POTW Show That the Curve is Straight

[Euge]

Fix points $p,q\in \mathbb{R}^n$, and let $\gamma : [a,b] \to \mathbb{R}^n$ be a continuously differentiable curve from $p$ to $q$ whose arclength equals the Euclidean distance between the points, $|q - p|$. Prove that $\gamma$ lies on the straight line passing through $p$ and $q$.

 

[anuttarasammyak]

It’s just a question on reading the problem, what do a and b mean?

 

[pasmith]

So the starting point is <br /> \|\gamma(b) - \gamma(a)\| = \int_a^b \|\gamma&#039;(t)\|\,dt if \gamma(a) = p and \gamma(b) = q, and we want to conclude that <br /> \gamma(t) = p + \gamma&#039;(a)G(t) for some G: [a,b] \to \mathbb{R} with \gamma&#039;(a) = K(q - p) for some K \in \mathbb{R}, G(a) = 0, and G(b) = K^{-1}.

I think that without loss of generality we can take a = 0 and b = 1.

 

[pasmith]

Spoiler: Proof by contradiction

Suppose there exists t_0 \in (0,1) such that r= \gamma(t_0) is not on the straight line from p to q, Then \begin{split}<br /> \int_a^b \|\gamma&#039;(t)\|\,dt &amp;= \int_a^{t_0} \|\gamma&#039;(t)\|\,dt + \int_{t_0}^b \|\gamma&#039;(t)\|\,dt \\<br /> &amp;\geq \|r - p\| + \|q - r\| \\<br /> &amp;&gt; \|q - p\| \end{split} since the straight line path gives the shortest possible path between p and q. But this a contradiction, so each point of \gamma lies on the straight line.

 

[Office_Shredder]

since the straight line path gives the shortest possible path between p and q

I believe the actual point of the challenge is to prove this is true.