Does the curve with the following equation holds a line segment?

Question

For a smooth curve $\alpha:I\to \mathbb{R}^3$ with $[a,b]\subset I$,

if $$|\alpha(b)-\alpha(a)| = \int_a^b |\alpha'(t)|dt$$ holds

Does the curve with the equation hold a line segment?

We can show for any smooth curve with point $\alpha(a),\alpha(b)$ fixed we have $|\alpha(b)-\alpha(a)| \le \int_a^b |\alpha'(t)|dt$ ,line segment between these two points are the one with lower bounded is attained.

Answer 1

If you have equality in the following inequality

$$ |\alpha(b) - \alpha(a)| = \left|\int_a^b \alpha '(t)dt\right| \leq \int_a^b |\alpha '(t)| dt $$

then since $\alpha$ is smooth there exists $\theta$ such that for all $t \in [a,b]$,

$$ \alpha'(t) = e^{i\theta}|\alpha'(t)| $$

(see the case of equality in the complex integral triangle inequality).

Therefore

$$ \alpha(b) - \alpha(a) = \int_a^b \alpha'(t)dt = e^{i\theta}\int_a^b |\alpha'(t)|dt $$

and for all $t \in [0,1]$

$$ \begin{aligned} \alpha(a + t(b-a)) &= \alpha(a) + \int_a^{a + t(b-a)} \alpha'(t)dt \\ &= \alpha(a) + e^{i\theta}\int_a^{a + t(b-a)} |\alpha'(t)|dt \\ &= \alpha(a) + \underbrace{\frac{\int_a^{a + t(b-a)} |\alpha'|}{\int_a^b |\alpha'|}}_{= \phi(t) \in [0,1]} (\alpha(b) - \alpha(a)) \end{aligned} $$

so $\alpha([a,b])$ is indeed the segment $ [\alpha(a),\alpha(b)] $ because $\phi : [0,1] \rightarrow [0,1]$ is continuous, non-decreasing and $\phi(0) = 0$, $\phi(1) = 1$.

Does that help you?

Answer 2

Let's use the well known equality

$$\tag 1\int_a^b |\alpha'(t)|\,dt = \sup\sum_{k=1}^{n}|\alpha (t_k)- \alpha (t_{k-1})|,$$

where the supremum is taken over all partitions $\{t_0,\dots, t_n\}$ of $[a,b].$ Suppose for some $t_0\in [a,b]$ that $\alpha(t_0)\notin [\alpha(a),\alpha(b)].$ Since the sum of lengths of two legs in a triangle is always greater than the length of the third leg, we then have

$$|\alpha(b)-\alpha(t_0)|+ |\alpha(t_0)-\alpha(a)|> |\alpha(b)-\alpha(a)|.$$

The hypothesis and $(1)$ then give a contradiction. Therefore $\alpha([a,b])\subset [\alpha (a),\alpha (b)],$ and by connectivity, it must be the whole segment.