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The Cauchy-Goursat theorem for a triangle contour states the following:

Let $\triangle=\triangle(a,b,c)$ be a triangle in an open set $\Omega \subseteq \mathbb{C},p\in \Omega,f:\Omega\rightarrow \mathbb{C}$ continuous and f analytical on $\Omega \setminus \{p\}$. Then:

$\int\limits_{\partial\triangle}f(z)dz=0$ , where $\partial\triangle$ means on and inside the triangle $\triangle$.

The proof then distinguishes a few cases around $p$... but why?:

1) $p \notin \triangle$

2) $p=a$

3) $p \in \triangle$ randomly.

I do understand the proof, but I don't understand what's the purposes of $p$. Why do we even mention it? What's the thought behind this? Some books don't mention it.

Edit:

Proof

Case 1: $ p\not\subset\triangle$. Let $a',\ b',\ c'$ be the midpoints of the segments $[b,\ c],\ [c,\ a],\ [a,\ b]$.

enter image description here

We denote with $\triangle^{j}(j=1,2,3,4)$ the triangles $(a,\ c',\ b'),\ (b,\ a',\ c'),\ (c,\ b',\ a'),\ (a',\ b',\ c')$. Let $I =\displaystyle \int_{\partial\triangle}f(z)dz$. We now have: $$ I\ =\sum_{j=1}^{4}\int_{\partial\triangle^{j}}f(z)\mathrm{d}z $$ There is a $j_{0}\in\{1,2,3,4\}$ so that $$ \left|\int_{\partial\triangle^{j_{0}}}f(z)\mathrm{d}z\right|\geq\frac{|I|}{4} $$ Let $\triangle_{1}:=\triangle^{j_{0}}$.

The same construction with $\triangle_{1}$ instead of $\triangle$ yields a triangle $\triangle_{2}$, etc. By induction we get a sequence $(\triangle_{n})$ of triangles with $\triangle\supseteq\triangle_{1}\supseteq\triangle_{2}\supseteq\ldots$, so that

$|I| \displaystyle \leq 4^{n}\left|\int_{\partial\triangle_{n}}f(z)\mathrm{d}z\right|,\ (n\in \mathbb{N})$

Further we have the length $L(\partial\triangle_{n})=2^{-n}L(\partial\triangle)$ and $\displaystyle \lim_{n\rightarrow\infty}$ diam $(\triangle_{n})=0$ because diam $(\triangle_{n})\leq L(\partial\triangle_{n})$. It follows that (Cantor) $$ \triangle \bigcap_{n=1}\triangle_{n}=\{z_{0}\}. $$ Because $f$ is in $z_{0}$ complex differentiable, there exists for $\varepsilon>0$ a $r>0$ with $$ |f(z)-f(z_{0})-f'(z_{0})(z-z_{0})|\leq \varepsilon|z-z_{0}|\ (z\in C(z_{0},\ r)\subseteq\Omega) $$ and there is a $n\in \mathbb{N}$ mit $\triangle_{n}\subseteq C(z_{0},\ r)$. For this $n$ we have: $$ |z-z_{0}|\leq 2^{-n}L(\partial\triangle)\ (z\in\triangle_{n}) $$ Furthermore: $$ \int_{\partial\triangle_{n}}f(z)\mathrm{d}z=\int_{\partial\triangle_{n}}f(z)-f(z_{0})-f'(z_{0})(z-z_{0})\mathrm{d}z $$

We have: $$ \left|\int_{\partial\triangle_{n}}f(z)\mathrm{d}z\right|\leq \varepsilon\cdot(2^{-n}L(\partial\triangle))^{2} $$ and thus $$ |I|\leq 4^{n}\left|\int_{\partial\triangle_{n}}f(z)\mathrm{d}z\right|\leq \varepsilon(L(\partial\triangle))^{2} $$ Because $\varepsilon>0$ was chosen arbitrarly, $I =0$ follows.

$\color{red}{\text{In my point of view, we are now completely done with the proof. But the proof continues, see below}}$

Case 2:

Now, let $p$ be a vertex of $\triangle$, for instance let $p=a$. If $a,\ b,\ c$ lie on the same line, then we have trivially $I=0$. If not, we choose $x\in[a,\ b],y\in[a,\ c]$ near $a$.

enter image description here

Then we have: $$ \displaystyle \int_{\partial\triangle}f(z)\mathrm{d}z=\int_{\partial\triangle(a,x,y)}f(z)\mathrm{d}z+\int_{\partial\triangle(x,b,y)}f(z)\mathrm{d}z+\int_{\partial\triangle(b,c,y)}f(z)\mathrm{d}z $$ $$ =\int_{\partial\triangle(a,x,y)}f(z)dz. $$ Because we can make $L(\partial\triangle(a,x,\ y))$ arbitrarily small, $I =0$ follows.

Case 3: If $ p\in\triangle$ arbitrarily, $I =0$ follows from case 2:

$$ \displaystyle I\ =\int_{\partial\triangle(a,b,p)}f(z)\mathrm{d}z+\int_{\partial\triangle(b,c,p)}f(z)\mathrm{d}z+\int_{\partial\triangle(c,a,p)}f(z)\mathrm{d}z=0 $$

enter image description here

$\color{red}{\text{Why the hassle of introducing } p?}$

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These are not the usual assumptions in Goursat's theorem. I don't think I've ever seen it formulated like that. Normally you'd require that $f$ is analytic (in the sense that $f'$ exists everywhere) on $\Delta$. Continuity is then implied. Without knowing where your version comes from, I can't really guess why the author fomulate the theorem like this? Perhaps they want a quick proof of Riemann's theorem of removable singuliarities without first developing power series expansions of holomorphic functions? –  mrf Jun 5 '12 at 22:32
    
It's from my complex analysis course at my university. I will soon provide the complete proof. Maybe this will shed some light on the problem. –  Chris Jun 5 '12 at 22:37
    
haha okay yeah the point is basically $f$'s gonna be analytic at $p$ too - if it weren't, it'd blow up to $\infty$ at least as fast as $\frac{1}{z}$, so not continuous –  uncookedfalcon Jun 5 '12 at 23:06
    
I've added the proof! –  Chris Jun 6 '12 at 1:04

1 Answer 1

The introduction of $p$ weakens the hypotheses of the theorem: $f$ is not assumed to be differentiable at $p$. In effect, along with the usual Goursat's theorem (integral of complex differentiable function over a closed contour is zero) we are dealing with the removability of a point for continuous holomorphic functions. (A set $E$ is removable for continuous holomorphic functions in $\Omega$ if every continuous function on $\Omega$ that is holomorphic on $\Omega\setminus E$ is actually holomorphic in $\Omega$. See Which sets are removable for holomorphic functions?)

I find that combining these two issues within the same theorem makes its meaning less clear. You can restore some clarity by dividing it into two theorems. The case $p\notin \triangle$ does not involve $p$ at all: this is a standard theorem you normally find in a textbook. Parts 2 and 3 introduces the aforementioned aspect of removability: we obtain the same conclusion without assuming differentiability at a point $p$.

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