proof of Goursat's theorem
ProofOfGoursatsTheorem
2002-08-02 13:17:18
2003-07-15 22:42:33
Proof
Cauchy integral theorem
0
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We argue by contradiction. Set
$$\eta = \oint_{\partial R} f(z)\, dz,$$
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
$$\eta_i = \oint_{\partial R_i} f(z)\, dz.$$
\begin{center}
\includegraphics[scale=.7]{ProofGoursatTheorem.eps}
{\tiny Figure 1: subdivision of the rectangle contour.}
\end{center}
Now subdivide each of the four sub-rectangles, to get 16 congruent
sub-sub-rectangles $R_{i_1i_2},\; i_1,i_2=1\ldots 4$, 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
$$\eta_{i_1\ldots i_k} = \eta_{i_1\ldots i_k1}+\eta_{i_1\ldots i_k2}+\eta_{i_1\ldots i_k3}+\eta_{i_1\ldots i_k4}.$$
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 $\vert\eta_{j_1}\vert$ is the maximum of
$\vert \eta_i \vert,\, i=1,\ldots,4$. By the triangle inequality we have
$$\vert\eta_1\vert+\vert\eta_2\vert+\vert\eta_3\vert+\vert\eta_4\vert \geq \vert\eta\vert,$$
and hence
$$\vert\eta_{j_1}\vert\geq 1/4 \vert\eta\vert.$$
Continuing inductively, let $j_{k+1}$ be such that
$\vert\eta_{j_1\ldots j_k j_{k+1}}\vert$ is the maximum of
$\vert\eta_{j_1 \ldots j_k i}\vert,\, i=1,\ldots,4$. We then have
\begin{equation}
\label{eq:etaest}
\vert\eta_{j_1\ldots j_k j_{k+1}}\vert \geq 4^{-(k+1)} |\eta|.
\end{equation}
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
$f'(z_0)$ 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
$$\vert f(z)-f'(z_0)(z-z_0) \vert \leq \epsilon \vert z-z_0\vert.$$
Now we make use of the following.
\begin{lemma}
Let $Q\subset\cnums$ be a rectangle, let $a,b\in\cnums$, and let
$f(z)$ be a continuous, complex valued function defined and bounded
in a domain containing $Q$. Then,
\begin{eqnarray*}
&&\oint_{\partial Q} (az+b) dz = 0 \\
&&\left\vert \oint_{\partial Q} \!\!\!f(z) \right \vert \leq MP,
\end{eqnarray*}
where $M$ is an upper bound for $|f(z)|$ and where $P$ is the length
of $\partial Q$.
\end{lemma}
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
$$\eta_{j_1\ldots j_k} = \left\vert \oint_{\partial R_{j_1\ldots
j_k}}\hskip -2em f(z)\, dz \right\vert \leq \epsilon
\vert\partial R_{j_1\ldots j_k}\vert^2 = 4^{-k}\, \vert
\partial R\vert^2 \epsilon.$$
where $\vert\partial R\vert$ denotes the
length of perimeter of the rectangle $R$. This contradicts the
earlier estimate \eqref{eq:etaest}. Therefore
$\eta=0$.