The geometric and analytic definitions of quasiconformality agree

Assume (b). By [F3, F4], $f$ is differentiable almost everywhere. At any point $z$ where $Df(z)$ exists and is nonsingular, write $f(z+v)-f(z)=Av+o(\mathrm{\mid }v\mathrm{\mid })$ with $A=Df(z)$. If $c={\min }_{\mathrm{\mid }u\mathrm{\mid }=1}\mathrm{\mid }Au\mathrm{\mid }>0$, then for all sufficiently small $r$, the straight homotopy from $v\mapsto f(z+v)-f(z)$ to $v\mapsto Av$ stays nonzero on $\mathrm{\mid }v\mathrm{\mid }=r$, since the remainder is less than $cr\mathrm{/}2$. Hence the induced map on local homology has the same sign as $A$, namely $\mathrm{sgn}\det A$; this is the determinant orientation rule in [F7]. At singular differentiability points $J_f=0$. If $f$ were orientation-reversing, it would follow that $J_f\le 0$ almost everywhere. The analytic inequality gives $J_f=\mathrm{\mid }{f}_z{\mathrm{\mid }}^2-\mathrm{\mid }{f}_{\bar{z}}{\mathrm{\mid }}^2\ge (1-k^2)\mathrm{\mid }{f}_z{\mathrm{\mid }}^2\ge 0$, so $J_f=0$ and $Df=0$ almost everywhere. By ACL and the one-dimensional fundamental theorem, on almost every horizontal segment in any small rectangle the restriction of $f$ would be constant, contradicting injectivity. Thus $f$ preserves orientation. Finally [F2] gives the modulus bounds in (a).

Assume (a). Fix a rectangle $R⋐\mathrm{\Omega }$ and define the finite Borel measure $\nu (B)=\mathrm{\mid }f(\{(x,y)\in R:y\in B\})\mathrm{\mid }$ on $\mathbb{R}$. Let $A(y)=\nu ((-\mathrm{\infty },y))$. Applying the measure-differentiation theorem in [F5] to the forward and backward half-intervals, which shrink nicely to $y$, shows that $A^{\mathrm{\prime }}(y)$ exists and is finite for almost every $y$. At such a height choose finitely many disjoint intervals $(u_j,v_j)$, of total length $l$, and put $d_j=\mathrm{\mid }f(v_j,y)-f(u_j,y)\mathrm{\mid }$. For strips of height $t$ above these intervals, uniform continuity makes every path joining the image vertical sides have length at least $(d_j-\epsilon {)}_{+}$ when $t$ is sufficiently small. Constant density one and the geometric lower modulus bound give $(d_j-\epsilon {)}_{+}^2\le K(v_j-u_j)\mathrm{\mid }f(R_j)\mathrm{\mid }\mathrm{/}t$. Their open images are disjoint and lie in the full strip, so Cauchy–Schwarz gives $({\sum }_j(d_j-\epsilon {)}_{+}{)}^2\le Kl(A(y+t)-A(y))\mathrm{/}t$. Let $t\downarrow 0$ and then $\epsilon \downarrow 0$. The bound ${\sum }_j{d}_j\le \sqrt{Kl{A}^{\mathrm{\prime }}(y)}$ is exactly absolute continuity on the horizontal line. Repeat vertically and cover by countably many interior rectangles. Partial derivatives exist almost everywhere; [F4]'s fixed-constant rectangle argument gives total differentiability almost everywhere.

At a differentiability point with singular values $s_1\ge s_2>0$, use a small square aligned with the right singular vectors and rescale by its side length. The image lies in an $o(1)$ neighborhood of the linear rectangle of dimensions $s_1,s_2$, so its area is at most $s_1{s}_2+o(1)$. Every path joining the image sides perpendicular to the first singular vector has length at least $s_1-o(1)$, by endpoint separation. Constant density one therefore gives image modulus at most $(s_1{s}_2+o(1))\mathrm{/}(s_1-o(1){)}^2$. The source square modulus is one, so its geometric lower bound gives $1\mathrm{/}K\le s_2\mathrm{/}{s}_1$ in the limit. Hence $s_1\mathrm{/}{s}_2\le K$, equivalent to the Beltrami inequality in [F6]. If the derivative has rank one, the same rescaled image has area $o(1)$ while the joining length remains bounded below, contradicting that lower modulus bound. Rank zero satisfies the inequality directly. This proves the sharp differential bound without presuming continuity at a degenerate quadrilateral or a later inverse result.

By the derivative bound, $\mathrm{\parallel }Df{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}^2\le K{J}_f$ almost everywhere. The independently proved lower area inequality [F4]–[F5] gives ${\int }_Q\mathrm{\parallel }Df{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}^2\le K\mathrm{\mid }f(Q)\mathrm{\mid }$ on every interior square. The Hilbert–Schmidt square is at most twice this, so both partials are locally square integrable. The ACL characterization now identifies them as weak derivatives; the bounded continuous map is also locally square integrable. Thus $f\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$, proving (b).

The two implications hold for each $K\ge 1$, so the least geometric constant and the least analytic constant coincide. The geometric definition gives the same bounds and orientation for the inverse. Applying the implication just proved to that inverse gives its analytic K-quasiconformality with the same least constant. Finally, the Beltrami definition gives $K_f=1$ iff $\mathrm{\parallel }{\mu }_f{\mathrm{\parallel }}_{\mathrm{\infty }}=0$, which is equivalent to ${\mu }_f=0$ almost everywhere.