Show that if $f : G \to \mathbb{C}$ is analytic and $\gamma$ is a rectifiable curve in $G$ then $f \circ \gamma$ is also a recifiable curve.

Question

Show that if $f : G \to \mathbb{C}$ is analytic and $\gamma$ is a rectifiable curve in $G$, then $f \circ \gamma$ is also a rectifiable curve.

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How can I solve this problem?

Answer 1

It depends on your definition of rectifiable. Presumably $G$ is an open set.

If rectifiable implies continuity, then $\Gamma = \gamma([a,b]) \subset G$ is compact, and since $f$ is analytic, then so is $f'$, which is bounded on $\Gamma$ by some $L$. If $p_0 = a, p_1,...,p_n =b$ is a sufficiently small (so that each pair $p_i,p_{i+1}$ is contained in some open ball contained in $G$) partition of $[a,b]$, then $\sum |f(\gamma(p_{i+1}))-f(\gamma(p_i)))| \le L \sum |\gamma(p_{i+1})-\gamma(p_i))| \le L l(\gamma)$. It follows that $f \circ \gamma$ is rectifiable and $l(f \circ \gamma) \le L\ l(\gamma)$.

Addendum: The above argument needs a little refinement, thanks to LEY for point it out. For $z \in G$ let $r_z>0$ be such that $\overline{B}(z,r_z) \subset G$ and consider the open cover $B(z,{1 \over 2}r_z)$. Since $\Gamma$ is compact there is a finite open cover $B(z_k,{1 \over 2}r_{z_k})$. The set $\cup_k \overline{B}(z,r_z) \subset G$ is compact, so $|f'|$ is bounded by $L$. Choose the partition such that $|\gamma(p_{i+1})-\gamma(p_i))|< {1 \over 2}\min_k r_{z_k}$. It follows that any two successive points lie in the some $B(z_k,r_{z_k})$ (which is convex) and so the mean value theorem applies.

If rectifiable does not imply continuous, then it is not true. Take $G = B(0,1)$, $f(z) = \frac{1}{1-z}$, and $\gamma(t) = \begin{cases} t & t \in [0,1) \\ 0 & t = 1 \end{cases}$. An easy computation shows that $l(\gamma) = 2$ (hence rectifiable). Since $(f \circ \gamma)([0,1])$ is unbounded, it is clear that $f \circ \gamma$ is not rectifiable.

Answer 2

Using the anlogue of the Fundamental Theorem of Caculus for line integrals, for any partition $P$: \begin{align}\sum_i|f(\gamma(p_{i+1}))-f(\gamma(p_{i}))|&=\sum_i|\int_{\gamma_i}f'\mathrm d\gamma|\\&\leq \sum_{i}\int_{\gamma_i}|f'||\mathrm d\gamma|\\&\leq\sum_i\sup_{\{\gamma_i\}}|f'|\mathrm V(\gamma_i)\\&\leq\sup_{\{\gamma\}}|f'|\mathrm V(\gamma)\end{align} where $\gamma_i$ stands for the restriction of $\gamma$ on $[p_i,p_{i+1}]$, and $V(\gamma)$ the total variation of $\gamma$. Hence $$\mathrm V(f\circ\gamma)\leq \sup_{\{\gamma\}}|f'|\mathrm V(\gamma)$$

Answer 3

If on the image of $\gamma$, $f'$ is bounded, or $f$ is locally Lipschitz, you are done. Analyticity is too strong a condition.