Arc Length and Rectifiable Curves

If $u<v$, [L1] says the chord length is one member of the set whose supremum is the arc length, so it is at most that supremum.

If $u=v$, the chord is the zero vector and [L2] makes both sides zero.

If the whole path has length zero, applying steps 1.1--1.2 to every $u\le v$ makes every chord zero; separation for the Euclidean norm gives $\gamma (u)=\gamma (v)$, so $\gamma$ is constant.

If one point $s$ is inserted between consecutive points $u<v$ of $P$, [L2] gives $\parallel \gamma (v)-\gamma (u){\parallel }_2\le \parallel \gamma (s)-\gamma (u){\parallel }_2+\parallel \gamma (v)-\gamma (s){\parallel }_2$.

Every other summand is unchanged, so insertion of one point cannot decrease polygonal length.

Because $Q$ is finite and contains $P$, it is obtained by finitely many one-point insertions. Repeated use of step 2.1 gives ${\mathrm{\ell }}_P(\gamma )\le {\mathrm{\ell }}_Q(\gamma )$.

Cauchy--Schwarz in [L1] gives $\mathrm{\mid }{z}_j\mathrm{\mid }=\mathrm{\mid }⟨z,e_j⟩\mathrm{\mid }\le \parallel z{\parallel }_2$. Applying this to every chord of every partition gives $V({\gamma }_j,P)\le {\mathrm{\ell }}_P(\gamma )$.

Conversely, $z={\sum }_{j<n}{z}_j{e}_j$ and the norm axioms in [L1] give $\parallel z{\parallel }_2\le {\sum }_{j<n}\mathrm{\mid }{z}_j\mathrm{\mid }$. Applying this to every chord gives ${\mathrm{\ell }}_P(\gamma )\le {\sum }_{j<n}V({\gamma }_j,P)\le {\sum }_{j<n}\mathrm{Var}({\gamma }_j)$ for every partition.

If $\gamma$ is rectifiable, taking suprema in step 1.1 gives $\mathrm{Var}({\gamma }_j)\le L(\gamma )$ for every $j$, so all coordinates have bounded variation and the left displayed bound holds.

If every coordinate has bounded variation, the final real number in step 1.2 bounds all polygonal sums. Hence $\gamma$ is rectifiable, and taking the supremum gives the right displayed bound.

If $a=b$, the singleton conventions in [L2] and [L3] make all quantities zero, so both directions and both bounds remain valid.

Concatenating a partition of $[a,c]$ with one of $[c,b]$ gives a partition of $[a,b]$ whose polygonal length is the sum of the two polygonal lengths.

Given a partition $P$ of $[a,b]$, insert $c$ if necessary. By [L1] the refined length is at least ${\mathrm{\ell }}_P$, and splitting the refined sum at $c$ makes it at most $L_{[a,c]}+L_{[c,b]}$.

Taking independent suprema in step 1.1 gives $L_{[a,c]}+L_{[c,b]}\le L_{[a,b]}$; if either left summand is infinite, this already forces the total length to be infinite.

Taking the supremum over $P$ gives the reverse inequality. Together with step 2.1 this proves equality.

If $c$ is an endpoint, one summand is zero by [L3]. The equality also shows that the total is finite exactly when both summands are finite.

Suppose first that $\phi$ is nondecreasing. The image under $\phi$ of any partition of $[c,d]$ is a nondecreasing finite list from $a$ to $b$; deleting repetitions produces a partition of $[a,b]$ with the same polygonal sum for $\gamma$.

Conversely, for a partition $a=t_0<\cdots <t_m=b$, choose one $s_i\in {\phi }^{-1}(t_i)$ for each of its finitely many values. Monotonicity forces $s_0<\cdots <s_m$, after taking $s_0=c$ and $s_m=d$, and the resulting polygonal sum of $\gamma \circ \phi$ equals that of $\gamma$.

Hence every polygonal sum of $\gamma \circ \phi$ is at most $L(\gamma )$, so $L(\gamma \circ \phi )\le L(\gamma )$.

Taking the supremum over target partitions gives $L(\gamma )\le L(\gamma \circ \phi )$, proving equality in the nondecreasing case.

If $\phi$ is nonincreasing, reverse the order of every finite list in steps 1.1 and 1.2; Euclidean chord lengths are symmetric, so the same two inequalities hold.

If $[c,d]$ is a singleton, so is its image $[a,b]$, and the singleton convention in [L2] gives both lengths as zero. If instead $a=b$ while $c<d$, then $\gamma \circ \phi$ is constant, so every polygonal sum for it vanishes and $L_{[c,d]}(\gamma \circ \phi )=0$, while $L_{[a,b]}(\gamma )=0$ by the same convention.

For every partition $P$, applying [L1] to each chord and summing gives ${\mathrm{\ell }}_P(T\circ \gamma )\le K{\mathrm{\ell }}_P(\gamma )$.

Taking suprema proves the Lipschitz estimate when $K>0$ and also when $L(\gamma )$ is finite. If $K=0$, [L1] makes all images equal, so every polygonal sum is zero.

Under the similarity identity, every chordwise inequality in step 1.1 is an equality, so ${\mathrm{\ell }}_P(T\circ \gamma )=\lambda {\mathrm{\ell }}_P(\gamma )$ for every $P$.

Taking suprema gives exact scaling for $\lambda >0$; for $\lambda =0$ the map is constant on the trace and the length is zero. With $\lambda =1$, [L2] identifies the isometric case.

Fix a partition $P=(t_0,\dots ,t_m)$. By [L1], ${\gamma }_k(t_i)\to \gamma (t_i)$ for each of its finitely many points.

By [L2], every chord norm converges and hence ${\mathrm{\ell }}_P({\gamma }_k)\to {\mathrm{\ell }}_P(\gamma )$.

For every $k$, [L3] gives ${\mathrm{\ell }}_P({\gamma }_k)\le L({\gamma }_k)$. Passing to the limit inferior yields ${\mathrm{\ell }}_P(\gamma )\le {\mathrm{lim\hspace{0.17em}inf}}_{k}L({\gamma }_k)$.

The right side is independent of $P$. Taking the supremum of the left side over all partitions and using [L3] gives $L(\gamma )\le {\mathrm{lim\hspace{0.17em}inf}}_{k}L({\gamma }_k)$.

The argument also covers an infinite right side or an infinite $L(\gamma )$, because all suprema and the limit inferior are taken in the extended reals.

For each coordinate and $h>0$, [L2] gives $({\gamma }_j(a+h)-{\gamma }_j(a))\mathrm{/}h={\gamma }_j^{\mathrm{\prime }}({\xi }_h)$ for some ${\xi }_h\in (a,a+h)$. Continuity of $v_j$ makes this tend to $v_j(a)$; the analogous argument at $b$ gives the left derivative $v_j(b)$. Hence $\gamma$ is differentiable relative to $[a,b]$ with derivative $v$ everywhere.

Fix $\epsilon >0$. By uniform continuity in [L5], choose $\delta >0$ so that $\parallel v(t)-v(s){\parallel }_2<\epsilon \mathrm{/}(2(b-a))$ whenever $\mathrm{\mid }t-s\mathrm{\mid }<\delta$. Choose a tagged partition $P$ of mesh below $\delta$ whose Riemann sum for the continuous speed differs from its integral by less than $\epsilon \mathrm{/}2$.

Applying [L3] on every subinterval $[u,w]\subseteq [a,b]$ gives $\gamma (w)-\gamma (u)={\int }_u^{w}v$.

For any partition $P$, [L4] applied to each increment from step 2.1 gives ${\mathrm{\ell }}_P(\gamma )\le {\sum }_i{\int }_{t_i}^{t_{i+1}}\parallel v{\parallel }_2={\int }_a^b\parallel v{\parallel }_2$.

On a subinterval with tag ${\xi }_i$, step 2.1 gives $\gamma (t_{i+1})-\gamma (t_i)=v({\xi }_i)\mathrm{\Delta }{t}_i+{\int }_{t_i}^{t_{i+1}}(v-v({\xi }_i))$. By [L4] and the reverse triangle inequality, its norm is at least $\parallel v({\xi }_i){\parallel }_2\mathrm{\Delta }{t}_i-\epsilon \mathrm{\Delta }{t}_i\mathrm{/}(2(b-a))$.

Taking the supremum over $P$ gives $L(\gamma )\le {\int }_a^b\parallel v{\parallel }_2$, so in particular $\gamma$ is rectifiable.

Summing step 3.2 and using the tagged-sum choice gives ${\mathrm{\ell }}_P(\gamma )>{\int }_a^b\parallel v{\parallel }_2-\epsilon$. Since $L(\gamma )\ge {\mathrm{\ell }}_P(\gamma )$ and $\epsilon$ is arbitrary, the reverse inequality follows.

Combining steps 4.1 and 4.2 proves equality. On $[a,a]$ the length and oriented integral are both zero by definition.

By [L1], the graph path has interior derivative $(1,f^{\mathrm{\prime }}(t))$ and continuous extension $v(t)=(1,q(t))$.

By [L2], $\parallel v(t){\parallel }_2=\sqrt{1+q(t{)}^2}$.

Apply [L3] and substitute the speed from step 2.1.

By [L1], the $i$-th restriction is rectifiable and has length ${\int }_{t_i}^{t_{i+1}}\parallel v_i{\parallel }_2$.

Repeated application of [L2] expresses the total length as the sum of the finitely many piece lengths.

Substituting step 1.1 into step 2.1 proves the formula and finiteness. Endpoint derivative extensions are local to each piece, so no matching condition is used.

On a singleton there are no nondegenerate pieces, and both the empty sum and the defined length are zero.

From [L1], $s(v)\ge s(u)$ whenever $u\le v$, so $s$ is nondecreasing.

Let $V_j(t)={\mathrm{Var}}_{[a,t]}({\gamma }_j)$. Each $V_j$ is continuous by [L2], [L3], and continuity of the path's coordinates.

If $s(u)=s(v)$ for some $u<v$, [L1] says the intervening path has length zero, and [L4] makes it constant on $[u,v]$.

For $u\le v$, [L1] and [L2] give $0\le s(v)-s(u)\le {\sum }_j(V_j(v)-V_j(u))$. The finite sum on the right tends to zero as $v\to u$, from either permitted side, so $s$ is continuous.

Conversely, if $\gamma$ is constant on $[u,v]$, every polygonal sum there is zero, so [L1] gives $s(u)=s(v)$. Thus equality at distinct arguments occurs exactly on a constant subinterval, proving the strictness equivalence.

By continuity and the endpoint values in [L1], $s([a,b])=[0,L]$. If $s(u)=s(v)$ with $u\le v$, [L1] makes the intervening length zero and [L2] gives $\gamma (u)=\gamma (v)$.

For $r\in [0,L]$, define $\bar{\gamma }(r)$ to be the unique common value $\gamma (t)$ of all $t$ with $s(t)=r$. Existence follows from surjectivity and well-definedness from step 1.1. This definition immediately gives $\gamma =\bar{\gamma }\circ s$ and uniqueness.

For $r<q$, take $u\le v$ with $s(u)=r$ and $s(v)=q$. The chord bound and [L1] give $\parallel \bar{\gamma }(q)-\bar{\gamma }(r){\parallel }_2\le L_{[u,v]}(\gamma )=q-r$. Hence $\bar{\gamma }$ is $1$-Lipschitz and continuous.

The restriction $s{\mathrm{\mid }}_{[u,v]}$ is a continuous surjective nondecreasing map onto $[r,q]$, and $\gamma {\mathrm{\mid }}_{[u,v]}=\bar{\gamma }{\mathrm{\mid }}_{[r,q]}\circ s{\mathrm{\mid }}_{[u,v]}$. By [L3], $L(\bar{\gamma }{\mathrm{\mid }}_{[r,q]})=L(\gamma {\mathrm{\mid }}_{[u,v]})=q-r$.

If $L=0$, [L2] makes $\gamma$ constant, $s$ has singleton image, and the construction gives the unique constant map on $[0,0]$; the subinterval formula is $0=0$.

By [L1]--[L2], $s^{\mathrm{\prime }}(t)=\parallel v(t){\parallel }_2>0$. By [L5], $s$ is increasing, and continuity plus its endpoint values makes it a bijection onto $[0,L]$; in particular $L>0$.

By [L3], the inverse $\tau :[0,L]\to [a,b]$ is differentiable and ${\tau }^{\mathrm{\prime }}(r)=1\mathrm{/}\parallel v(\tau (r)){\parallel }_2$. This derivative is continuous because $v$, the norm, $\tau$, and reciprocal on positive reals are continuous.

Apply [L4] componentwise to $\sigma =\gamma \circ \tau$ to get ${\sigma }^{\mathrm{\prime }}(r)=v(\tau (r))\mathrm{/}\parallel v(\tau (r)){\parallel }_2$.

Absolute homogeneity of the Euclidean norm gives $\parallel {\sigma }^{\mathrm{\prime }}(r){\parallel }_2=1$, and the formula is continuous in $r$, so $\sigma$ is $C^1$. The relative endpoint derivatives follow from the relative forms in [L2] and [L3].