Let $\gamma : [0,1] \rightarrow \mathbb R^n$ be s.t. $\gamma(0)=a, \gamma(1)=b$ and $\|\gamma' \|\in L^1$. How can I show that $$ \mathscr L (\gamma) = \int _0^1 \| \gamma'(t) \| dt \geq \|a-b\| \qquad ?$$
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You mean bigger in the title? – user10444 Apr 07 '14 at 14:44
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@user10444 It seems he means "how would I exclude the possibility that the arc described by $\gamma$ is shorter than the straight line segment between $a$ and $b$," i.e. prove that it is longer or equal. – Jeppe Stig Nielsen Apr 07 '14 at 15:06
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Somewhat related: The shortest distance between any two distinct points is the line segment joining them. How can I see why this is true?. – Jeppe Stig Nielsen Apr 07 '14 at 15:17
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@JeppeStigNielsen Thank you, I already knew the variational approach and was looking for an easier way to prove this fact. It seems a little bit odd to me there isn't a straightforward way to solve this using something like Lagrange theorem – user139928 Apr 07 '14 at 15:26
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This is false as stated. For example, let $\gamma(t)=(t,C(t))$ where $C$ is the Cantor function. Then $\|\gamma(1)-\gamma(0)\|=\sqrt{2}$ but since $\|\gamma'\|=1$ almost everywhere on $[0,1]$, the integral of $\|\gamma'\|=1$ is $1$.
We must assume that $\gamma$ is absolutely continuous. Then the following works: introduce the function $\varphi(t) = \langle \gamma(t),b-a \rangle$ and note that $\varphi(1)-\varphi(0)=\|a-b\|^2$. Since $\varphi$ is absolutely continuous, the fundamental theorem of calculus applies to it: $$ \|a-b\|^2 = \int_0^1 \varphi'(t)\,dt = \int_0^1 \langle \gamma'(t),b-a\rangle \,dt \le \|a-b\|\int_0^1 \| \gamma'(t)\| \,dt $$ which yields the desired inequality.
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