\right)  \ \mathrm dz$.

This contour $C'$ has the property that for all $k \in \left\{ {1, \ldots, n-1} \right\}$, the intersection of the images of $C_k$ and $C_{k+1}$ is equal to their common endpoint $\gamma_k \left({b_k}\right)$.

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
As induction basis, if $C$ consists of just one directed smooth curve, we can simply put $C' = C$.

Then, we have to check a statement for all $k \in \emptyset$, so it is trivially true.

Induction Hypothesis
As induction hypothesis, suppose that if $C$ is a concatenation of $n$ directed smooth curves, we can find a contour $C'$ with the properties above.

Induction Step
In the induction step, let $C$ be a concatenation of $n+1$ directed smooth curves.

Suppose that there exists $k \in \left\{ {1, \ldots, n}\right\}$ such that the intersection of the images of $C_k$ and $C_{k+1}$ contain more elements than their common endpoint $\gamma_k \left({b_k}\right)$, otherwise we can put $C' = C$.

Then, one of the two line segments must contain the other line segment, so:


 * $\operatorname{Im} \left({C_k}\right) \cap \operatorname{Im} \left({C_{k+1} }\right) = \gamma_k \left({ \left[{a'\,.\,.\,b_k }\right] }\right) \cup \gamma_{k+1} \left({ \left[{a_{k+1} \,.\,.\,b' }\right] }\right)$

where either $a' = a_k$, or $b' = b_{k+1}$.

Define $C_k'$ as the contour that is parameterized as a line segment that goes from $\gamma_k \left({a_k}\right)$ to $\gamma_k \left({a'}\right)$, and then to $\gamma_{k+1} \left({b_{k+1} }\right)$.

Then, the image of $C_k'$ is equal to the part of the line segments that does not overlap.

Define $\tilde C_k$ as the contour with the parameterization $\gamma_k \restriction_{ \left[{a'\,.\,.\,b_k }\right] }$, and $\tilde C_{k+1}$ as the contour with the parameterization $\gamma_{k+1} \restriction_{ \left[{a_{k+1} \,.\,.\,b' }\right] }$.

Here, $\gamma_k \restriction_{ \left[{a'\,.\,.\,b_k }\right] }$ denotes the restriction of $\gamma_k$ to $\left[{a'\,.\,.\,b_k }\right]$.

Now $\tilde C_k$ is the reversed contour of $\tilde C_{k+1}$, so:

which shows that the contour integal of $f$ along $C$ is equal to the the integral of $f$ along $C'' = C_1 \cup \ldots \cup C_{k-1} \cup \tilde C_k \cup C_{k+2} \ldots \cup C_{n+1}$.

This new contour $C''$ is a staircase contour which is a concatenation of $n$ directed smooth curves, so by the induction hypothesis, there exists a contour $C'$ such that:


 * $\displaystyle \oint_{C'} f \left({z}\right) \ \mathrm dz = \oint_{C''} f \left({z}\right)  \ \mathrm dz$

where $C'$ has the desired property.

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 endpoint $\gamma_k \left({b_k}\right)$.

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, $\displaystyle \oint_{C} f \left({z}\right) \ \mathrm dz = 0$ 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:


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

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: