Conformal invariance, monotonicity, and the series and parallel laws for extremal length

Fix a rectifiable $\gamma :[a,b]\to \mathrm{\Omega }$ with positive length $L$, let $\alpha :[0,L]\to \mathrm{\Omega }$ be its unit-speed arc-length parametrization, and put $w(s)=\mathrm{\mid }{f}^{\mathrm{\prime }}(\alpha (s))\mathrm{\mid }$. By [F2], $w$ is continuous and strictly positive. Define $T(s)$ to be the arc length of $f\circ \alpha$ on $[0,s]$. It is finite: locally on a convex disk around a point of the compact trace of $\alpha$, boundedness of $f^{\mathrm{\prime }}$ makes $f$ Lipschitz by the real one-variable mean-value theorem; finitely many such disks and a subdivision of $[0,L]$ then give finite length. Arc-length additivity makes $T(v)-T(u)$ the length of $f\circ \alpha$ on $[u,v]$.

Fix $s_0\in (0,L)$ and $\epsilon >0$. Choose a convex disk $D$ about $\alpha (s_0)$ on which $\mathrm{\mid }{f}^{\mathrm{\prime }}(z)-f^{\mathrm{\prime }}(\alpha (s_0))\mathrm{\mid }<\epsilon$. On $D$, the map $g(z)=f(z)-f^{\mathrm{\prime }}(\alpha (s_0))z$ is $\epsilon$-Lipschitz by the real mean-value estimate applied along line segments. For $u<v$ sufficiently close to $s_0$, $\alpha ([u,v])\subset D$; polygonal sums and [F5] then give ${\displaystyle \mid T(v)-T(u)-w(s_0)(v-u)\mid \le \epsilon (v-u),}$ since $\alpha$ has length $v-u$ on $[u,v]$. Hence $T^{\mathrm{\prime }}(s_0)=w(s_0)$; as $s_0$ varies this derivative is continuous.

On every interval $[u,v]\subseteq [0,L]$, apply the real mean-value theorem to $T$ on each interval of a partition. The resulting sums for $T(v)-T(u)$ are Riemann sums for the continuous function $w$, so refinement gives ${\displaystyle T(v)-T(u)={\int }_u^{v}w(s)ds\mathrm{.}}$ The positive continuous function $w$ has a positive minimum on $[0,L]$, so $T$ is strictly increasing. Its range is $[0,T(L)]$, and the arc-length parametrization of $f\circ \gamma$ is $\beta =f\circ \alpha \circ T^{-1}$.

If $\mathrm{\Gamma }\subseteq {\mathrm{\Gamma }}^{\mathrm{\prime }}$, then for each Borel density $\rho$, ${\mathrm{\ell }}_{\rho }({\mathrm{\Gamma }}^{\mathrm{\prime }})\le {\mathrm{\ell }}_{\rho }(\mathrm{\Gamma })$, so each quotient for ${\mathrm{\Gamma }}^{\mathrm{\prime }}$ is at most the corresponding quotient for $\mathrm{\Gamma }$; taking suprema gives $\lambda ({\mathrm{\Gamma }}^{\mathrm{\prime }})\le \lambda (\mathrm{\Gamma })$. If every path of ${\mathrm{\Gamma }}_0$ contains a subpath in ${\mathrm{\Gamma }}_1$, [F1] gives ${\mathrm{\ell }}_{\rho }({\mathrm{\Gamma }}_0)\ge {\mathrm{\ell }}_{\rho }({\mathrm{\Gamma }}_1)$ for every $\rho$, hence $\lambda ({\mathrm{\Gamma }}_0)\ge \lambda ({\mathrm{\Gamma }}_1)$. Reciprocal order gives $\mu ({\mathrm{\Gamma }}_0)\le \mu ({\mathrm{\Gamma }}_1)$, including zero and infinite values.

For a path family $\mathrm{\Gamma }$, call a Borel density $\sigma$ width-admissible when ${\mathrm{\ell }}_{\sigma }(\mathrm{\Gamma })\ge 1$, and let $W(\mathrm{\Gamma })$ be the infimum of $A(\sigma )$ over such densities. Then $W(\mathrm{\Gamma })=\mu (\mathrm{\Gamma })$. Indeed, if $A(\rho )\in (0,\mathrm{\infty })$ and $0<{\mathrm{\ell }}_{\rho }(\mathrm{\Gamma })<\mathrm{\infty }$, scaling by $1\mathrm{/}{\mathrm{\ell }}_{\rho }(\mathrm{\Gamma })$ gives a width-admissible density with area the reciprocal of its extremal-length quotient; if ${\mathrm{\ell }}_{\rho }(\mathrm{\Gamma })=+\mathrm{\infty }$, arbitrary positive rescalings have areas tending to zero. Taking the supremum over $\rho$ proves $W(\mathrm{\Gamma })\le \mu (\mathrm{\Gamma })$. Conversely, any width-admissible $\sigma$ with finite positive area gives $\lambda (\mathrm{\Gamma })\ge 1\mathrm{/}A(\sigma )$, so $\mu (\mathrm{\Gamma })\le A(\sigma )$. If $A(\sigma )=0$, [F1] makes $\sigma =0$ almost everywhere; adding $\epsilon {\mathbf{1}}_Q$ for the rectangle in [F7] preserves width-admissibility and has area ${\epsilon }^2\mathrm{\mid }Q\mathrm{\mid }$, forcing $\lambda (\mathrm{\Gamma })=+\mathrm{\infty }$ and again $\mu (\mathrm{\Gamma })\le W(\mathrm{\Gamma })$. If $\lambda (\mathrm{\Gamma })=0$, no width-admissible density can have finite area by these same implications; thus $W(\mathrm{\Gamma })=+\mathrm{\infty }=\mu (\mathrm{\Gamma })$. These cases establish the claimed equality with all extended values.

For a nonnegative Borel function $q$ on $[0,T(L)]$, define $\nu (B)={\int }_{T^{-1}(B)}w(s)ds$. This is a finite Borel measure by [F4]. For $0\le u<v\le L$, step 1.3 gives $\nu ((T(u),T(v)])=T(v)-T(u)$; endpoint singletons have zero measure because $w$ is bounded. The measure is supported on $[0,T(L)]$; clipping any half-open interval to this range and using the interval identity and the zero endpoint atoms shows that $\nu$ agrees with Lebesgue measure restricted to $[0,T(L)]$ on every half-open interval in $\mathbb{R}$. Hence [F6] gives equality of these two restricted measures on Borel sets. Indicators, simple functions and increasing simple approximations now yield ${\displaystyle {\int }_0^{T(L)}q(t)dt={\int }_0^{L}q(T(s))w(s)ds\mathrm{.}}$ Taking $q={\rho }^{\mathrm{\prime }}\circ \beta$ and using [F1] proves for every Borel ${\rho }^{\mathrm{\prime }}:{\mathrm{\Omega }}^{\mathrm{\prime }}\to [0,+\mathrm{\infty }]$ that ${\displaystyle {\mathrm{\ell }}_{{\rho }^{\mathrm{\prime }}}(f\circ \gamma )={\mathrm{\ell }}_{({\rho }^{\mathrm{\prime }}\circ f)\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }}(\gamma )\mathrm{.}}$

For the series law, if either $\lambda ({\mathrm{\Gamma }}_j)=0$, overflow in step 1.4 gives the desired lower bound by the other term. If either value is $+\mathrm{\infty }$, the same overflow gives $\lambda (\mathrm{\Gamma })=+\mathrm{\infty }$. It remains to consider $0<\lambda ({\mathrm{\Gamma }}_0),\lambda ({\mathrm{\Gamma }}_1)<\mathrm{\infty }$. Choose for each $j$ a Borel density with finite positive area whose quotient $q_j$ is arbitrarily close from below to $\lambda ({\mathrm{\Gamma }}_j)$. Restrict it to $E_j$; its path infimum on ${\mathrm{\Gamma }}_j$ is unchanged, and its area can only decrease. The restricted area is positive, since zero area together with a positive path infimum would, by [F1] and the rectangle perturbation of step 1.5, force $\lambda ({\mathrm{\Gamma }}_j)=+\mathrm{\infty }$. Thus the restricted quotient remains positive and finite.

For the parallel law, take any width-admissible density $\sigma$ for ${\mathrm{\Gamma }}_0\cup {\mathrm{\Gamma }}_1$. Its restrictions ${\sigma }_j=\sigma {\mathbf{1}}_{E_j}$ remain width-admissible for ${\mathrm{\Gamma }}_j$, since every path of ${\mathrm{\Gamma }}_j$ lies in $E_j$. By nonnegative-integral monotonicity and additivity on the disjoint sets [F1, F4], ${\displaystyle A(\sigma )\ge A({\sigma }_0)+A({\sigma }_1)\ge \mu ({\mathrm{\Gamma }}_0)+\mu ({\mathrm{\Gamma }}_1),}$ using step 1.5. Taking the infimum over $\sigma$ gives $\mu ({\mathrm{\Gamma }}_0\cup {\mathrm{\Gamma }}_1)\ge \mu ({\mathrm{\Gamma }}_0)+\mu ({\mathrm{\Gamma }}_1)$; if there is no width-admissible density the left side is $+\mathrm{\infty }$ and the inequality still holds.

The same local Lipschitz estimate applies to $f^{-1}$ on compact subsets of ${\mathrm{\Omega }}^{\mathrm{\prime }}$. Consequently $f\circ \gamma$ is rectifiable exactly when $\gamma$ is rectifiable: each direction follows by covering the compact path trace with finitely many convex disks and subdividing its parameter interval. Nonrectifiable paths have infinite length by definition on both sides of the last identity, and a zero-length path is constant, for which both integrals vanish because the arc-length measure is zero. Thus the length identity holds for every path.

Write $L_j={\mathrm{\ell }}_{{\rho }_j}({\mathrm{\Gamma }}_j)$ and $A_j=A({\rho }_j)$ for the restricted densities in step 2.2, and replace ${\rho }_j$ by $(L_j\mathrm{/}{A}_j){\rho }_j$. The scaling law [F1] makes both its path infimum and its area equal to $q_j=L_j^2\mathrm{/}{A}_j$. Put $\rho ={\rho }_0+{\rho }_1$; its supports are disjoint, so [F1] gives $A(\rho )=q_0+q_1$. Each $\gamma \in \mathrm{\Gamma }$ contains subpaths from ${\mathrm{\Gamma }}_0,{\mathrm{\Gamma }}_1$ on disjoint parameter intervals. Subpath additivity and nonnegativity show ${\mathrm{\ell }}_{\rho }(\gamma )\ge q_0+q_1$, hence ${\mathrm{\ell }}_{\rho }(\mathrm{\Gamma }{)}^2\mathrm{/}A(\rho )\ge q_0+q_1$. Letting the two quotients approach their suprema proves $\lambda (\mathrm{\Gamma })\ge \lambda ({\mathrm{\Gamma }}_0)+\lambda ({\mathrm{\Gamma }}_1)$.

If both $\mu ({\mathrm{\Gamma }}_j)$ are finite, choose width-admissible densities ${\sigma }_j$ with areas arbitrarily close above their infima, restrict them to $E_j$, and put $\sigma ={\sigma }_0+{\sigma }_1$. Every path of either family has $\sigma$-length at least one, while disjoint area additivity gives $A(\sigma )=A({\sigma }_0)+A({\sigma }_1)$. Therefore $\mu ({\mathrm{\Gamma }}_0\cup {\mathrm{\Gamma }}_1)\le \mu ({\mathrm{\Gamma }}_0)+\mu ({\mathrm{\Gamma }}_1)$ by step 1.5 and passage to arbitrarily small errors. If either summand is infinite this upper bound is automatic in the extended order. Combined with step 2.3 this proves equality.

Given a Borel ${\rho }^{\mathrm{\prime }}$ of finite positive area on ${\mathrm{\Omega }}^{\mathrm{\prime }}$, put $\rho =({\rho }^{\mathrm{\prime }}\circ f)\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }$ on $\mathrm{\Omega }$. The function is Borel by [F2, F4], and [F3] with $\det Df=\mathrm{\mid }{f}^{\mathrm{\prime }}{\mathrm{\mid }}^2$ (The Jacobian determinant of a holomorphic map is $\mathrm{\mid }{f}^{\mathrm{\prime }}{\mathrm{\mid }}^2$ and is positive exactly where $f^{\mathrm{\prime }}\mathrm{\ne }0$) gives $A_{\mathrm{\Omega }}(\rho )=A_{{\mathrm{\Omega }}^{\mathrm{\prime }}}({\rho }^{\mathrm{\prime }})$. The two areas are therefore finite and positive, while step 3.1 gives ${\mathrm{\ell }}_{\rho }(\mathrm{\Gamma })={\mathrm{\ell }}_{{\rho }^{\mathrm{\prime }}}(f\mathrm{\Gamma })$. Hence the corresponding extremal-length quotients agree. Applying the same construction to $f^{-1}$ shows $\lambda (f\mathrm{\Gamma })=\lambda (\mathrm{\Gamma })$; taking reciprocals gives $\mu (f\mathrm{\Gamma })=\mu (\mathrm{\Gamma })$. The definition of conformal equivalence then gives the stated invariance.

By definition $\lambda =1\mathrm{/}\mu$ with $1\mathrm{/}0=+\mathrm{\infty }$ and $1\mathrm{/}(+\mathrm{\infty })=0$. Thus the reciprocal of $\mu ({\mathrm{\Gamma }}_0)+\mu ({\mathrm{\Gamma }}_1)$ is the harmonic sum of $\lambda ({\mathrm{\Gamma }}_0),\lambda ({\mathrm{\Gamma }}_1)$, including when either modulus is zero or infinite. Steps 4.1, 1.4, 3.2, 2.3 and 3.3 establish (a), (b), (c) and (d), respectively.