Arc Length and Rectifiable Curves: Examples and Counterexamples

The segment derivative is the constant vector $q-p$, so [L1] gives $L(\gamma )={\int }_0^1\parallel q-p{\parallel }_{2}dt=\parallel q-p{\parallel }_2$.

On a polygonal path, the $i$-th affine piece has length $\parallel p_{i+1}-p_i{\parallel }_2$ by step 1.1, independently of its positive traversal time.

Add the piece lengths using [L1] or [L2] to obtain the displayed sum. If $m=0$, the path is constant and both sides are zero.

If $\theta >0$, [L1]--[L2] give ${\gamma }_{\theta }^{\mathrm{\prime }}(t)=(-\sin t,\cos t)$.

By [L3], $\parallel {\gamma }_{\theta }^{\mathrm{\prime }}(t){\parallel }_2=\sqrt{{\sin }^{2}t+{\cos }^{2}t}=1$.

Apply [L4] and [L5] to get $L({\gamma }_{\theta })={\int }_0^{\theta }1dt=\theta$.

If $\theta =0$, the domain is a singleton and the defined length is $0=\theta$.

Since $\mathrm{\mid }f(x)\mathrm{\mid }\le x$ for $x>0$, $f(x)\to 0=f(0)$ as $x\downarrow 0$; away from zero it is continuous. Thus $\gamma$ is a path.

Put $x_k=2\mathrm{/}((2k+1)\pi )$. Positivity of $\pi$ and [L5] give $x_k\downarrow 0$, so choose $K$ with $x_K\le 1$; and [L1] gives $f(x_k)=(-1{)}^k{x}_k$.

For $N>K$, take the partition whose points are $0,x_N,x_{N-1},\dots ,x_K,1$, omitting a repeated endpoint if $x_K=1$. Its variation contribution from consecutive $x_k$ is $\mathrm{\mid }f(x_k)-f(x_{k+1})\mathrm{\mid }=x_k+x_{k+1}\ge x_k$.

Since $x_k=2\mathrm{/}((2k+1)\pi )\ge 1\mathrm{/}(\pi (k+1))$, [L2] says the tails ${\sum }_{k=K}^{N-1}{x}_k$ are unbounded. Hence the variation sums in step 2.1 are unbounded and $f$ is not of bounded variation.

The first coordinate $x\mapsto x$ has bounded variation, but the second does not by step 3.1. Therefore [L4] says the graph path is not rectifiable.

Define $R(x,y):=(x\mathrm{/}2-\sqrt{3}y\mathrm{/}2,\sqrt{3}x\mathrm{/}2+y\mathrm{/}2)$. Since $(\sqrt{3}{)}^2=3$, direct expansion gives $\parallel Rv{\parallel }_2=\parallel v{\parallel }_2$ and $\parallel v-Rv{\parallel }_2=\parallel v{\parallel }_2$ for every $v\in {\mathbb{R}}^2$.

Put ${\kappa }_0(t)=(t,0)$. Recursively, suppose ${\kappa }_n$ is affine between consecutive points of ${\sigma }_n:=\{j\mathrm{/}{4}^n:0\le j\le 4^n\}$. For an old edge from $A={\kappa }_n(j\mathrm{/}{4}^n)$ to $A+v={\kappa }_n((j+1)\mathrm{/}{4}^n)$, prescribe the five successive values of ${\kappa }_{n+1}$ at parameters $(4j+r)\mathrm{/}{4}^{n+1}$, $0\le r\le 4$, to be $A$, $A+v\mathrm{/}3$, $A+v\mathrm{/}3+Rv\mathrm{/}3$, $A+2v\mathrm{/}3$, and $A+v$, and make ${\kappa }_{n+1}$ affine between them. The endpoint prescriptions agree on adjacent old edges, so [L1] gives the sequence.

Every old vertex is retained. By step 1.1, each old edge of length $\mathrm{\ell }$ is replaced by four edges of length $\mathrm{\ell }\mathrm{/}3$. Induction therefore gives $4^n$ edges of length $3^{-n}$ in ${\kappa }_n$, and [L5] gives $L({\kappa }_n)=4^n{3}^{-n}=(4\mathrm{/}3{)}^n$.

On an old edge of length $3^{-n}$, compare ${\kappa }_{n+1}$ with the affine chord ${\kappa }_n$. At the five subdivision parameters their differences have norms at most $0,3^{-n}\mathrm{/}12,3^{-n}\mathrm{/}2,3^{-n}\mathrm{/}12,0$. On each intervening interval the difference is affine, so the triangle inequality gives ${\sup }_t\parallel {\kappa }_{n+1}(t)-{\kappa }_n(t){\parallel }_2\le 1\mathrm{/}(2\cdot 3^n)$. Consequently, for $m>n$, telescoping and the finite geometric-sum identity give ${\sup }_t\parallel {\kappa }_m(t)-{\kappa }_n(t){\parallel }_2<3\mathrm{/}(4\cdot 3^n)$.

Each coordinate sequence is uniformly Cauchy by step 4.1, so [L3] supplies uniform coordinate limits. Let $\kappa$ be the resulting vector-valued limit. The inequality $\parallel z{\parallel }_2\le \mathrm{\mid }{z}_1\mathrm{\mid }+\mathrm{\mid }{z}_2\mathrm{\mid }$, obtained from the coordinate decomposition and norm axioms in [L2], makes the convergence ${\kappa }_n\to \kappa$ uniform. Each ${\kappa }_n$ is continuous, and [L3] makes $\kappa$ continuous, hence a path.

Fix $n$. Every vertex of ${\sigma }_n$ remains unchanged in every later path, so uniform convergence gives $\kappa (j\mathrm{/}{4}^n)={\kappa }_n(j\mathrm{/}{4}^n)$ for all $0\le j\le 4^n$. Thus the inscribed polygonal sum of $\kappa$ on ${\sigma }_n$ is ${\mathrm{\ell }}_{{\sigma }_n}(\kappa )=4^n{3}^{-n}=(4\mathrm{/}3{)}^n$. These sums are unbounded by [L4]. The defining supremum in [L5] is therefore $+\mathrm{\infty }$, so $\kappa$ is not rectifiable.

The classical Koch snowflake boundary is the concatenation of three isometric copies of $\kappa$. By [L7], each copy is nonrectifiable, and a rectifiable concatenation would have rectifiable restrictions. Hence the snowflake boundary is nonrectifiable.

Lower semicontinuity is consistent with the result but cannot prove it: [L6] yields only $L(\kappa )\le {\mathrm{lim\hspace{0.17em}inf}}_n(4\mathrm{/}3{)}^n=+\mathrm{\infty }$, a vacuous upper bound. The retained-vertex partitions in step 6.1 supply the necessary lower bounds.

On $[-1,0]$ the derivative is $(1,-1)$, and on $[0,1]$ it is $(1,1)$; each has Euclidean norm $\sqrt{2}$ by [L3].

The right difference quotients at $0$ equal $(1,1)$ and the left difference quotients equal $(1,-1)$. They have different limits, so the vector derivative at $0$ does not exist.

By [L2], the two pieces have lengths $\sqrt{2}$ and $\sqrt{2}$, so the total length is $2\sqrt{2}$.

Thus the path is rectifiable by step 2.1 but not differentiable, and therefore not continuously differentiable, at its corner.

The identity path has constant derivative and speed $1$, so [L1]--[L2] give $L(\gamma )=1$.

On the three parameter thirds, $\eta$ has derivatives $3,-3,3$, hence speed $3$. Each speed integral is $3(1\mathrm{/}3)=1$, so [L1]--[L2] give $L(\eta )=3$.

Every value of either path lies in $[0,1]$, and each path traverses the whole segment, so their traces agree even though steps 1.1--1.2 give different lengths.

Every ${\gamma }_k(t)$ has first coordinate $t$ and second coordinate between $0$ and $1\mathrm{/}(2k)$, so ${\sup }_t\parallel {\gamma }_k(t)-\gamma (t){\parallel }_2\le 1\mathrm{/}(2k)\to 0$ by [L3].

On each of the $2k$ parameter intervals, ${\gamma }_k$ has derivative $(1,1)$ or $(1,-1)$ and hence constant speed $\sqrt{2}$. By [L1]--[L2], each piece contributes $\sqrt{2}\mathrm{/}(2k)$ and $L({\gamma }_k)=\sqrt{2}$.

The limit path has constant derivative $(1,0)$ and speed $1$, so [L1]--[L2] give $L(\gamma )=1$.

Thus lengths do not converge to the length of the uniform limit. The valid inequality [L4] reads $1\le \sqrt{2}$, as expected.

The identity coordinate and $c$ are nondecreasing by [L1], hence have bounded variation by [L2]. Their pairing is continuous by componentwise continuity.

The $2^n$ retained stage intervals have pairwise disjoint interiors and total length $2^n{3}^{-n}=(2\mathrm{/}3{)}^n$, which tends to zero by [L4] and [L6].

By [L5], the sum of the absolute increments of $c$ over those same intervals is $2^n{2}^{-n}=1$ for every $n$.

Therefore [L2] makes $\gamma$ rectifiable; quantitatively its length is at most the sum of the two coordinate variations, namely $2$.

Taking $\epsilon =1\mathrm{/}2$, every proposed $\delta >0$ is defeated by a sufficiently large $n$: step 1.2 makes the total interval length below $\delta$, while step 1.3 leaves total image increment $1$. This contradicts [L3], so $c$ is not absolutely continuous.