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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.

Scale the disk to the unit disc, so that the starting point is $a=1/r\in (0,1)$. Use the Möbius transformationwhich maps the disk onto itself, sends $a$ to $0$, and sends $1$ to $1$. By conformal invariance of planar Brownian motion, the exit image is the circular exit of a Brownian motion starting at $0$. Rotational invariance of planar Brownian motion makes that exit uniform in angle.

The endpoints of the right semicircle satisfyIts image is the arc through $1$ between these points, of angular length $2\phi$. Thus the Möbius calculation of circular Brownian exit givesSince $E_{+}$ and $E_{-}$ partition the exit almost surely, part (b) yieldsThe last equality uses $2\arctan (1/r)\in (0,\pi /2)$, so the double-angle tangent identity uses the stated principal branch. As a check, the boundary angle derivative of ${\psi }_a$ is $(1-a^2)/|e^{i\theta }-a{|}^2$; integrating this circular exit density over the right semicircle gives exactly the integral supplied in the question.

Here the otherwise undefined printed domain $H$ must mean $D=\mathbb{H}\setminus K$. Let the Brownian motion start at $z\in D$, and let ${\tau }_R$ stop it on reaching height $R$ or leaving $D$. Take $R>\max (\mathrm{Im}z,{\sup }_K\mathrm{Im}w)$. This stopping time is finite almost surely, since it is no later than exit of its imaginary coordinate from $(0,R)$.

By part (d), $0\le v(w)\le R$ in the stopped domain. Its finite-boundary values vanish except on the top boundary, by the argument of part (d). Localization and the optional stopping theorem for this bounded harmonic martingale therefore giveWrite $p_R={\mathbb{P}}_z(\mathrm{Im}{B}_{{\tau }_R}=R)$ and use the uniform displacement bound $M$ from part (c). On the top boundary, $|v(w)-R|\le M$, henceThe stopped imaginary coordinate is itself a bounded martingale. Since its exit height is nonnegative,Consequently $|v(z)-R{p}_R|\le M\mathrm{Im}z/R\to 0$, proving the high-level escape representation of a mapping-out height:If $H$ instead meant the whole upper half-plane, the right side would always be $\mathrm{Im}z$. For example, the mapping-out function of a vertical slit $K=(0,ia]$ is $g_K(w)=\sqrt{w^2+a^2}$ with branch asymptotic to $w$; at $z=iy$, $y>a$, its height is $\sqrt{y^2-a^2}<y$. This demonstrates why killing on the hull is essential to the printed formula.

Use $U_t=2W_t$ and write $Z_t=X_t+i{Y}_t$ up to its interior-point swallowing time for a Loewner chain $T_z$. The Chordal Loewner equation givesThe branch of the complex logarithm with argument in $(0,\pi )$ is well-defined while $t<T_z$. The Itô formula yields the crucial cancellationThe $2/Z_t^2$ drift is cancelled by the quadratic variation term, because $\kappa =4$. Thus both $L_t=\log |Z_t|$ and ${\theta }_t=\arg Z_t$ are continuous local martingales, withThis is the logarithmic martingale for SLE4.

We justify absence of a finite swallowing time rather than presuming that the logarithm survives forever. On a finite horizon $H$, $0<Y_t\le \mathrm{Im}z$. Also $|X_t|$ is bounded pathwise before $T_z\wedge H$. Indeed, while $|X_t|\ge 1$ its drift $2X_t/|Z_t{|}^2$ has absolute value at most $2$; on each excursion outside $[-1,1]$, integrate from its starting point and bound the Brownian oscillation on $[0,H]$. For example,Therefore $L_t$ is bounded above pathwise on this interval.

Suppose $T_z\le H$. By the Dambis-Dubins-Schwarz theorem, $L$ is a Brownian motion run at its own quadratic variation. If that clock diverged as $t\uparrow T_z$, Brownian oscillation would make $L$ unbounded above, contradicting the preceding bound. The one-sided bound criterion for a martingale clock therefore gives a finite clock limit and a finite real limit for $L_t$. Hence $|Z_t|$ is bounded away from zero near $T_z$.

Now $Y_t=Y_0\exp (-2{\int }_0^t|Z_s{|}^{-2}ds)$ has a strictly positive limit at $T_z$. The drift in $X$ is integrable there, so continuity of $W$ gives a finite limit for $X$ as well. The limiting point is in $\mathbb{H}$ and away from the Loewner driver singularity, and the differential equation extends past $T_z$, a contradiction. Thus $T_z=\infty$ almost surely. A point of the Loewner trace at a finite time belongs to that time's hull, soThis proves the fixed-interior-point avoidance of SLE4 without using simplicity as an input.

The angle remains in $(0,\pi )$ at all finite times. The bounded local martingale criterion upgrades its local martingale equation to a genuine martingale:In particular this is the SLE4 angle martingale, and it converges almost surely and in $L^1$ by bounded Martingale convergence theorem.

It remains to identify the limiting angle using the assumed simple path tending to infinity. Orient that path from $0$ to infinity. Its left component is the one adjacent to the negative real half-axis. Under $g_t$, the left boundary of the slit domain maps to $(-\infty ,U_t)$ and the right boundary to $(U_t,\infty )$. The harmonic measure of the former as seen from $z$ isThis follows by conformal invariance of planar Brownian motion: in the upper half-plane, $\arg (w-U_t)/\pi$ is the bounded harmonic function with values $1$ on the left half-axis and $0$ on the right.

To justify the limiting boundary classification, condition on a simple proper realization of the path and use an independent planar Brownian motion from $z$. It exits the upper half-plane in finite time almost surely, so its path up to that time is compact. The curve tends to infinity, so its intersection with this compact set is contained in a finite initial curve segment. Once that segment has been drawn, the Brownian path exits the slit domain through its left boundary exactly when $z$ is in the final left component: a path from that component cannot reach the right boundary without crossing the curve, and the reverse assertion holds on the right. Endpoints have zero harmonic measure. Bounded convergence of these exit indicators provesTaking expectations in the bounded angle martingale gives the SLE4 left-passage probabilityFor the imaginary axis it is $1/2$; near the negative real axis it tends to $1$, fixing the orientation of “left”.

First prove nowhere monotonicity of Brownian motion. On a fixed interval $[a,b]$ with rational endpoints and $a<b$, the $2^m$ increments over its equal subdivision are independent centered normal random variables. If the Brownian motion path were nondecreasing, all these increments would be nonnegative, an event with probability $2^{-2^m}$. Letting $m\to \infty$ gives probability zero. The same argument excludes nonincreasing paths. A countable union over rational intervals shows that, almost surely, no nontrivial interval supports a monotone function restriction of the path, since every such interval contains one with rational endpoints.

Work on this event and on the event of continuous paths. Inside any open interval $I\subset (0,\infty )$, choose two separated smaller intervals, the first to the left of the second. The first contains $r<s$ with $B_r<B_s$, because its restriction is not nonincreasing. The second contains $u<v$ with $B_u>B_v$, because its restriction is not nondecreasing. Thus $r<s<u<v$, all in $I$.

By the extreme value theorem, the path attains its maximum on $[r,v]$. This value exceeds $B_r$, since it is at least $B_s$, and exceeds $B_v$, since it is at least $B_u$. A maximizing time therefore lies in $(r,v)$ and is a local maximum of Brownian motion. Every open interval in the half-line contains a positive-time interval of this kind. Consequently the set of local maxima of Brownian motion is dense in $[0,\infty )$ almost surely.

The construction is well defined. Since $D$ is contained in a ball, its Brownian exit time is finite almost surely: at successive integer times there is a fixed positive probability that the next independent unit-time Brownian motion increment has length greater than the ball's diameter, forcing an exit. The survival probability is therefore bounded by a geometric sequence. Path continuity gives $B_{{\tau }_D}\in \partial D$. Thus the bounded Dirichlet boundary data give a bounded Borel function $u$ on $D$.

For interior harmonicity, fix a ball with closure in $D$, centred at $x$, and let $\sigma$ be its Brownian exit time. The Strong Markov property, followed by the tower property of conditional expectation, givesThe orthogonal invariance of Brownian motion makes $B_{\sigma }$ uniform on the boundary sphere. Hence $u$ has the spherical mean value property for harmonic functions for every such ball. A locally bounded Borel function with this property is smooth and harmonic: integrating the spherical averages against any smooth radial mollifier gives $u=u\ast \rho$ locally, which first proves smoothness; the mean value property for harmonic functions then implies $\Delta u=0$.

For the boundary limit at $\xi$, choose the harmonic barrier for the Dirichlet problem $h_{\xi }$ from (i). The Itô formula, localized inside $D$, makes $h_{\xi }(B_{t\wedge {\tau }_D})$ a bounded martingale. Compact localization and the bounded convergence theorem, first to the exit and then as $t\to \infty$, yieldFor $\epsilon >0$, if $\partial D\setminus B(\xi ,\epsilon )$ is nonempty, compactness and barrier positivity giveIf that boundary subset is empty, the probability is already zero. ThereforeFirst let $x\to \xi$ and then $\epsilon \downarrow 0$. The continuity of the Dirichlet boundary data proves $u(x)\to f(\xi )$.

Finally, the difference of two continuous solutions is a harmonic function vanishing on the boundary. The maximum principle for harmonic functions on the bounded domain gives that it is zero. Thus the Kakutani solution of the Dirichlet problem exists, attains every prescribed boundary value, and is unique.

A standard form of the Kakutani solution of the Dirichlet problem uses a bounded domain $D\subset {\mathbb{R}}^d$, continuous Dirichlet boundary data $f\in C(\partial D)$, and regularity of every boundary point for the Dirichlet problem. The boundedness of $D$ makes $\partial D$ a compact set, so $f$ is a bounded function and a uniformly continuous function. A bounded domain with $C^2$ boundary is a sufficient geometric case; one must not omit boundary regularity for an arbitrary bounded domain.

One standard analytic characterization of a regular boundary point on a bounded domain is the existence of a positive harmonic barrier: for each $\xi \in \partial D$ there is a harmonic function $h_{\xi }\in C(\overline{D})$ with $h_{\xi }(\xi )=0$ and $h_{\xi }(x)>0$ for $x\in \overline{D}\setminus \{\xi \}$. This is the harmonic form of a barrier for the Dirichlet problem; the barrier characterization of regularity is a standard fact of potential theory.

Let $B$ be $d$-dimensional Brownian motion started at $x\in D$ and let ${\tau }_D=\inf \{t\ge 0:B_t\notin D\}$ be its Brownian exit time. Then the unique solution in $C(\overline{D})\cap C^2(D)$ isEquivalently, $u(x)={\int }_{\partial D}f(\xi ){\omega }_D(x,d\xi )$, where ${\omega }_D$ is harmonic measure, the exit probability distribution of Brownian motion. The harmonic function equation is $\Delta u=0$, with the Laplace operator convention $\Delta ={\sum }_j{\partial }_j^2$.

Fix the initial state $x$, or a prescribed initial distribution. A weak stochastic solution consists of a filtered probability space, a Brownian motion $W$ in that filtration, and a continuous adapted process $X$, with the prescribed initial law, such that the integrals exist andalmost surely for every $t$. The integrability conditions on each finite interval are ${\int }_0^t|b(X_s)|ds<\infty$ and ${\int }_0^t\sigma (X_s{)}^{2}ds<\infty$ almost surely. The probability space and driving Brownian motion are part of what may be chosen.

A strong stochastic solution is constructed on a space carrying a specified driving Brownian motion and specified initial variable. It satisfies the same equation and is adapted to the completed filtration generated by that initial variable and the Brownian motion. Equivalently, it is a nonanticipating measurable function of those data, requiring no additional randomness. For a random initial variable, it is independent of future Brownian increments, as required by the Brownian property of the enlarged filtration.

Uniqueness in law means that any two weak stochastic solutions with the same prescribed initial distribution have the same distribution as random elements of the continuous-path space. Their probability spaces and driving Brownian motions may differ.

Pathwise uniqueness means that if two solutions are defined on the same filtered probability space, driven by the same Brownian motion, and have the same initial variable almost surely, thenThis is indistinguishability of stochastic processes, a statement about coupled sample paths. It is stronger than requiring only equality of their distributions.

On $I=g(\mathbb{R})$ define $h(y)=g^{\prime }(g^{-1}(y))$. Differentiation givesThe mean value theorem proves Lipschitz continuity of $h$ on $I$, with constant $2B$.

If an endpoint of $I$ is finite, the Lipschitz bound gives a finite limiting value of $h$ there. That value must be zero. Otherwise $h$ would be bounded below by a positive number near the endpoint, andwould make $g^{-1}$ approach a finite limit, contradicting its tending to $+\infty$ or $-\infty$. The zero extension of a scale diffusion coefficient at finite endpoints therefore gives a globally Lipschitz function $\bar{h}$ on $\mathbb{R}$: set it to zero beyond each finite endpoint and retain $h$ on $I$.

Use the following standard global existence theorem for stochastic differential equations with Lipschitz coefficients: globally Lipschitz drift and diffusion coefficients give a nonexplosive, pathwise unique strong stochastic solution for every prescribed initial state and driving Brownian motion. Global Lipschitz continuity also gives the required linear-growth bound. Apply it toIt remains to check that $Y$ stays in $I$, so the inverse transform is defined.

Before the first boundary time, put $X=g^{-1}(Y)$. Since $(g^{-1}{)}^{\prime }=1/h$ and $(g^{-1}{)}^{\prime \prime }=-h^{\prime }/h^2$, the Itô formula, stopped inside compact subintervals of $I$, yieldsFor each finite $T$, up to that boundary time,A finite endpoint of $I$ would require $X$ to diverge, which this bound excludes. Therefore $Y$ stays in $I$ at every finite time, and $X=g^{-1}(Y)$ is a global strong stochastic solution.

Finally, any two solutions driven by the same $W$ transform into solutions of the globally Lipschitz $Y$ equation. Its pathwise uniqueness makes the transforms, and hence their inverses, indistinguishable. Thus

A scalar process is a Brownian motion in $({\mathcal{F}}_t)$ when it is adapted, has continuous sample paths, starts at zero almost surely, and, for every $0\le s<t$,Independence from the whole past sigma-field is the essential filtration requirement. It implies independent increments over disjoint intervals; the variance $t-s$ fixes the standard Brownian time scale.

The scalar Lévy characterization of Brownian motion states that an adapted process $X$ starting at zero is a Brownian motion if and only if it is a continuous local martingale with

For necessity, the centered independent increments make a Brownian motion a martingale. Their conditional second moments show that $X_t^2-t$ is also a martingale. The defining uniqueness of the quadratic variation compensator gives $⟨X{⟩}_t=t$.

For sufficiency, fix $\theta \in \mathbb{R}$ and apply the Itô formula toThe time drift cancels the second-order Itô term, leaving $d{Z}_t=i\theta Z_{t}d{X}_t$. Its real and imaginary parts are local martingales. On any deterministic interval $[0,T]$, $|Z_t|=e^{{\theta }^{2}t/2}\le e^{{\theta }^{2}T/2}$, so the bounded local martingale criterion makes them true martingales. ThereforePart (b) proves the Brownian property. This also supplies a proof of Lévy's characterization of Brownian motion through conditional Fourier transforms.

Because $K$ is continuous, adapted, and strictly positive, $K^{-1/2}$ is predictable and locally integrable against $M$. DefineThe quadratic variation of a stochastic integral isThus $W$ starts at zero and is a continuous local martingale with bracket $t$. The Lévy characterization of Brownian motion makes it a Brownian motion in the original filtration.

The process $H$ is continuous and adapted, and ${\int }_0^t{H}_s^{2}ds={\int }_0^t{K}_{s}ds<\infty$ almost surely. The associativity of stochastic integration givesThis is the stochastic-integral representation from absolutely continuous quadratic variation.

Put $A_t=e^{-W_t+t/2}$ and $J_t={\int }_0^t{A}_s^{-1}d{B}_s$. The Itô formula gives $d{A}_t=-A_{t}d{W}_t+A_{t}dt$. The independence of the two Brownian motions gives $[W,B]=0$, so the product formula yieldsFor $f(x)=\arctan x$, $f^{\prime }(x)=(1+x^2{)}^{-1}$ and $f^{\prime \prime }(x)=-2x(1+x^2{)}^{-2}$. The drift terms cancel in the Itô formula:Define the rotated stochastic integralIt is a continuous local martingale withThe Lévy characterization of Brownian motion makes $Z$ a Brownian motion. Since ${\Theta }_t\in (-\pi /2,\pi /2)$ and $\cos {\Theta }_t=(1+X_t^2{)}^{-1/2}$,This is the arctangent transform of a two-noise affine diffusion.