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If $(X, d)$ is a metric space, and $p \colon [0,1] \to X$ is a path (continous function).

Then, can we define the length of $p$? Should we assume somehow differentiable?

The only I know needs $X = \mathbb{R}^n$

somenxavier
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  • What is your notion of differentiability for maps with values in metric spaces? What is your motivation? Do you have a specific problem that requires such a generalization, or are you interested in generalization for sake of generalization? In normed spaces, the notion of arc length is similar to that of $\mathbb{R}^n$. – Mittens Jun 12 '21 at 20:28

1 Answers1

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No, we do not need to assume differentiability, the length can be defined for any path at all. The only problem that can happen is that the length is infinite.

Here is the basic definition.

Consider a partition $P$ of the unit interval into subintervals (just as in the definition of the Riemann integral): $$P = (s_i) = (0 = s_0 < s_1 < \cdots < s_N = 1) $$ Define $$L_p(P) = \sum_{i=1}^N d_X(p(s_{i-1}),p(s_i)) $$ Now define $$\text{Length}(p) = \sup_P L_p(P) $$ It is possible for $\text{Length}(p)$ to be infinite, even for paths $p$ in Euclidean space. A path for which $\text{Length}(p)$ is finite is called rectifiable, and a path of infinite length is nonrectifiable.

It's not too hard to prove that if $X$ is the Euclidean space $\mathbb R^n$ and if $p$ is piecewise continuously differentiable then $p$ is rectifiable, and the above definition of length is in agreement with the integration formula $$\text{Length}(p) = \int_0^1 |p'(t)| \, dt $$ You can find some analysis books which devote space to the study of rectifiable curves.

Lee Mosher
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