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> [!proposition] Proposition. ([[straight lines as shortest]])
> Let $I \subset \mathbb{R}$ be an [[open interval]] and let $\alpha:I \to \mathbb{R}^{n}$ be a smooth [[parameterized curve]]. Let $[a,b] \subset I$. We have $\|\alpha(b) - \alpha(b)\|_{2}\leq \int_{a}^{b} \|\alpha'(t)\|_{2} \ dt, $
> that is, the curve of shortest length from $\alpha(a)$ to $\alpha(b)$ is the straight line joining these points (see [[arc length of a path|arc length]]).
^b767d3
> [!proof]- Proof. ([[straight lines as shortest]])
> ~
>
**a.** $\alpha(a):=p, \alpha(b):=q$. We first show that, for any unit-length constant vector $v \in \mathbb{R}^{n}$, $(q-p)\cdot v = \int_{a}^{b} \big( \alpha'(t) \cdot v \big)\, dt \leq \int_{a}^{b} \|\alpha'(t)\|_{2}\, dt. $
Indeed, for the first equality, let $\alpha(t)=(\alpha_{1}(t), \dots, \alpha_{n}(t)), v=(v_{1},\dots,v_{n})$ and observe via [[linearity of the integral]] and the [[fundamental theorem of calculus]] that $\begin{align}
\int_{a}^{b} \big( \alpha'(t) \cdot v \big) \, dt = & \int_{a}^{b} v_{i} \sum_{i=1}^{n} \alpha_{i}'(t) \\
= & \sum_{i=1}^{n} v_{i} \int_{a}^{b}\, \alpha_{i}'(t) \\
= & \sum_{i=1}^{n} v_{i} (\alpha_{i}(b)- \alpha_{i}(a)) \\
= & v \cdot (q-p).
\end{align}$
For the second inequality, we apply the [[triangle inequality for integrals]] and [[Cauchy-Schwarz inequality|Cauchy-Schwarz]] + [[comparison property of the integral]] thus: $\begin{align}
| \int_{a}^{b} \big( \alpha'(t) \cdot v \big) \, dt | \leq & \int_{a}^{b} | \alpha'(t) \cdot v |\ , dt \\
\leq & \int_{a}^{b} \|\alpha'(t)\|_{2} \ \cancel{\|v\|_{2}}^{1}\, dt
\end{align}$
**b.** Next, set $v:=\frac{q-p}{\|q-p\|_{2}}$. This is a constant of unit length, hence we apply **a** thus: $\begin{align} \|\alpha(b) - \alpha(a)\|_{2} = \|q-p\|_{2} =& \frac{\overbrace{(q-p) \cdot \textcolor{Skyblue}{(q-p)}}^{\|q-p\|_{2}^{2}}{}{}}{\textcolor{Skyblue}{\|q-p\|_{2}}}\\ = &
(q-p) \cdot \textcolor{Skyblue}{v} \\
\leq & \int_{a}^{b} \|\alpha'(t)\|_{2}\,dt.
\end{align}$
^8a7dbb
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#### References
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