A path $\gamma:[a,b]\to \mathbb C$ which is not rectifiable.

Question

In Conway's Functions of one complex variables (part 1) book , the definition of a path is written as : A continuous function $\gamma:[a,b]\to \mathbb C$ where $a,b \in \mathbb R, a<b$ is called a path and it is rectifiable if $\gamma $ is of bounded variation , i.e., $\gamma $ has finite length.

Now my question is : As $\gamma$ is continuous , $\gamma[a,b]$ is compact , hence closed and bounded so any path must be of finite length. But why there is another definition of rectifiable path.

Answer 1

Simply because it's not enough for a function to bounded & continuous to have that sum be finite, it must be of bounded variation. Recall that you are not changing the interval $[a,b]$ in the definition of $L_{[a,b],\gamma} ( \mathcal{P})$, it is the partition $\mathcal{P}$ i.e even though $\gamma$ is bounded, continuous, the variational sum i.e,

$$L_{[a,b],\gamma} ( \mathcal{P})=\sum_{\mathcal{P}} \|\gamma(t_{j-1})- \gamma(t_j)\|$$

may blow up as $\|\mathcal{P}^*\| \to 0$, where $\mathcal{P}^*$ is the supremum of $\|\gamma(t_{j-1})-\gamma_(t_j)\|$ (making the partition larger). For example, if we take any continuous curve $\gamma$ connecting the points,

$$ \left\{\left(\frac{1}{2n},\frac{1}{n}\right), \left(\frac{1}{2n+1},0\right) \right\}$$

then its variational sum diverges since $\sum 1/n \to \infty$. By definition the original function we defined is bounded, cts, but not rectifiable.

Answer 2

Consider a sequence of squares:

where each square has its side half as long as the previous one.

Let $\gamma : [0, 1] \to \mathbb{C}$ be defined as follows: $\gamma(0)$ is the lower left corner. Then:

• let $\gamma \upharpoonright \left[ 0, \frac{1}{2} \right]$ traverse the biggest square once at constant speed;
• let $\gamma \upharpoonright \left[ \frac{1}{2}, \frac{3}{4} \right]$ traverse the second biggest square twice at constant speed;
• let $\gamma \upharpoonright \left[ \frac{3}{4}, \frac{7}{8} \right]$ traverse the third biggest square four times at constant speed;
• ...and so on.

Then $\gamma(t)$ is clearly continuous except at $t = 1$, where it is also continuous, since the square side tends to zero. But on each interval described above $\gamma$ traverses the path of length equal to the circumference of the biggest square, and this happens infinitely many times, therefore it's length is infinite.