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We use the following standard planar Brownian motion facts: its exit probability distribution is harmonic measure; conformal maps preserve that probability distribution after an increasing time change; and the Poisson kernel for the upper half-plane gives its real-boundary exit density. For rough hulls the boundary is interpreted through prime ends, with boundary correspondence at harmonic-measure-almost every exit point. This avoids imposing a smoothness or local-connectedness assumption on $K$.

Under $g_K$, the exit event through $K$ corresponds to a measurable set $E_K\subset \mathbb{R}$ of boundary images. It is bounded: the mapping-out function extends by Schwarz reflection along the real line outside a sufficiently large interval enclosing $K$, and its boundary near infinity corresponds to the original real boundary near infinity. All prime ends representing $K$ therefore have images in a bounded interval. The boundary correspondence assertion can equivalently be formulated using the almost-everywhere limits of $g_K^{-1}$; changing null sets does not change $E_K$'s measure.

Write $g_K(iy)=u_y+i{v}_y$. The hydrodynamic normalization at infinity gives $u_y=O(y^{-2})$ and $v_y=y-a_K/y+O(y^{-2})$. By conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane,On the bounded set $E_K$, the integrand multiplied by $y$ tends uniformly to one. Therefore the limit exists and is finite:This is the harmonic capacity from infinity in the upper half-plane with the normalization $c(K)=\mathrm{cap}(K)/\pi$; it is different from half-plane capacity. If real boundary points are retained in the definition of $K$, include their boundary images in $E_K$ as well; the argument is unchanged.

For the scaling statement, Brownian scaling says that $r^{-1}{B}_{r^{2}t}$ started from $iy/r$ is again standard planar Brownian motion. The exit event scales with the hull, soSet $v=y/r$, multiply by $y=rv$ and take the limit. It follows thatThe same conclusion follows geometrically from $E_{rK}=r{E}_K$ and the formula above.

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 ${\gamma }_t=2i\sqrt{t}$.

The Brownian reflection principle and Brownian scaling givewhere $Z$ has the standard normal distribution. Combine part (c) with the allowed Gaussian tail bound to obtainThis is the Brownian martingale proof of Gaussian concentration. It proves the intended Gaussian concentration inequality. The denominator printed in the initial bound is $2K$, whereas the gradient hypothesis and part (c) give $2K^2$. The stronger printed bound is false in general. For a concrete counterexample, take $d=1$, $K=2$, and $f(x)=2x$. At $r=10$,where the first strict inequality follows by integrating the decreasing normal density over $[5,6]$. The rightmost term is the printed bound. Thus the square on $K$ is a necessary source correction, rather than a change of proof technique.

Put $L_r={\theta }_{T_r}$. The hitting times $T_r$ are finite almost surely by recurrence of planar Brownian motion, and increase with $r$. At $T_r$, the position is $e^{-r}{e}^{i{L}_r}$. By the Strong Markov property, Brownian scaling, and rotational invariance of planar Brownian motion, the future processis a standard planar Brownian motion started at $1$, independent of ${\mathcal{F}}_{T_r}$. The random rotation does not change its conditional law, since that law is rotationally invariant.

The first time this rescaled process reaches radius $e^{-h}$ corresponds to the original time $T_{r+h}$. Its continuous argument starts at zero and equals the increment of the original winding angle. HenceFor any ordered finite collection of levels, apply this assertion successively. Earlier winding values are measurable at the current hitting time, so all increments are independent and their laws depend only on the level differences. This proves the Lévy process property of winding at logarithmic radial passage levels.

One can also verify stochastic continuity and identify this process. The logarithmic-radius and argument martingales have the common clockThe same Cauchy-Riemann equations calculation as in part (a), applied locally to the logarithm and joined along the continuous argument, gives independent Brownian motions $U,V$ such that $\log R_t=U_{H_t}$ and ${\theta }_t=V_{H_t}$. This clock diverges: if it had a finite limit, convergence of martingales with finite total quadratic variation would make $\log R_t$ converge to a finite value, forcing the clock integral itself to diverge.

Thus, with ${\eta }_r=\inf \{u:U_u=-r\}$, we have $L_r=V_{{\eta }_r}$. The Brownian first-passage Laplace transform is $\mathbb{E}{e}^{-q{\eta }_r}=e^{-r\sqrt{2q}}$. It follows either from the Exponential martingale for Brownian motion stopped at ${\eta }_r\wedge n$, or from the one-dimensional exit equation. Conditioning on the independent process $U$ givesSo this is a symmetric Cauchy process, with level parameter $r$. The formula implies stochastic continuity at zero and, by stationary increments, everywhere. Under the usual definition requiring càdlàg paths, use the right-continuous passage clock ${\eta }_r^{+}=\inf \{u:U_u<-r\}$. It agrees with ${\eta }_r$ for each fixed level almost surely, and $V_{{\eta }_r^{+}}$ is the càdlàg modification. This distinction concerns simultaneous passage levels and does not alter any increment law.

For $\lambda >0$, define ${\tilde{g}}_t(z)={\lambda }^{-1/2}{g}_{\lambda t}(\sqrt{\lambda }z)$. Differentiating the Chordal Loewner equation givesBrownian scaling makes $\tilde{W}$ have the original driver law. The corresponding hull is ${\lambda }^{-1/2}{K}_{\lambda t}$, soThe time scaling applies to the whole coupled hull process. In particular, the joint law of swallowing times satisfies $(T(cx),T(cy))\stackrel{d}{=}{c}^2(T(x),T(y))$.

Thus $F(cx,cy)=F(x,y)$. A pair $0<x<y$ is determined up to dilation by $z=y/(y-x)>1$: dilating by $(y-x{)}^{-1}$ turns it into $(z-1,z)$. Define $f(z)=F(z-1,z)$. Then

A compact H-hull is a bounded set $K\subset \mathbb{H}$ that is closed relative to the complex upper half-plane and whose complement $D=\mathbb{H}\setminus K$ is a simply connected domain. Its Euclidean closure is compact in $\overline{\mathbb{H}}$; “compact” here does not require separation from the real axis. One may equivalently describe the closed hull in $\overline{\mathbb{H}}$, with its real-boundary convention understood. Its mapping-out function is the unique conformal map to $\mathbb{H}$ with hydrodynamic normalization at infinity.

Use the capacity convention $\mathrm{hcap}(K_t)=2t$. A chordal Schramm–Loewner evolution from $0$ to infinity is the hull family of the Chordal Loewner equationwhere $W$ is a real standard Brownian motion and $\kappa \ge 0$. The points with interior-point swallowing time for a Loewner chain at most $t$ constitute $K_t$. For each $t$, $g_t$ maps $\mathbb{H}\setminus K_t$ conformally onto $\mathbb{H}$ and has expansion $g_t(z)=z+2t/z+O(z^{-2})$. The case $\kappa =0$ is the deterministic vertical-slit chain.

First identify the relevant transformed Loewner driving functions directly. For $\lambda >0$, defineDifferentiation gives ${\partial }_t{\tilde{g}}_t=2/({\tilde{g}}_t-{\lambda }^{-1}{U}_{{\lambda }^{2}t})$. Its initial value is $z$, its domain is $\mathbb{H}\setminus {\lambda }^{-1}{K}_{{\lambda }^{2}t}$, and its expansion is $z+2t/z+O(z^{-2})$. Thus its driving function is ${\lambda }^{-1}{U}_{{\lambda }^{2}t}$, by uniqueness of the differential equation and of the hydrodynamically normalized map. Brownian scaling provesThis is capacity-parametrized scale invariance of a Loewner chain.

For a fixed time $t$, map the future remaining domains by $z\mapsto g_t(z)-U_t$. The corresponding hulls ${\hat{K}}_s$ are specified without any boundary-image ambiguity byTheir mapping-out functions areAt $s=0$ this is the identity. Differentiating at a point in its domain yieldsThe Laurent series is $z+2s/z+O(z^{-2})$. Hence the transformed Loewner driver is exactly $U_{t+s}-U_t$, with the original capacity clock unchanged. This proves the composition rule for chordal Loewner driving functions, rather than assuming an identification of transforms.

The Brownian motion increments after $t$ are independent of its past and have the original law. Since past hulls are measurable functions of the past Loewner driver, conditionally on the past, the centered mapped future is an independent copy of the original chain. This is the domain Markov property of a chordal Loewner chain. At an almost surely finite stopping time it follows likewise from the Strong Markov property.

For the converse, a precise class is essential: take capacity-parametrized Loewner chains generated by a continuous real Loewner driving function $U$, with $U_0=0$, and require the two displayed hull properties, including independence from the entire hull past in the domain Markov property. Within this class they characterize ${\mathrm{SLE}}_{\kappa }$, for some $\kappa \ge 0$.

To see why the hull assertions determine the Loewner driver assertions, the hull domains determine their normalized maps uniquely. For any surviving point,The left time derivative suffices for $t>0$, since $U$ is continuous. A sufficiently high point survives up to any given finite time, so points $in$, $n\in \mathbb{N}$, suffice to reconstruct the Loewner driver locally and measurably from the hull past. Thus the driving-function reconstruction for a chordal Loewner chain identifies the two past filtrations and makes the preceding scale and composition computations reversible.

The domain Markov property therefore gives stationary independent increments for $U$. Its continuity makes it a continuous Lévy process. Its classification gives $U_t=bt+\sqrt{\kappa }{W}_t$: in the Lévy–Khintchine formula continuity removes the jump measure, leaving characteristic function $\exp (t(ibu-\kappa u^2/2))$. Driver scaling would turn the drift $b$ into $\lambda b$, so invariance for every $\lambda$ forces $b=0$. The diffusion coefficient is unchanged. This proves the scale-and-domain-Markov characterization of SLE with its regularity and parametrization hypotheses stated explicitly.

Let $\mu$ be the given Brownian motion law on $C[0,1]$. Choose $\alpha <\beta <1/2$ and an integer $p$ with $p(1/2-\beta )>1$. For the largest adjacent level-$n$ increment $A_n$, Brownian scaling and the Markov inequality givewhere $c_p=\mathbb{E}|\mathcal{N}(0,1){|}^p<\infty$. These probabilities are summable, so the Borel-Cantelli lemma gives a finite random $K$ such that $A_n\le K{2}^{-n\beta }$ for every $n$, after absorbing the finitely many exceptional levels.

For $s<t$, choose $m$ with $2^{-m}\le t-s<2^{1-m}$. Their left dyadic approximations at level $m$ differ by at most two grid steps; each finer approximation adds at most one adjacent increment. Continuity and dyadic increment chaining therefore giveThus paths are $\beta$-Hölder and hence $\alpha$-Hölder on $[0,1]$. This direct proof supplies the relevant Brownian Hölder regularity, also obtainable from the Kolmogorov continuity theorem.

The full-measure subset $E_{\alpha }=C^{\alpha }[0,1]$ is measurable in $C[0,1]$, because continuity identifies its seminorm with a countable supremum:Its finiteness event is a countable union of countable intersections of coordinate-measurable events. The sigma-algebra specified on $E_{\alpha }$ is precisely the trace of the coordinate sigma-algebra on $C[0,1]$. Define $\nu (A)=\mu (A)$ for trace-measurable $A\subseteq E_{\alpha }$; equivalently restrict the Brownian law to its full-measure subset. Since $\mu (E_{\alpha })=1$, no renormalization is needed, and all finite-dimensional coordinate laws are unchanged. This is the required probability measure:The construction uses the requested coordinate sigma-algebra and makes no assumption about a separable Hölder-norm topology.

For $c>0$, Brownian scaling and the bijection $t\mapsto c^{2}t$ of $[0,\infty )$ give $Z\stackrel{d}{=}{c}^{-1}Z$, or $\boxed{cZ\stackrel{d}{=}Z}$. The supremum is a measurable extended random variable because continuity allows the supremum over rational times, and $Z\ge 0$.

For $z>0$, the function $F(z)=\mathbb{P}(Z\le z)$ must satisfy $F(z)=F(z/c)$ for every $c>0$, so it is constant on $(0,\infty )$. Letting $z\downarrow 0$ and $z\uparrow \infty$ shows $\mathbb{P}(Z=0)=\mathbb{P}(Z<\infty )$, hence $\boxed{Z\in \{0,+\infty \}\text{ almost surely}}$. This scaling argument alone does not decide the mass at zero; part (d) does.

Taking $x=y=a$ in (c) and using that $B_t$ has no atom at $a$ givesThis is the distribution function of the Brownian first-passage time; (b) ensures that it is a proper probability distribution. Comparing these distribution functions gives the Brownian scaling identity $T_{ca}\stackrel{d}{=}{c}^2{T}_a$ for $c>0$.

For an integer $n\ge 1$, set $T_0=0$ and successively hit the levels $a,2a,\dots ,na$. The Strong Markov property at these finite stopping times and spatial translation imply thatare independent random variables, each with the probability distribution of $T_a$. Thus, for independent copies $T_a^{(j)}$,ThereforeThe paper's definition is the strictly stable distribution convention, with no centering term, a special case of a stable distribution.

Put $\Delta B_h=B_{t+h}-B_t$ and $R_h={\int }_t^{t+h}(H_s-H_t)d{B}_s$. The constant-in-time coefficient $H_t$ on $(t,t+h]$ is previsible, so the ratio minus $H_t$ is $R_h/\Delta B_h$. Since $|H|\le C$, continuity and the dominated convergence theorem implyThe Itô isometry gives $\mathbb{E}{R}_h^2\le h{\eta }_h$. For $\epsilon ,\delta >0$, use the Chebyshev inequality and the normal distribution of $\Delta B_h/\sqrt{h}$ to obtainFirst let $h\downarrow 0$, then $\delta \downarrow 0$; the normal distribution has no atom at zero. This provesNo independence between $R_h$ and the denominator is needed.

The first estimate suggested by the hint is valid: Brownian scaling and the Gamma function give $\mathbb{E}|\Delta B_h{|}^{-1/2}=h^{-1/4}{2}^{-1/4}\Gamma (1/4)/\sqrt{\pi }$. The second hint requires care about the placement of its expectation: with $V_h={\int }_t^{t+h}(H_s-H_t{)}^{2}ds$, the interpretation $\mathbb{E}|R_h{|}^{1/2}\le (\mathbb{E}{V}_h{)}^{1/4}$ is correct by the Jensen inequality and the Itô isometry, whereas the interpretation $\mathbb{E}|R_h{|}^{1/2}\le \mathbb{E}(V_h^{1/4})$ is false with constant $1$. For the latter interpretation, the Burkholder-Davis-Gundy inequality supplies a constant $C_{1/2}$ instead. One can see the obstruction with a Brownian motion stopped on exiting $(-1,1)$: if $\tau$ is that Brownian exit time, then $|B_{\tau }|=1$, $\mathbb{E}\tau =1$, and strict Jensen inequality gives $\mathbb{E}{\tau }^{1/4}<1=\mathbb{E}|B_{\tau }{|}^{1/2}$.

This obstruction also survives the continuity condition on $H$. On a long finite interval approximate $1_{\{s<\tau \}}$ by a continuous adapted process equal to $1$ until $\tau$ and decreasing linearly to $0$ during the next $ϵ$ units of time, and take its initial value to be $0$ by inserting a short continuous ramp before starting the exit experiment. The Itô isometry controls these approximations, and letting the interval length tend to infinity recovers the strict inequality above; hence some bounded continuous adapted example also violates the latter interpretation with constant $1$. The direct proof avoids relying on this ambiguous hint.

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.

If $g_t$ is driven by $U_t=\sqrt{\kappa }{B}_t$, then the mapping-out functions ofareand their driver is ${\tilde{U}}_t=r{U}_{t/r^2}$. By Brownian scaling, $(r{B}_{t/r^2}{)}_{t\ge 0}$ is Brownian. Thus $\tilde{U}$ has the same law as $U$, and uniqueness of the Chordal Loewner equation provesThis is the Scaling invariance of SLE.

Let ${\hat{B}}_t=r{B}_{t/r^2}$. By Brownian scaling, $\hat{B}$ is standard Brownian motion. Substituting $Y_t=r{X}_{t/r^2}$ into the Bessel equation givesWeak uniqueness for the Bessel equation shows that $Y$ is a Bessel process of dimension $d$ started at $rx$. This is the Scaling invariance of a Bessel process.

For Schramm–Loewner evolution in $(\mathbb{H},0,\infty )$, the Scaling invariance of SLE states that, for every $a>0$,has the same law as $K_t$. The scaled Loewner driving function is ${\tilde{U}}_t=a^{-1}{U}_{a^{2}t}$. Since $U_t=\sqrt{\kappa }{B}_t$, the Brownian scaling identity $a^{-1}{B}_{a^{2}t}\stackrel{d}{=}{B}_t$ proves the claim.

The Conformal Markov property of SLE states that, conditionally on the hull through time $t$, the future hull mapped by $g_t-U_t$ is an independent ${\mathrm{SLE}}_{\kappa }$ in $(\mathbb{H},0,\infty )$. More precisely,has driving function ${\tilde{U}}_s=U_{t+s}-U_t$. The stationary increments and independent increments of Brownian motion show that $\tilde{U}$ is independent of ${\mathcal{F}}_t$ and has the same law as $\sqrt{\kappa }B$. The deterministic correspondence between continuous drivers and Loewner chains completes the proof.

For every $a>0$, defineThe Brownian scaling theorem makes $\hat{B}$ a standard Brownian motion, and substitution shows that $(\hat{X},\hat{Y})$ satisfies the same coupled Bessel process equations from initial values $(x/a,y/a)$. Both hitting times are divided by $a^2$, so their order is unchanged. Taking $a=x$ or comparing any two pairs with the same ratio proves that $\mathbb{P}({\tau }_x>{\tau }_y)$ depends only on $x/y$.

By Brownian scaling,where $Z$ has the standard three-dimensional multivariate normal distribution. The right-hand side tends to zero in probability, since $Z$ has no atom at the origin. Part (d) gives almost-sure convergence to $Y$, which also implies convergence in probability to $Y$. Uniqueness of a limit in probability therefore gives $Y=0$ almost surely. Hence $|B_t|\to \infty$ almost surely, proving the transience of Brownian motion in dimension at least three in dimension three.

If $X$ is a Bessel process of dimension $d$ started at $x$ and $r>0$, thenis a Bessel process of dimension $d$ started at $rx$. This follows from Brownian scaling in the stochastic differential equation and uniqueness in law.

For every $a>0$, if $(K_t)$ is ${\mathrm{SLE}}_{\kappa }$, then $(a^{-1}{K}_{a^{2}t})$ has the same law. Its driving function is $a^{-1}{U}_{a^{2}t}$, which has the same law as $U_t$ by Brownian scaling.

Brownian motion in ${\mathbb{R}}^d$ is transient for $d\ge 3$: its distance from the origin tends to infinity almost surely. In dimension three, the positive local martingale $|B_t{|}^{-1}$ and Brownian scaling give a short proof.

It is a dimension-dependent long-time property of Brownian motion.