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Let $\gamma:[a,b] \to \mathbb{R}^2$ be a smooth path, and denote by $\| \cdot\|$ the standard Euclidean norm.

It is easy to see "geometrically" that $ \| \gamma(a)-\gamma(b)\| \le \int_a^b \| \gamma '\|$.

Indeed, $\int_a^b \| \gamma '\|=L(\gamma)$ and straight lines are the shortest paths between points; Since Euclidean distance is intrinsic\Riemannian it follows that

$$ \| \gamma(a)-\gamma(b)\| =d(\gamma(a),\gamma(b)) \le L(\gamma)=\int_a^b \| \gamma '\|.$$

However, we used here the quite strong fact that $d(\gamma(a),\gamma(b)) \le L(\alpha)$ for any path $\alpha$ which connects $\gamma(a),\gamma(b)$.

We only used it for the speciic path $\alpha=\gamma$ though.

Is there a way to prove more directly (without using the fact straight lines are length minimizers) $\| \gamma(a)-\gamma(b)\| \le \int_a^b \| \gamma '\|$ ?

I was thinking on a more calculs-ish style proof.

However, the naive attempted proof amounts to:

$$ \| \gamma(a)-\gamma(b)\|=\sqrt{(\gamma_1(a)-\gamma_1(b))^2+(\gamma_2(a)-\gamma_2(b))^2} $$

$$= \sqrt{\big(\int_a^b \gamma_1'\big)^2+\big(\int_a^b \gamma_2'\big)^2} \stackrel{?}{\le} \int_a^b \sqrt{\gamma_1'^2+\gamma_2'^2}=\int_a^b \| \gamma '\|$$

I don't see any reason why inequality $(?)$ should hold.

2 Answers 2

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Use $$\eqalign{\|\gamma(b)-\gamma(a)\|^2&=\bigl(\gamma(b)-\gamma(a)\bigr)\cdot\bigl(\gamma(b)-\gamma(a)\bigr)=\bigl(\gamma(b)-\gamma(a)\bigr)\cdot\int_a^b\gamma'(t)\>dt\cr&=\int_a^b \bigl(\gamma(b)-\gamma(a)\bigr)\cdot\gamma'(t)\>dt\leq\int_a^b \bigl|\gamma(b)-\gamma(a)\bigr|\>\bigl|\gamma'(t)\bigr|\>dt\cr&= \bigl|\gamma(b)-\gamma(a)\bigr|\int_a^b\bigl|\gamma'(t)\bigr|\>dt\ .\cr}$$

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It turns out this also follows from Minkowski inequality for $0

$$ \left(\int_0^1|u|^p\right)^{1/p}+\left(\int_0^1|v|^p\right)^{1/p}\leq \left(\int_0^1|u+v|^p\right)^{1/p},$$

Taking $u=\gamma_1'^2,v=\gamma_2'^2, p=\frac{1}{2}$ we get

$$ \sqrt{\big(\int_a^b \gamma_1'\big)^2+\big(\int_a^b \gamma_2'\big)^2} \le \sqrt{\big(\int_a^b |\gamma_1'|\big)^2+\big(\int_a^b |\gamma_2'|\big)^2} {\le} \int_a^b \sqrt{\gamma_1'^2+\gamma_2'^2}$$