Here is Prob. 18, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Let $\gamma_1, \gamma_2, \gamma_3$ be curves in the complex plane, defined on $[0, 2 \pi ]$ by $$ \gamma_1 (t) = e^{\iota t}, \qquad \gamma_2(t) = e^{2 \iota t}, \qquad e^{2\pi \iota t \sin (1/t) }. $$ Show that these three curves have the same range, that $\gamma_1$ and $\gamma_2$ are rectifiable, that the length of $\gamma_1$ is $2 \pi$, that the length of $\gamma_2$ is $4 \pi$, and that $\gamma_3$ is not rectifiable.

Here is Definition 6.26 in Baby Rudin, 3rd edition:

A continuous mappaing $\gamma$ of an interval $[a, b]$ into $\mathbb{R}^k$ is called a curve in $\mathbb{R}^k$. To emphasize the parameter interval $[a, b]$, we may also say that $\gamma$ is a curve on $[a, b]$.

If $\gamma$ is one-to-one, $\gamma$ is called an arc.

If $\gamma(a) = \gamma(b)$, $\gamma$ is said to be a closed curve.

It should be noted that we define a curve to be a mapping, not a point set. Of course, with each curve $\gamma$ in $\mathbb{R}^k$ there is associated a subset of $\mathbb{R}^k$, namely the range of $\gamma$, but different curves may have the same range.

We associate to each partition $P = \left\{ \ x_0, \ldots, x_n \ \right\}$ of $[a, b]$ and to each curve $\gamma$ on $[a, b]$ the number $$ \Lambda (P, \gamma) = \sum_{i=1}^n \left\lvert \gamma \left( x_i \right) - \gamma \left( x_{i-1} \right) \right\rvert. $$ The $i$th term in this sum is the distance (in $\mathbb{R}^k$) between the points $\gamma \left( x_{i-1} \right)$ and $\gamma \left( x_i \right)$. Hence $\Lambda (P, \gamma)$ is the length of a polygonal path with vertices at $\gamma \left(x_0\right), \gamma \left( x_1 \right), \ldots, \gamma \left( x_n \right)$, in this order. As our partition becomes finer and finer, this polygon approaches the range of $\gamma$ more and more closely. This makes it seem reasonable to define the length of $\gamma$ as $$ \Lambda (\gamma) = \sup \Lambda (P, \gamma), $$ where the supremum is taken over all partitions of $[a, b]$.

If $\Lambda (\gamma) < \infty$, we say that $\gamma$ is rectifiable.

In certain cases, $\Lambda(\gamma)$ is given by a Riemann integral. We shall prove this for continuously differentiable curves, i.e., for curves $\gamma$ whose derivative $\gamma^\prime$ is continuous.

Here is Theorem 6.27 in Baby Rudin, 3rd edition:

If $\gamma^\prime$ is continuous on $[a, b]$, then $\gamma$ is rectifiable, and $$ \Lambda ( \gamma) = \int_a^b \left\lvert \gamma^\prime(t) \right\rvert \ \mathrm{d} t. $$

My first question is, how to define $\gamma_3(0)$? The function $\sin (1/t)$ is not defined at $t=0$.

My Attempt:

We note that $$ \begin{align} \mathrm{Range} \ \gamma_1 &= \left\{ \ e^{\iota t} \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ \cos t + \iota \sin t \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ \left( \cos t, \sin t \right) \in \mathbb{R}^2 \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ z \in \mathbb{C} \ \colon \ \lvert z \rvert = 1 \ \right\}. \end{align} $$ Similarly, $$ \begin{align} \mathrm{Range} \ \gamma_2 &= \left\{ \ e^{2 \iota t} \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ \cos 2t + \iota \sin 2t \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ \left( \cos 2t, \sin 2t \right) \in \mathbb{R}^2 \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ z \in \mathbb{C} \ \colon \ \lvert z \rvert = 1 \ \right\}. \end{align} $$

What about the range of $\gamma_3$?

As $\gamma_1^\prime(t) = \iota e^{\iota t}$ and $\gamma_2^\prime (t) = 2 \iota e^{2 \iota t}$ on $[0, 2 \pi ]$, and as both of $\gamma_1^\prime$ and $\gamma_2^\prime$ are continuous on $[a, b]$, so $\gamma_1$ and $\gamma_2$ both are rectifiable, by Theorem 6.27 in Rudin, and by the same theorem we obtain $$ \begin{align} \Lambda \left( \gamma_1 \right) &= \int_0^{2\pi} \left\lvert \gamma_1^\prime(t) \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} \left\lvert \iota e^{\iota t} \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} \lvert \iota \rvert \left\lvert e^{\iota t} \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} 1 \ \mathrm{d} t \\ &= 2 \pi, \end{align} $$ and $$ \begin{align} \Lambda \left( \gamma_2 \right) &= \int_0^{2\pi} \left\lvert \gamma_2^\prime(t) \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} \left\lvert 2\iota e^{2 \iota t} \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} 2 \lvert \iota \rvert \left\lvert e^{2\iota t} \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} 2 \ \mathrm{d} t \\ &= 4 \pi. \end{align} $$

Is there any mistake in what I've done so far? If not, then how to handle $\gamma_3$?


You need to look at the graph of $t\sin(1/t)$. It tends to zero as $t\to0$. So take $\gamma_3(0)=\exp(0)=1$. It will have a maximum $M$ and minimum $m$ on $[0,2\pi]$. If you can show $M-m\ge1$ then $\exp(2\pi i t\sin(1/t))$ will cover the whole unit circle. To prove $M-m\ge1$ all you need are values $a$ and $b$ with $f(a)-f(b)\ge1$.

To show the path is not rectifiable, all you need is $\int|\gamma_3'|=\infty$.

Everything else is fine so far.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.