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Use the unitary normalization of the Fourier transform,The Schwartz space $\mathcal{S}({\mathbb{R}}^n)$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }f(x)|$ is finite. Such functions and all their derivatives are integrable. Differentiating under the integral and integrating by parts giveConsequentlyExpand the derivative by the product rule and insert the integrable weight $(1+|x|{)}^{-n-1}$; the right side is bounded by finitely many Schwartz seminorms of $f$. This proves continuity $\mathcal{F}:\mathcal{S}\to \mathcal{S}$. The identical argument with the opposite sign proves continuity of $\mathcal{G}$.

For surjectivity and injectivity an inversion proof is needed. If $f\in \mathcal{S}$, multiply $\mathcal{F}f(\xi )$ by $e^{-\epsilon |\xi {|}^2/2}$. Fubini's theorem and the elementary Gaussian integral giveThe Gaussian has total integral one and its mass outside any fixed ball tends to zero. The right side therefore tends to $f(x)$, by uniform continuity of $f$. On the left, $\mathcal{F}f\in \mathcal{S}\subset L^1$, so dominated convergence gives $\mathcal{GF}f(x)$. This proves $\mathcal{GF}=I$. Reversing the signs in the same calculation proves $\mathcal{FG}=I$. ThusIts inverse is continuous by the seminorm estimate already proved. This establishes the requested bijection without assuming Fourier inversion as an unexplained theorem.

Inversion gives ${\mathcal{F}}^{2}f(x)=f(-x)$ and the Parseval identity. Indeed, inserting the integral for $\mathcal{F}f$ into $\int \mathcal{F}f\overline{\mathcal{F}g}$ and using $f,\mathcal{F}g\in L^1$ gives $\int f\overline{g}$. Density of Schwartz functions in $L^2$ extends $\mathcal{F}$ to a unitary operator on $L^2$, the Plancherel theorem. Translation of $f$ becomes multiplication by a phase, modulation becomes frequency translation, andThese formulas explain why the Fourier transform diagonalizes constant-coefficient differential operators. By transpose it also defines a bijection of the tempered distributions: with the complex-linear distributional pairing, $⟨\mathcal{F}T,\varphi ⟩=⟨T,\mathcal{F}\varphi ⟩$, since the Fourier kernel is symmetric in its variables.

The connection with creation and annihilation operators is especially transparent for the harmonic oscillator. On the common invariant dense domain $\mathcal{S}$, setIntegration by parts shows that these are formal Hilbert-space adjoints. They are closable unbounded operators; commutators below are computed on $\mathcal{S}$, avoiding undefined products on arbitrary $L^2$ vectors. The identity $[{\partial }_j,x_k]={\delta }_{jk}I$ gives the canonical commutation relationsIt also givesThe normalized vacuum $h_0(x)={\pi }^{-n/4}{e}^{-|x{|}^2/2}$ satisfies $a_j{h}_0=0$. Define the Hermite functionsMoving annihilation operators past creation operators with the commutation relations provesThese same relations show that the Hermite functions are orthonormal: unequal occupations are orthogonal eigenvectors of one of the commuting symmetric number operators $a_j^{†}{a}_j$, and the ladder relations recursively give norm one.

For completeness, let $g\in L^2$ be orthogonal to every $h_{\alpha }$. The Hermite polynomials have nonzero leading monomials and triangular lower-degree terms, so these orthogonality relations say that all polynomial moments of $h(x)=g(x)e^{-|x{|}^2/2}\in L^1$ vanish. Its Fourier integral extends to an entire function of $\zeta \in {\mathbb{C}}^n$:The Cauchy-Schwarz inequality, with the Gaussian absorbing every exponential linear weight and every polynomial, justifies differentiation uniformly on compact sets of $\zeta$. Every Taylor coefficient at zero is a vanishing moment, so the entire function is zero. Fourier uniqueness on $L^1$ follows here without circularity: if an $L^1$ function has zero Fourier transform, convolving it with ${\rho }_{\epsilon }$ gives zero by the Gaussian integral and Fubini, and the approximate-identity limit in $L^1$ gives the original function zero. Thus $h=0$ and $g=0$. The Hermite functions form an orthonormal basis of $L^2$.

Finally, the elementary Gaussian integral gives $\mathcal{F}{h}_0=h_0$. The derivative and multiplication identities above implywhere the operators on the right act on the frequency variable. HenceThus creation and annihilation organize both the oscillator's spectral decomposition and the four possible Fourier eigenvalues. On Schwartz functions the oscillator powers control weighted derivatives, and Hermite coefficients decay faster than every power of $1+|\alpha |$; conversely this rapid coefficient decay implies all Schwartz seminorms are finite. The Fourier phases preserve this decay, providing another explanation of the Fourier automorphism of Schwartz space.

The loop group here is $LU(1)=C^{\infty }(S^1,U(1))$, with pointwise multiplication. Although it is abelian, its nontrivial positive-energy quantum representations are projective: implementers can commute only up to a phase. A positive energy representation of the loop group of U(1) has rotations $R_t$ whose self-adjoint generator is bounded below and which implement $g(z)\mapsto g(e^{it}z)$ in the projective unitary group. The fermionic construction gives both that phase and an explicit bosonic description.

Choose the one-particle Hilbert space $H=L^2(S^1)$ with modes $e_r(z)=z^r$, $r\in \mathbb{Z}$, and projection $P$ onto $r\ge 0$. Use the polarized fermionic Fock spacewhere the exterior powers are Hilbert-completed and the tensor product is graded. Its vacuum fills all modes $r<0$ and leaves $r\ge 0$ empty; excitations are particles in the empty modes and holes in the filled modes. The fermionic annihilation operators $c_r$ and fermionic creation operators $c_r^{\ast }$ satisfy the canonical anticommutation relationsFinite particle-hole configurations form an orthonormal basis. As in the finite-dimensional exterior algebra construction, their creation and annihilation operators act irreducibly: an operator in the commutant sends the vacuum to a scalar vacuum, since the joint kernel of its vacuum annihilators is one-dimensional, and then acts by that scalar on the dense configuration basis.

Segal's quantization criterion for fermions says that a unitary one-particle operator $u$ has a unitary implementer $\Gamma (u)$ on this Fock space, satisfying $\Gamma (u)c^{\ast }(f)\Gamma (u{)}^{\ast }=c^{\ast }(uf)$, exactly whenEquivalently $P-uP{u}^{\ast }$ is Hilbert-Schmidt. Implementers are unique up to phase, hence form a projective unitary representation of the restricted unitary group. This is the fermionic, polarized criterion; it is not a condition that the entire operator $u-I$ be Hilbert-Schmidt. The criterion and the corresponding continuity topology are stated in Theorem III.2.1.1 of the fermionic construction of loop-group representations. For the compact principal-angle part of a pair of polarizations, the threshold has a concrete interpretation. Its two-dimensional blocks are rotations through angles ${\theta }_j$; finitely many unmatched modes can be handled separately. Each angle ${\theta }_j$ rotates the vacuum into $\cos {\theta }_j$ times its original component plus $\sin {\theta }_j$ times a particle-hole pair. After discarding finitely many exceptional blocks, the successive tensor products are Cauchy precisely when ${\prod }_j\cos {\theta }_j$ converges to a nonzero value, equivalently ${\sum }_j{\sin }^2{\theta }_j<\infty$. In this block model the sum is exactly half the squared Hilbert-Schmidt norm of the projection difference; finitely many unmatched modes account for a change of charge. Infinitely many unmatched modes already violate the criterion. This explains why infinitely many arbitrarily small mixings can fail to be implementable.

For $g(z)={\sum }_n{g}_n{z}^n\in LU(1)$, multiplication $M_g$ is a unitary one-particle operator. Its matrix entry from $e_s$ to $e_r$ is $g_{r-s}$. ThereforeFor a fixed difference $n$, exactly $|n|$ pairs cross the polarization boundary. Smoothness makes the sum finite, so Segal's quantization criterion for fermions constructs a continuous projective unitary representation of every smooth loop. Smooth convergence gives operator-norm convergence of $M_g$ and Hilbert-Schmidt convergence of its commutator, which is sufficient for the strong projective continuity of the implementers.

Normal ordering fixes the charge and rotation energy. On the dense finite-configuration domain putThe sums are finite on each configuration. A particle at $r\ge 0$ adds charge one and energy $r$; a hole at $r<0$ adds charge minus one and energy $-r$. Consequently $L_0$ is a nonnegative self-adjoint diagonal operator with integer spectrum and finite-dimensional eigenspaces. The unitaries $R_t=e^{it{L}_0}$ are strongly continuous, have period $2\pi$, and implement $e_r\mapsto e^{itr}{e}_r$. Conjugating multiplication by this one-particle rotation gives $M_{g(e^{it}\cdot )}$, so their Fock implementers rotate the loop representation projectively. This is an actual positive-energy circle action.

The scalar anomaly is visible in the Heisenberg current algebra of a fermion on the circle. Define on finite-energy vectorsFor fixed $n$ the sum has only finitely many nonzero actions on any configuration. Direct use of the canonical anticommutation relations givesOn summing, the two noncentral sums cancel. If $m=-n$ the central sum is ${\sum }_r(1_{r<0}-1_{r+n<0})=n$: for $n>0$ the terms are precisely $r=-n,\dots ,-1$. ThusIn particular $J_n\Omega =0$ for $n>0$. The coefficient one in the central term is the level of this basic representation.

Let $h(\theta )={\sum }_n{h}_n{e}^{in\theta }$ be smooth and real. The normal-ordered infinitesimal implementer of multiplication by $h$ isThe sign of the current index matters: multiplication by $z^n$ raises the one-particle mode by $n$, so it corresponds to $J_{-n}$. It follows either directly from the bilinear commutator or from the fermionic currents on the circle that $[A(h),c_r^{\ast }]={\sum }_n{h}_n{c}_{r+n}^{\ast }$. For trigonometric $h$ these are oscillator fields on the finite-energy domain; their self-adjoint closures exponentiate to the implementers of $e^{ih}$. For general smooth $h$, their extensions are obtained by the oscillator-field construction below: ${\sum }_{n>0}n|h_n{|}^2<\infty$ guarantees the square-summable field coefficients, and Fourier truncation converges in the corresponding Weyl relations implementers. Thus no unjustified infinite operator product is required.

SetIt is real and antisymmetric for real $h,k$. The Heisenberg current algebra of a fermion on the circle relations give $[A(h),A(k)]=i\omega (h,k)I$. The Weyl relations, or the finite-mode Baker-Campbell-Hausdorff computation followed by the square-summable limit, yieldThis is an explicit two-cocycle on the identity component of the loop group. A different phase choice changes it by a group coboundary, but its antisymmetric commutator remains $e^{-i\omega (h,k)}$, and hence cannot be removed.

The winding components also matter. Every smooth loop is $z^w{e}^{ih}$ with $w\in \mathbb{Z}$, real periodic $h$, and mean $h_0$ determined modulo $2\pi$. Let $S$ implement multiplication by $z$. It moves the filled boundary one place upwards, so it raises charge by one; phases of charge ground states can be chosen so that $S{\Omega }_q={\Omega }_{q+1}$. Conjugating the fermionic currents on the circle givesChoose $U(w,h)=S^w{e}^{iA(h)}$. Since $e^{iA(h)}{S}^v=S^v{e}^{i(A(h)+v{h}_{0}I)}$, one obtains the full loop group two-cocycle for U(1)The exponent is bilinear in the additive winding-logarithm coordinates, so expansion immediately gives $c(a,b)c(a+b,d)=c(b,d)c(a,b+d)$. Replacing $h$ by $h+2\pi \ell$ changes $A(h)$ by $2\pi \ell Q$, whose exponential is identity since charges are integers; the scalar cocycle also stays unchanged since $v\in \mathbb{Z}$. Thus the expression is well-defined on loops, not just on chosen logarithms. It records both the derivative anomaly and the commutator of winding with constant loops.

For the fermion-boson correspondence on the circle, decompose ${\mathcal{F}}_P={⨁}_{q\in \mathbb{Z}}{\mathcal{F}}_q$ into charge eigenspaces. The charge-$q$ ground state ${\Omega }_q$ fills all integer modes $r<q$. Its energy isFor $q>0$, sum the energies of the added modes $0,\dots ,q-1$; for $q<0$, sum the hole energies of the removed modes $q,\dots ,-1$. Every other charge-$q$ configuration is obtained by moving finitely many occupied levels upwards, so this is the lowest energy. Since $J_n$ preserves charge and lowers energy by $n>0$, $J_n{\Omega }_q=0$ for $n>0$.

Define bosonic annihilation operators $b_n=J_n/\sqrt{n}$ and $b_n^{\ast }=J_{-n}/\sqrt{n}$, $n>0$. They obey $[b_n,b_m^{\ast }]={\delta }_{nm}I$. Their occupation vectorsare orthonormal by commuting positive fermionic currents on the circle through negative fermionic currents on the circle. Their energies are $q(q-1)/2+{\sum }_{n>0}n{m}_n$. It remains to prove that they span, rather than assume the boson correspondence. Count configurations by formal charge $u$ and energy $t$. Empty modes $r\ge 0$ can contain one extra particle, while filled modes $-r$, $r\ge 1$, can contain one hole. Therefore the fermionic character isThe charge form of the Jacobi triple product proved in Question 4 identifies this withThe reciprocal product counts bosonic occupations, or partitions of an integer. For fixed charge and energy, the finite-dimensional fermion space has exactly as many basis configurations as the independent fermionic current on the circle monomials already constructed. Hence those monomials exhaust every such space, proving their dense completeness. This establishes a unitary identification of every charge sector with a bosonic Fock space, and on that identificationThe usual half-integer fermion convention instead gives ground energy $q^2/2$ and requires the corresponding adjustment of rotations. Here integer modes were deliberately chosen to obtain a genuine circle action with integral nonnegative energies. The two ground states of energy zero, at charges zero and one, are therefore consistent, not a negative-energy defect.

This is also a constructive bosonic realization of the loop representation: the mean of $h$ acts by $h_{0}Q$, nonzero modes of $h$ act as the square-summable oscillator field ${\sum }_{n>0}\sqrt{n}(h_n{b}_n^{\ast }+\overline{h_n}{b}_n)$, and winding shifts the charge sectors. Taking $k$ tensor copies adds the fermionic current on the circle central terms and gives level $k\in \mathbb{N}$. Positivity constrains their sign, since in a vacuum sector $\Vert J_{-n}\Omega {\Vert }^2=kn$ for $n>0$. The level-one construction, its explicit loop cocycle, and its complete bosonic occupation basis show how an abelian classical loop group acquires noncommuting quantum currents while retaining positive rotation energy.

The infinite product converges locally uniformly in $\mathbb{H}$, because ${\sum }_{n\ge 1}|q{|}^n<\infty$ uniformly on compact subsets. None of its factors vanishes there, and the convergent logarithm of its tail shows that the product is nonzero. Logarithmic differentiation, justified by locally uniform convergence, givesThe middle equality groups the absolutely convergent terms $n{q}^{nr}$ by $m=nr$. Thus $F^{\prime }/F=2\pi i{E}_2$.

Let $H(\tau )={\tau }^{-12}F(-1/\tau )/F(\tau )$. It is nonzero and holomorphic on the simply connected upper half-plane. Using the printed transformation $E_2(-1/\tau )={\tau }^2{E}_2(\tau )+6\tau /(\pi i)$, its logarithmic derivative isSo $H$ is constant. At $\tau =i$ we have $-1/i=i$ and $i^{-12}=1$, giving $H(i)=1$. Therefore $F(-1/\tau )={\tau }^{12}F(\tau )$. Also $F(\tau +1)=F(\tau )$ directly from $q=e^{2\pi i\tau }$. Since these two transformations generate $S{L}_2(\mathbb{Z})$, $F$ has the weight-twelve modular form transformation law.

At the unique modular cusp its expansion is $F=q+O(q^2)$, so it extends holomorphically with zero value. HenceThis is the modular discriminant. The transformed argument contributes $6\tau /(\pi i)$; the converted TeX's numerator $67$ is a transcription error, not the printed relation.

Use $\Theta =(2\pi i{)}^{-1}d/d\tau$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, differentiating the weight-$k$ law givesThe transformations of $E_2$ under the generators $S,T$, and the chain rule for their compositions, giveEquivalently, one can check just $S,T$ in the next cancellation, since they generate the group. Substituting these formulas intocancels the extra term and gives $(D_{k}f)(\gamma \tau )=(c\tau +d{)}^{k+2}{D}_{k}f(\tau )$. Both constituents are holomorphic on $\mathbb{H}$ and have Fourier expansions of a modular form with nonnegative exponents, so $D_{k}f$ is holomorphic at the modular cusp. ThusThis operator is the Serre derivative.

If $f={\sum }_{n\ge 0}{a}_n{q}^n$, then its derivative has constant coefficient zero, while $E_2$ has constant coefficient one. ThereforeFor $k\ne 0$ this vanishes exactly when $a_0(f)=0$, proving the claimed cusp form equivalence in the intended positive-weight setting. At weight zero the printed equivalence is false: $f=1$ is a noncuspidal weight-zero modular form, but $D_{0}f=0$ belongs to $S_2$. Indeed all weight-zero forms are constants by compactness of $X(1)$. Negative-weight level-one modular forms are zero by the valence formula for the modular group, so there is no further nonzero exception. The transformation and holomorphy conclusion for $D_{k}f$ remains true at weight zero.

Let ${\xi }_t=\sqrt{\kappa }{B}_t$ and $0<\kappa <4$. We use these precise Bessel process facts, which the question permits: for dimension $\delta >2$ a process started at $r>0$ exists for all time, never reaches zero, and has a strictly positive all-time minimum. More quantitatively,No assertion about SLE simplicity or transience is being assumed.

For a real point $x>0$ before its Loewner lifetime, set $Z_t^x=g_t(x)-{\xi }_t$. Its equation isThus $Z^x/\sqrt{\kappa }$ is a Bessel process of dimension $1+4/\kappa >2$. For each rational $x>0$ it stays positive on every finite interval, almost surely. These countably many events hold together. They cover every real $y>0$ as well: choose rational $0<x<y$ and observe that the noise cancels in the difference,Consequently $Z^y$ cannot reach zero while $Z^x$ remains positive. The negative side has the identical argument with the reflected Brownian motion.

This excludes Loewner trace contact with every real point other than zero, rather than just with each fixed rational point. On any finite interval a positive gap at a real point stays bounded away from zero. Continuous dependence of the Chordal Loewner equation then solves it on a complex neighborhood of that point, and the analytic flow extends the map conformally across a real interval there. Its real derivative is $\exp (-2{\int }_0^t(Z_s^x{)}^{-2}ds)>0$, and its image stays away from the driving point. The Loewner trace tip therefore cannot be there. We have proved $\gamma [0,\infty )\cap \mathbb{R}\subseteq \{0\}$.

Now use Brownian motion independent increments to restart at every deterministic rational time $s>0$. The centered future maps have Loewner driver ${\xi }_{s+t}-{\xi }_s$ and therefore the original SLE law, independently of the past. Its Loewner trace stays in $\mathbb{H}\cup \{0\}$ by the preceding argument. The inverse map $g_s^{-1}$ extends continuously to the real boundary for a continuous trace-generated compact H-hull: the finite Loewner trace gives a locally connected boundary, and the Caratheodory boundary extension theorem applies. Its value at ${\xi }_s$ is $\gamma (s)$. Thus the future in the original plane can meet $K_s\cup \mathbb{R}$ only at $\gamma (s)$. All these restart conclusions hold simultaneously for rational $s$.

Suppose $\gamma (u)=\gamma (v)$ with $u<v$. The Loewner trace cannot be constant throughout $(u,v)$, since its compact H-hull capacity increases strictly. By continuity there is a rational $s\in (u,v)$ with $\gamma (s)\ne \gamma (u)$. But $\gamma (v)=\gamma (u)$ lies in $K_s\cup \mathbb{R}$, contradicting the restart conclusion at $s$. This also excludes any return to the starting point. Hence the Loewner trace is simple and lies in the open complex upper half-plane at every positive time.

For transience, first strengthen the boundary conclusion to avoidance of the closure at each fixed real point. Fix $x>0$ and $0<r<x/4$. If the now-simple curve first enters the closed half-disc of radius $r$ about $x$ at time $\sigma$, join its tip to $x$ by the segment inside that disc. Before $\sigma$ the segment misses the Loewner trace. Together with the Loewner trace and the real interval from zero to $x$, it bounds a pocket whose boundary includes the positive side of the Loewner trace and $[0,x/2]$. A Brownian motion path from high on the imaginary axis reaching that boundary arc must enter the pocket through the segment. The maximum principle for harmonic functions bounds its harmonic measure by the chance of hitting the entire half-disc.

The boundary arc maps under $g_{\sigma }$ to $[{\xi }_{\sigma },g_{\sigma }(x/2)]$. Its harmonic measure from $iY$, as $Y\to \infty$, isThe half-disc maps out by $z\mapsto z+r^2/(z-x)$; its semicircle becomes an interval of length $4r$. Its harmonic measure has asymptotic $4r/(\pi Y)$. Comparing and letting $Y\to \infty$ yieldsBut $m_x={\inf }_{t\ge 0}(g_t(x/2)-{\xi }_t)>0$ by the all-time minimum of a transient Bessel process fact. The curve cannot enter any such disc with $4r<m_x$. Therefore it stays a strictly positive distance from $x$ for all time. Reflection proves the same for fixed $x<0$. This is the boundary approach forces a small SLE Bessel gap argument; the crosscut comparison is also described in the boundary-closure lemma in the primary SLE paper.

Map out the simple initial slit at time one. The starting point zero has two distinct real prime end images $b_{-},b_{+}$ under $g_1$, lying on opposite sides of ${\xi }_1$. Conditionally on the initial slit, the image future is a fresh independent SLE translated by ${\xi }_1$. The just-proved fixed-point closure avoidance applies to each of $b_{-}-{\xi }_1$ and $b_{+}-{\xi }_1$. Therefore neither $b_{-}$ nor $b_{+}$ lies in the closure of the mapped future. If a sequence of future Loewner trace points approached zero in the original domain, their images would approach one of these two prime ends. This is impossible. ThusFinally Scaling invariance of SLE gives $R_T:={\inf }_{t\ge T}|\gamma (t)|\stackrel{d}{=}\sqrt{T}{R}_1$. For every fixed $R>0$,The events of returning to the closed radius-$R$ disc after time $T$ decrease as $T$ increases. Continuity of probability on decreasing events implies that the probability of returns at arbitrarily late times is zero. Taking the countable intersection over positive integer $R$ provesThis proves actual eventual escape, not merely unboundedness of the growing compact H-hulls.

Each marginal is a Bessel process of dimension $\delta =1+2a\in (1,2)$, but finite lifetime can also be proved directly. Apply the Itô formula before $X$ reaches zero:With clock $u(t)={\int }_0^t{X}_s^{-2}ds$, the Itô integral is a Brownian motion $W_u$. The inverse clock satisfiesThe clock runs through all $u\ge 0$ up to the maximal lifetime: a finite terminal clock would leave $X$ at a finite positive limit, allowing continuation, or, at an infinite physical lifetime, would make its clock grow at a positive rate, a contradiction. There is no finite-time explosion to infinity; for example $d{X}_t^2=2X_{t}d{B}_t+(1+2a)dt$ and localization bounds the probability of reaching arbitrarily large levels on a fixed time interval. Since $W_u/u\to 0$ almost surely and $1-2a>0$, the last exponential integral has finite limit, while $X_{t(u)}\to 0$. ThereforeThe same argument applies to $Y$, whose noise $-B$ is also Brownian motion, so its lifetime is finite too. No independence between the two lifetimes is being claimed.

Set $S=X+Y$ and $U=Y/S$ before $T=\zeta \wedge \tau$. Brownian motion terms cancel in the sum:Thus simultaneous extinction is impossible. At the finite first lifetime, precisely one coordinate is zero and the other remains positive. Because $S$ has finite variation, differentiating the ratio givesAfter clock change $v={\int }_0^t{S}_s^{-2}ds$, its generator isA scale function of a one-dimensional diffusion $H$ solves $\mathcal{L}H=0$, soBoth endpoint singularities are integrable since $a<1/2$. Normalize $H(0)=0,H(1)=1$ to obtainThe Itô formula makes $H(U_{t\wedge T})$ a bounded martingale, first by localization away from the endpoints and then by bounded convergence theorem. Its terminal value is one on $\{\zeta <\tau \}$, when $U_T=1$, and zero on the opposite event, when $U_T=0$. optional stopping theorem therefore proves the oppositely driven Bessel exit probability

Here $\mathrm{B}$ is the beta function and $\Gamma$ is the Gamma function.

The integral printed in the question is incorrect for general $a$. Its first exponent is $2-4a$, whereas the two coupled equations give $2a$. At $a=1/4$, which lies in the allowed interval, its integral from zero diverges for every positive upper limit; no finite normalizing constant gives a probability. Moreover exchanging $X$ and $Y$ while changing $B$ to $-B$ shows that at $x=y$ the probability must be $1/2$. The corrected symmetric integral has this value for every $a\in (0,1/2)$, whereas the printed asymmetric normalized integral generally does not. For $a>1/4$, even when the printed integral converges, its derivative density $H_{\mathrm{print}}^{\prime }(u)\propto u^{-(2-4a)}(1-u{)}^{-2a}$ satisfiesso it is harmonic for the required exit problem only at $a=1/3$. The two exponents agree exactly at that value.

To relate that value to percolation, write the two SLE boundary gaps as $g_t(x)-{\xi }_t$ and ${\xi }_t-g_t(-y)$. In Brownian-time units $s=\kappa t$, they have the displayed coupled equations with $a=2/\kappa$, after reversing the Brownian motion sign. Hence $a=1/3$ corresponds to $\kappa =6$. The event is that the positive marked point $x$ is swallowed before the negative marked point $-y$.

For a quadrilateral with boundary points in the order $(-y,0,x,\infty )$, its conformal cross-ratio coordinate is $u=y/(x+y)$. In the critical percolation exploration, wire the arcs $(-y,0)$ and $(x,\infty )$ blue and the other two arcs yellow. Exploration from zero, stopped when one of the two marked sides is swallowed, detects a blue crossing between those two blue arcs exactly in the event just computed. The continuum exploration has the SLE6 law; its Locality property of SLE lets one stop the ordinary chordal exploration before the distant boundary colors affect it. Thus the crossing probability isthe Cardy boundary crossing formula. This is the rigorous scaling-limit formula for critical site percolation on the triangular lattice; applying it to other critical planar models requires the corresponding universality assumption. At this percolation value of $a$, the printed integral is already the correct one.

Work in the prime end closure of the marked domain, so that its two conformal boundary points are distinct even if their physical boundary representatives coincide. A chordal filling is a closed connected set joining those points, meeting the conformal boundary nowhere else, and full relative to the two boundary arcs between them: complementary pockets that do not reach either boundary arc have been added. In the normalized complex upper half-plane, its two complementary sides are simply connected and abut the negative and positive real axes. For a proper boundary-to-boundary curve, chordal filling means adjoining all such trapped pockets to its range. Independent curves are combined by taking their union and then applying this same chordal filling operation.

An admissible marked subdomain $V\subset U$ contains neighborhoods of the two marked boundary points and has a simply connected component joining them. The restriction property says that, conditional on the random chordal filling $K$ being contained in $V$, its law is the chordal filling law for $(V,z_0,z_1)$. Equivalently, mapping the conditional chordal filling conformally back to $(U,z_0,z_1)$ recovers the original law. The contained-set events are understood away from the shared marked endpoints.

By transporting the family under conformal maps, it suffices to work in $(\mathbb{H},0,\infty )$. Let $A$ be an admissible compact H-hull whose closure avoids zero, and normalizeIn terms of the single normalized measure $\mu$, restriction is preciselyThe Schwarz reflection principle extends ${\Phi }_A$ analytically near zero, with ${\Phi }_A^{\prime }(0)>0$, so the stated conditioning events have positive probability for the two laws considered below. This is the chordal restriction property for filled sets. An unbounded marked subdomain can be approximated by such compact-hull removals; conformal maps transport the formulation to general domains.

We need a determination principle for filled sets, since a probability of one avoidance event alone would not establish equality of laws. Choose countably many thin polygonal compact H-hulls with rational data, attached to real boundary intervals away from zero. For each chordal filling, every point of its complement has a path within one of the two complementary sides to the appropriate real boundary interval. A sufficiently thin tube along such a path avoids the chordal filling and contains a neighborhood of that point. Hence the union of interiors of all avoided test tubes is exactly the complement.

Their joint avoidance probabilities are also determined by single-hull probabilities. For a finite union of test compact H-hulls, either the union separates zero from infinity, in which case a connected chordal filling cannot avoid it, or its unbounded complementary component contains both marked endpoints. In the latter case fill the union's bounded pockets to get an admissible compact H-hull $C$. A connected set joining zero to infinity and avoiding the union cannot enter one of these separated pockets, so it avoids $C$ as well. Thus joint avoidance is either impossible or is the avoidance event for $C$. The inclusion-exclusion principle gives every finite joint probability distribution of the tube-avoidance indicators, and these indicators determine the random complement and chordal filling. This proves avoidance probabilities determine a chordal filling law.

In the given avoidance formulas, the argument of the conformal map must be the retained open domain, rather than the random closed filling. We express that domain as $\mathbb{H}\setminus A$ to keep the two objects distinct. The formulas then imply for the two filled processesThe chordal filling operation does not change avoidance of an admissible compact H-hull: every piece of that compact H-hull is attached to the real boundary away from the endpoints, and so cannot be inside a trapped pocket while its generating curve avoids the compact H-hull. These laws are conformally invariant as unparameterized chordal filling laws. For SLE this follows from Question 1; for the Brownian half-plane excursion, Brownian scaling of its independent one- and three-dimensional coordinates shows that its unparameterized law is unchanged by positive dilations, and mapping the Brownian half-plane excursion to another marked domain gives the conformally natural law. Conformal invariance of planar Brownian motion describes the corresponding change of time, which does not alter the chordal filling.

Here is the conditional-law calculation for either exponent $\alpha \in \{5/8,1\}$. For another admissible compact H-hull $B$ in the image complex upper half-plane, setThe complement of $C$ is mapped to $\mathbb{H}$ by this composition, so $C$ is again admissible; both maps fix zero and have derivative one at infinity. The chain rule gives ${\Phi }_C^{\prime }(0)={\Phi }_B^{\prime }(0){\Phi }_A^{\prime }(0)$. ThereforeThese are exactly the original chordal filling's avoidance probabilities for every $B$. The determination principle proves equality of the conditional image law with the original law. Hence both generated chordal fillings have restriction, with exponents $5/8$ and $1$ respectively. This is a complete implication from the two allowed avoidance formulas; the restriction conclusion was not assumed.

For independent chordal fillings, avoidance of $A$ by their filled union is equivalent to avoidance by every component. Independence multiplies the probabilities, and thus adds their restriction exponents of a chordal filling. ConsequentlyThe determination principle applies once more, givingThe equality concerns the filled, unparameterized closed sets. The numerical reason is $8\cdot (5/8)=5\cdot 1=5$.

Let $I=(U^{\prime }{)}^{-1}$ be the inverse marginal utility. The derivative $U^{\prime }$ is continuous and strictly decreasing: concavity makes it nonincreasing, and equality at two distinct points would make $U$ affine between them, contradicting strict concavity. The Inada conditions therefore make $U^{\prime }$ a bijection from $(0,\infty )$ onto $(0,\infty )$. For each $y>0$, the derivative of $U(x)-xy$ is positive below $I(y)$ and negative above it. Consequently its unique global maximum is finite andThere is no need to differentiate $I$ to find the first derivative of this utility conjugate. Comparing the maxima at $y$ and $y+h$ gives, for $h>0$,Continuity of $I$, together with the corresponding inequalities for $h<0$, givesThus $V$ is strictly decreasing and strictly convex, since its derivative is strictly increasing. This proves all the asserted properties except the second differentiability, which does not follow from the printed hypotheses.

Here is an explicit Inada utility with vanishing curvature. PutThe positive integral defining $p$ makes $U$ strictly increasing. Also $p^{\prime }=-w$ is negative except at one point, so $p$ is strictly decreasing and $U$ is strictly concave. Near zero $w(s)$ is asymptotic to $s^{-2}$, and at infinity it is asymptotic to $s^{-2}$; thus $p(0+)=\infty$ and $p(\infty )=0$. This smooth utility function satisfies every printed positive-domain assumption, but $U^{\prime \prime }(1)=0$. At $y_0=p(1)=3-4\log 2$, differentiability of $I$ would imply, by differentiating $p(I(y))=y$,which is impossible. Hence $V^{\prime }=-I$ has no finite derivative there. More explicitly, $p(x)-p(1)=-(x-1{)}^3/12+o((x-1{)}^3)$, so its inverse function has a cube-root singularity. The twice-differentiable conclusion is false as printed. With the additional assumption $U^{\prime \prime }(x)<0$ everywhere, the inverse function theorem proves the intended dual differentiability with nonvanishing utility curvature:

For the financial assertions, use the positive-wealth admissible trading strategy class appropriate to the positive-domain utility function, and expectations for which the expressions are defined. A normalized terminal state-price density satisfies $Z_N>0$, $\mathbb{E}{Z}_N=1$, and defines an equivalent martingale measure $Q$ by $dQ=Z_{N}dP$. A self-financing portfolio obeyswhere the holdings are measurable before the next price increment. Under $Q$ the gains are a martingale when integrable; for the usual lower-bounded admissible class, localization gives a supermartingale. In particular the state-price budget constraint is $\mathbb{E}[Z_N{X}_N]\le x$. The positive-domain qualification matters: assumptions only on $U$ at positive arguments do not control its values at nonpositive wealth.