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If a compact H-hull lies in a disc of radius $r$ centred at $\xi \in \mathbb{R}$, the displayed estimate holds for $|z-\xi |>2r$, with an absolute constant $C$. It refines the Laurent expansion uniformly in the hull and supplies the first-order increment needed to derive the Chordal Loewner equation from the Loewner local growth property.

An increasing family of compact H-hulls of half-plane capacity $2t$ with the Loewner local growth property corresponds uniquely to a continuous real Loewner driving function. Its mapping-out functions solve the Chordal Loewner equation up to the maximal lifetime of each point. Conversely each continuous driver generates that hull family. A continuous Loewner trace is an additional property, not a consequence for every continuous driver. For SLE it is supplied by the trace-existence theorem.

A compact H-hull is a bounded relatively closed subset $K$ of $\mathbb{H}$ with simply connected complement, considered together with its compact closure in $\overline{\mathbb{H}}$. Its unique mapping-out function has hydrodynamic normalization at infinityThe nonnegative coefficient $a_K$ is its half-plane capacity. Thus the half-plane-capacity parameterization here gives $g_t(z)=z+2t/z+O(z^{-2})$. Strict increase means $K_s⊊K_t$ whenever $s<t$; the strictly increasing capacity already excludes equality.

The Loewner local growth property means that the new growth becomes small after removing the old compact H-hull. More precisely set $K_{t,t+h}=\overline{g_t(K_{t+h}\setminus K_t)}$, taking the closure from the complex upper half-plane. On each bounded time interval, the diameters of these image compact H-hulls tend uniformly to zero as $h\downarrow 0$. Equivalently the new growth can be separated from infinity by crosscuts of vanishing diameter in the mapped domain. It is this single-point local growth, together with capacity continuity, that gives a continuous real Loewner driver.

The Loewner transform ${\xi }_t$ is the limiting point at which this mapped growth occurs:The limit is independent of the chosen new-growth points. The Loewner correspondence theorem says that the resulting mapping-out functions satisfyConversely, given a continuous real $\xi$, solve this ordinary differential equation until the maximal lifetime $T_z$ of each $z\in \mathbb{H}$. The surviving set $H_t=\{z:T_z>t\}$ is mapped conformally onto $\mathbb{H}$ by $g_t$; its complement, with the appropriate boundary closure, is $K_t$. The Laurent coefficient is $2t$, and the composition law reconstructs the local growth. A continuous Loewner driver always determines the compact H-hulls; generation by a continuous Loewner trace is an additional property, not a conclusion for every continuous Loewner driver.

A chordal SLE Loewner trace in the complex upper half-plane with parameter $\kappa \ge 0$ is the continuous curve associated with the Loewner driver ${\xi }_t=\sqrt{\kappa }{B}_t$, where $B$ is standard real Brownian motion started at zero. It starts at zero, its compact H-hull at time $t$ is the curve's past with all pockets cut off from infinity filled, andThe capacity clock has $\mathrm{hcap}(K_t)=2t$. For $\kappa =0$ this gives the vertical slit $\gamma (t)=2i\sqrt{t}$.

To prove Scaling invariance of SLE, fix $r>0$ and putThe Laurent series of ${\hat{g}}_t$ has coefficient $2t$, and direct differentiation givesBrownian scaling says $(r^{-1}{B}_{r^{2}t}{)}_{t\ge 0}$ has the same law as $B$. The Loewner drivers therefore agree in law, and uniqueness of the Chordal Loewner equation gives equality in law of the maps, compact H-hulls, and continuous Loewner traces:This identifies the time change $t\mapsto r^{2}t$ as well as the spatial scaling.

For a simply connected Jordan domain $D$ with distinct marked boundary points, choose a conformal bijection $f:\mathbb{H}\to D$ taking zero to the initial point and infinity to the target, and use $f(\gamma )$ as an unparameterized curve. The Caratheodory boundary extension theorem supplies the boundary values of $f$. Any second such map is $f\circ (z\mapsto rz)$ for $r>0$, because the automorphisms of $\mathbb{H}$ fixing zero and infinity are positive dilations. The Scaling invariance of SLE just proved makes the two image laws identical, up to the corresponding deterministic change of capacity clock. Thus this defines SLE consistently in marked Jordan domains. The assumed $|\gamma (t)|\to \infty$ and the continuous boundary extension give $f(\gamma (t))\to f(\infty )$, the specified target. The law is conformally natural and does not impose a distinguished half-plane-capacity parameterization on an arbitrary domain.

Write $g_t=g_{K_t}$ and, for $s<t$, let $K_{s,t}$ be the filled image hull $g_s(K_t\setminus K_s)$. The half-plane-capacity composition rule givesThe Loewner local growth property says that these mapped increments shrink uniformly on bounded time intervals: for every $A<\infty$,Using diameter instead of radius is equivalent, since a nonempty compact H-hull has its closure meeting the real axis. This is a condition on the increments after mapping out the past, not merely on the Euclidean size of $K_t\setminus K_s$ in the original domain.

For fixed $t$, the nonempty compact sets $\overline{K_{t,t+h}}$, $h>0$, are nested as $h$ decreases. Their diameters tend to zero by Loewner local growth property. The nested-compact-set property gives a single intersection point, and its imaginary part is zero because $K_{t,t+h}$ lies in a real-centred disc of radius tending to zero. Define the Loewner transform byOne may restrict the intersection to sufficiently small $h$. Every point of $K_{t,t+h}$ is within $2{\omega }_A(h)$ of ${\xi }_t$ when $t+h\le A$.

Here is a direct proof of continuity, using the permitted displacement estimate. Fix $s<t$ and put $\delta =t-s$. Take $0<h\le \delta$ and a point $v\in K_{t,t+h}$. Set $z=g_{K_{s,t}}^{-1}(v)$. The hull composition rule puts $z\in K_{s,t+h}$, while the displacement estimate givesSince $z$ is within $2{\omega }_A(\delta +h)$ of ${\xi }_s$, let $h\downarrow 0$, taking $v\to {\xi }_t$, to obtainUse a slightly larger fixed time interval if $t$ is its endpoint. Both terms tend to zero with $\delta$. This proves continuity from both sides, and indeed uniform continuity on every compact time interval. The real boundary value is obtained as a limit; evaluating the displacement estimate directly on a possibly rough boundary is unnecessary.

The Loewner correspondence theorem states that a capacity-$2t$ increasing hull family with Loewner local growth property has the unique continuous real Loewner driving function just defined, and its mapping-out functions solve the Chordal Loewner equationFor each $z\in \mathbb{H}$, this solution is taken up to its maximal lifetime; $\mathbb{H}\setminus K_t$ is precisely the set of points whose lifetimes exceed $t$. Conversely a continuous real driver gives such a hull family through this ordinary differential equation. A continuous driver alone need not have a continuous Loewner trace; that additional assertion is part of the trace theory of SLE, not of the deterministic correspondence theorem.

A continuous process $\gamma :[0,\infty )\to \overline{\mathbb{H}}$ is chordal Schramm–Loewner evolution with parameter $\kappa$ from $0$ to infinity when ${\gamma }_0=0$, its filled past generates these capacity-$2t$ hulls, and their Loewner driving function iswith $B$ standard real Brownian motion. The filling includes bounded components cut off from infinity. Equivalently its Loewner trace is the continuous extension of ${\gamma }_t={\lim }_{y\downarrow 0}{g}_t^{-1}({\xi }_t+iy)$. This last boundary limit is the trace-existence input for SLE.

Fix $r>0$ and scale space and time together:The half-plane capacity of ${\hat{K}}_t$ is $2t$ by part (a). Differentiation givesBy Brownian scaling, $r^{-1}{B}_{r^{2}t}$ is again standard Brownian motion, so the new driver has the same law as the old one. Uniqueness of the Chordal Loewner equation transfers this equality to the hulls and their continuous Loewner traces. ThusThis is capacity-parametrized scale invariance of a Loewner chain. It includes $\kappa =0$, when the driver is zero and the trace is the deterministic vertical slit ${\gamma }_t=2i\sqrt{t}$.

Take increasing to mean strict growth on every nonempty time interval; otherwise a constant family has no uniquely determined driving point. The half-plane-capacity parameterization imposed below guarantees strict growth. For $s<t$, define the increment hull in the mapped domain byThis definition includes filling and boundary conventions automatically. Its mapping-out function is $g_{s,t}=g_t\circ g_s^{-1}$. The Loewner local growth property meansfor every finite $T$ within the parameter range. Equivalently, the mapped increments have uniformly vanishing diameter on compact time intervals.

For fixed $t$, the nonempty compact Euclidean closures ${\overline{K}}_{t,t+h}$ are nested as $h$ decreases. Their diameters tend to zero, so their intersection is a singleton. Its point lies on the real axis: the imaginary part of any point in a hull is at most its enclosing radius. Define the Loewner transform bySince ${\xi }_t$ lies in each closure, every point of $K_{t,t+h}$ is at distance at most $2\mathrm{rad}(K_{t,t+h})$ from it.

To prove continuity, choose $z\in K_{t+2h}\setminus K_{t+h}$ and write $w=g_t(z)$, $w^{\prime }=g_{t+h}(z)$. Then $w\in K_{t,t+2h}$, $w^{\prime }\in K_{t+h,t+2h}$ and $w^{\prime }=g_{t,t+h}(w)$. The continuity estimate from part (i) givesThe right-hand side tends to zero uniformly on compact time intervals. Applying the same inequality with the earlier time as the base proves left continuity as well. Thus the Loewner transform is continuous.

Now impose $\mathrm{hcap}(K_t)=2t$. The half-plane-capacity composition rule follows by composing Laurent expansions at infinity and givesFor a point not yet swallowed, put $z_s=g_s(z)$ and $r_{s,t}=2\mathrm{rad}(K_{s,t})$. The increment is contained in the half-disc of radius $r_{s,t}$ centred at ${\xi }_s$. The continuity estimate first proves continuity of $s\mapsto g_s(z)$: its increment is bounded by $3\mathrm{rad}(K_{s,t})$. The differentiability estimate for a mapping-out function then gives, when $t-s$ is small enough,On a compact interval before swallowing the denominator stays away from zero. Divide by $t-s$ and let $t\downarrow s$. The local-growth property makes the error tend to zero. The analogous backward quotient has the same limit, using continuity of $g_s(z)$ and ${\xi }_s$. Thus the Chordal Loewner equation follows:The initial value follows from $\mathrm{hcap}(K_0)=0$, hence $K_0=\varnothing$. Without the capacity parameterization, the same argument gives $d{g}_t(z)=d\mathrm{hcap}(K_t)/(g_t(z)-{\xi }_t)$, interpreted with the capacity clock.

Put ${\tilde{\xi }}_t={\varphi }_t({\xi }_t)$ and $p_t={\varphi }_t^{\prime }({\xi }_t)$. The given conformal change of half-plane capacity makes the image mapping-out equationIndeed its capacity speed is $2p_t^2$, and the stated Loewner local growth property identifies the indicated single driver. For $t<T$, differentiate the conjugacy identityat a fixed original point. The Chordal Loewner equation then gives the conformal conjugacy derivative for the chordal Loewner equationThe change of variable $z=g_t(w)$ covers the whole upper half-plane. The conformal automorphism of the upper half-plane ${\varphi }_t$ is a real Möbius transformation. Its pole is the image of the still-unswallowed boundary point $-1$, so it is holomorphic in a neighbourhood of ${\xi }_t$ before $T$.

To evaluate the apparent singularity, write $h=z-{\xi }_t$, $p={\varphi }_t^{\prime }({\xi }_t)>0$ and $q={\varphi }_t^{\prime \prime }({\xi }_t)$. The Taylor series isThus the first term in the derivative is $2p/h-q+O(h)$, while the second is $2p/h+2q+O(h)$. Their poles cancel and their constant terms differ by $-3q$. Therefore

Write $g_s=g_{K_s}$ and let $K_{s,t}$ be the relatively closed hull obtained by mapping the increment $K_t\setminus K_s$ through $g_s$. For a bounded compact H-hull $L$, defineThe Loewner local growth property isOne can equivalently use diameter: for a nonempty compact H-hull, $\mathrm{diam}(L)\le 2\mathrm{rad}(L)\le 2\mathrm{diam}(L)$, using a point of its closure on $\mathbb{R}$ for the second inequality. The condition concerns the mapped increment, not the original difference $K_{t+h}\setminus K_t$, which can be large when a narrow opening closes. It makes growth occur near one continuous Loewner driving function rather than by a sudden addition at separated boundary points.

The Loewner local growth property gives a continuous Loewner driving function, and we normalize $U_0=0$. Mapping out an initial hull transforms the future driver to $U_{t+s}-U_t$. The Conformal Markov property of SLE therefore gives stationary increments and independent increments.

Consequently $U$ is a continuous Lévy process. In the Lévy–Khintchine formula, continuity excludes its jump measure, so its only possible components are a deterministic linear drift and Brownian motion:Conformal invariance under dilations gives $U_{r^{2}t}/r\stackrel{d}{=}{U}_t$. Comparing the expectations of these Gaussian random variables yields $\mu rt=\mu t$ for every $r>0$, hence $\mu =0$. The Brownian term already has the required law by Brownian scaling. Therefore the characterization of the SLE driving function isThe case $\kappa =0$ is the constant zero driver. No reflection-symmetry assumption is needed to remove the drift; scaling suffices.

Interpret the printed $A_0=0$ as the empty initial hull. Write $a(t)=\mathrm{hcap}({\tilde{A}}_t)$ and $F_t={\psi }_t$. Conformal maps preserve inclusion and simple connectedness of the complementary domains, so the image sets form an increasing hull family. Boundedness of $\psi$ on bounded sets ensures that these image hulls are bounded. The Loewner local growth property is preserved under conformal transport: after mapping out the hull at time $t$, the small new hull is transported by $F_t$ near its single boundary growth point. The Schwarz reflection principle extends $F_t$ analytically across that point, with real positive derivative. The image diameters therefore tend to zero as the original ones do. These statements use the usual hull closures and prime ends; the initial image hull is empty.

For completeness, the infinitesimal capacity rule underlying this argument is that a shrinking hull $J$ attached near $u$, transported by a map $F$ analytic there, hasOne obtains this by rescaling at $u$: the transported map tends to its linear part, and half-plane capacity scales by the square of the dilation. Uniform analytic distortion near the growth point controls the error. Applying it to the mapped-out increments gives local absolute continuity of $a$ and $a^{\prime }(t)=2F_t^{\prime }(U_t{)}^2$.

The coefficient can also be read directly from the Chordal Loewner equation. The image driver is ${\tilde{U}}_t=F_t(U_t)$. Differentiate $F_t={\tilde{g}}_t\circ \psi \circ g_t^{-1}$, at a fixed point $w$, to obtainThe left side is regular at $w=U_t$. On the right, the coefficient of $(w-U_t{)}^{-1}$ is $a^{\prime }(t)/F_t^{\prime }(U_t)-2F_t^{\prime }(U_t)$, so it must vanish. This proves the conformal change of half-plane capacity rule and its integrated version:If the derivative at the initial boundary point is singular, the formula is interpreted by integrating from a positive time and taking the lower limit to zero; finite image capacity and kernel continuity give this limit. The duplicated incomplete normalization sentence in the TeX is absent from the PDF.

It has the Loewner local growth property when, after mapping out the old hull, each short new increment is small: for every $T,ϵ>0$ there is $\delta >0$ such thatEquivalent formulations use a crosscut of diameter below $ϵ$ separating the new increment from infinity.

The parameterization in part (c) gives the exact self-similarityfor every $\lambda >0$. The Loewner local growth property supplies a continuous Loewner driving function $U$. Under this scaling of the hulls, the deterministic scaling rule for the Chordal Loewner equation givesTaking $t=1$ and $\lambda =\sqrt{t}$ shows thatfor the real constant $c=U_1$.