proof of Goursat’s theorem

We argue by contradiction. Set

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

and suppose that $\eta \ne 0$. 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 $R_{i_1{i}_2},i_1,i_2=1\mathrm{\dots }4$, and then continue ad infinitum to obtain a sequence of nested families of rectangles $R_{i_1\mathrm{\dots }{i}_k}$, with ${\eta }_{i_1\mathrm{\dots }{i}_k}$ 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

$${\eta }_{i_1\mathrm{\dots }{i}_k}={\eta }_{i_1\mathrm{\dots }{i}_{k}1}+{\eta }_{i_1\mathrm{\dots }{i}_{k}2}+{\eta }_{i_1\mathrm{\dots }{i}_{k}3}+{\eta }_{i_1\mathrm{\dots }{i}_{k}4}.$$

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 $|{\eta }_i|,i=1,\mathrm{\dots },4$. By the triangle inequality we have

$$|{\eta }_1|+|{\eta }_2|+|{\eta }_3|+|{\eta }_4|\ge |\eta |,$$

and hence

$$|{\eta }_{j_1}|\ge 1/4|\eta |.$$

Continuing inductively, let $j_{k+1}$ be such that $|{\eta }_{j_1\mathrm{\dots }{j}_k{j}_{k+1}}|$ is the maximum of $|{\eta }_{j_1\mathrm{\dots }{j}_{k}i}|,i=1,\mathrm{\dots },4$. We then have

$$|{\eta }_{j_1\mathrm{\dots }{j}_k{j}_{k+1}}|\ge 4^{-(k+1)}|\eta |.$$ (1)

Now the sequence of nested rectangles $R_{j_1\mathrm{\dots }{j}_k}$ converges to some point $z_0\in R$; more formally

$$\{z_0\}=\bigcap _{k=1}^{\mathrm{\infty }}{R}_{j_1\mathrm{\dots }{j}_k}.$$

The derivative $f^{\prime }(z_0)$ is assumed to exist, and hence for every $ϵ>0$ there exists a $k$ sufficiently large, so that for all $z\in R_{j_1\mathrm{\dots }{j}_k}$ we have

$$|f(z)-f^{\prime }(z_0)(z-z_0)|\le ϵ|z-z_0|.$$

Now we make use of the following.

Lemma 1

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

${\displaystyle {\oint }_{\partial Q}}(az+b)𝑑z=0$

$\left|{\displaystyle {\oint }_{\partial Q}}f(z)\right|\le MP,$

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

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 $ϵ>0$ there exists a $k$ sufficiently large that

$${\eta }_{j_1\mathrm{\dots }{j}_k}=\left|{\oint }_{\partial R_{j_1\mathrm{\dots }{j}_k}}f(z)𝑑z\right|\le ϵ{|\partial R_{j_1\mathrm{\dots }{j}_k}|}^2=4^{-k}{|\partial R|}^2ϵ.$$

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
Canonical name	ProofOfGoursatsTheorem
Date of creation	2013-03-22 12:54:37
Last modified on	2013-03-22 12:54:37
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	13
Author	rmilson (146)
Entry type	Proof
Classification	msc 30E20