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Let $Z_t=g_t(z)-U_t$. For ${\mathrm{SLE}}_4$, $d{U}_t=2d{B}_t$, and the Chordal Loewner equation givesThe complex Itô formula yieldsTherefore

Fix $z$ and writeBy assumption, $L(z)$ is a continuous local martingale, so $Z_t(z)=e^{L_t(z)}$ is a semimartingale. The Chordal Loewner equation giveswhich has finite variation. Thereforeis a semimartingale. Thus the Loewner driver $U$ is a continuous semimartingale.

Fix $x\in \mathbb{R}\setminus \{0\}$ and, before $x$ is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt{\kappa }$. The Chordal Loewner equation and $U_t=\sqrt{\kappa }{B}_t$ show, after changing the sign of the Brownian motion, thatThus $X$ is a Bessel process of dimensionA Bessel process hits zero exactly when $\delta <2$, which here is equivalent to $\kappa >4$. Hitting zero is precisely the swallowing of a nonzero boundary point by the SLE hull. For $\kappa \le 4$ the trace is simple and swallows no such point; for $\kappa >4$ swallowing occurs through a boundary contact. Hence the trace intersects $\partial \mathbb{H}\setminus \{0\}$ exactly when $\kappa >4$.

For a continuous real driving function $U$, solve the Chordal Loewner equationup to the first time $T_z$ at which $g_t(z)-U_t$ tends to zero. The generated hulls areThey form an increasing Loewner chain of compact H-hulls, satisfy $\mathrm{hcap}(K_t)=2t$, and $g_t$ is the mapping-out function of $K_t$.

For $x>0$, the Chordal Loewner equation and $U_t=\sqrt{\kappa }{B}_t$ givePut $D_t=V_t^r-V_t^1$. The common Brownian term cancels, soThe Itô formula applied to $Z_t=\log D_t-\log V_t^1$ givesSincethe clock $q(u)={\int }_0^u(V_s^1{)}^{-2}ds$ and its inverse $\sigma$ turn the local-martingale term into Brownian motion by the Dambis-Dubins-Schwarz theorem. ThereforeThis is the SLE boundary-point logarithmic separation diffusion.

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 $c=U_1$.

Put $w=g_T^{-1}(U_T+z)$ and defineThen $f_0(z)=z$. Differentiating with the Chordal Loewner equation givesUniqueness for this ordinary differential equation shows that $f_s=h_s$. At $s=T$,which is the endpoint identity for the Reverse Loewner flow.

For $0<s<T$, the pathwise identity $h_s(z)=g_s^{-1}(U_s+z)-U_s$ is generally false. The left side is built from the reversed final driver segment $(U_{T-r}-U_T{)}_{0\le r\le s}$, whereas the right side is built from the initial segment $(U_r{)}_{0\le r\le s}$. By time reversal and symmetry of Brownian motion they have the same probability distribution, but they are not equal for the given Brownian path.

With the upper-half-plane branch, ${\theta }_t\in (0,\pi )$ before the Loewner swallowing time. Write $Z_t=X_t+i{Y}_t$. The Itô formula and the Chordal Loewner equation giveAt parameter four its drift vanishes. At other parameters the drift describes the competition between the deterministic conformal evolution and the Brownian driver.

For $\kappa >8$, the Loewner trace visits every point of the complex upper half-plane almost surely. Choose $\rho =\kappa -8$ in the SLE interior-point martingale to obtain $M_t={{\rm Y}}_t^{(\kappa -8)/8}{S}_t^{-(\kappa -8)/\kappa }$. The maximal inequality for a nonnegative supermartingale bounds its running supremum. If a fixed interior ball were avoided, its center would have ${{\rm Y}}_t\ge c_1>0$ before swallowing, forcing $S_t\ge c_2>0$. But the Chordal Loewner equation gives $d{Y}_t^2/dt=-4S_t^2$, forcing a finite lifetime with $Y_t\to 0$, and $d\log {{\rm Y}}_t/d\log Y_t=2S_t^2$ then forces ${{\rm Y}}_t\to 0$, a contradiction. Countably many rational balls give a dense trace; continuity and Transience of chordal SLE make its image relatively closed, so it is all of the complex upper half-plane. This argument does not cover the critical parameter eight.

For the Chordal Loewner equation driven by $U_t=a\sqrt{t}$, the hull is a straight slit whose endpoint is proportional to $\sqrt{t}$. Put $r_{\pm }=(a\pm \sqrt{a^2+16})/2$ and $\alpha =-r_{-}/(r_{+}-r_{-})\in (0,1)$. Its inverse map isusing upper-half-plane branches. The interval between the two real roots maps onto the two sides of a segment of argument $\pi \alpha$.