Probabilistic Liouville three-point identity and its primary boundary value

First DOZZ supplier: exact GMC three-point normalization. Research theorem with a complete reduction, not an assumed DOZZ identity. Use the actual spherical covariance $K(z,w)=\log (1\mathrm{/}\mathrm{\mid }z-w\mathrm{\mid })+\log \mathrm{\mid }z{\mathrm{\mid }}_{+}+\log \mathrm{\mid }w{\mathrm{\mid }}_{+}$, background $dA0=\mathrm{\mid }z{\mathrm{\mid }}_{+}^{(-4)}{d}^{2}z$, variance-normalized GMC Mγ, $V\alpha =e^{(\alpha \phi )}$, and zero-mode measure dc. Here $0<\gamma <\mathrm{sqrt}(2)$, $\mu >0$, $Q=2\mathrm{/}\gamma +\gamma \mathrm{/}2$. This is the full γ interval of the current bootstrap target. The same argument applies at any $0<\gamma <2$ once the stated local GMC moment interface is supplied there; it does not assert the missing construction outside its currently proved range.

The actual local GMC input is the coordinate owner's completed disk-gmc-local-moments-and-uniform-inverse-moments.md in ../conformal-field-theory-string-theory-all-97-2026-10-26-owner-step1-liouville-spectral/: for $0<q<1$ and small disks, $E[M\gamma (B_r{)}^q]\le C_q{r}^{\xi }(q)$, $\xi (q)=(2+{\gamma }^2\mathrm{/}2)q-{\gamma }^2{q}^2\mathrm{/}2$, uniformly for centers in a fixed chart; every fixed positive-area anchor disk has all inverse moments. These two properties, together with the admitted mollifier-independent measure convergence, are the ONLY GMC estimates used below. They have an explicit local covariance/comparison proof in that supplier, not a DOZZ input.

Statement. For real $\alpha 1,\alpha 2,\alpha 3<Q$ and $A=\alpha 1+\alpha 2+\alpha 3>2Q$, set $s=(A-2Q)\mathrm{/}\gamma$ and

$\begin{array}{c}{\displaystyle \rho \text{(}\alpha \text{1,}\alpha \text{2,}\alpha \text{3)=}\int \text{\_C |x|\_+^}\text{(}\gamma \text{A)}}\\ {\displaystyle \text{|x|^}\text{(}-\gamma \alpha \text{1)|x}-\text{1|^}\text{(}-\gamma \alpha \text{2) M}\gamma \text{(d}{}^2\text{x).}}\end{array}$

Then $0<\rho <\mathrm{\infty }$ almost surely, $E[{\rho }^{(-s)}]<\mathrm{\infty }$, and the three-point coefficient of the sphere model with zero-mode measure dc is

$\begin{array}{c}{\displaystyle \text{C\_sph(}\alpha \text{1,}\alpha \text{2,}\alpha \text{3)=(1/}\gamma \text{) }\mu \text{^}\text{(}-\text{s) }\mathrm{\Gamma }\text{(s) E[}\rho \text{^}\text{(}-\text{s)]. (1)}}\end{array}$

Here $C_{sph}$ is the limit $\mathrm{\mid }z3{\mathrm{\mid }}^{(4\mathrm{\Delta }\alpha 3)}⟨V\alpha 1(0)V\alpha 2(1)V\alpha 3(z3)⟩$ as $z3\to \mathrm{\infty }$, with $\mathrm{\Delta }\alpha =\alpha (Q-\alpha \mathrm{/}2)\mathrm{/}2$. The same correlation definition with 2dc instead has $C_{KRV}=2C_{sph}$. This factor is mathematical normalization, not a change of μ or a renormalization of individual vertex operators.

1. Singular integrability and uniform insertion cutoff. Near an insertion zi with positive charge αi, divide a fixed small disk into annuli $r\mathrm{/}2<\mathrm{\mid }x-zi\mathrm{\mid }\le r$. Its singular integral has qth moment bounded by C $r^{[\xi (q)-\gamma \alpha iq]}$, since (sum $a_j{)}^q\le \sum a_j^q$ for $q\le 1$. Choose

$\begin{array}{c}{\displaystyle \text{0<q<min(1,2(Q}-\alpha \text{i)/}\gamma \text{).}}\end{array}$

Then ξ(q)−γαi $q=q[\gamma (Q-\alpha i)-{\gamma }^{2}q\mathrm{/}2]>0$, so the dyadic sum converges. The integral is finite almost surely and its mass in radius r tends to zero in probability with a strictly positive power of r. Negative charges only improve the local singularity and can be replaced by zero for this upper bound. The same bound holds for weights truncated at ε: the innermost disk gives ${\epsilon }^{(-\gamma \alpha iq)}E[M\gamma (B_{\epsilon }{)}^q]$, with the identical exponent. All other smooth weight factors are bounded above and below on the chosen disk.

At infinity use $w=1\mathrm{/}x$. The covariance and dA0 are invariant under this inversion, so the transformed GMC has the same law. The integrand in ρ behaves as $\mathrm{\mid }x{\mathrm{\mid }}^{(\gamma \alpha 3)}$; thus in w it is a bounded positive factor times $\mathrm{\mid }w{\mathrm{\mid }}^{(-\gamma \alpha 3)}$. The preceding proof with $\alpha 3<Q$ handles infinity. Away from the three singularities the weight is bounded and Mγ is finite. Therefore $\rho <\mathrm{\infty }$ almost surely.

Choose an anchor disk avoiding $0,1,\mathrm{\infty }$. Its weight has a positive lower bound c, so $\rho \ge cM\gamma (anchor)>0$ and every inverse moment of ρ is finite. In particular $E[{\rho }^{(-s)}]<\mathrm{\infty }$. These estimates also give uniform inverse-moment bounds when the three weights are replaced by their covariance-smoothed versions, because they remain uniformly bounded below on the same anchor disk.

2. Finite-cutoff Cameron–Martin identity, including every background factor. First take n distinct finite insertion points zi, real $\alpha i<Q$, and $A=\sum \alpha i>2Q$. At a fixed insertion smoothing ε, the zero-mode lemma applies pathwise. It gives

$\begin{array}{c}{\displaystyle \text{(1/}\gamma \text{)}\mu \text{^}\text{(}-\text{s)}\mathrm{\Gamma }\text{(s) }\prod \text{|zi|\_+^}\text{(}-\text{4}\mathrm{\Delta }\text{i)}}\\ {\displaystyle \text{E[}\prod \text{exp(}\alpha \text{iX}\epsilon \text{(zi)}-\alpha \text{i}{}^2\text{Var(X}\epsilon \text{(zi))/2) M}\gamma \text{(S)^}\text{(}-\text{s)].}}\end{array}$

For clarity, the Gaussian tilt used here needs no imported infinite-dimensional formula. For a finite Gaussian vector its identity follows by completing the square in its density. Apply it to finitely many Gaussian coordinates and the finite insertion vector, approximate bounded measurable functionals by cylinder functions in L1, and truncate the nonnegative functional before monotone convergence. This proves the same identity for a measurable functional of the Gaussian field.

Let $h\epsilon (x)={\sum }_i$ αi Cov(Xε(zi),X(x)). This is a deterministic continuous bounded function for fixed ε (the $\log \mathrm{\mid }x{\mathrm{\mid }}_{+}$ gauge term need not be smooth across the unit circle). The GMC shift law is exactly $M\gamma (X+h\epsilon )=e^{(\gamma h\epsilon )}M\gamma (X)$: it holds first for the normalized exponential regularizations, and then follows from their measure convergence and uniform convergence of the regularized continuous deterministic shifts. Completing the square therefore gives the preceding expression as

$\begin{array}{c}{\displaystyle \text{(1/}\gamma \text{)}\mu \text{^}\text{(}-\text{s)}\mathrm{\Gamma }\text{(s)}}\\ {\displaystyle \text{exp(sum\_(i<j) }\alpha \text{i}\alpha \text{j K}\epsilon \epsilon \text{(zi,zj))}}\\ {\displaystyle \prod \text{|zi|\_+^}\text{(}-\text{4}\mathrm{\Delta }\text{i)}}\\ {\displaystyle \text{E[(}\int \text{exp(}\gamma \text{h}\epsilon \text{(x))M}\gamma \text{(dx))^}\text{(}-\text{s)]. (2)}}\end{array}$

As $\epsilon \to 0$, exp(γhε(x)) converges uniformly away from the insertion disks to

$\begin{array}{c}{\displaystyle \prod \text{|zi|\_+^}\text{(}\gamma \alpha \text{i) }\cdot \text{ |x|\_+^}\text{(}\gamma \text{A) }\prod \text{|x}-\text{zi|^}\text{(}-\gamma \alpha \text{i).}}\end{array}$

The uniform cutoff/annular estimate in step 1 makes the omitted disks uniformly negligible in probability. Thus the integrals in (2) converge in probability. Their negative powers are dominated by a constant times $M\gamma (anchor{)}^{(-s)}$, which is integrable. Every subsequence has an almost-surely convergent further subsequence, so dominated convergence on those subsequences proves convergence of the full expectations.

Now compute the deterministic factors, rather than suppressing them. The off-diagonal Gaussian prefactor is

$\begin{array}{c}{\displaystyle \prod \text{\_(i<j)|zi}-\text{zj|^}\text{(}-\alpha \text{i}\alpha \text{j)}}\\ {\displaystyle \prod \text{\_i |zi|\_+^}\text{[}\alpha \text{i(A}-\alpha \text{i)].}}\end{array}$

Multiplying by the vertex backgrounds gives $\prod \mathrm{\mid }zi{\mathrm{\mid }}_{+}^{[\alpha i(A-2Q)]}$, because $-4\mathrm{\Delta }i=-2Q\alpha i+\alpha i^2$. The constant $\prod \mathrm{\mid }zi{\mathrm{\mid }}_{+}^{(\gamma \alpha i)}$ in the interaction integral, raised to −s, gives $\prod \mathrm{\mid }zi{\mathrm{\mid }}_{+}^{(-\gamma s\alpha i)}$. These cancel exactly since $\gamma s=A-2Q$. Hence the actual sphere-model n-point function is

$\begin{array}{c}{\displaystyle ⟨\prod \text{V}\alpha \text{i(zi)}⟩\text{=(1/}\gamma \text{)}\mu \text{^}\text{(}-\text{s)}\mathrm{\Gamma }\text{(s)}}\\ {\displaystyle \prod \text{\_(i<j)|zi}-\text{zj|^}\text{(}-\alpha \text{i}\alpha \text{j)}}\\ {\displaystyle \text{E[(}\int \text{ |x|\_+^}\text{(}\gamma \text{A)}\prod \text{|x}-\text{zi|^}\text{(}-\gamma \alpha \text{i) M}\gamma \text{(dx))^}\text{(}-\text{s)]. (3)}}\end{array}$

It is finite and strictly positive under the displayed strict Seiberg conditions. Necessity at the boundary is not claimed by this lemma; only the domain needed here has been proved.

3. The insertion at infinity. Put $z1=0,z2=1$ and let $z3\to \mathrm{\infty }$. Multiply the interaction integral in (3) by $\mathrm{\mid }z3{\mathrm{\mid }}^{(\gamma \alpha 3)}$. On compact sets away from 0,1 its weight converges to the weight in ρ. The same annular estimate proves uniform negligibility at 0 and 1. For the infinity chart the exact transformed scaled weight is $\mathrm{\mid }w{\mathrm{\mid }}_{+}^{(\gamma A)}\mathrm{\mid }1-w{\mathrm{\mid }}^{(-\gamma \alpha 2)}\mathrm{\mid }w-w3{\mathrm{\mid }}^{(-\gamma \alpha 3)}$, where $w3=1\mathrm{/}z3\to 0$. Thus it has one singularity of charge α3 at w3 times uniformly bounded factors near $w=0$, while the limiting singularity is at 0. Annuli centered at w3 and the uniform local mass estimate give a bound C $r^{[\xi (q)-\gamma \alpha 3q]}$ on the integral near $w=0$, uniformly for $\mathrm{\mid }w3\mathrm{\mid }\le r\mathrm{/}4$, after increasing the radius by a fixed factor. Thus those tails also tend to zero uniformly in probability. An anchor disk again dominates negative powers. This proves convergence of the scaled negative moments to $E[{\rho }^{(-s)}]$.

The remaining deterministic power of |z3| is

$\begin{array}{c}{\displaystyle \text{4}\mathrm{\Delta }\alpha \text{3}-\alpha \text{3(}\alpha \text{1+}\alpha \text{2)+}\gamma \alpha \text{3s}}\\ {\displaystyle \text{=2Q}\alpha \text{3}-\alpha \text{3}{}^2-\alpha \text{3(A}-\alpha \text{3)+}\alpha \text{3(A}-\text{2Q)=0.}}\end{array}$

Therefore the limit is exactly (1), with no additional constant.

4. Normalization crosswalk and actual source limit. Kupiainen–Rhodes–Vargas, arXiv:1707.08785, saved krv-dozz.tex, label liouvlaw1 and its footnote, defines the correlation functional with an explicit overall factor 2, “included to match with the physics literature normalization which is based on the DOZZ formula.” Its labels Z00, Z1 and expression3pointstruct consequently have 2/γ. Guillarmou–Kupiainen–Rhodes–Vargas, arXiv:2005.11530v1, (1.3)–(1.4), uses dc without this factor; footnote 5 explicitly records KRV's factor 2. The current production sphere definition uses the latter normalization. Hence once the independent DOZZ proof establishes $C_{KRV}=C_{DOZZ}$, our coefficient will be $C_{sph}=(1\mathrm{/}2)C_{DOZZ}$. That equality has NOT been assumed to prove (1).

This supplier establishes the exact probabilistic quantity that the next reflection/tail and BPZ/shift suppliers must evaluate. It does not evaluate its fractional GMC moment, assume a reflection amplitude, assume self-duality, or claim analytic continuation to complex charges.

Second DOZZ supplier: radial reflection tail and its Mellin normalization. Research theorem. No DOZZ formula, reflection relation, duality, or BPZ equation is assumed. Use the same actual spherical covariance and variance-normalized GMC as in three-point-gmc-normalization.md, with $0<\gamma <\mathrm{sqrt}(2)$. The argument is on the full bootstrap interval. Put $\alpha \in (\gamma \mathrm{/}2,Q)$, $\nu =Q-\alpha >0$ and $p=2\nu \mathrm{/}\gamma <4\mathrm{/}{\gamma }^2$.

Statement. Let $I\alpha ={\int }_{(\mathrm{\mid }z\mathrm{\mid }<1)}\mathrm{\mid }z{\mathrm{\mid }}^{(-\gamma \alpha )}M\gamma (d^{2}z)$. There is a specifically constructed positive random variable Sν below, independent of an exponential random variable M of rate 2ν, such that

$\begin{array}{c}{\displaystyle \text{P(I}\alpha \text{>t)=Rbar}\gamma \text{(}\alpha \text{)t^}\text{(}-\text{p)+O(t^}\text{(}-\text{p}-\delta \text{)),}}\\ {\displaystyle \text{Rbar}\gamma \text{(}\alpha \text{)=E[S}\nu \text{^}\text{p]}\in \text{(0,}\mathrm{\infty }\text{), (1)}}\end{array}$

for some $\delta >0$ depending on the strict parameters. Every positive moment of Sν below 4/γ² is finite. The Mellin transform $E[I{\alpha }^z]$ is holomorphic for Re $z<p$ and extends meromorphically to Re $z<p+\delta$, with its only pole in that added strip at $z=p$ and

$\begin{array}{c}{\displaystyle \text{lim\_(z}\to \text{p)(p}-\text{z)E[I}\alpha \text{^}\text{z]=p Rbar}\gamma \text{(}\alpha \text{). (2)}}\end{array}$

Define the full reflection normalization, away from $p\in N$, by

$\begin{array}{c}{\displaystyle \text{R}\gamma \text{,}\mu \text{(}\alpha \text{)=}\mu \text{^}\text{p }\mathrm{\Gamma }\text{(}-\text{p) p Rbar}\gamma \text{(}\alpha \text{). (3)}}\end{array}$

Then ${\lim }_{(\alpha \uparrow Q)}R\gamma ,\mu (\alpha )=-1$. This fixes a probabilistic normalization at Q; it does not identify (3) with the explicit DOZZ reflection function or prove the reflection relation for vertex operators.

1. Positive local GMC moments beyond L². The existing local GMC supplier proves construction, fractional scaling and all anchored inverse moments for ${\gamma }^2<2$. Its positive moments can be upgraded to every $r<4\mathrm{/}{\gamma }^2$ without assuming an integrable formula. Here are the steps used later.

For a bounded planar chart and its regularized mass Tε, Gaussian square completion gives

$\begin{array}{c}{\displaystyle \text{E[T}\epsilon \text{^}\text{r]=}\int \text{ E[(}\int \text{exp(}\gamma {}^2\text{K}\epsilon \text{(x,y))M}\epsilon \text{(dy))^}\text{(r}-\text{1)]dx.}}\end{array}$

All harmless bounded area densities are inserted in the integrals. Set $q=r-1$. Induct on successive intervals $r\in (k,k+1]$, starting with $q\in (0,1]$. The covariance comparison under rescaling gives

$\begin{array}{c}{\displaystyle \text{E[M}\epsilon \text{(B\_a)^}\text{q]}\le \text{C a^}\xi \text{(q),}}\\ {\displaystyle \xi \text{(q)=(2+}\gamma {}^2\text{/2)q}-\gamma {}^2\text{q}{}^2\text{/2, }\epsilon \le \text{a,}}\end{array}$

once the qth moment on the unit chart is uniformly bounded. For $q\le 1$ the bound follows from concave comparison and Jensen, as in the actual local supplier. For $q>1$, convex comparison with an independent common Gaussian of variance log(1/a)+C and a fresh unit-chart field gives the same scaling bound, now using the previously proved uniform qth moment. Thus this inductive step does not invoke an unproved higher moment.

Decompose the inner tilted mass into dyadic shells about x, down to the ε ball. Its tilt is bounded by C $a^{(-{\gamma }^2)}$. If $q\le 1$, use (sum $b_j{)}^q\le \sum b_j^q$. The shell sum converges because

$\begin{array}{c}{\displaystyle \xi \text{(q)}-\gamma {}^2\text{q=q(2}-\gamma {}^2\text{r/2)>0.}}\end{array}$

If $q>1$, use the Lq triangle inequality; the shell norms sum because $\xi (q)\mathrm{/}q-{\gamma }^2=2-{\gamma }^{2}r\mathrm{/}2>0$. The innermost ε ball has the same bound. This gives supε $E[T{\epsilon }^r]<\mathrm{\infty }$ whenever $r<4\mathrm{/}{\gamma }^2$. Fatou gives the limit moment. The same comparison proves local scaling at every such r. A finite chart cover handles compact cylinders and seams. This proof uses only Gaussian square completion, the covariance bound, and lower moments.

2. Exact radial decomposition and stationary cylinder chaos. For $z=e^{(-s+i\theta )}$, $s\ge 0$, the spherical covariance inside the unit disk decomposes as

$\begin{array}{c}{\displaystyle \text{K(z,w)=min(s,t)+C\_Y(s}-\text{t,}\theta -\theta {}^{\mathrm{\prime }}\text{),}}\\ {\displaystyle \text{C\_Y(u,v)=}-\text{log|1}-\text{e^}\text{(}-\text{|u|+iv)|.}}\end{array}$

The circular average is therefore a standard Brownian motion $B_s$ starting at zero. Its Gaussian orthogonal complement Y is independent. A concrete stationary extension of Y to all $s\in R$ is obtained from independent stationary Ornstein–Uhlenbeck modes with covariance $e^{(-n\mathrm{\mid }s-t\mathrm{\mid })}$, multiplied by cos(nθ)/sqrt(n), sin(nθ)/sqrt(n), and summed over $n\ge 1$. The resulting covariance is the displayed $C_Y$. Finite spectral cutoffs followed by the local GMC limit construct its cylinder chaos Nγ(ds dθ). It is stationary in s and independent of B. Write $Z_{s}ds=\int N\gamma (dsd\theta )$; this notation denotes the integrated chaos, not a new independence assertion between different s.

Its mass on a unit strip has all positive moments below 4/γ² by step 1, uniformly in the strip location, and all anchored inverse moments by the same cone comparison in the local GMC supplier: $C_Y$ is log(1/distance)+a bounded remainder in every fixed cylinder chart. That proof depends only on these covariance bounds and hence applies to Y as well.

The polar area Jacobian is $e^{(-2s)}$, and Brownian variance normalization contributes $e^{(-{\gamma }^{2}s\mathrm{/}2)}$. Consequently the radial formula is EXACTLY

$\begin{array}{c}{\displaystyle \text{I}\alpha \text{ law=}\int \text{\_0^}\mathrm{\infty }\text{ e^}\text{[}\gamma \text{(B\_s}-\nu \text{s)] Z\_s ds, (4)}}\end{array}$

since $\gamma \alpha -2-{\gamma }^2\mathrm{/}2=-\gamma (Q-\alpha )$. There is no extra radial constant or cosmological factor in (4).

3. Conditioned Brownian motion, constructed from its kernel. Let $h\nu (x)=1-e^{(2\nu x)}$, $x<0$. A Brownian motion of drift −ν killed at zero has transition kernel

$\begin{array}{c}{\displaystyle \text{k\_t(x,y)=e^}\text{[}-\nu \text{(y}-\text{x)}-\nu {}^2\text{t/2]}}\\ {\displaystyle \text{[g\_t(y}-\text{x)}-\text{g\_t(y+x)],}}\\ {\displaystyle \text{g\_t(u)=(2}\pi \text{t)^}\text{(}-\text{1/2)e^}\text{(}-\text{u}{}^2\text{/(2t)).}}\end{array}$

Reflection at zero proves this formula. The positive harmonic function hν is precisely the probability of never hitting zero, from the elementary exponential martingale. Conditioning on that event gives kernel $k_t(x,y)h\nu (y)\mathrm{/}h\nu (x)$. Taking x↑0 gives the entrance density, in $r=-y>0$,

$\begin{array}{c}{\displaystyle \text{q\_t}\nu \text{(r)=[2r sinh(}\nu \text{r)/(}\nu \text{t)]e^}\text{(}-\nu {}^2\text{t/2)g\_t(r). (5)}}\end{array}$

This defines the negative conditioned path Yν starting at zero. Existence and continuity at that entrance can be constructed without a conditioning-on-probability-zero assertion: start with the norm $R_t$ of three independent standard Brownian motions and tilt its law up to time t by

$\begin{array}{c}{\displaystyle \text{e^}\text{(}-\nu {}^2\text{t/2) sinh(}\nu \text{R\_t)/(}\nu \text{R\_t).}}\end{array}$

The numerator divided by νR is the spherical average of $e^{(\nu u\cdot B_t)}$, so this is a positive mean-one martingale. Its consistent path laws have density (5), continuous paths, and generator for $R\nu =-Y\nu$ equal to (1/2)∂rr+νcoth(νr)∂r. Thus Yν has generator (1/2)∂xx−ν∂x+(hν′/hν)∂x and stays negative after time zero.

For every $a>0$, (5), by completing the square in r, implies

$\begin{array}{c}{\displaystyle \text{E[e^}\text{(aY}\nu \text{\_t)]}\le \text{C\_a e^}\text{(}-\text{c\_a t), t}\ge \text{1, c\_a>0. (6)}}\end{array}$

Indeed its integral is bounded by $t^{(-3\mathrm{/}2)}{e}^{(-{\nu }^{2}t\mathrm{/}2)}$ times a polynomially weighted integral of exp[(ν−a)r−r²/(2t)]. If $a<\nu$ the exponent gains (ν−a)²t/2, leaving decay $\nu a-a^2\mathrm{/}2>0$; if $a\ge \nu$ the remaining integral grows at most polynomially. Absorb that polynomial into a smaller positive exponential rate. On a unit future interval the drift of Yν is at most −ν. Coupling its SDE to a Brownian motion of drift −ν and localizing away from zero gives

$\begin{array}{c}{\displaystyle \text{E[e^}\text{(a sup\_(n}\le \text{t}\le \text{n+1)Y}\nu \text{\_t)]}\le \text{C\_a e^}\text{(}-\text{c\_a n). (7)}}\end{array}$

The independent Brownian increment supremum has all exponential moments; this proves (7) from (6). The localization limit is allowed because the conditioned process never returns to zero.

Take two independent copies of Yν, one for each half of R, forming a continuous two-sided path $B{\nu }_s$ with maximum zero. Independently take the stationary cylinder chaos above, and define

$\begin{array}{c}{\displaystyle \text{S}\nu \text{=}\int \text{\_R e^}\text{(}\gamma \text{B}\nu \text{\_s) Z\_s ds. (8)}}\end{array}$

On each unit strip the integrand mass is bounded by exp(γ sup Bν) times the chaos mass. Their factors are independent. Equations (7) and step 1 give summable exponentially decaying rth moments for every $r<4\mathrm{/}{\gamma }^2$. For $r\le 1$ sum those moments; for $r>1$ sum Lr norms. Therefore Sν has every positive moment claimed and is finite. It is strictly positive because a fixed strip has positive chaos mass and the Brownian weight there has a positive minimum.

4. Maximum decomposition and its time reversal, with the kernel calculation. For $X_t=B_t-\nu t$, the exponential martingale stopped between −a and m, followed by $a\to \mathrm{\infty }$, gives

$\begin{array}{c}{\displaystyle \text{P(sup X}\ge \text{m)=e^}\text{(}-\text{2}\nu \text{m), m}\ge \text{0.}}\end{array}$

Conditional on hitting m, the pre-hit process is Brownian motion of drift +ν stopped at its hit of m: the Doob factor $e^{(2\nu x)}$ changes drift −ν to +ν. Condition the maximum to lie in [m,m+ε]. By the strong Markov property at the hit of m, the post-hit process is an independent drift −ν process conditioned not to hit ε. Its transition kernels converge, as ε↓0, to m+Yν. The band probability is $e^{(-2\nu m)}(1-e^{(-2\nu \epsilon )})$; dividing by ε gives the exponential density. These kernel limits determine bounded cylinder functionals and then path laws by continuity. This proves the pre/post maximum decomposition with a maximum M of rate 2ν.

The needed reverse pre-hit law also has a direct density proof. Translate the drift +ν process to start at −m and stop at zero. Its killed Green kernel from −m is

$\begin{array}{c}{\displaystyle \text{u\_m(x)=h}\nu \text{(x)/}\nu \text{ for }-\text{m}\le \text{x<0,}}\\ {\displaystyle \text{u\_m(x)=e^}\text{(2}\nu \text{x)(e^}\text{(2}\nu \text{m)}-\text{1)/}\nu \text{ for x<}-\text{m.}}\end{array}$

This follows by integrating the reflected Gaussian kernel; use ${\int }_0^{\mathrm{\infty }}{e}^{(-{\nu }^{2}t\mathrm{/}2)}{g}_t(a)dt=e^{(-\nu \mathrm{\mid }a\mathrm{\mid })}\mathrm{/}\nu$. The factor

$\begin{array}{c}{\displaystyle \pi \text{\_m(x)=}\nu \text{u\_m(x)/h}\nu \text{(x)}}\end{array}$

is exactly the probability that Yν started at x will visit −m in the future: it is one above −m, and below −m it follows from the exponential hit probability and the h transform. Let $L_{-}m$ be Yν's last visit to −m.

For reverse observation times $0<t1<\mathrm{.}\mathrm{.}\mathrm{.}<tk$, integrate the original pre-hit segment length beyond tk. Its density is $u_m(xk)$ times the reversed killed transition kernels and the final first-hit-zero density. Lebesgue duality changes the reversed +ν kernels to −ν kernels. The first-hit density at time t1 equals ν/hν(x1) times the entrance density of Yν. Telescoping the h factors therefore gives exactly

$\begin{array}{c}{\displaystyle \text{entrance\_(t1)(x1) }\cdot \text{ product\_j k^}\text{h\_(t[j+1]}-\text{t[j])(xj,x[j+1]) }\cdot \pi \text{\_m(xk).}}\end{array}$

This is the density of Yν observed before its last visit to −m: ${\pi }_m$ is the probability of a later visit. Hence the time-reversed pre-maximum segment is Yν stopped at $L_{-}m$. This establishes the specific reversal used here rather than citing a decomposition as an unproved input.

We may therefore construct independent M, two-sided Bν, and cylinder noise so that (4) has the law

$\begin{array}{c}{\displaystyle \text{I}\alpha \text{=e^}\text{(}\gamma \text{M)}\int \text{\_(-L\_}-\text{M)^}\mathrm{\infty }\text{ e^}\text{(}\gamma \text{B}\nu \text{\_s)Z\_s ds. (9)}}\end{array}$

The random time translation of Z used to center the maximum changes no law: Z is stationary and independent of the whole Brownian path, so even a Brownian-measurable random translation has the same conditional law. This is not independence of disjoint cylinder strips.

5. The omitted past has a higher moment. Define $J=e^{(\gamma M)}S\nu$ and $D=J-I\alpha$ in the coupling (9). Conditional on $M=m$, D is the weighted chaos AFTER the last passage of −m on the negative branch, shifted upward by m. This shifted path has the same law as Yν.

Here is the last-passage conditioning that justifies that assertion. For $x<-m$, the probability under the original hν-process of never returning to −m is

$\begin{array}{c}{\displaystyle \text{1}-\text{h}\nu \text{(}-\text{m)e^}\text{[2}\nu \text{(x+m)]/h}\nu \text{(x)=h}\nu \text{(x+m)/h}\nu \text{(x).}}\end{array}$

Applying this further Doob transform makes the combined harmonic factor hν(x+m); its generator after translating x by m is precisely the original Yν generator. To obtain the post-last-visit entrance, approximate the successful final departure by a downcrossing from −m to −m−ε followed by no return. Strong Markov at successive such downcrossings gives this same conditional future kernel, independently of the earlier failed excursions. Let ε↓0. Continuity and transience of Yν make the final departure times tend to $L_{-}m$, and the kernels tend to the same zero entrance constructed in step 3. This proves the asserted post-last-passage law. Stationarity of the independent cylinder field handles its random translation as above.

Consequently, for every $r<4\mathrm{/}{\gamma }^2$,

$\begin{array}{c}{\displaystyle \text{sup\_(m}\ge \text{0) E[D^}\text{r | M=m] <}\mathrm{\infty }\text{. (10)}}\end{array}$

The bound is the one-sided version of (8). In particular choose $p<r<4\mathrm{/}{\gamma }^2$; this is possible exactly because $\alpha >\gamma \mathrm{/}2$. Thus $E{D}^r<\mathrm{\infty }$. No moment of the heavy Pareto factor $e^{(\gamma M)}$ has been incorrectly assumed here: it is cancelled by translating the path after its last visit to −M.

6. Tail comparison, coefficient, and Mellin pole. Since M is independent of Sν and $P(e^{(\gamma M)}>u)=u^{(-p)}$ for $u\ge 1$,

$\begin{array}{c}{\displaystyle \text{P(J>t)=t^}\text{(}-\text{p)E[S}\nu \text{^}\text{p;S}\nu \le \text{t]+P(S}\nu \text{>t).}}\end{array}$

Using $E[S{\nu }^r]<\mathrm{\infty }$ for $r>p$ shows

$\begin{array}{c}{\displaystyle \text{P(J>t)=E[S}\nu \text{^}\text{p]t^}\text{(}-\text{p)+O(t^}\text{(}-\text{r)).}}\end{array}$

Its conditional density at $t>S\nu$ is $pS{\nu }^p{t}^{(-p-1)}$, so for $0<a<1$,

$\begin{array}{c}{\displaystyle \text{P(t<J}\le \text{t+t^}\text{a)}\le \text{pE[S}\nu \text{^}\text{p]t^}\text{(}-\text{p}-\text{1+a).}}\end{array}$

Choose $p\mathrm{/}r<a<1$. Since $I\alpha \le J$ and $J-I\alpha =D$,

$\begin{array}{c}{\displaystyle \text{0}\le \text{P(J>t)}-\text{P(I}\alpha \text{>t)}}\\ {\displaystyle \le \text{P(D>t^}\text{a)+P(t<J}\le \text{t+t^}\text{a)}}\\ {\displaystyle \le \text{C t^}\text{(}-\text{ar)+C t^}\text{(}-\text{p}-\text{1+a).}}\end{array}$

Together these prove (1), for any δ below min(r−p,ar−p,1−a). All quantities are strictly positive and finite by step 3.

Iα has all inverse moments: on a fixed annulus away from zero its weight is bounded below and the local anchored inverse GMC moments apply. Hence its Mellin transform is holomorphic for Re $z<p$, by dominated difference quotients using slightly larger positive/negative powers to control logarithms. Split at $I\alpha =1$ and integrate its upper tail:

$\begin{array}{c}{\displaystyle \text{E[I}\alpha \text{^}\text{z]=E[I}\alpha \text{^}\text{z;I}\alpha \le \text{1]+P(I}\alpha \text{>1)}}\\ {\displaystyle \text{+ z Rbar}\gamma \text{(}\alpha \text{)/(p}-\text{z)}}\\ {\displaystyle \text{+ z}\int \text{\_1^}\mathrm{\infty }\text{t^}\text{(z}-\text{1)[P(I}\alpha \text{>t)}-\text{Rbar}\gamma \text{(}\alpha \text{)t^}\text{(}-\text{p)]dt.}}\end{array}$

The first term is entire because all negative moments exist. The last integral is holomorphic for Re $z<p+\delta$, locally uniformly by the error in (1). This proves the meromorphic continuation and (2), with no DOZZ value supplied for Rbar.

**7. Normalization at $\alpha =Q\mathrm{.}\ast \ast$ It remains to prove $Rbar\gamma (\alpha )\to 1$ as ν↓0, rather than substitute an assumed reflection constant. At $\nu =0$ the negative conditioned path is minus a three-dimensional Bessel process. Its entrance density is $2r^2{g}_t(r)\mathrm{/}t$, so $E[e^{(aY{0}_t)}]\le C_a(1+t{)}^{(-3\mathrm{/}2)}$. The unit-strip supremum version follows by the same Brownian comparison. Since $\nu coth(\nu r)\ge 1\mathrm{/}r$, couple Rν and R0 so $R\nu \ge R0$. Thus these polynomial bounds dominate all $\nu \in [0,\nu 0]$. Choosing any fixed $0<q<\min (1,4\mathrm{/}{\gamma }^2)$, summing strip qth moments gives $su{p}_{(0\le \nu \le \nu 0)}E[S{\nu }^q]<\mathrm{\infty }$.

There is also a uniform negative moment. On one fixed strip, $S\nu \ge e^{[-\gamma su{p}_{(0\le t\le 1)}R{\nu }_t]}$ Nγ(strip). The inequality $\nu coth(\nu r)\le \nu +1\mathrm{/}r$ and one-dimensional comparison give $R{\nu }_t\le R{0}_t+\nu t$ under a suitable common Brownian coupling: the latter has drift $\nu +1\mathrm{/}R{0}_t$, which is at least $\nu +1\mathrm{/}(R{0}_t+\nu t)$. The supremum of R0 over a unit interval has Gaussian tails, since it is the norm of three Brownian motions. All its exponential moments are finite. Independence of radial path and cylinder chaos, and the anchored inverse moments from step 2, therefore give $su{p}_{(0\le \nu \le \nu 0)}E[S{\nu }^{(-q)}]<\mathrm{\infty }$.

With $p=2\nu \mathrm{/}\gamma \to 0$, for $p\le q\mathrm{/}2$ the elementary inequality $\mathrm{\mid }{x}^p-1\mathrm{\mid }\le C_{q}p(x^q+x^{(-q)})$ yields $E[S{\nu }^p]\to 1$. Finally ${\mu }^p\to 1$ and $p\mathrm{\Gamma }(-p)\to -1$ by the elementary Gamma pole at zero. Equation (3) therefore tends to −1. This is the actual probabilistic reflection normalization used by the later shift argument, not its closed-form evaluation.

Remaining use boundary. This supplier proves a one-singularity tail with a power error, its exact radial coefficient, the associated Mellin pole, and $R(Q)=-1$. It does not yet prove the multi-singularity $3\to 2$ residue bridge, analyticity in the charge parameters, degenerate fusion/BPZ shifts, inversion $R(\alpha )R(2Q-\alpha )=1$, the dual shift, or the DOZZ expression. Those are subsequent obligations; the present theorem does not assume any of them. The source KRV paper motivates the decomposition, but its tail/reflection theorems are not used as black-box premises in the proof above.

Ninth DOZZ supplier: a smooth weighted local tail with uniform power error. Research lemma needed for the two-pole reflection continuation. Fix $0<\gamma <\mathrm{sqrt}(2)$. Let $\mathrm{\mid }z0\mathrm{\mid }>2$ and let F be strictly positive C1 on the closed disk B(z0,1). Define, with the ACTUAL sphere-normalized chaos including dA0,

$\begin{array}{c}{\displaystyle \text{X\_F,}\alpha \text{=}\int \text{\_(B(z0,1)) |x}-\text{z0|^}\text{(}-\gamma \alpha \text{)F(x)M}\gamma \text{(dx),}}\\ {\displaystyle \gamma \text{/2<}\alpha \text{<Q, }\nu \text{=Q}-\alpha \text{, p=2}\nu \text{/}\gamma \text{.}}\end{array}$

Then

$\begin{array}{c}{\displaystyle \text{P(X\_F,}\alpha \text{>t)=c\_F,}\alpha \text{ t^}\text{(}-\text{p)+O(t^}\text{(}-\text{p}-\delta \text{)),}}\\ {\displaystyle \text{c\_F,}\alpha \text{=F(z0)^}\text{p |z0|^}\text{[4}\alpha \text{(}\alpha -\text{Q)]Rbar}\gamma \text{(}\alpha \text{), (1)}}\end{array}$

for some $\delta >0$. If α varies in a compact subinterval of (γ/2,Q), and F and its inverse and C1 norm vary in fixed compact bounds, there is a uniform positive δ and uniform error constant. In particular there is a $\delta \ast >0$ valid for α near 2/γ BEFORE choosing the small two-pole parameter η. This lemma is derived from the centered tail supplier and an explicit Gaussian/OU coupling; it is not an assumed general GMC tail theorem.

1. The constant circle average and its exact correlations. Write $u=x-z0$ and $g(x)=\log \mathrm{\mid }x\mathrm{\mid }$, harmonic on B(z0,1). The covariance of X(z0+u) is log(1/|u−v|)+g(z0+u)+g(z0+v). Let S be its average over $\mathrm{\mid }u\mathrm{\mid }=1$ and subtract S. The remaining field has the PURE-LOG covariance log(1/|u−v|) inside the unit disk. Thus it is $B_s+Y(s,\theta )$, with $s=\log (1\mathrm{/}\mathrm{\mid }u\mathrm{\mid })$, B standard Brownian, and the same stationary lateral OU field as in the centered tail supplier. Direct covariance averaging gives

$\begin{array}{c}{\displaystyle \text{Var S=2g(z0), Cov(B\_s,S)=0,}}\\ {\displaystyle \text{Cov(Y(s,}\theta \text{),S)=g(z0+e^}\text{(}-\text{s+i}\theta \text{))}-\text{g(z0). (2)}}\end{array}$

The second equality uses harmonic mean value and implies independence of B from (S,Y). More explicitly, expand $\log (1+u\mathrm{/}z0)={\sum }_{(n\ge 1)}(-1{)}^{(n+1)}{u}^n\mathrm{/}(nz{0}^n)$. The right side of (2) is its real part. In the OU mode representation of Y, S is therefore a linear combination of the modes at time zero plus an independent Gaussian. The variance of that projection is ${\sum }_{(n\ge 1)}\mathrm{\mid }z0{\mathrm{\mid }}^{(-2n)}\mathrm{/}n=-\log (1-\mathrm{\mid }z0{\mathrm{\mid }}^{(-2)})$; the independent Gaussian has variance $\log (\mathrm{\mid }z0{\mathrm{\mid }}^2-1)>0$. In particular S is independent of every OU innovation after time zero.

Wick normalization and the actual background area density must both be kept. In polar coordinates the exact local mass is

$\begin{array}{c}{\displaystyle \text{X\_F,}\alpha \text{=e^}\text{(}\gamma \text{S)}\int \text{\_0^}\mathrm{\infty }\text{ e^}\text{[}\gamma \text{(B\_s}-\nu \text{s)]a(s,}\theta \text{)N}\gamma \text{(ds d}\theta \text{),}}\\ {\displaystyle \text{a(s,}\theta \text{)=F(z0+e^}\text{(}-\text{s+i}\theta \text{))}}\\ {\displaystyle \cdot \text{exp[}-\text{(4+}\gamma {}^2\text{)g(z0+e^}\text{(}-\text{s+i}\theta \text{))]. (3)}}\end{array}$

Here the factor −γ²g comes from the difference of Wick variances, and −4g is dA0. Thus $a(s,\theta )=a0(1+O(e^{(-s)}))$ uniformly in θ, with $a0=F(z0)e^{[-(4+{\gamma }^2)g(z0)]}$. These factors cannot be omitted when computing the tail constant.

2. Early scales have a sufficiently high moment. Fix a compact α range, hence p in [pmin,pmax] with $pmax<4\mathrm{/}{\gamma }^2$. Choose $R>\max (1,pmax)$ and $R<4\mathrm{/}{\gamma }^2$. The higher local chaos moments proved in the centered tail supplier allow a Hölder exponent $q>1$ with $Rq<4\mathrm{/}{\gamma }^2$. Gaussian exponential moments of S and Hölder give a uniform bound on $E[e^{(\gamma RS)}N\gamma ([n,n+1]\times T{)}^R]$, despite their correlation. Brownian motion is independent of those variables.

Let $E_L$ be the integral in (3) over $0\le s\le L$. Brownian supremum on the nth unit interval has exponential moment bounded by C $\exp [A_{R}n]$, where

$\begin{array}{c}{\displaystyle \text{A\_r=}\gamma {}^2\text{r(r}-\text{p)/2.}}\end{array}$

This is just the Gaussian exponential moment for $B_n-\nu n$ followed by the bounded-time Brownian supremum. The LR triangle inequality therefore gives

$\begin{array}{c}{\displaystyle \text{E[E\_L^}\text{R]}\le \text{C(1+L)^}\text{R exp[A\_R L]. (4)}}\end{array}$

For a fixed α the same statement applies to every $r\in (p,4\mathrm{/}{\gamma }^2)$. All constants in (4) can be chosen uniformly on the stated compact parameter set.

3. Late scales couple to an independent centered singular mass. For $s\ge L$, replace the initial OU values of Y at time zero by independent stationary Gaussian copies, keeping the innovations after zero. This gives a stationary Y′ independent of S. The explicit OU solution gives

$\begin{array}{c}{\displaystyle \text{sup\_(s}\ge \text{L,}\theta \text{)|Y(s,}\theta \text{)}-\text{Y}{}^{\mathrm{\prime }}\text{(s,}\theta \text{)|}\le \text{e^}\text{(}-\text{L)W, (5)}}\end{array}$

where W is a sum of absolute Gaussian coordinates with weights bounded by C $e^{(}-(n-1))\mathrm{/}\mathrm{sqrt}(n)$. It has E $\exp (c{W}^2)<\mathrm{\infty }$ for some $c>0$. To verify this directly use weighted Cauchy–Schwarz and the product of the one-dimensional Gaussian square moment-generating functions; the weights and their sum are finite. No independence of the actual early and late field is asserted.

The variance normalization of Y and Y′ is the same. Equations (3), (5) and the C1 bound on a therefore sandwich the late mass $T_L$ by

$\begin{array}{c}{\displaystyle \text{A\_L I}\alpha {}^{\mathrm{\prime }}\cdot \text{exp[}\pm \text{C e^}\text{(}-\text{L)(1+W)],}}\\ {\displaystyle \text{A\_L=a0 exp[}\gamma \text{S+}\gamma \text{(B\_L}-\nu \text{L)], (6)}}\end{array}$

where Iα′ is an independent copy of the centered singular mass. It is independent of $A_L$: the replaced stationary OU process is independent of S, and the fresh Brownian increments after L are independent of $B_L$ and of all lateral variables. The possible correlation between W and Iα′ is retained in (6).

By the proved centered tail there is a uniform $\delta c>0$ such that $P(I{\alpha }^{\mathrm{\prime }}>u)=Rbar\gamma (\alpha )u^{(-p)}+O(u^{(-p-\delta c)})$. Decrease δc so $p+\delta c<4\mathrm{/}{\gamma }^2$ uniformly. Extending the error bound to $u<1$ by increasing its constant permits averaging over all $A_L>0$. Therefore

$\begin{array}{c}{\displaystyle \text{P(A\_L I}\alpha {}^{\mathrm{\prime }}\text{>t)=Rbar}\gamma \text{(}\alpha \text{)t^}\text{(}-\text{p)E[A\_L^}\text{p]}}\\ {\displaystyle \text{+O(t^}\text{(}-\text{p}-\delta \text{c)E[A\_L^}\text{(p+}\delta \text{c)]).}}\end{array}$

Gaussian calculation, using the independence of S and B, gives

$\begin{array}{c}{\displaystyle \text{E[A\_L^}\text{p]=a0^}\text{p exp[}\gamma {}^2\text{p}{}^2\text{g(z0)],}}\\ {\displaystyle \text{E[A\_L^}\text{r]=a0^}\text{r exp[}\gamma {}^2\text{r}{}^2\text{g(z0)]exp[A\_r L]. (7)}}\end{array}$

The first equality has no L factor: $p=2\nu \mathrm{/}\gamma$ is exactly the Brownian exponential-martingale exponent. No arbitrary volume normalization is involved.

4. Choose $L=c$ log t and keep every error below the leading tail. Choose $c>0$ sufficiently small that c sup $A_{(p+\delta c)}<\delta c\mathrm{/}2$ and c sup $A_R<(R-pmax)\mathrm{/}4$. Set $L=c$ log t, and choose $d=(pmax+R)\mathrm{/}(2R)<1$. Markov and (4) give

$\begin{array}{c}{\displaystyle \text{P(E\_L>t^}\text{d)}\le \text{C(log t)^}\text{R t^}\text{[}-\text{dR+cA\_R]}}\\ {\displaystyle \text{=O(t^}\text{(}-\text{pmax}-\delta \text{E))}}\end{array}$

for some $\delta E>0$, after a harmless reduction of δE. The model tail from (7) has a uniform power error at least δc/2.

On $W\le t^{(c\mathrm{/}2)}$, the multiplicative error in (6) is $1+O(t^{(-c\mathrm{/}2)})$. Its complement has probability $\le Cexp(-c{t}^c)$, which is smaller than every inverse power. Thus the late tail has the same coefficient as its model and a power error at least min(c/2,δc/2). Finally

$\begin{array}{c}{\displaystyle \text{P(T\_L>t)}\le \text{P(X\_F,}\alpha \text{>t)}}\\ {\displaystyle \le \text{P(T\_L>t}-\text{t^}\text{d)+P(E\_L>t^}\text{d).}}\end{array}$

Replacing t by $t-t^d$ changes the leading model power by $O(t^{(-p-1+d)})$. Combining these estimates proves a uniform positive error exponent, for example any number strictly below min(δE,c/2,δc/2,1−d). This supplies an ACTUAL uniform δ*, not a limit comparison without a rate.

5. Compute the coefficient and the two-pole margin. From (7), the coefficient is

$\begin{array}{c}{\displaystyle \text{a0^}\text{p exp[}\gamma {}^2\text{p}{}^2\text{g(z0)]Rbar}\gamma \text{(}\alpha \text{)}}\\ {\displaystyle \text{=F(z0)^}\text{p exp{[}\gamma {}^2\text{(p}{}^2-\text{p)}-\text{4p]g(z0)}Rbar}\gamma \text{(}\alpha \text{).}}\end{array}$

Since $\gamma p=2(Q-\alpha )$ and $2\gamma Q=4+{\gamma }^2$, the exponent in braces is 4α(α−Q). This proves exactly (1), including the background-density contribution.

Apply the compact-parameter version near $\alpha =2\mathrm{/}\gamma$, whose p value is one and remains strictly below 4/γ². First fix a small compact neighborhood there and obtain $\delta \ast >0$ from step 4. Then choose η smaller if necessary so $4\eta \mathrm{/}\gamma <\delta \ast \mathrm{/}2$. The two local exponents 1−2η/γ and 1+2η/γ are closer than the available uniform local power error. Together with the already-proved separated mixed moments, this is the strict margin needed for the two-term sum tail in the analytic two-pole helper.

No reflection formula or special-function value of Rbar is assumed. This extends the centered radial tail precisely to the smooth weighted outside-circle disks used in the two-pole argument; it does not assert an unrelated arbitrary-covariance tail theorem.

Third DOZZ supplier: the real three-to-two residue bridge. Research theorem using the two preceding constructed suppliers, not a reflection axiom. Fix $0<\gamma <\mathrm{sqrt}(2)$, $\mu >0$ and $\gamma \mathrm{/}2<\alpha 2\le \alpha 3<Q$. Put $p=2(Q-\alpha 3)\mathrm{/}\gamma$ and assume p is not a positive integer. For $\epsilon >0$ small set $\alpha 1=\alpha 3-\alpha 2+\epsilon$, $A=2\alpha 3+\epsilon$, and $q=p-\epsilon \mathrm{/}\gamma$. Let ρε be the EXACT three-point random integral of three-point-gmc-normalization.md with these charges. Its qth moment is finite. Define the Mellin-form auxiliary three-point expression

$\begin{array}{c}{\displaystyle \text{C\_aux,KRV(}\epsilon \text{)= (2/}\gamma \text{) }\mu \text{^}\text{q }\mathrm{\Gamma }\text{(}-\text{q) E[}\rho \epsilon \text{^}\text{q].}}\end{array}$

Then

$\begin{array}{c}{\displaystyle \text{lim\_(}\epsilon \downarrow \text{0) }\epsilon \text{ C\_aux,KRV(}\epsilon \text{)}}\\ {\displaystyle \text{=2R}\gamma \text{,}\mu \text{(}\alpha \text{3) if }\alpha \text{2<}\alpha \text{3,}}\\ {\displaystyle \text{=4R}\gamma \text{,}\mu \text{(}\alpha \text{3) if }\alpha \text{2=}\alpha \text{3. (1)}}\end{array}$

For the sphere model normalized by dc, the auxiliary expression and both residues are divided by two. This theorem does NOT yet assert charge analyticity or identify $C_{aux}$ with analytic continuation of the convergent zero-mode correlation; that is the next separate prerequisite. It proves the actual positive-moment boundary limit needed for that identification.

1. Singular positive moments with a strict margin. For a local charge $a<Q$, annular scaling and the positive moments established in reflection-tail-and-mellin-normalization.md give

$\begin{array}{c}{\displaystyle \text{E[(}\int \text{\_(|x}-\text{z|<r)|x}-\text{z|^}\text{(}-\gamma \text{a)M}\gamma \text{(dx))^}\text{t]<}\mathrm{\infty }}\end{array}$

whenever $0<t<\min (4\mathrm{/}{\gamma }^2,2(Q-a)\mathrm{/}\gamma )$. For $t\le 1$ use the subadditivity of the tth power on annuli; for $t>1$ use its Lt triangle inequality. The summability exponent is ξ(t)−γat, or divided by t, respectively; it is positive exactly in the stated range. Constants are uniform when the parameters stay inside strict inequalities. Consequently ρε has a finite qth moment: its three charges are all $<Q$ and q is strictly below each singular threshold. All regions away from these singularities have moments up to 4/γ².

2. A mixed-moment estimate for separated singularities. Let X1,X2 be the weighted masses of two disjoint small disks, each with its own insertion singularity. Suppose $ti>0$ lies strictly below that disk's singular moment threshold. Then $E[X{1}^{t}1X{2}^{t}2]<\mathrm{\infty }$, uniformly in strict compact parameter ranges.

Here is the Gaussian argument, which avoids assuming actual independence of the disk fields. First truncate each deterministic singular weight at a fixed insertion cutoff, and at finite FIELD cutoff compare the actual Gaussian field on the two disks with independent copies of its two restrictions. Covariances WITHIN each disk are identical in the two models. The only interpolation derivative is the cross covariance, bounded in absolute value by a fixed C because the disks are separated. Gaussian square completion/differentiation for $F(x,y)=x^{t}1y^{t}2$ gives

$\begin{array}{c}{\displaystyle \text{|d/du E[F(M1,u,M2,u)]|}}\\ {\displaystyle \le \gamma {}^2\text{ C t1t2 E[F(M1,u,M2,u)].}}\end{array}$

Indeed $F_{xy}=t1t2x^{(t1-1)}{y}^{(t2-1)}$, and the two weighted integration sums in the cross term equal xy. There are no within-disk $F_{xx}$ or $F_{yy}$ terms to control. One may first use x+δ,y+δ and bounded truncations; finite-cutoff masses have all moments. Gronwall gives an upper bound $e^{({\gamma }^{2}Ct1t2)}$ times the product of the marginal moments of the independent model. First pass to the field limit with the bounded truncated weights. Then remove the insertion truncation, using the strict singular-mass moment bounds of step 1 and Fatou. No uniform bound is claimed when a smooth field cutoff is kept fixed while an unintegrable deterministic insertion weight is untruncated. This ordered limit gives the assertion for the actual singular masses. Thus correlation between disks is retained and bounded, not replaced by an independence identity.

3. Moment residues add over the heaviest singularities. For $x,y\ge 0$, q in a small fixed neighborhood of $p>0$, and a sufficiently small fixed $\eta >0$, elementary integration of $(x+y{)}^q$ along one variable gives

$\begin{array}{c}{\displaystyle \text{|(x+y)^}\text{q}-\text{x^}\text{q}-\text{y^}\text{q|}}\\ {\displaystyle \le \text{C[x^}\text{(q}-\eta \text{)y^}\eta \text{+x^}\eta \text{ y^}\text{(q}-\eta \text{)]. (2)}}\end{array}$

Choose $\eta <\min (p\mathrm{/}4,1\mathrm{/}2)$ and restrict q sufficiently close to p that all displayed powers are positive. If x,y are masses of separated singularities with threshold p, each mixed expectation on the right is uniformly finite by step 2: both powers lie a fixed distance below p. The same assertion holds if one mass has a strictly larger moment threshold, using Hölder with strict margins. Therefore the difference in (2) is bounded uniformly as q↑p. Multiplication by p−q makes it vanish.

Partition ρε into small neighborhoods of its heaviest singularities and the remaining pieces. When $\alpha 2<\alpha 3$ the unique heaviest singularity is at infinity. When $\alpha 2=\alpha 3$ the two heaviest singularities are at 1 and infinity. The charge $\alpha 1=\alpha 3-\alpha 2+\epsilon$ stays strictly below α3, uniformly as $\epsilon \to 0$; its neighborhood and the bounded rest have a moment of order p+δ for a fixed $\delta >0$. To avoid touching-support issues, the mixed terms between a heaviest mass and its remaining bounded annulus can alternatively be bounded by Hölder: $(q-\eta )\mathrm{/}p+\eta \mathrm{/}(p+\delta )<1$ uniformly near $q=p$. Hence every non-heaviest contribution and every cross term is bounded after subtracting the heaviest qth powers. It suffices to compute one local moment residue at each heaviest point.

4. A local residue at infinity and at 1. Under $w=1\mathrm{/}x$ the exact ρε weight is

$\begin{array}{c}{\displaystyle \text{|w|\_+^}\text{(}\gamma \text{A)|w|^}\text{(}-\gamma \alpha \text{3)|1}-\text{w|^}\text{(}-\gamma \alpha \text{2).}}\end{array}$

Thus near $w=0$ it is $\mathrm{\mid }w{\mathrm{\mid }}^{(-\gamma \alpha 3)}$ times a positive factor tending to one, uniformly in ε. The covariance there is exactly log(1/|w−w′|). Let Iα,r denote the pure-log singular integral in a disk of radius r. Exact Gaussian scaling gives

$\begin{array}{c}{\displaystyle \text{I}\alpha \text{,r law= r^}\text{[}\gamma \text{(Q}-\alpha \text{)] e^}\text{(}\gamma \text{Z\_r) I}\alpha \text{,1,}}\\ {\displaystyle \text{Var(Z\_r)=log(1/r), Z\_r independent of I}\alpha \text{,1.}}\end{array}$

Therefore, with $p\alpha =2(Q-\alpha )\mathrm{/}\gamma$,

$\begin{array}{c}{\displaystyle \text{E[I}\alpha \text{,r^}\text{q]=r^}\text{[}\gamma {}^2\text{q(p}\alpha -\text{q)/2]E[I}\alpha \text{,1^}\text{q],}}\\ {\displaystyle \text{lim\_(q}\uparrow \text{p}\alpha \text{)(p}\alpha -\text{q)E[I}\alpha \text{,r^}\text{q]=p}\alpha \text{ Rbar}\gamma \text{(}\alpha \text{), (3)}}\end{array}$