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Let $D$ be a region in $\hat{\mathbb C}$, the Riemann surface, $I=[T_0,T_1]$ be a bounded closed interval in $\mathbb R$ and $\phi: I\rightarrow D$ be a function whose real part and imaginary part are continuous as real functions. I want to show that there exists a partition $T_0=t_0<t_1<\dots<t_n=T_1$ of $I$ such that for every $1\le j\le n$ there exists an open disk $\Delta_j\subset D$ that contains $\phi([t_{j-1},t_j])$, and if $\infty\in\Delta_j$ either $\phi(t_{j-1})$ or $\phi(t_j)$ is $\infty$.

How do you prove this?

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1 Answer 1

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As the real and the imaginary part of $\phi$ are continuous on the compact interval $I$ the point $\infty$ is not part of the problem.

The essential ingredient in the following proof is the compactness of $I$ resp. its image in $D$.

For $t\in I$ put $\rho(t):=\sup\{r>0\ |\ D_r\bigl(\phi(t)\bigr)\subset D\}$. We may assume $\rho(t)<\infty$ for all $t\in I$, or there would be nothing to prove. The function $t\mapsto \rho(t)$ is positive and continuous on $I$. To prove the latter assume that an $\epsilon>0$ is given. Then there is a $\delta>0$ with $|\phi(t)-\phi(t')|<{\epsilon\over 2}$ as soon as $|t-t'|<\delta$. It follows that $\rho(t')\geq \rho(t)-{\epsilon\over 2}$ when $|t-t'|<\delta$, whence by symmetry $|\rho(t')- \rho(t)|\leq{\epsilon\over 2}$ when $|t-t'|<\delta$.

Therefore we may conclude that there is a $\delta>0$ with $\rho(t)>2\delta$ for all $t\in I$. Choose $N$ so large that $|\phi(t)-\phi(t')|<\delta$ when $|t-t'|<h:=(T_1-T_0)/N$, and put $t_j:=T_0+ jh$ $\ (0\leq j\leq N)$. Then the open disks $\Delta_j$ $\ (1\leq j\leq N)$ with center $\phi(t_j)$ and radius $\delta$ are subsets of $D$ and contain the respective arcs $\phi\bigl([t_{j-1},t_j]\bigr)$.

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Before I asked this, I thought something like your $\rho$ should be used to prove this, but I couldn't prove its continuity. How do you do this? –  Pteromys Apr 3 '12 at 0:05
    
@Pteromys: See my edit. –  Christian Blatter Apr 3 '12 at 8:13

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