[Math] Prove that a straight line is the shortest curve between two points in $R^n$.

differential-geometry

Let $p,q∈R^n$ and let $\gamma$ be a curve such that $\gamma(a) = p, \gamma(b) = q$, where $a$ < $b$.

(a) Show that, if $\mathbf u$ is a unit vector, then
$$\dot\gamma \cdot \mathbf u\leq \|\dot\gamma\|$$

(b) Show that
$$(q – p)\cdot\mathbf u ≤ \int_b^a\|\dot\gamma\|\,dt$$
(c) Show that the arc length of $\gamma$ from $\gamma(a)$ to $\gamma(b)$ is at least $\|q – p\|$, with equality when $\gamma$ is a straight line.

This is what I have worked out so far, for (b): $$(q – p)·\mathbf u = (\gamma(a) – \gamma(b))·\mathbf u$$
$$=\int_b^a\dot\gamma\cdot\mathbf u\,dt$$
and thus, using part (a):
$$\int_b^a\dot\gamma\cdot\mathbf u\,dt\leq \int_b^a\|\dot\gamma\|\,dt$$

and for (c), as $\mathbf u$ is a unit vector we use the equation: $$\mathbf u = \frac{(q – p)}{\|q – p\|}$$
from here you can see that arc length of $\gamma$ from $\gamma(a)$ to $\gamma(b)$ is at least $\|q – p\|$, but I'm not sure what working to show or if I'm even doing this right.

Best Answer

The first parts look good (up to Thomas Andrews' comment). Are you asking if you're done? If so, you should just notice that the length of the line from $\gamma(a)$ to $\gamma(b)$ is exactly $\lVert \gamma(a) - \gamma(b) \rVert$. You've shown that the length of any curve has to be at least this number so in particular, any curve has to be at least as long as the line.