Ex. 16.2

Ex. 16.2

Let \(\alpha(t)\in\mathbb{R}^p\) be a piecewise-differentiable and continuous coefficient profile, with \(\alpha(0)=0\). The \(L_1\) arc-length of \(\alpha\) from time 0 to \(t\) is defined by

\[\begin{equation} \Lambda(t) = \int_0^t |\dot{\alpha}(s)|_1ds.\non \end{equation}\]

Show that \(\Lambda(t)\ge |\alpha(t)|_1\), with equality iff \(\alpha(t)\) is monotone.

Soln. 16.2

The inequality follows directly by noting

\[\begin{equation} \alpha(t) = \int_0^t\dot{\alpha}(s)ds.\non \end{equation}\]

When \(\alpha(t)\) is monotone, we know that \(\dot{\alpha(s)}\) is either non-positive or non-negative, so that the equality holds.

On the other hand, suppose that equality holds but \(\alpha(t)\) is not monotone. Without loss of generality, we may assume that there exist \(0 < t_1 < t\) such that \(\alpha(t_1) > \alpha(0)\) and \(\alpha(t_1) > \alpha(t)\). It is then not difficult (via classical real analysis alike arguments) to show that \(\Lambda(t) > |\alpha(t)|_1\), which is a contradiction.