Zero Staircase Integral Condition for Primitive
Theorem
Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.
Let $z_0 \in D$.
Suppose that $\ds \oint_C \map f z \rd z = 0$ for all closed staircase contours $C$ in $D$.
Then $f$ has a primitive $F: D \to \C$ defined by:
- $\ds \map F w = \int_{C_w} \map f z \rd z$
where $C_w$ is any staircase contour in $D$ with start point $z_0$ and end point $w$.
Proof
From Connected Domain is Connected by Staircase Contours, it follows that there exists a staircase contour $C_w$ in $D$ with start point $z_0$ and end point $w$.
If $C_w'$ is another staircase contour with the same endpoints as $C_w$, then $C_w' \cup \paren {-C_w}$ is a closed staircase contour.
Then the definition of $F$ is independent of the choice of contour, as:
\(\ds \int_{C_w} \map f z \rd z\) | \(=\) | \(\ds \int_{C_w} \map f z \rd z + \int_{C_w' \cup \paren {-C_w} } \map f z \rd z\) | by assumption | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_{C_w} \map f z \rd z + \int_{C_w'} \map f z \rd z + \int_{-C_w} \map f z \rd z\) | Contour Integral of Concatenation of Contours | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_{C_w} \map f z \rd z + \int_{C_w'} \map f z \rd z - \int_{C_w} \map f z \rd z\) | Contour Integral along Reversed Contour | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_{C_w'} \map f z \rd z\) |
We now show that $F$ is the primitive of $f$.
Let $\epsilon \in \R_{>0}$.
By definition of continuity, there exists $r \in \R_{>0}$ such that the open ball $\map {B_r} w \subseteq D$, and for all $z \in \map {B_r} w$:
- $\size {\map f z - \map f w} < \dfrac \epsilon 2$
Let $h = x+iy \in \C \setminus \set 0$ with $x, y \in \R$ such that $\size h < r$.
Let $\LL$ be the staircase contour that goes in a horizontal line from $w$ to $w + x$, and continues in a vertical line from $w + x$ to $w + h$.
As $w + x, w + h \in \map {B_r} w$, it follows from Open Ball is Convex Set that $\LL$ is a contour in $\map {B_r} w$.
Then $C_w \cup \LL$ is a staircase contour from $z_0$ to $w + h$, so:
\(\ds \map F {w + h} - \map F w\) | \(=\) | \(\ds \int_{C_w \cup \LL} \map f z \rd z - \int_{C_w} \map f z \rd z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_\LL \map f z \rd z\) | Contour Integral of Concatenation of Contours |
From Derivative of Complex Polynomial, it follows that $\dfrac \d {\d z} \map f w z = \map f w$, so:
\(\ds \int_\LL \map f w \rd z\) | \(=\) | \(\ds \map f w \paren {w + h} - \map f w w\) | Fundamental Theorem of Calculus for Contour Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds h \map f w\) |
We can now show that $\map {F'} w = \map f w$, as:
\(\ds \size {\dfrac {\map F {w + h} - \map F w} h - \map f w}\) | \(=\) | \(\ds \size {\dfrac 1 h \int_\LL \map f z \rd z - \dfrac 1 h h \map f w}\) | a priori | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {\dfrac 1 h} \size {\int_\LL \paren {\map f z - \map f w} \rd z}\) | Linear Combination of Contour Integrals | |||||||||||
\(\ds \) | \(<\) | \(\ds \size {\dfrac 1 h} \dfrac \epsilon 2 \map L \LL\) | Estimation Lemma for Contour Integrals, as $z \in \map {B_r} w$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\size x + \size y} {\size h} \dfrac \epsilon 2\) | the lengths of the line segments are $\size x$ and $\size y$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \epsilon\) | Modulus Larger than Real Part and Imaginary Part |
When $h$ tends to $0$, we have $\map {F'} w = \map f w$ by definition of differentiability.
$\blacksquare$
Sources
- 2001: Christian Berg: Kompleks funktionsteori: $\S 2.3$