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Take the principal square root $w=u+iv$, which has $u>0$ on the slit domain. The inverse relation $z=w^2$ gives $x=u^2-v^2$ and $y=2uv$. Hence the defining inequality for the parabolic domain becomesThus the image is the strip $\boxed{0<\mathrm{Re}w<1}$. Conversely, every point of this strip squares to a point of the parabolic domain outside the removed nonpositive real axis, and its principal square root is the original $w$. This proves that the map is onto the strip and one-to-one. Its derivative $1/(2\sqrt{z})$ is nonzero there, so it is a conformal map.

It helps to separate the tangent and the squaring map. Put $t=\tan (z/2)$, with $z=x+iy$. The identityshows that $0<x<\pi /2$ gives $\mathrm{Re}t>0$ and $|t|<1$. To see that every point of this right half-disc occurs, setThen $z=-i\mathrm{Log}R$ has $0<\mathrm{Re}z<\pi /2$ and satisfies $\tan (z/2)=t$. The principal complex logarithm is unambiguous because $R$ is in the first quadrant.

Squaring maps this right half-disc bijectively onto the unit disc with the interval $(-1,0]$ removed: the argument of $t$ lies between $-\pi /2$ and $\pi /2$, and doubles into $(-\pi ,\pi )$. Thus the desired image isNeither derivative vanishes in the domain, so the composition is also a conformal map onto that slit disc.

Use the Green function of killed planar Brownian motion, normalized as the density of expected occupation with respect to area:for nonnegative measurable $f$. Equivalently, $G_D(z,w)={\int }_0^{\infty }{p}_D(t,z,w)dt$, where $p_D$ is the killed Brownian transition density. With generator $\frac{1}{2}\Delta$, the distributional normalization is $-\frac{1}{2}{\Delta }_w{G}_D(z,w)={\delta }_z(w)$, and the singularity is $-{\pi }^{-1}\log |w-z|$ plus a locally harmonic function. This fixes the normalization of the Dirichlet Green function explicitly.

By conformal invariance of planar Brownian motion and its conformal Brownian clock,The Jacobian determinant of a conformal map is $|{\varphi }^{\prime }(w){|}^2$. Changing the area variable to $w^{\prime }=\varphi (w)$ givesUniqueness of the occupation density proves the desired equality almost everywhere. Both functions are continuous and harmonic away from their pole, so it holds at every $w\ne z$. The Green function is conformally invariant:If the Dirichlet Green function is instead normalized for $-\Delta$, both kernels are divided by two and the invariance statement is unchanged.

The scale is precisely the inverse conformal map:By definition, a chordal SLE in a specified scale in $(\mathbb{H},0,1)$ is a curve whose image under $\sigma$ is the half-plane-capacity-parameterized SLE from zero to infinity. Hereand the right-hand side is the given ${\mathrm{SLE}}_6$. Equivalently, this is Conformal invariance of SLE, with the target point transported from infinity to $1$. Thus $\tilde{\gamma }$ is ${\mathrm{SLE}}_6$ in $(\mathbb{H},0,1)$ in scale $\sigma$, using the original parameter $t$. This scaled parameter is distinguished from the ordinary half-plane-capacity parameterization of its image hulls used next.

The PDF specifies decay as $y\to \infty$; the local TeX drops this limit from its displayed boundary data. Take $a=4\pi$, so the conformal map is $w=\sin (z/4)$. Its components areFor $-2\pi <x<2\pi$ and $y>0$, $\eta >0$. The bottom edge maps monotonically to $(-1,1)$, and the two vertical sides map to the real rays $(-\infty ,-1)$ and $(1,\infty )$. The derivative $\cos (z/4)/4$ never vanishes in the interior. Injectivity follows from the identities for equal sines: the alternatives $z_1/4=z_2/4+2\pi n$ and $z_1/4=\pi -z_2/4+2\pi n$ give only the identical point within this strip. Surjectivity can be seen directly: for fixed $\xi$ and $v=y/4>0$, solve $\sin (x/4)=\xi /\cosh v$; thenOn $v>\max (0,\mathrm{arcosh}|\xi |)$, with the lower endpoint read as $0$ when $|\xi |\le 1$, the right side increases strictly from zero to infinity. Thus every point of the upper half-plane has a unique preimage.

By conformal invariance of harmonicity, write the transformed harmonic function as $\Phi (\xi ,\eta )$. Its real-axis boundary data areThe inverse sine on $(-1,1)$ is its real principal branch. For bounded integrable data, take the Fourier transform convention $\hat{\Phi }(k,\eta )={\int }_{\mathbb{R}}\Phi (\xi ,\eta )e^{-ik\xi }d\xi$. The Laplace equation becomes ${\hat{\Phi }}_{\eta \eta }-k^2\hat{\Phi }=0$, with the bounded decaying solution $\hat{\Phi }=\hat{F}(k)e^{-|k|\eta }$. Inverting the Fourier transform givesConsequently the Poisson kernel for the upper half-plane yieldsThe decay is uniform across the strip: $|w{|}^2={\sin }^2(x/4)+{\sinh }^2(y/4)\ge {\sinh }^2(y/4)$ and, for $|w|>1$, $|\Phi |\le \Vert F{\Vert }_{L^1}|w|/(\pi (|w|-1{)}^2)\to 0$. This verifies the printed decay condition. Boundary limits give the prescribed values at continuity points away from the corners.

For the given sine data,Values assigned at $s=\pm 1$ do not alter the integral. There is no solution continuous on the entire closed half-strip for these data: the bottom edge has corner limits $\pm 1$, but the side edges have corner limits zero. The formula is the bounded Dirichlet problem solution with the specified limits away from those two corners. This is the usual interpretation of the printed problem; boundedness excludes additional boundary-singular harmonic functions, while decay alone without a regularity or boundedness condition does not establish uniqueness.

For $r>0$, the conformal map $h(z)=r{g}_A(z/r)$ maps $\mathbb{H}\setminus rA$ onto $\mathbb{H}$ and has Laurent series $h(z)=z+r^{2}a/z+O(z^{-2})$. The uniqueness in hydrodynamic normalization at infinity givesAt $r=0$ the scaled closure is contained in $\mathbb{R}$, so the interior compact H-hull is empty and both sides are zero. This is the scaling part of scaling and translation of half-plane capacity.

For $x\in \mathbb{R}$, the conformal mapmaps $\mathbb{H}\setminus (A+x)$ onto $\mathbb{H}$ and has hydrodynamic normalization at infinity. Its Laurent series is $h(z)=z+a/z+O(z^{-2})$. Uniqueness of the mapping-out function of a compact H-hull therefore givesThis is the translation part of scaling and translation of half-plane capacity.

For the vertical slit, choose the branch of $\sqrt{z^2+1}$ asymptotic to $z$ at infinity. It is the mapping-out function of a compact H-hull for $(0,i]$, andThus half-plane capacity of a vertical slit givesThe endpoint $0$ is merely part of the boundary closure convention.

For the filled upper half-disc $K=\{z\in \mathbb{H}:|z|\le 1\}$, the conformal map $g_K(z)=z+1/z$ maps the exterior half-disc onto $\mathbb{H}$. Indeed, it sends the semicircle to $[-2,2]$, the remaining real boundary to the complementary intervals, and its inverse is the branch of $(w+\sqrt{w^2-4})/2$ asymptotic to $w$. Hence half-plane capacity of a half-disc givesIf $\mathbb{D}$ denotes the open unit disc, the printed $\mathbb{H}\cap \mathbb{D}$ is not relatively closed and is literally not a compact H-hull. The intended value is the one for its filled relative closure $K$. With a closed-disc convention the displayed notation already represents that hull.

For a nonempty compact H-hull $A$, its closure meets $\mathbb{R}$: otherwise a nonempty compact set strictly inside $\mathbb{H}$ would be separated from the lower half-plane in the complement, contradicting the simply connected domain condition for $\mathbb{H}\setminus A$. Choose $x\in \overline{A}\cap \mathbb{R}$ and put $d=\mathrm{diam}(A)$. Taking limits in the definition of diameter shows $|z-x|\le d$ for every $z\in A$. Thus $A$ lies in the filled half-disc of radius $d$ centered at $x$. Its half-plane capacity is $d^2$ by scaling and translation of half-plane capacity. Using monotonicity of half-plane capacity,For the empty compact H-hull, use diameter zero. This proves the unit-constant version of half-plane capacity is bounded by squared diameter; no claim of optimality of the constant is needed.

Assume the domains are proper, so $d$ and $\tilde{d}$ are finite and positive, and interpret the conformal transformation as a bijection. Apply the Koebe quarter theorem toThe open ball $B(z,d)$ lies in $D$, so $F$ is a normalized univalent function. Its image inclusion shows that $B(\tilde{z},d|f^{\prime }(z)|/4)\subseteq \tilde{D}$. Therefore $d|f^{\prime }(z)|/4\le \tilde{d}$, or $|f^{\prime }(z)|\le 4\tilde{d}/d$.

Apply the same argument to the inverse conformal map, whose derivative at $\tilde{z}$ is $1/f^{\prime }(z)$. It gives $\tilde{d}/(4|f^{\prime }(z)|)\le d$. Thus the boundary-distance derivative bound for a conformal bijection isThe full domains need not be simply connected domains: only the two interior discs are used. If the whole plane is allowed as a domain, a conformal bijection onto another plane domain is affine and both domains are the whole plane. For example, $D=\tilde{D}=\mathbb{C}$ and $f(z)=z$ make both boundary distances infinite and the printed ratios $\infty /\infty$ undefined. This is a necessary proper-domain convention. Bijectivity also matters: the injective conformal map $f(w)=w/8$ from the unit disc into itself has $d=\tilde{d}=1$ at zero but $|f^{\prime }(0)|=1/8<1/4$. Thus an interpretation allowing a map into, rather than onto, the second domain makes the lower bound false.

The chordal restriction property concerns the unparameterized trace law. Let $A$ be a compact H-hull with $0\notin \overline{A}$, and put ${\psi }_A(z)=g_A(z)-g_A(0)$. This conformal map takes $\mathbb{H}\setminus A$ onto $\mathbb{H}$, fixes $0$ and infinity, and has derivative one at infinity. On the event that the curve avoids $A$, its image is a curve from $0$ to infinity in $\mathbb{H}$. Restriction requires positive avoidance probability for every admissible $A$, and the conditional identityup to increasing changes of the curve parameter. Equivalently, conditioning an SLE in a marked simply connected domain to stay in an admissible simply connected domain gives the same SLE law in that subdomain. The property does not assert that the original half-plane-capacity parameterization survives an arbitrary conformal map.

Interpret the malformed printed domain inequality as the first quadrant $x>0,y>0$, as indicated by the two axis boundary conditions. There is also a genuine missing hypothesis: decay of the boundary data does not require the harmonic function itself to be bounded. For zero data, both $q=0$ and $q=xy$ satisfy the Laplace equation and both edge values, but the Wirtinger derivative of the latter isThus the data do not determine an arbitrary solution's derivative without a growth restriction. We derive the integral representation for the bounded solution, which is also the standard decaying solution when the continuous edge data decay. Any other admissible unbounded solution can differ by a zero-boundary harmonic function.

The conformal map $w=z^2$ sends the quadrant bijectively to the complex upper half-plane: the positive real axis maps to $s>0$, and the positive imaginary axis maps to $s<0$. Define its real-axis boundary function byThe compatibility at the corner makes $f$ continuous there. The Dirichlet Poisson integral in a quadrant follows from the upper-half-plane Poisson kernel:It is bounded by the supremum of the bounded edge data, and the kernel's approximate identity property supplies both Dirichlet boundary conditions. More explicitly, for $z^2=a+ic$, where $a=x^2-y^2$ and $c=2xy>0$, this integral isIt displays directly which part of the real-line boundary corresponds to each original axis.

To obtain its derivative, first take real data and use the Schwarz integral formula. A holomorphic function whose real part is the Poisson solution in the $w$-plane isThe regularizing subtraction is independent of $w$ and affects only the imaginary normalization; it permits bounded data without an artificial integrability assumption. Then $q=\mathrm{Re}F(z^2)$. By the chain rule and the Wirtinger derivative,The factor is $z$, not $2z$: taking the real part contributes the factor $1/2$ in the complex derivative. Differentiation under the integral sign is justified uniformly on compact interior subsets because the denominator stays away from the real boundary and decays quadratically at infinity. Splitting the real integral at zero and substituting $s=r^2$ or $s=-r^2$ gives the holomorphic derivative of a quadrant Dirichlet solution:This converges for bounded data, and in particular for the stated sufficiently decaying data. No interior pole is encountered since $\mathrm{Im}(z^2)=2xy>0$. For complex-valued data the same formula holds by applying the real-data construction separately to their real and imaginary parts and using linearity.

An equivalent form, often convenient when the data's derivatives decay, comes from integration by parts. Sincethe two endpoint terms are $-g_2(0)/z^2$ and $+g_1(0)/z^2$. They cancel by the specified compatibility, leavingThis is the harmonic quadrant solution from tangential boundary derivatives applied to the actual derivatives $g_1^{\prime },g_2^{\prime }$; the supplied $g_1,g_2$ in this question are boundary values, not derivatives. Omitting those primes would solve a different boundary problem.

For completeness, bounded Dirichlet uniqueness in a quadrant selects the solution just constructed. The difference of two bounded solutions is zero on both axes. Odd reflection across each axis gives a bounded harmonic extension to the punctured plane; the removable singularity for a bounded harmonic function fills the corner. The harmonic Liouville theorem makes the extension constant, and oddness makes it zero. Decaying data give a decaying Poisson extension: split the real-axis integral into a fixed compact interval and a tail with arbitrarily small data; the compact contribution tends to zero as $|z^2|\to \infty$, while the tail is bounded by that small supremum.

As literally printed without boundedness or a growth condition, the most one can assert isThe zero-boundary harmonic growth in a quadrant includes $v=cxy$ and more generally $v=\mathrm{Im}(z^{2m})$. This is an actual nonuniqueness of the printed problem, not a defect in the integral for the normalized bounded solution.

The convention used for planar Gaussian free field theory is the Dirichlet energy space:It is a real Hilbert space with the extended Dirichlet inner product. A conformal map from the unit disc onto $D$ identifies the two energy completions, so local representatives can be understood through the disc's zero-boundary space.

For an arbitrary unbounded domain this is the homogeneous energy completion; it should be distinguished from completion in the inhomogeneous norm $\Vert f{\Vert }_{L^2}+\Vert \nabla f{\Vert }_{L^2}$ used in another standard definition of the zero-boundary Sobolev space. On bounded domains with a Poincare inequality, the two norms are equivalent.

For real test functions $f,g\in C_c^{\infty }(D)$, the Dirichlet inner product in the normalization compatible with the logarithmic Green kernel isIts norm measures gradient energy rather than the values of the functions. A change of variables shows that this energy is invariant under planar conformal maps: the squared derivative in the transformation of the gradient cancels the area Jacobian determinant.

Set $F(w)=(\varphi (w)-\varphi (0))/w$ for $w\ne 0$. The singularity is removable, with $F(0)={\varphi }^{\prime }(0)$. Injectivity of the conformal map makes $F(w)\ne 0$ away from zero, and conformality makes $F(0)\ne 0$ as well. Since the disc is simply connected, $F$ has a holomorphic logarithm. Therefore $\psi =\log |F|$ is a harmonic function, including at zero, and $\psi (0)=\log |{\varphi }^{\prime }(0)|$.

The mean value property for harmonic functions gives, for every $0<r<1$,For general $D$, boundary values mean radial limits, rather than a continuous extension of $\varphi$ to every point of the circle. The boundary logarithmic mean of a univalent function justifies taking $r\uparrow 1$: the Koebe distortion theorem bounds $|F|$ below by $|{\varphi }^{\prime }(0)|/4$, while the standard integral-mean bound for a univalent function, ${\sup }_{r<1}\int |\varphi (r{e}^{i\theta }){|}^{p}d\theta <\infty$ for $0<p<1/2$, gives uniform integrability of its positive logarithm. Radial limits exist almost everywhere, and passage to the integral follows. Thus, writing $dw=d\theta$ for arc length on the unit circle,The expression on the left is the logarithm of the conformal radius of $D$ at $z=\varphi (0)$.

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.

Let $v_1,v_2\ne 0$ be tangent vectors to the two smooth curves at $p$. complex differentiability at a point givesWhen $f^{\prime }(p)\ne 0$, its real derivative is multiplication by a nonzero complex number, which is a rotation followed by a positive scaling. It therefore preserves the angle between $v_1$ and $v_2$, so $f$ is conformal at $p$. The condition is essential: $f(z)=z^2$ has $f^{\prime }(0)=0$ and sends rays making angle $\pi /4$ at zero to rays making angle $\pi /2$.

For $J(z)=z+z^{-1}$,If $z,w$ both satisfy $|z|>1$, or both satisfy $0<|z|<1$, the alternative $zw=1$ is impossible unless the points lie on the omitted unit circle. Hence $J$ is one-to-one on each region. Also $J^{\prime }(z)=1-z^{-2}$ has no zero there, so the restrictions are conformal. Solving $z^2-wz+1=0$ shows that $w\in [-2,2]$ exactly when both roots lie on the unit circle; otherwise one root is inside and the other outside. Thus each region has image

Taking the reciprocal of the interior restriction and filling its removable value at zero givesThe reciprocal sends $\mathbb{C}\setminus [-2,2]$ ontoand $g(0)=0$. Therefore $g$ is the required one-to-one conformal map from the unit disc.

A compact H-hull is a bounded, relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityThe half-plane capacity is

Let $D$ be simply connected domain with distinct marked boundary points $a,b$, and choose a conformal map $f:D\to \mathbb{H}$ with $f(a)=0$ and $f(b)=\infty$. Chordal ${\mathrm{SLE}}_{\kappa }$ from $a$ to $b$ is the unparameterized curve $f^{-1}(\gamma )$, where $\gamma$ is chordal SLE in $(\mathbb{H},0,\infty )$.

Any other such map is $\tilde{f}=rf$ for some $r>0$. The Scaling invariance of SLE says that $r^{-1}\gamma$ has the same unparameterized law as $\gamma$; only its capacity clock changes. Hence the pullback law is independent of $f$. This proves the Conformal invariance of SLE definition is well-defined.