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I have a few questions about the proof of the following theorem (outlined in red).

Lemma Cauchy-Goursat

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 $\triangle$.

Cauchy Theorem for convex sets

Let $\Omega\subseteq \mathbb{C}$ open and convex, $ p\in\Omega,\ f$ : $\Omega\rightarrow \mathbb{C}$ continuous, $f\in H(\Omega\backslash \{p\})$ ($f$ analytic on $\Omega \backslash \{p\}$. Then there is a $F\in H(\Omega)$ (F analytic on $\Omega$) with $F'(z)=f(z)$ for all $ z\in\Omega$.

It also folows that $\displaystyle \int\limits_{\gamma}f(z)\mathrm{d}z=0$ for every piecewise continuously differentiable and closed path $\gamma\in\Omega$. (e)

Proof

Fix $ a\in\Omega$ . Because of the convexity of $\Omega$, we have $$ F(z):=\int\limits_{[a,z]}f(\xi)\mathrm{d}\xi\ (z\in\Omega).\ (a) $$

$\color{red}{\text{(1) Why (a) follow from }\Omega \text{ being convex ? }}$

Also, because $\Omega$ is convex, we have every triangle $\triangle(a,\ z_{0},\ z)$ in $\Omega\ (z,\ z_{0}\in\Omega). \ \ \ (b)$

triangle

$\color{red}{\text{(2) Why is (b) true ?}}$

From the Cauchy-Goursat lemma presented above, we have: $$ F(z)-F(z_{0})=\int\limits_{[a,z]}f(\xi)\mathrm{d}\xi+\int_{[z_{0},a]}f(\xi)\mathrm{d}\xi=\int_{[z_{0},z]}f(\xi)\mathrm{d}\xi. \ (c) $$

$\color{red}{\text{(3) How does (c) follow from the Cauchy-Goursat lemma?}}$

For a fixed $ z_{0}\in\Omega$ it follows for $z\in\Omega\backslash \{z_{0}\}$: $$ \frac{F(z)-F(z_{0})}{z-z_{0}}-f(z_{0})=\frac{1}{z-z_{0}}\int\limits_{[z_{0},z]}(f(\xi)-f(z_{0}))\mathrm{d}\xi \ \ (d) $$

$\color{red}{\text{(4) Why is in (d): }\displaystyle f(z_0)=\frac{1}{z-z_{0}}\int\limits_{[z_{0},z]}f(z_{0})\mathrm{d}\xi}$

Let $\varepsilon>0$. Because $f$ is in $z_{0}$ continuous, we have a $\delta>0$ with $|f(\xi)-f(z_{0})|<\varepsilon$, if $|\xi-z_{0}|<\delta$. For $ 0<|z-z_{0}|<\delta$ we have: $$ \left|\frac{F(z)-F(z_{0})}{z-z_{0}}-f(z_{0})\right|<\frac{1}{|z-z_{0}|}L([z_{0},z])\varepsilon=\varepsilon. \ (d) $$ $L[z_0,z]$ is the length from $z_0$ to $z$.

$\color{red}{\text{(5) In (d): This is the ML-inequality, right? Where does the }\frac{1}{|z-z_{0}|} \text{come from?}}$

So $F'(z_{0})=f(z_{0})$. Because $ z_{0}\in\Omega$ was chosen arbitrarily, we have $F$ analytic in $\Omega$.

$\color{red}{\text{(6) But we still have not shown }\\ \displaystyle \int\limits_{\gamma}f(z)\mathrm{d}z=0. \text{ This is appears in the statement of the theorem, the last paragraph (see (e) ). How does this follow?}}$

As you can see, I am utterly lost here.

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2 Answers 2

Here is a start.

  1. Because the domain is convex, for any pair of points in the domain, the line segment connecting them is in the domain, hence the triangle property.
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I shall address $\,(4)\,$ as is the only one that the question is in red. In all the other ones I can't be sure whether you're asking about something below or above the red question...

Anyway: $$\int_{[z_0,z]}f(z_0)d\xi=f(z_0)\int_{[z_0,z]}d\xi=f(z_0)|z_0-z|$$

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The questions in red always refer to the line above in black. –  Chris Jun 6 '12 at 3:18
    
I've added references for the remaining questions. –  Chris Jun 6 '12 at 14:17
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