# proof of Goursat’s theorem

 $\eta=\oint_{\partial R}f(z)\,dz,$

and suppose that $\eta\neq 0$. Divide $R$ into four congruent rectangles $R_{1},R_{2},R_{3},R_{4}$ (see Figure 1), and set

 $\eta_{i}=\oint_{\partial R_{i}}f(z)\,dz.$

Figure 1: subdivision of the rectangle contour.

Now subdivide each of the four sub-rectangles, to get 16 congruent sub-sub-rectangles $R_{i_{1}i_{2}},\;i_{1},i_{2}=1\ldots 4$, and then continue ad infinitum to obtain a sequence of nested families of rectangles $R_{i_{1}\ldots i_{k}}$, with $\eta_{i_{1}\ldots i_{k}}$ the values of $f(z)$ integrated along the corresponding contour.

Orienting the boundary of $R$ and all the sub-rectangles in the usual counter-clockwise fashion we have

 $\eta=\eta_{1}+\eta_{2}+\eta_{3}+\eta_{4},$

and more generally

 $\eta_{i_{1}\ldots i_{k}}=\eta_{i_{1}\ldots i_{k}1}+\eta_{i_{1}\ldots i_{k}2}+% \eta_{i_{1}\ldots i_{k}3}+\eta_{i_{1}\ldots i_{k}4}.$

In as much as the integrals along oppositely oriented line segments cancel, the contributions from the interior segments cancel, and that is why the right-hand side reduces to the integrals along the segments at the boundary of the composite rectangle.

Let $j_{1}\in\{1,2,3,4\}$ be such that $|\eta_{j_{1}}|$ is the maximum of $|\eta_{i}|,\,i=1,\ldots,4$. By the triangle inequality we have

 $|\eta_{1}|+|\eta_{2}|+|\eta_{3}|+|\eta_{4}|\geq|\eta|,$

and hence

 $|\eta_{j_{1}}|\geq 1/4|\eta|.$

Continuing inductively, let $j_{k+1}$ be such that $|\eta_{j_{1}\ldots j_{k}j_{k+1}}|$ is the maximum of $|\eta_{j_{1}\ldots j_{k}i}|,\,i=1,\ldots,4$. We then have

 $|\eta_{j_{1}\ldots j_{k}j_{k+1}}|\geq 4^{-(k+1)}|\eta|.$ (1)

Now the sequence of nested rectangles $R_{j_{1}\ldots j_{k}}$ converges to some point $z_{0}\in R$; more formally

 $\{z_{0}\}=\bigcap_{k=1}^{\infty}R_{j_{1}\ldots j_{k}}.$

The derivative $f^{\prime}(z_{0})$ is assumed to exist, and hence for every $\epsilon>0$ there exists a $k$ sufficiently large, so that for all $z\in R_{j_{1}\ldots j_{k}}$ we have

 $|f(z)-f^{\prime}(z_{0})(z-z_{0})|\leq\epsilon|z-z_{0}|.$

Now we make use of the following.

###### Lemma 1

Let $Q\subset\mathbb{C}$ be a rectangle, let $a,b\in\mathbb{C}$, and let $f(z)$ be a continuous, complex valued function defined and bounded in a domain containing $Q$. Then,

 $\displaystyle\oint_{\partial Q}(az+b)dz=0$ $\displaystyle\left|\oint_{\partial Q}\!\!\!f(z)\right|\leq MP,$

where $M$ is an upper bound for $|f(z)|$ and where $P$ is the length of $\partial Q$.

The first of these assertions follows by the Fundamental Theorem of Calculus; after all the function $az+b$ has an anti-derivative. The second assertion follows from the fact that the absolute value of an integral is smaller than the integral of the absolute value of the integrand — a standard result in integration theory.

Using the Lemma and the fact that the perimeter of a rectangle is greater than its diameter we infer that for every $\epsilon>0$ there exists a $k$ sufficiently large that

 $\eta_{j_{1}\ldots j_{k}}=\left|\oint_{\partial R_{j_{1}\ldots j_{k}}}\hskip-20% .0ptf(z)\,dz\right|\leq\epsilon|\partial R_{j_{1}\ldots j_{k}}|^{2}=4^{-k}\,|% \partial R|^{2}\epsilon.$

where $|\partial R|$ denotes the length of perimeter of the rectangle $R$. This contradicts the earlier estimate (1). Therefore $\eta=0$.

Title proof of Goursat’s theorem ProofOfGoursatsTheorem 2013-03-22 12:54:37 2013-03-22 12:54:37 rmilson (146) rmilson (146) 13 rmilson (146) Proof msc 30E20