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For fixed $u$, write $h(t)={\phi }_{X_t}(u)$. Independent increments and stationary increments giveThe preceding part gives continuity. Also $h$ never vanishes: if $h(t_0)=0$ with $t_0>0$, then $h(t_0/m{)}^m=0$ for every positive integer $m$, contradicting $h(t_0/m)\to 1$.

Here is a direct proof of the exponential form of Lévy characteristic functions. Choose $a>0$ small enough that $|h(t)-1|<1/2$ on $[0,a]$. The principal complex logarithm gives a continuous $L(t)=\log h(t)$ there, with $L(0)=0$. For $s,t\ge 0$ with $s+t\le a$, the multiplicative identity impliesThis difference is continuous on the connected triangle of allowed $(s,t)$ and equals zero at $(0,0)$, so it is identically zero. Thus $L$ satisfies the additive Cauchy functional equation locally. Subdivision gives $L(a/m)=L(a)/m$ and $L(ka/m)=kL(a)/m$; continuity then gives $L(t)=tL(a)/a$ for every $t\in [0,a]$.

Define $\eta (u)=L(a)/a$. For any $t\ge 0$, choose an integer $m$ with $t/m\le a$; thenThe coefficient is unique: if two coefficients give the same exponential for every $t\ge 0$, their derivatives at zero agree. In particular $\eta (0)=0$, and $\mathrm{Re}\eta (u)\le 0$ follows from $|h(t)|\le 1$. The characteristic exponent of a Lévy process has therefore been obtained from first principles, without invoking the Lévy–Khintchine 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”.

Use the principal complex logarithm in the upper half-plane. Deform the interval upward into a half-strip: the vertical side at zero is traversed upward, and the side at one downward. The top side vanishes exponentially; a small indentation at zero contributes $O(r|\log r|)$ and vanishes. Thus the contour deformation gives the exact representation, for $x>0$,Since $\log (is)=\log s+i\pi /2$, scaling $u=xs$ and using the integral defining Euler's constant givesFor the second endpoint, $\log (1+is)=is+s^2/2+O(s^3)$. Watson's lemma, or integration of these powers against $e^{-xs}$ with an exponentially small tail, yieldsIn particular the required first terms areThe two different endpoint scales come from the logarithmic singularity at zero and the simple zero of the amplitude at one.

Analytic continuation extends a holomorphic function by holomorphic elements agreeing on overlaps; along a path, a chain of overlapping neighbourhoods transports the initial element. The identity theorem makes compatible continuation unique on connected overlaps, although different paths can lead to different branches.

Choose the principal complex logarithm and the slit domain $D=\mathbb{C}\setminus [1,\infty )$. On $D$, the functionis holomorphic, with the removable value $h(0)=1$. The domain $D$ is simply connected, so $h$ has a primitive and its integral from $0$ to $z$ is independent of the path in $D$. For $|z|<1$, the uniformly convergent power series on compact subdiscs givesTherefore $f$ is a single-valued analytic continuation of the dilogarithm to $\mathbb{C}\setminus [1,\infty )$. The apparent problem at $0$ is removable; $1$ is a genuine branch point, even though the function has a finite one-sided limit there, because its derivative has a logarithmic singularity. Other cuts from $1$ to infinity give corresponding branches.

ChooseIt is holomorphic on the right half-plane with derivative $1/z\ne 0$, and its inverse $z=e^w$ has positive real part precisely for $|\mathrm{Im}w|<\pi /2$. Thus this branch of the complex logarithm is a conformal bijection onto $S$.

Write $A_n=n^{-1}{\sum }_{j=1}^n{X}_j$ and $\phi (t)=\mathbb{E}{e}^{it{X}_1}$. The given two-sided limit $|f(u)|/u\to 0$ implies $|f(u)|/|u|\to 0$, so $f(u)=o(|u|)$. For each fixed real $t$, independence and the characteristic function of a sum of independent variables giveSet $z_n=iat/n+f(t/n)$. For $t\ne 0$, $n{z}_n\to iat$ and $z_n=O(n^{-1})$. The local Taylor expansion of the complex logarithm at $1$ givesThus $\mathbb{E}{e}^{it{A}_n}\to e^{iat}$; at $t=0$ the identity is immediate. The limit is continuous at zero and is the characteristic function of the constant random variable $a$. The Lévy continuity theorem proves the weak law from a characteristic-function expansion:No integrability hypothesis has been used. The conclusion is also convergence in probability by (d)'s convergence in distribution to a constant implies convergence in probability. If the word “constant” were to allow complex $a$, the characteristic function symmetry $\phi (-t)=\overline{\phi (t)}$ forces $a=\overline{a}$, so it is necessarily real.

Write $z=r{e}^{i\theta }$ with $r>0$ and $0\le \theta <2\pi$, and write $w=u+iv$. Sincewe haveThus the inverse of the exponential is the infinitely valued complex logarithmFor the angle convention in the question, its principal value is

All values of the complex logarithm areBecause $x>0$, one may take the complex argument $\arg z=\arctan (y/x)$. Consequently

Rouché's theorem states that if $g$ and $h$ are holomorphic functions on a neighbourhood of the closure of a bounded domain andon its positively oriented boundary, then $g$ and $g+h$ have the same number of zeros in the domain, counted with multiplicity.

The Open mapping theorem states that a nonconstant holomorphic function on a domain maps every open subset to an open subset. To prove it, fix $z_0$ and put $w_0=f(z_0)$. By the identity theorem, the zeros of $f-w_0$ are isolated. Choose $r>0$ so that the closed disc $\overline{D(z_0,r)}$ lies in the domain and $z_0$ is the only zero of $f-w_0$ in that disc. ThenWhenever $|w-w_0|<\delta$, the constant $w_0-w$ has modulus smaller than $f-w_0$ on the circle. The Rouche theorem therefore says thathas the same positive number of zeros in the disc as $f-w_0$. Thus every $w\in D(w_0,\delta )$ lies in the image of $f$, proving that the image is open.

If $|f|$ had a local maximum at $z_0$, take a small open disc $D$ on which $|f(z)|\le |f(z_0)|$. The restriction of $f$ to $D$ cannot be constant, since the identity theorem would then make $f$ constant on the whole domain. The complex open mapping theorem says that $f(D)$ is an open neighbourhood of $f(z_0)$, and such a neighbourhood contains points of modulus greater than $|f(z_0)|$, a contradiction. This proves the maximum modulus principle from the complex open mapping theorem.

Now let $\Omega$ be bounded. Its closure is compact, so the continuous function $|f|$ attains a maximum there. If $f$ is nonconstant, the maximum modulus principle excludes an interior maximum; if $f$ is constant, the claimed bound is immediate. Hencewhich is the maximum modulus principle on a bounded domain.

Finally let $\Omega =\{z:\mathrm{Re}z>1\}$ and suppose $|f|\le K$ throughout $\overline{\Omega }$. Fix $z_0\in \Omega$. The principal complex logarithm is holomorphic on the right half-plane, so for each positive integer $n$ the functionis holomorphic on $\Omega$ and continuous on its closure. Given $\epsilon >0$, choose $R>|z_0|$ so large thatApply the bounded-domain result to $g_n$ on $\Omega \cap D(0,R)$. On the vertical part of the boundary, $|z|\ge 1$ and hence $|g_n(z)|\le M$. On the circular part,It follows thatLetting $n\to \infty$ and then $\epsilon \to 0$ gives $|f(z_0)|\le M$. Since $z_0$ was arbitrary, this proves the bounded half-plane maximum principle.

The boundedness assumption is necessary. The functionis holomorphic on $\Omega$ and continuous on its closure, and $|f(1+iy)|=e$ on the boundary, but $|f(x)|=e^x$ is unbounded for real $x>1$. Thus the boundary estimate does not control an unbounded holomorphic function on this unbounded domain.

Let ${\wp }_{\Lambda }$ be the Weierstrass elliptic function forBecause $2\pi i$ is a period, the formulais independent of the branch of the complex logarithm. It is a nonconstant harmonic function onand its other period gives $u(qz)=u(z)$.