Proof. Denote \(\hat{B}_t = (1/r) B_{tr^2}\), then \((\hat{B}_t)_{t\ge 0}\) has the law of a Brownian motion starting from \(z\) which is independent of \(h\). Let \(\hat{h}\) denote the image of \(h^{2r}\) under the maps \(w \mapsto (w-z)/r\) (which by conformal invariance has the law of a GFF with Dirichlet boundary conditions in \(B_2(0)\)). Note that, by 12 , we have for \(w\in B_{2r}(z)\) \(G(z,w) = - \log |z-w|.\) Hence \[\begin{align} \nu(\tau_r) &= \lim_{\epsilon\to0}\;\int_0^{\tau_r}e^{\gamma h_{\epsilon}(B_s)}\epsilon^{\gamma^2/2}ds\\ &= \lim_{r\epsilon\to 0}\;\int_0^{\tau_{r}}e^{\gamma (h^{2r}_{r\epsilon}(B_s)+u^{2r}_{r\epsilon}(B_s))}(r\epsilon)^{\gamma^2/2} e^{\gamma^2 G(0, B_s)}ds\\ & \le \limsup_{\epsilon\rightarrow0} \int_0^{r^2\hat{\tau}_1}e^{\gamma (h^{2r}_{r\epsilon}(r\hat{B}_{s/r^2})+ \bar u_r)}(r\epsilon)^{\gamma^2/2}\vert rB_{s/r^2}\vert^{-\gamma^2} ds\\ & \le \limsup_{\epsilon\rightarrow0} e^{\gamma \bar u_r} \int_0^{\tau_1}r^2e^{\gamma h^{2r}_{r\epsilon}(r \hat{B}_t)}(r\epsilon^{\gamma^2/2})\vert r \hat{B}_t\vert^{-\gamma^2}dt\\ & = e^{\gamma \bar u_r} \;r^{2-\gamma^2/2} \limsup_{\epsilon\rightarrow0} \int_0^{\hat{\tau}_1}e^{\gamma \hat{h}_{\epsilon}( \hat{B}_{t})}\epsilon^{\gamma^2/2}\vert \hat{B}_{t}\vert^{-\gamma^2}dt\\ &= e^{\gamma \overline{u}_r}r^{2-\gamma^2/2}\hat{\nu}(\hat{\tau}_1), \end{align}\] where \(\hat{\nu}(\cdot)\) denotes the clock process associated to \(\hat{B}\) in the environment \(\hat{h}\). This gives the upper bound, the lower bound follows by a similar argument. ◻

Proof. Under \(\mathbb{P}^*_z\), we have \(h=h'+\gamma \log\vert z-\cdot\vert\), so conditioned on \({\mathcal{E} }_{r,z}(\alpha)\), we get by 12 \[\begin{align} h'_{2r}(z)=(\alpha-\gamma)\log1/2r. \end{align}\] Plugging that into ?? , we obtain ?? .

Now suppose that \(\delta>0\) and \(\alpha <Q\) so that \(\kappa(\alpha)>0\). Then, we get \[\begin{align} \mathbb{P}^*_z\otimes P_z[\nu(\tau_{r}) \leq t| {\mathcal{E} }_{z,r} (\alpha)]&\le\mathbb{P}^*_z\otimes P_z[t^{\kappa(\alpha)(1-\delta)(2+\gamma^2/2-\gamma\alpha)} e^{\gamma(\underline{u}_r-u(z))} \hat{\sigma}_1\le t | {\mathcal{E} }_{z,r} (\alpha)]\\ &\le\mathbb{P}^*_z\otimes P_z[t^{(1-\delta)} e^{\gamma(\underline{u}_r-u(z))} \hat{\sigma}_1\le t | {\mathcal{E} }_{z,r} (\alpha)]\\ &\le\mathbb{P}^*_z\otimes P_z[e^{\gamma(\underline{u}_r-u(z))} \hat{\sigma}_1\le t^{\delta} | {\mathcal{E} }_{z,r} (\alpha)]\\ &\le\mathbb{P}^*_z\otimes P_z[\hat{\sigma}_1\le t^{\delta/2}| {\mathcal{E} }_{z,r} (\alpha)]+\mathbb{P}^*_z[e^{\gamma(\underline{u}_r-u(z))}\le t^{\delta/2} | {\mathcal{E} }_{z,r} (\alpha)]. \end{align}\] Now, \(\hat{\sigma}_1\) is independent of \({\mathcal{E} }_{z,r} ( \alpha)\) so the first term vanishes FTP by Lemma 8, uniformly in \(\alpha<Q\). The exact same is true for the second term, as by Lemma 9 \(u - u(z)\) and \(u(z)\) are independent and the tail can be bounded by ?? . The second case \(r<t^{\kappa(\alpha)(1+\delta)}\) follows in an analogous manner.

Now suppose \(\alpha \ge Q\) so \(\kappa(\alpha ) \le 0\), and let \(\delta>0\) and \(r\le 1\). Then \(r^{2+\gamma^2/2-\gamma\alpha}\ge1\) so we have \[\begin{align} \mathbb{P}^*_z\otimes P_z[\nu(\tau_r)\le t\vert {\mathcal{E} }_{r,z}(\alpha)]&\le\mathbb{P}^*_z\otimes P_z[ e^{\gamma(\underline{u}_r-u(z))} \hat{\sigma}_1\le t | {\mathcal{E} }_{z,r} (\alpha)]\\ &\le\mathbb{P}^*_z\otimes P_z[\hat{\sigma}_1\le t^{1/2}| {\mathcal{E} }_{z,r} (\alpha)]+\mathbb{P}^*_z[e^{\gamma(\underline{u}_r-u(z))}\le t^{1/2} | {\mathcal{E} }_{z,r} (\alpha)]. \end{align}\] Again, we conclude by Lemma 8 and Lemma 9 that the right hand side tends to zero FTP in \(t\), uniform in \(\alpha\ge Q\). ◻

Proof of Proposition 6. As already indicated at the beginning of this section, we start applying Girsanov’s theorem, to exchange the expectation and the integral. For every point \(z\in \Omega\), under \(\mathbb{P}^*_z\) our field can be written as \(h=h'+\gamma G ( z, \cdot)\) where \(h'\) has again the law \(\mathbb{P}\) (see 12 ). Note that this is equal to \(h(x) = h'(x) - \gamma \log\vert x-z\vert\), if \(x \in B_1(0)\). We will show that \(\gamma\)-thick points do not contribute uniformly to the integral (cf. Lemma 10). The contribution of a point is governed by the circle average of \(h'\) at the scale of its distance to the boundary. Writing \(r=d(z,\partial\Omega)\), we therefore condition on the value of \(h_{2r}(z)\) and rewrite the integral accordingly.

We have to show that \[\begin{align} I=\int_{\Omega}\mathbb{P}^*_z\otimes P_z[\nu(\tau_\Omega)<t]dz=t^{\Delta+o(1)}. \end{align}\] We start with the upper bound. We condition with respect to the value of \(h_r(z)/ \log (1/r)\), a value which we will denote by \(\alpha\) (so we will integrate with respect to \(\alpha\)). As \[\begin{align} \label{eq:compare32h32to32h39} \frac{h_r(z) }{\log (1/r)} = \gamma + \frac{h'_r (z)}{\log (1/r)} \end{align}\tag{15}\] and as \(h'_r(z)/ \log (1/r)\) is a centered Gaussian random variable under \(\mathbb{P}^*_z\) and variance \(1/\log (1/r)\), we deduce (after lower-bounding \(\tau_\Omega\) crudely by \(\tau_r\)) the following: \[\begin{align} \label{eq:boundI} I&\le\int_{\alpha\in \mathbb{R}}\int_{z\in \Omega}\mathbb{P}^*_z\otimes P_z\Bigl[\nu(\tau_r)<t\Big\vert h_{2r}(z)=\alpha\log1/r\Bigr]f_{r,z}(\alpha)dz d\alpha ;\\ & \text{ with } f_{r,z}(\alpha) = r^{(\alpha-\gamma)^2/2+o(1)}; \text{ and } \int_\alpha f_{r,z} (\alpha) d\alpha = 1 \text{ for all } z \in \Omega\nonumber \end{align}\tag{16}\]

As we will see, the dominant contribution (even for a given \(\alpha \in \mathbb{R}\)) to the second integral on the right hand side turns out to come from a thin “strip” of points at a certain distance from the boundary of \(\Omega\), namely points whose distance to the boundary is of order \(t^{\kappa (\alpha)}\), where \(\kappa(\alpha)\) is given by ?? .

Let us first comment on the case \(\alpha\ge Q\) (i.e. \(\kappa(\alpha)\le 0\)), and show that the portion of the first integral coming from such values of \(\alpha\) can be neglected. By Corollary 10, the probability in 16 can be bounded by \(Ct^m\) for any \(m>0\) (in particular any \(m>\Delta\)), for some \(C=C(m,\gamma)>0\). Thus, we get \[\begin{align} \int_Q^\infty&\int_{z\in \Omega}\mathbb{P}^*_z\otimes P_z\Bigl[\nu(\tau_r)<t\Big\vert h_{2r}(z)=\alpha\log1/r\Bigr]f_{r,z}(\alpha) d\alpha\\ &\le\int_Q^\infty\int_{z\in \Omega}Ct^m f_{r,z}(\alpha)dz d\alpha\\ &\le Ct^m \int_{z\in \Omega}\int_Q^\infty f_{r,z}(\alpha)d\alpha dz\\ &\le C\vert\Omega\vert t^m \end{align}\] and obtain that this case contributes less than \(t^{\Delta+o(1)}\) to the integral.

For \(\alpha< Q\), we pick a small parameter \(\delta>0\), let \(\rho = t^{\kappa(\alpha)(1-\delta)}\) and decompose the domain \(\Omega\) by setting \[\begin{align} \Omega_{\alpha, n}^\delta&: = \{ z \in \Omega: \rho^{n+1} < d ( z, \partial \Omega) \le \rho^{n}\},\qquad n\ge1\\ \Omega^\delta_{\alpha,\mathrm{in}}&:=\{z\in\Omega:d(z,\partial\Omega)>\rho\} \end{align}\] so that \[\Omega=\Omega_{\alpha,\mathrm{in}}^\delta\cup\bigcup_{n\ge0}\Omega_{\alpha,n}^\delta.\] Let us first show that \(\Omega_{\alpha,\mathrm{in}}^\delta\) only contributes a negligible amount to the integral. Note that for \(z\in \Omega_{\alpha,\mathrm{in}}^\delta\), we have \(r = d(z, \partial \Omega) \ge \rho = t^{\kappa (\alpha) ( 1- \delta)}\). So by ?? in Corollary 10, we see that for any \(m>0\), there is a constant \(C = C(m, \delta, \gamma)\) (but independent of \(\alpha<Q\)) such that \(\mathbb{P}^*_z\otimes P_z[\nu(\tau_\Omega)<t] \le C t^m\). Hence \[\begin{align} \label{eq:32upper32estimate32first32moment32of32the32heat32content} \int_{\alpha< Q} \int_{\Omega_{\alpha,\mathrm{in}}^\delta} \mathbb{P}^*_z\otimes P_z[\nu(\tau_r)<t] f_{r,z} (\alpha) dz d\alpha \le & C t^m \int_{\Omega_{\alpha,\mathrm{in}}^\delta} \int_{\alpha<Q} f_{r,z} (\alpha) d\alpha \le C t^m |\Omega|. \end{align}\tag{17}\] Since \(m>0\) is arbitrary this shows the left hand side is in particular less than \(t^{\Delta + o(1)}\) by picking \(m>\Delta\).

Now to the case \(\alpha<Q\) and \(z\in\Omega^\delta_{\alpha,n}\) for some \(n\ge1\). It will turn out to be convenient to first show that the values of \(\alpha\) outside of a compact set (say \(\alpha< -2\)) contribute a negligible amount overall to the integral. To see this, we bound crudely \(\mathbb{P}^*_z\otimes P_z[\nu(\tau_r)<t] \le 1\), and recall that \[\int_{\alpha< -2} f_{r, z} ( \alpha) d\alpha \le C \exp \Bigl( - \frac{1}{2} (\gamma+2)^2 \log (1/r) \Bigr)\] using standard results for the tail of a standard Gaussian random variable (where \(C\) is a universal constant). Furthermore, if \(\alpha< -2\), \[\bigcup_{n\ge0} \Omega_{\alpha,n}^\delta \subseteq \{ z \in \Omega: d( z, \partial \Omega) \le t^{\eta}\}, \text{ with } \eta = \kappa(-2) /2\] so that uniformly for \(z \in \cup_{n\ge1} \Omega_{\alpha,n}^\delta\), we have \(r \le t^{\eta}\). Therefore, using Fubini’s theorem, \[\int_{\alpha< -2} \int_{z\in \cup_{n\ge0} \Omega_{\alpha,n}^\delta } f_{r,z} (\alpha) d\alpha \le |\Omega| \exp \Bigl( - \frac{1}{2} (\gamma+2)^2 \eta \log (1/t)\Bigr).\] One can easily check that \(\eta ( \gamma+2)^2/2 > \Delta\), so that the integral on the left hand side is \(t^{\Delta + o(1)}\).

We will thus henceforth be able to assume without loss of generality that \(-2 \le \alpha <Q\). Let us call \(I_n(\alpha)\) the portion of the integral in the right hand side of 16 on \(\Omega_{\alpha, n}^\delta\). By the assumption on the inner Minkowski dimension of the boundary 2, we have \[\begin{align} \vert\Omega^\delta_{\alpha,n}\vert=( \rho^n)^{2\mathsf{x} (1+ o(1))} \end{align}\] Since \(\kappa(\alpha) \ge \kappa (-2)\) uniformly over \(-2 \le \alpha \le Q\), and since \(n\ge 1\) we deduce that \[\begin{align} \vert\Omega^\delta_{\alpha,n}\vert\le t^{2\mathsf{x}n\kappa(\alpha)(1-\delta)+o(1)}, \end{align}\] where now \(o(1)\) is uniform in \(\alpha\in[-2,Q)\) and \(0<\delta<1/2\) and only depends on \(t\). Again, on \(\Omega^\delta_{\alpha,n}\), we can crudely bound \(\mathbb{P}^*_z\otimes P_z\Bigl[\nu(\tau_r)<t\Big\vert h_r(z)=\alpha\log1/r\Bigr]\le 1\), and we estimate \(f_{r,z}(\alpha)\) by \[r^{(\alpha-\gamma)^2/2} \le \rho^{(n+1)(\alpha- \gamma)^2/2} = t^{(n+1)\kappa(\alpha)(1-\delta)(\alpha-\gamma)^2/2}.\] Thus, we get the following upper bound for \(I_n(\alpha)\) when \(-2\le \alpha<Q\): \[I_n(\alpha) \le t^{2\kappa(\alpha)(1-\delta)(n+1)\mathsf{x}}t^{(n+1)\kappa(\alpha)(1-\delta)(\alpha-\gamma)^2/2+o(1)},\] Where the \(o(1)\)-term is uniform in \(z,\alpha,\delta\) and \(n\). Therefore, \[\begin{align} \int_{\alpha<Q}\sum_{n\ge0} I_n(\alpha) d\alpha&\le t^{\Delta+o(1)}+\int_{-2}^Q\sum_{n\ge0}\int_{\Omega_{\alpha,n}^\delta}t^{(n+1)\kappa(\alpha)(1-\delta)(\alpha-\gamma)^2/2+o(1)}dzd\alpha\\ &\le t^{\Delta+o(1)}+\int_{-2}^Q\sum_{n\ge0}t^{2\kappa(\alpha)(1-\delta)(n+1)\mathsf{x}}t^{(n+1)\kappa(\alpha)(1-\delta)(\alpha-\gamma)^2/2+o(1)}d\alpha\\ &\le t^{\Delta+o(1)}+\int_{-2}^Q t^{o(1)}\sum_{n\ge 0} \left(t^{(1-\delta) f(\alpha)}\right)^n d\alpha, \end{align}\] where \[f(\alpha )= \kappa(\alpha)(2\mathsf{x}+(\alpha-\gamma)^2/2).\] Some elementary but tedious calculations show that the minimum of \(\alpha\mapsto f(\alpha)\) is attained at \[\label{eq:alpha0} \alpha_0=Q-\sqrt{Q^2-4(1-\mathsf{x})}.\tag{18}\] The corresponding minimum is the value \[\begin{align} f(\alpha_0) = \frac{2\mathsf{x}+(\alpha_0-\gamma)^2/2}{2+\gamma^2-\alpha_0\gamma}=\Delta, \label{eq:Delta95alt} \end{align}\tag{19}\] where one can check that \(\Delta\) agrees with 5 .

As a consequence, the sum in the integral above can be bounded (uniformly over \(-2 \le \alpha<Q\)) by \[\begin{align} \sum_{n\ge 1}\Bigl(t^{(1-\delta) \Delta}\Bigr)^n \le \frac{t^{(1-\delta)\Delta}}{1- t^{\Delta/2}}. \end{align}\] if we assume without loss of generality that \(\delta <1/2\) and \(0<t<1\). Integrating, we thus obtain \[I\le C t^{(1-\delta) \Delta + o(1)}\] with \(C = 2(Q+2)\) and \(t\) sufficiently small. Since \(0<\delta<1/2\) is arbitrarily small, this shows \(I \le t^{\Delta+o(1)}\), as desired.

The lower bound is easier. We want to restrict to an event and a region which together carry the main contribution. Let \(\alpha_0\) be as in 18 be the optimal value, \(\kappa(\alpha_0)\) given by ?? (note that \(\alpha_0 < Q\), so \(\kappa(\alpha_0)>0\)) and let \(\delta>0\). Under the rooted law \(\mathbb{P}^*_z\) we have by 12 , \(h=h'-\gamma\log\vert\cdot-z\vert\), where \(h'\) is again a full plane GFF normalized to have zero average on the unit circle. Let \[\begin{align} {\mathcal{E} }_{z,r}(\alpha_0,\delta)&:=\{h_{2r}(z)\in[\alpha_0\log1/r,(\alpha_0+\delta)\log1/r]\}\\ \Omega^*_{\delta,t}&:=\Bigl\{z\in\Omega\Big\vert t^{\kappa(\alpha_0)(1+2\delta)}\le d(z, \partial^*\Omega)\le t^{\kappa(\alpha_0)(1+\delta)}\Bigr\}. \end{align}\] (We recall that \(\partial^*\Omega\) denotes a fixed connected component of positive diameter of the boundary). Let also \(z\in\Omega^*_{\delta,t}\). Let \(n\ge1\). By scaling (for ordinary planar Brownian motion) and Beurling’s estimate, since \(\partial^* \Omega\) is connected, we have for \(t>0\) small enough \[\begin{align} P_z[\tau_\Omega<\tau_{r\log1/t}]\ge\frac{n-1}{n}. \end{align}\] On \(\Omega^*_{\delta,t}\) we have \(r\le t^{\kappa(\alpha_0)(1+\delta)}\). Thus, we get by Corollary 10 \[\begin{align} \mathbb{P}^*_z\otimes P_z[\nu(\tau_{r\log1/t})<t\vert {\mathcal{E} }_{z,r}(\alpha_0,\delta)]\ge\frac{n-1}{n}, \end{align}\] for \(t>0\) small enough. Combining these estimates, we get \[\begin{align} \mathbb{P}^*_z\otimes P_z[\nu(\tau_\Omega)<t\vert {\mathcal{E} }_{z,r}(\alpha_0,\delta)]&\ge\mathbb{P}^*_z\otimes P_z[\nu(\tau_\Omega)<\nu(\tau_{Cr})<t\vert {\mathcal{E} }_{z,r}(\alpha_0,\delta)]\\ &=\mathbb{P}^*_z\otimes P_z[\nu(\tau_\Omega)<\nu(\tau_{Cr})\vert {\mathcal{E} }_{z,r}(\alpha_0,\delta)]\\ &\qquad-\mathbb{P}^*_z\otimes P_z[\nu(\tau_\Omega)<\nu(\tau_{Cr}),\nu(\tau_{Cr})\ge t\vert {\mathcal{E} }_{z,r}(\alpha_0,\delta)]\\ &\ge P_z[\tau_\Omega<\tau_{Cr}]-\mathbb{P}^*_z\otimes P_z[\nu(\tau_{Cr})\ge t\vert {\mathcal{E} }_{z,r}(\alpha_0,\delta)]\\ &\ge1-\frac{1}{n}-\frac{1}{n}=\frac{n-2}{n}, \end{align}\] so the probability tends to \(1\) uniformly in \(z\in\Omega^*_{\delta,t}\), when \(t\to0\). Next, we estimate the size of \(\Omega^*_{\delta,t}\). By ?? and Assumption 1, we have \[\begin{align} \vert\Omega^*_{\delta,t}\vert \ge t^{2\mathsf{x}\kappa(\alpha_0)(1+\delta)+o(1)}-t^{2\mathsf{x}\kappa(\alpha_0)(1+2\delta)+o(1)}=t^{2\mathsf{x}\kappa(\alpha_0)(1+\delta)+o(1)}, \end{align}\] where \(\mathsf{x}\) is the Euclidean scaling exponent.

Finally, we bound the probability of \({\mathcal{E} }_{z,r}(\alpha_0,\delta)\) from below for a point \(z\in\Omega^*_{\delta,t}\). Note that \(h'_r(z)\) (where we recall that \(h=h'-\gamma\log\vert\cdot-z\vert\) under \(\mathbb{P}^*_z\)) is a centered Gaussian random variable with variance \(\mathrm{Var}(h'_r(z))= \mathrm{Var}(h_r(z))=\log(1/r)\). Thus, using 15 , we get \[\begin{align} \mathbb{P}^*_z[ {\mathcal{E} }_{z,r}(\alpha_0,\delta)]&\ge\int_{\alpha_0\log1/r}^{(\alpha_0+\delta)\log1/r}r^{(\gamma-\alpha)^2/2+o(1)}d\alpha\\ &\ge r^{(\gamma-\alpha_0)^2/2+o(1)}\\ &\ge t^{\kappa(\alpha_0)(1+2\delta)(\gamma-\alpha_0)^2/2+o(1)}, \end{align}\] where again the \(o(1)\)-term is uniform in \(\alpha,z\) and \(\delta\). So we get in total \[\begin{align} I&\ge\int_{\Omega^*_{\delta,t}}\mathbb{P}^*_z\otimes P_z[\nu(\tau_\Omega)<t\cap {\mathcal{E} }_{z,r}(\alpha_0, \epsilon)]dz\\ &=\int_{\Omega^*_{\delta,t}}\mathbb{P}^*_z\otimes P_z[\nu(\tau_\Omega)<t\vert {\mathcal{E} }_{z,r}(\alpha_0, \delta)]\mathbb{P}^*_z[ {\mathcal{E} }_{z,r}(\alpha_0, \delta)]dz\\ &\ge \frac{\vert\Omega^*_{\delta,t}\vert}{2} t^{\kappa(\alpha_0)(1+2\delta)(\gamma-\alpha_0)^2/2+o(1)}\\ &=t^{2\mathsf{x}(\kappa(\alpha_0)(1+\delta)+\kappa(\alpha_0)(1+2\delta)(\gamma-\alpha_0)^2/2+o(1)}. \end{align}\] As \(\delta>0\) was arbitrary, we have in total, using 19 , \[\begin{align} I_1&\ge t^{\kappa(\alpha_0)(2\mathsf{x}+(\gamma-\alpha_0)^2/2)+o(1)}=t^{\Delta+o(1)}. \end{align}\]

All together, we proved \[\begin{align} \mathbb{E}[ {\mathbf{K} }_\Omega(t)]=t^{\Delta+o(1)}, \end{align}\] which completes the proof of Proposition 6 assuming Lemmas 8 and 9. ◻

Proof of Lemma 9. The independence of \(h'\) and \(u\) is just given by the spatial Markov property of the GFF. As \(u\) is harmonic on \(B_{Nr}(z_0)\), we can write for all \(z,w\in B_{r}(z_0)\) \[\begin{align} u(z)-u(w)=(h,\nu_z)-(h,\nu_w)=\frac{1}{2\pi}\int_0^{2\pi}h(z_0+Nr e^{i\phi})(\underbrace{f_z(\phi)-f_w(\phi)}_{=:f_{z,w}(\phi)})d\phi, \end{align}\] where \(\nu_z,\nu_w\) are the harmonic measures on \(\partial B_{Nr}(z_0)\) of \(z,w\) respectively and \(f_z,f_w\) are the densities of \(\nu_z,\nu_w\) to the uniform measure on \(\partial B_{Nr}(z_0)\). In particular \(f_{z,w}\) induces a signed measure on \(\partial B_{Nr}(z_0)\) with total mass \(0\). By estimating the first derivative of \(f_{z,w}\) for \(z,w\in B_{r}(z_0)\), it can be easily shown that \[\begin{align} \label{eq:estimate32on32f95123z44w125} \vert f_{z,w}(\phi)\vert\le L\frac{\vert z-w\vert}{(N-1)r}, \end{align}\tag{20}\] where \(L>0\) is a universal constant (note that for this bound \(z,w\in B_r(z_0)\) is crucial); this corresponds to Harnack’s inequality. We will write \(L_N = L / (N-1)\); the only property we will need is that \(L_N\) depends only on \(N\) and tends to 0 as \(N\to \infty\). By 12 , and since \(\bar B_{Nr}(z_0) \subset \mathbb{D}\), we have \(G(z',w')=\log (1/{\vert z'-w'\vert})\) for any \(z', w' \in \bar B_{Nr} (z_0)\). Since for any fixed \(x\in\mathbb{C}\) the function \(z\mapsto G(z,x)\) is harmonic on \(\mathbb{C}\setminus\{x\}\), we get for \(z,w\in B_r(z_0)\), \[\begin{align} \mathrm{Cov}(u(z)-u(w),u(z_0))&=\frac{1}{4\pi^2}\int_0^{2\pi}\int_0^{2\pi}G(z_0+Nre^{i\phi},z_0+Nre^{i\psi})f_{z,w}(\phi)d\psi d\phi\\ &=\frac{1}{2\pi}\int_0^{2\pi}G(z_0,z_0+N re^{i\phi})f_{z,w}(\phi)d\phi\\ &=\frac{1}{2\pi}\int_0^{2\pi}\log\frac{1}{N r}f_{z,w}(\phi)d\phi=0. \end{align}\] As \(u\) is centered Gaussian, we obtain independence of \(u(z)-u(w)\) and \(u(z_0)\). We also want to estimate the variance of \(u(z)-u(w)\). Using that \(f_{z,w}\) integrates to \(0\) and 20 , we get \[\begin{align} \mathrm{Var}(u(z)-u(w))&=\frac{1}{4\pi^2}\int_0^{2\pi}\int_0^{2\pi}f_{z,w}(\phi)G(z_0+ Nre^{i\phi},z_0+ Nre^{i\psi})f_{z,w}(\psi)d\psi d\phi\\ &=\frac{1}{4\pi^2}\int_0^{2\pi}\int_0^{2\pi}f_{z,w}(\phi)\log\frac{1}{Nr\vert e^{i\phi}-e^{i\psi}\vert}f_{z,w}(\psi)d\psi d\phi\\ &\le\frac{1}{4\pi^2}\int_0^{2\pi}\int_0^{2\pi}\vert f_{z,w}(\phi)\vert \Big\vert\log\frac{1}{\vert e^{i\phi}-e^{i\psi}\vert}\Big\vert\vert f_{z,w}(\psi)\vert d\psi d\phi\\ &\qquad+\frac{1}{4\pi^2}\int_0^{2\pi}\int_0^{2\pi}\log\frac{1}{N r}f_{z,w}(\phi)f_{z,w}(\psi)d\psi d\phi\\ &\le L_N^2\frac{\vert z-v\vert}{4\pi^2r^2}\int_0^{2\pi}\int_0^{2\pi} \Big\vert\log\frac{1}{\vert e^{i\phi}-e^{i\psi}\vert}\Big\vert d\psi d\phi\\ &\le CL_N^2 \frac{\vert z-w\vert^2}{r^2}, \end{align}\] where \(C>0\) is a universal constant.

Let us use this to deduce the desired Gaussian tail for the supremum of \(\vert u(z)-u(z_0)\vert\) over \(z\in B_r(0)\). For \(n\ge1\), we define the lattice \[\begin{align} A_n:=\{z\in B_r(z_0):z=(r2^{-n}k,r2^{-n}l), k,l\in\mathbb{Z}\} \end{align}\] Then we have \(\vert A_n\vert\le 4r^22^{2n}\). Furthermore, for any given \(z\in B_r(z_0)\) we define a sequence \((z_n)_{n\in\mathbb{N}}\), by setting \(z_n\) to being the closest point in \(A_n\) to \(z\) Then note that \(\vert z_n-z_{n-1}\vert\le \sqrt{2}r2^{-(n-1)}\) and thus \[\begin{align} \label{eq:variance32estimate32u} \mathrm{Var}(u(z_n)-u(z_{n-1}))\le CL_N^2 2^{-2n}, \end{align}\tag{21}\] for some universal constant \(C>0\). As \(u\) is harmonic and therefore also continuous, we can write \[\begin{align} u(z)-u(z_0)=\sum_{n\ge1}u(z_n)-u(z_{n-1}). \end{align}\] Let \(\lambda\ge\lambda_0\) for some fixed \(\lambda_0>0\). Then using 21 and the estimate on the variance of \(u(z)-u(w)\) above, we have \[\begin{align} \label{eq:estimate32in32proof32of32decay32mean} \mathbb{P}[\sup_{z\in B_r(z_0)}\vert u(z)-u(z_0)\vert\ge\lambda]&\le\mathbb{P}[\sup_{z\in B_r(z_0)}\vert \sum_{n\ge1}u(z_n)-u(z_{n-1})\vert\ge\frac{6\lambda}{\pi^2}\sum_{n\ge1}n^{-2}]\nonumber\\ &\le\sum_{n\ge1}\mathbb{P}[\sup_{z\in B_r(z_0)}\vert u(z_n)-u(z_{n-1})\vert\ge\frac{6\lambda}{\pi^2}n^{-2}]\nonumber\\ &\le\sum_{n\ge1}8\sum_{z_n\in A_n}\mathbb{P}[\vert u(z_n)-u(z_{n-1})\vert\ge\frac{6\lambda}{\pi^2}n^{-2}]\nonumber\\ &\le C_1\sum_{n\ge1}2^{2n}\exp\Bigl(-\frac{C_2 L_N^{-2} \lambda^22^{2n}}{n^4}\Bigr), \end{align}\tag{22}\] for some universal constant \(C_1,C_2>0\). To conclude, we first notice that there is a universal constant \(C>0\) such that \[\begin{align} \frac{C_2L_N^{-2} \lambda^2 2^{2n}}{n^4} & \ge CL_N^{-2} \lambda^2(n+2^n)\\ & \ge C L_N^{-2}\lambda^2 n + C \lambda^2 2^{2n} . \end{align}\]

Using this, we obtain \[\begin{align} \sum_{n\ge1}2^{2n}\exp\Bigl(-\frac{C_2\lambda^22^{2n}}{n^4}\Bigr)\le\max_{n\ge1}\exp\Bigl(2\log(2)n-C\lambda^22^n\Bigr)\sum_{n\ge1}e^{-CL_N^{-2}\lambda^2n}. \end{align}\] By maximizing over any \(n\in[1,\infty)\), we get that the maximum can be estimated by \[\begin{align} \label{eq:proof32of32decaymean321} \max_{n\ge1}\exp\Bigl(2\log(2)n-C\lambda^22^n\Bigr)\le\Bigl(\frac{2}{C\lambda^2}\Bigr)^2e^{-2}\le C' <\infty, \end{align}\tag{23}\] assuming \(\lambda\ge\lambda_0\), for some constant \(C'\) only depending on \(\lambda_0\). Furthermore, the sum gives \[\begin{align} \label{eq:proof32of32decaymean322} \sum_{n\ge1}e^{-C\lambda^2n}=\frac{e^{-C L_N^{-2}\lambda^2}}{1-e^{-C L_N^{-2} \lambda^2}}<\frac{1}{1- e^{-C}}e^{-CL_N^{-2}\lambda^2}. \end{align}\tag{24}\] By putting 23 and 24 together, we obtain \[\begin{align} \mathbb{P}[\sup_{z\in B_r(z_0)}\vert u(z)-u(z_0)\vert\ge\lambda]\le ce^{-c'\lambda^2},\qquad\lambda\ge\lambda_0 \end{align}\] for constants \(c,c'>0\), where \(c\) depends only on \(N\) and \(\lambda_0\) and where \(c' = C L_N^{-2}\) uniform in \(\lambda_0\) can be made arbitrarily big by choosing \(N\) large enough since \(L_N \to 0\) as \(N\to \infty\). This completes the proof of Lemma 9. ◻

Proof of Lemma 8. First we note that we can assume without loss of generality that \(\alpha = 0\), as when \(\alpha>0\) the exit time becomes larger (since \(\log (1/ |z|) \ge 0\) in \(\mathbb{D}\)). We therefore assume \(\alpha = 0\) in the rest of the proof. The strategy is inspired by Duplantier and Sheffield’s approach in [6] (where the Liouville area measure of a ball is used in place of the quantum clock accumulated by an independent Brownian motion until its exit time from the unit disc). Roughly, the idea is as follows. We estimate the exit time from below by two smaller copies of itself, by considering two balls of a slightly smaller radius (but still same scale overall). For each of them, by the spatial Markov property, the GFF splits into an independent part plus a harmonic remainder, and Lemma 9 controls the harmonic contribution. This yields a lower bound on the original exit time by the maximum of two independent, smaller copies of itself, up to a small error term. By the change of coordinate formula of the Liouville measure, these smaller exit times get transported back to the original scale, resulting in a recursive inequality for the distribution function of the exit time. Estimating this recursion gives the desired bound.
Setup. To make things rigorous, fix a path of the underlying (ordinary) Brownian motion \((B_t, t \ge 0)\) up to its exit from \(B_2(0)\). Let \[\begin{align} \tau_{1/2}=\inf\{t>0:\vert B_t\vert=1/2\},\qquad \text{ and let } z_0=B_{\tau_{1/2}} \end{align}\] be the first point where the (Euclidean) Brownian motion attains modulus 1/2. By Lemma 9, by choosing \(N\) sufficiently large we suppose without loss of generality that the constant \(c'\) in this Gaussian bound is such that \(c'>\log2\).

For notational convenience, set \[\begin{align} D^+&=B_{1/4}(0),\qquad D^-=B_{1/4}(z_0),\qquad D=D^+\cup D^-,\\ B^+&=B_{1/(4N)}(0)\subseteq D^+,\qquad B^-=B_{1/(4N)}(z_0)\subseteq D^-. \end{align}\] The argument runs as follows. Given the path of the ordinary Brownian motion \((B_t, t \ge 0)\), we bound \(\sigma_\mathbb{D}= \nu(\tau_\mathbb{D})\) from below by two ingredients: – The exit time \(\nu(\tau_{B^+})\) of the smaller ball \(B^+\) – and the clock accumulated by \(B\) between \(\tau_{1/2}\) and \(\tau_{B^-}\), defined as the first time after \(\tau_{1/2}\) when \(B\) leaves the ball \(B^-\). Equivalently, this can be viewed as the clock \(\nu^{(h, \tilde{B})}(\tilde{\tau}_{B^-})\) where the underlying planar Brownian motion is \(\tilde{B}_s = B_{s+\tau_{1/2}}, s\ge 0\), and \(\tilde{\tau}_{B^-} = \inf \{ s > 0: \tilde{B}_s \notin B^-\}\).

To make them independent, we use the spatial Markov property of the GFF \(h^{2\mathbb{D}}\) to get \(h^{2\mathbb{D}}=h^{D^+}+h^{D^-}+u\), where \(h^{D^+}, h^{D^-}\) are independent Dirichlet GFFs in \(D^+\) and \(D^-\) respectively and \(u\) is independent of the two other terms and harmonic in \(D^+\cup D^-\). We define \[\begin{align} A &= \log\nu^{2\mathbb{D}}(\tau_{\mathbb{D}}),\\ A^+ &= \log\nu^{(h^{D^+}, B)}(\tau_{B^+}),\\ A^- &= \log\nu^{(h^{D^-}, \tilde{B})}(\tilde{\tau}_{B^-}), \end{align}\] where \(\nu^{D^+},\nu^{D^-}\) are the quantum clocks associated to \(h^{D^+}\) and \(h^{D^-}\) respectively. Now, by the strong Markov property of planar Brownian motion, given \(\mathcal{F}_{\tau_{1/2}} = \sigma( B_s: 0 \le s \le \tau_{1/2})\), \((\tilde{B}_{s}, s \ge 0)\) is a Brownian motion starting in \(z_0\). Since \(h^{D^+}\) and \(h^{D^-}\) are also independent, we deduce that \(A^+\) and \(A^-\) are independent.
Construction of the recursion. By the change of coordinate formula of LQG, we have \[\begin{align} A^+, A^- \overset{(d)}{=} A - (2+\gamma^2/2)\log(8N). \end{align}\] We will write \(b = ( 2+ \gamma^2/2) \log (8N)\). Its value for the rest of the argument is of little relevance but we note that \(b >0\). Moreover, we have that \[\begin{align} \nu^{2\mathbb{D}}(\tau_{\mathbb{D}})\geq \nu^{2\mathbb{D}}(\tau_{B^+})\geq e^{\gamma\underline{u}}\nu^{D^+}(\tau_{B^+}), \end{align}\] and similarly for \(A^-\), so we conclude that \[\begin{align} \label{eq:boundAs} A\geq\max\{A^-,A^+\}+\gamma\underline{u}, \end{align}\tag{25}\] with \(\underline{u}=\inf_{z\in B^+\cup B^-}u(z)\). Our goal is to show that, for sufficiently negative \(x\), \[\begin{align} p_A(x):=\mathbb{P}\otimes P_0[A\leq x]\leq e^{-c'x^2}. \end{align}\]

For \(\rho\in(0,1)\) that we will specify later, we split this probability into two terms, \[\begin{align} p_A(x)=\underbrace{\mathbb{P}\otimes P_0[A\leq x,\underline{u}\leq\rho x/\gamma]}_{=:p_1(x)}+\underbrace{\mathbb{P}\otimes P_0[A\leq x,\underline{u}>\rho x/\gamma]}_{=:p_2(x)}. \end{align}\] Now, Lemma 9 allows us to immediately bound the first term, so that for sufficiently negative \(x\), \[\begin{align} \label{eq:sufficiently32negative} p_1(x)\leq e^{-c'x^2}, \end{align}\tag{26}\] where we made \(c'\) smaller to get rid of the factor \(\rho^2/\gamma^2\) in the exponent and also the constant \(c>0\). Using 25 , and the independence of \(A^+\) and \(A^-\), we get \[\begin{align} p_2(x) \le \mathbb{P}(\max (A^-, A^+) \le (1- \rho) x) \le p_A\Bigl((1-\rho)x+\underbrace{(2+\gamma^2/2)\log8N}_{=b}\Bigr)^2, \end{align}\] which in total gives \[\begin{align} p_A(x)\le p_A\Bigl((1-\rho)x+b\Bigr)^2+e^{-c'x^2}, \end{align}\] for some constant \(c'>0\). From here the idea is to use a recursive argument to obtain the desired Gaussian bound. Such an idea is sketched in [6]; however, we could not follow the details of their argument, and indeed our numerical simulations suggest that the induction alone as used in [6] is not enough to guarantee a Gaussian bound without additional nontrivial information on the start of this recursion. The same question is also studied by Aru in [37], following a similar idea but with a cleaner approach, and we follow this instead. While some of the details of the argument there were not entirely written (and indeed the crucial iteration step cannot be carried out all the way up to \(\nu_0\) in Aru’s notations) we were nevertheless able to carry out the argument and provide a complete proof following his outline. The details are as follows.

Note that if we choose \(N=2\) so \(b = (2+ \gamma^2/2) \log (8N)\) then \(b\le 16 \log 2\), so if \(x \le -1000\) and \(\rho =0.15\) then \(x' := ((1- \rho) x + b) \le 0.8 x = (4/5) x\). Using monotonicity of \(p_A\) we deduce \[\label{eq:Induction95Aru} p_A(x) \le e^{ - c' x^2} + p_A( 4x/5)^2; \quad \text{ for all } x \le - 1000.\tag{27}\] Fix \(a = c'/2\) and let \(x_0 \le -1000\) be such that if \(x \le x_0\) then 26 holds true and also \[\label{eq:x0} e^{ - a x^2 } - e^{ - c' x^2} \ge e^{ - 0.9 a x^2}.\tag{28}\]

Consider \(x_m = -x_0 \cdot (5/4)^m\). By monotonicity of \(p_A\) again it suffices to show that \(p_A(x_m) \le e^{ - a x_m^2}\) for all sufficiently large \(m\ge 0\). Suppose for contradiction that: \[\label{eq:contrad} p_A(x_m) > e^{ - a x_m^2}, \quad \text{ for infinitely many values of m.}\tag{29}\] If \(m\ge 1\) is any such value we also have from 27 \[e^{ - a x_m^2} \le p_A(x_m) \le e^{ - c' x_m^2} + p_A(x_{m-1})^2\] so by 28 (since \(x_m \le x_0\)), \[p_A(x_{m-1})^2 \ge e^{ - a x_m^2} - e^{ - c' x_m^2} \ge e^{ - 0.9 a x_m^2}.\] This implies (since \(0.45 \cdot (5/4)^2 \approx 0.7 \le 0.9\)) \[\label{induction95k611} p_A(x_{m-1}) \ge e^{ - 0.45 a x_m^2} \ge e^{ - 0.9 a x_{m-1}^2}.\tag{30}\]

This can be iterated by induction: for \(0 \le k \le m\), let \[y_k = 0.9^{k/2} x_{m-k}.\] Thus \(y_0 = x_m\), and as \(k\) increases, \(y_k\) gradually shrinks closer to 0 at a geometric rate which is faster than \(x_{m-k}\) itself; in particular \(y_k \ge x_{m-k}\). As a consequence, if \[k_m = \min \{ k \ge 0: y_k \ge x_0\}\] is the first time for which \(y_k\) gets above the value \(x_0\), then \(k_m< m\), and in fact \(m - k_m \to \infty\) as \(m\to \infty\) (linearly with \(m\), in fact). We will show by induction on \(k\ge 0\) that for all \(0 \le k < k_m\), \[\label{eq:iterated} p_A (x_{m-k}) \ge e^{- 0.9^k a x_{m-k}^2} = e^{ - a y_k^2}.\tag{31}\] For completeness we write out the induction carefully. The case \(k =0\) was the assumption on \(m\), and we have just done by hand the case \(k = 1\) in 30 . Let \(k \ge 1\) and suppose that 31 holds. Then from 27 , \[\begin{align} p_A(x_{m-k-1})^2 & \ge e^{ - a y_{k}^2 } - e^{ - c' x_{m-k}^2} \\ &\ge e^{ - a y_{k}^2 } - e^{ - c' y_{k}^2} \quad \text{ (since x_{m-k} \le y_k)}\\ & \ge e^{ - 0.9 a y_{k}^2} = e^{ - 0.9^{k+1} a x_{m-k}^2}, \end{align}\] where we used 28 , thanks to the fact that \(k< k_m\) and thus \(y_k \le x_0\). Therefore, taking square roots, \[\begin{align} p_A(x_{m-k-1})& \ge e^{ - (1/2) 0.9^{k+1} a x_{m-k}^2} \\ & = e^{ - 0.45 \cdot (5/4)^2 \cdot 0.9^k a x_{m-k-1}^2} \\ & \ge e^{ - 0.9^{k+1} a x_{m-k-1}^2} = e^{ - 0.9^{k+1} y_{k+1}^2}. \end{align}\] This completes the induction.

Now let us evaluate 31 at \(k = k_m-1\). This reads: \[\label{eq:contrad2} p_A ( x_{m- k_m+1} ) \ge e^{ - a y_{k_m-1}^2} \ge \rho: = e^{ - a \frac{x_0^2}{ 0.9 (4/5)^2}}.\tag{32}\] Note that 32 holds for the infinitely many values of \(m\) for which 29 holds. In particular, since \(m-k_m \to \infty\) as \(m\to \infty\) we have a subsequence \(n_m \to \infty\) such that \(p_A ( x_{n_m}) \ge \rho\). This contradicts the fact that \(p_A (x) \to 0\) as \(x \to - \infty\) (which follows from the fact that \(\nu^{2\mathbb{D}}( \tau_\mathbb{D}) >0\) a.s., see, e.g., Theorem 3.42 in [11]). Therefore, 29 cannot hold, and this concludes the proof of Lemma 8. ◻

Proof. Clearly, we have \(\mathbb{E}[ {\mathbf{K} }_{\Omega,z_i}(t)]\le\mathbb{E}[ {\mathbf{K} }_\Omega(t)]\), so it suffices to establish the lower bound. Our goal will be to establish that \[\label{eq:1st95trunc95goal} \mathbb{P}^* \otimes P_z ( \nu (\tau_\Omega)< t ; G_t) \ge c \mathbb{P}^*(G_t)\tag{40}\] for some constant \(c>0\), uniformly in \(z \in \Omega^t \cap B_{R(t)} (z_i)\) and uniformly in \(z_i \in \partial_{R(t)}\). Note that this implies the strong lower bound ?? . Indeed, given this, we can write by the Girsanov–Cameron–Martin lemma (more specifically 38 ): \[\begin{align} \mathbb{E}[ {\mathbf{K} }_{\Omega,z_i}(t)]&\ge\int_{\Omega^t \cap B_{R(t)} (z_i)}\mathbb{P}^*_z\otimes P_z[\nu(\tau_\Omega)<t; G_t]dz\ge c\int_{\Omega^t\cap B_{R(t)}(z_i)}\mathbb{P}^*_z[G_t]dz \end{align}\] Recall that \(G_t = (C1)' \cap (C2)'\). We can estimate the probability of \(G_t\) simply by that of \((C1)'\): more precisely, there exists \(p_0\) depending solely on \(\beta\), \(\gamma\) and \(M\) such that \[\label{eq:p0} \mathbb{P}^*_z ( (C2)' (z) | (C1)'(z) ) \ge \mathbb{P}[ B_t \le M + ( \beta - \gamma) t, \forall t \ge 0] \ge p_0,\tag{41}\] where \((B_t)_{t\ge 0}\) is a standard one-dimensional Brownian motion starting from \(B_0 = 0\). Furthermore, if \(\beta\) and \(\gamma\) are fixed, by choosing \(M\) large enough we can ensure that \(p_0 \ge 0.9\).

Thus, \[\mathbb{E}[ {\mathbf{K} }_{\Omega,z_i}(t)] \ge c p_0 \int_{\Omega^t}\mathbb{P}^*_z [ (C1)'(z)] dz.\] As \(X'_s\) is a centered Gaussian random variable with variance \(\log1/r\) and by using \(r=Nd(z,\partial\Omega)\) and 34 , we have \[\begin{align} \mathbb{P}^*_z[(C1)'(z)]\ge C\frac{t^{\kappa(\alpha_0)(\alpha_0-\gamma)^2/2}}{\log1/t}, \end{align}\] where \(C>0\) is a constant which is uniform in \(R(t),z\) and \(z_i\). We deduce that \[\begin{align} \mathbb{E}[ {\mathbf{K} }_{\Omega,z_i}(t)]&\ge cp_0 \left|\Omega^t\cap B_{R(t)}(z_0)\right| \mathbb{P}^*_z[(C1)'(z)]\\ &\ge c_\partial t^{2\mathsf{x}\kappa(\alpha_0)}R(t)^{2(1-\mathsf{x})}\frac{C}{\log1/t}t^{\kappa(\alpha_0)(\alpha_0-\gamma)^2/2}\\ &\ge c\frac{t^{\Delta} R(t)^{2(1-\mathsf{x})}}{\log1/t}=t^{\Delta+\epsilon(t)}R(t)^{2(1-\mathsf{x})}. \end{align}\] where \(c>0\) is a constant which is uniform in \(R(t)\) and \(z_i\) and \(\epsilon(t)\) is tending to 0 as \(t\to 0\), uniformly in \(z_i \in \partial_{R(t)}\), and uniformly in the choice of \(R(t)\).

It therefore suffices to prove 40 . Now, observe that \[\begin{align} \mathbb{P}^*_z \otimes P_z [ \nu (\tau_\Omega) < t ; G_t] & \ge \mathbb{P}^*_z \otimes P_z [ \nu (\tau_\Omega) < t ; (C1)'] - \mathbb{P}^*_z \otimes {\mathbf{P} }_z [ \nu (\tau_\Omega) < t; (C1)' ; \overline{(C2)'}] \nonumber \\ & \ge \mathbb{P}^*_z \otimes P_z [ \nu (\tau_\Omega) < t ; (C1)'] - \mathbb{P}^*_z \otimes P_z [ (C1)' ; \overline{(C2)'}] \nonumber \\ & \ge \mathbb{P}^*_z \otimes P_z [ \nu (\tau_\Omega) < t ; (C1)'] - (1-p_0) \mathbb{P}^*_z [ (C1)'] \label{eq:GtC1} \end{align}\tag{42}\] where \(\overline{(C2)'}\) denotes the complement of the event \((C2)'\), and \(p_0\) is as in 41 .

To estimate the first term in the right hand side, we will use a Beurling estimate to compare \(\nu (\tau_\Omega)\) with \(\nu (\tau_r)\) (cf. 35 ) as well as a scaling argument, where we recall that \(r=Nd(z,\partial\Omega)\). More precisely, \[\begin{align} \mathbb{P}^*_z \otimes P_z [ \nu (\tau_\Omega) < t ; (C1)'] & \ge \mathbb{P}^*_z \otimes P_z [ \nu (\tau_r) < t ; (C1)'] - \mathbb{P}^*_z \otimes P_z [ \nu (\tau_r) < \nu (\tau_\Omega) ; (C1)'] \nonumber \\ & \ge \mathbb{P}^*_z \otimes P_z [ \nu (\tau_r) < t ; (C1)'] - \mathbb{P}^*_z \otimes P_z [ \tau_r < \tau_\Omega ; (C1)'] \nonumber \\ & \ge \mathbb{P}^*_z \otimes P_z [ \nu (\tau_r) < t ; (C1)'] - p_1 \mathbb{P}^*_z [ (C1)']. \label{eq:C1tau95r} \end{align}\tag{43}\] Now, by Lemma 7, we have for \(z \in \Omega^t \cap B_{R(t)} (z_i)\), \[\begin{align} \nu(\tau_{r})\le r^{2-\gamma^2/2}e^{\gamma\bar u_r}\hat{\sigma}_1=r^{2-\gamma^2/2}e^{\gamma u(z)}e^{\gamma(\bar u_r-u(z))}\hat{\sigma}_1, \end{align}\] where all three factors \(e^{\gamma u(z)}\), \(e^{\gamma(\bar u_r-u(z))}\) and \(\hat{\sigma}_1\) are mutually independent. When \((C1)'\) holds, we therefore deduce (after some algebra) that \[\begin{align} \nu(\tau_{r})& \le r^{1/\kappa(\alpha_0)} e^{ - \gamma M} e^{\gamma(\bar u_r-u(z))} \hat{\sigma}_1\\ & = N^{1/\kappa ( \alpha_0)} t e^{ - \gamma M} e^{\gamma(\bar u_r-u(z))} \hat{\sigma}_1. \end{align}\] We may assume without loss of generality that \[\label{eq:MN} N^{ 1/ \kappa (\alpha_0)} e^{ - \gamma M/2} \le 1,\tag{44}\] so that in total, canceling the factors of \(t\) on either side of the inequality, \[\begin{align} \mathbb{P}^*_z \otimes P_z [ \nu (\tau_r) < t ; (C1)'] \ge \mathbb{P}^*_z \otimes P_z [e^{\gamma(\bar u_r-u(z))} \hat{\sigma}_1 < e^{ \gamma M /2} ; (C1)']. \end{align}\] Now, observe that \((C1)'\) depends only on \(u(z)\) and is thus independent of both random variables appearing in the right hand side. We deduce \[\begin{align} \mathbb{P}^*_z \otimes P_z [ \nu (\tau_r) < t ; (C1)'] & \ge \mathbb{P}^*_z [ (C1)'] \mathbb{P}^*_z \otimes P_z [e^{\gamma(\bar u_r-u(z))} \hat{\sigma}_1 < e^{ \gamma M /2}]. \end{align}\] The two random variables above are furthermore independent of each other, and their law does not depend on \(r\) or \(z\) (by scaling invariance of the Gaussian free field in the plane modulo constants). By choosing \(M\) sufficiently large and using the monotone convergence theorem, we can therefore ensure that the second term in the right hand side is as close to 1 as desired, say at least \(0.9\). Hence \[\mathbb{P}^*_z \otimes P_z [ \nu (\tau_r) < t ; (C1)'] \ge 0.9\; \mathbb{P}^*_z [ (C1)']\] Plugging this into 43 and recalling that \(p_1 =0.1\), we deduce \[\mathbb{P}^*_z \otimes P_z [ \nu (\tau_\Omega) < t ; (C1)'] \ge 0.8 \; \mathbb{P}^*_z [ (C1)'] .\] In turn, plugging this into 42 , and recalling that \(p_0 \ge 0.9\), we get \[\mathbb{P}^*_z \otimes P_z [ \nu (\tau_\Omega) < t ; G_t] \ge 0.7 \; \mathbb{P}^*_z[ (C1)'] \ge 0.7 \;\mathbb{P}^*_z [ G_t].\] Let us comment quickly on the order in which we need to choose constants in order to make this argument work: we first pick \(N\) large enough that 35 holds with \(p_1 = 0.1\). Then we choose \(M\) large enough so that 44 holds and 41 holds with \(p_0=0.9\).

Altogether, this proves 40 with \(c = 0.7\), and thus concludes the proof of Lemma 12. ◻

Proof. The lower bound is obvious from Jensen’s inequality and Lemma 12. The hard part is to prove the corresponding upper bound. Using Girsanov again, we get \[\begin{align} \mathbb{E}[ {\mathbf{K} }_{\Omega,z_i}(t)^2]= \displaystyle\iint_{(\Omega^t \cap B_{R(t)} (z_i))^2} \mathbb{E}^{**}\Bigl[ {\mathbf{P} }_z[\sigma_\Omega<t] {\mathbf{P} }_w[\sigma_\Omega<t]\boldsymbol{1}_{G_t(z)}\boldsymbol{1}_{G_t(w)}\Bigr]\vert z-w\vert^{-\gamma^2}dzdw, \end{align}\] where under the tilted measure \(\mathbb{P}^{**}\) we have \(h=h'+\gamma G(z,\cdot)+\gamma G(w,\cdot)\) with \(h'\) being a whole plane Gaussian free field, normalised to have zero unit circle average. (Indeed, recall that on \(B_1(0)\) we have \(G(z,w)=-\log\vert z-w\vert\) and \(\Omega \subset B(0,1/2)\)). To make the computation easier, we split the integral in the variable \(w\) in the following way: we fix \(z\in \Omega^t \cap B_{R(t)}(z_i)\) and define for \(n\in\mathbb{N}\) \[\begin{align} A_n(z):=\Bigl\{w\in \Omega^t \cap B_R(z_i):\vert z-w\vert\in[e^{-(n+1)},e^{-n})\Bigr\}. \end{align}\] Thus, we get, bounding both \({\mathbf{P} }_z[\sigma_\Omega<t], {\mathbf{P} }_w[\sigma_\Omega<t]\) crudely by \(1\), \[\begin{align} \label{eq:proof32of32second32moment} \mathbb{E}[ {\mathbf{K} }_{\Omega,z_i}(t)^2]\lesssim\int_{\Omega^t\cap B_{R(t)(z_i)}}\sum_{n\ge n(t)}\int_{A_n(z)}\mathbb{E}^{**}\Bigl[\boldsymbol{1}_{G_t(z)}\boldsymbol{1}_{G_t(w)}\Bigr]e^{n\gamma^2}dwdz \end{align}\tag{45}\] where \(n(t)\) is such that \(e^{ - n(t)} \asymp R(t)\). The key step is to control the probability of the good events. To this end, we reformulate the conditions 36 once more; now in the presence of two logarithmic insertions rather than a single one.

The approach is as follows. Let us write \(r_0:=t^{\kappa(\alpha_0)}>0\). We decompose the double integral into two regimes.

• First we look at the mesoscopic case, where \(\vert z-w\vert\ge 100r_0\). In this regime the two balls \(B_{r_0}(z),B_{r_0}(w)\) are disjoint. The objective is to show that, at bigger separations, the restriction of the GFF to the two balls are roughly independent, so the \(\mathbb{P}^{**}\)-probability of \(G_t(z)\) and \(G_t(w)\) (which concern the behaviour of the GFF inside balls of radius roughly \(r_0\) around both points) factorises, yielding the correct exponent. As the distance between the two points gets closer to the bound \(100 r_0\), the contribution to the integral coming from points at a given distance depends in a delicate way on the geometry of the domain near the boundary, in particular the size of balls of a given radius. This can be controlled by using Assumption 3.

• The second case is the microscopic one, where \(\vert z-w\vert<100r_0\). Here the two balls \(B_{r_0}(z),B_{r_0}(w)\) may overlap, and the two logarithmic insertions from the Girsanov transform are close, yielding (almost) a \(2\gamma\)-log singularity. This proximity makes it very unlikely for both events \(G_t(z)\) and \(G_t(w)\) to hold - this is for similar reasons as in the GMC case ([12]).

This mesoscopic/microscopic splitting corresponds to partitioning the sum over \(n\) into \(n\le N_t:=\lfloor\log(1/r_0)-\log(100)\rfloor -1\) and \(n>N_t\). Let us first comment on the size of \(A_n(z)\). If \(z \in \Omega^t\), there exists \(\tilde{z}\in\partial\Omega\) with \(d(\tilde{z},z)<e^{-n}\). So we can estimate \(A_n(z)\) using Assumption 2: \[\begin{align} \label{eq:estimate32on32A95n32mezoscopic32case} \vert A_n(z)\vert\le\vert \Omega_{t^{\kappa(\alpha_0)}}\cap B_{2e^{-n}}(\tilde{z})\vert\le 4C_\partial t^{2\mathsf{x}\kappa(\alpha_0)}e^{-2n(1-\mathsf{x})}. \end{align}\tag{46}\] This will be useful in the mesoscopic case. On the other hand, in the microscopic case we will simply use the trivial bound \(\vert A_n(z)\vert\le\vert B_{e^{-n}}(z)\vert=\pi e^{-2n}\).

Now let us start with the estimate of 45 . We begin with the mesoscopic case.
Mesoscopic case. Let \(n \le N_t\), and suppose \(w\in A_n(z)\). Under \(\mathbb{P}^{**}\), we may write by the Girsanov–Cameron–Martin lemma: \[\begin{align} \label{eq:h32with32two32singularities} h=h'+\gamma G(z,\cdot)+\gamma G(w,\cdot), \end{align}\tag{47}\] where \(h'\) is a full plane GFF, normalised to have zero unit circle average. Let \(s_z=\log(1/Nd(z,\partial\Omega))\) and \(s_w=\log(1/Nd(w,\partial\Omega))\) (where \(N\) was chosen in 35 ), and let \(X_{s_z}':=h'_{e^{-s_z}}(z)\) and \(Y_{s_w}':=h'_{e^{-s_w}}(w)\). So \((X_{s_z}',Y'_{s_w})\) is a centered Gaussian vector, with respective variances and covariance given by \[\begin{align} \label{eq:variance32and32covariance32of3240X3944Y3941} \sigma_z^2:=\mathbb{E}[(X_{s_z}')^2]=s_z+O(1),\quad\sigma_w^2:=\mathbb{E}[(Y_{s_w}')^2]=s_w+O(1),\quad \rho:=\mathbb{E}[X_{s_z}'Y_{s_w}']=n+O(1), \end{align}\tag{48}\] where all \(O(1)\) terms are bounded by a constant \(C>0\) which is uniform in \(z,w,n,t\) and \(z_i\). (Note that the estimate on the covariance relies on the fact that we are in the mesoscopic regime).

Because of 45 , our goal is to estimate \(\mathbb{P}^{**} [G_t(z), G_t(w)].\) In fact we will ignore the condition \((C2)\) and simply estimate these events using the conditions \((C1)\). Note that the conditions \((C1)(z) \cap (C1)(w)\) imply \[\begin{align} X_{s_z}'&\in\Bigl[(\alpha_0-\gamma)s_z-\gamma n-C,(\alpha_0-\gamma)s_z-\gamma n+C\Bigr],\\ Y_{s_w}'&\in\Bigl[(\alpha_0-\gamma)s_w-\gamma n-C,(\alpha_0-\gamma)s_w-\gamma n+C\Bigr], \end{align}\] where \(C=2M\) (and where \(M\) was chosen in the proof of Lemma 12, and hence is uniform in \(z,w,n\) and \(t\)). Note that \(s_z = \log ( 1/ r_0) + O(1)\), and analogously for \(s_w\). For simplicity, we write \(\lambda = \lambda_n:=n/(\log1/r_0)\) (note that \(\lambda \in [0,1]\) since \(n \le N_t\)). Thus, it suffices to compute the joint density of \((X'_{s_z}, Y'_{s_w})\) in the box \[\begin{align} \label{eq:box} \Bigl[(\alpha_0-\gamma(1+\lambda))\log(1/r_0)-C,(\alpha_0-\gamma(1+\lambda))\log(1/r_0)+C\Bigr]^2, \end{align}\tag{49}\] for some \(C>0\) which is uniform in \(z,w\) and \(t\). To describe the joint law of \((X'_{s_z}, Y'_{s_w})\) we use the fact that we can explicitly describe the conditional law of \(Y'_{s_w}\) given \(X'_{s_w}\), from which it is not hard to see that there exists an independent centered normal random variable \(Z\) such that \[\begin{align} \left(X'_{s_z}, Y'_{s_w}\right)\overset{d}{=}\left(X'_{s_z},\frac{\rho}{\sigma_z^2}X'_{s_z}+Z\right). \end{align}\] To make this work, by 48 we have to choose \[\begin{align} \mathrm{Var}(Z)&=\sigma_w^2-\frac{\rho^2}{\sigma_z^2}=\log\frac{1}{r_0}-\frac{n^2}{\log1/r_0}+O(1) =\log\frac{1}{r_0}(1-\lambda^2+o(1))\nonumber \\ &\in\Bigl[\log\frac{1}{r_0}\Bigl(1-\lambda^2\frac{\log1/r_0}{\log1/r_0-C}\Bigr)-C,\log\frac{1}{r_0}\Bigl(1-\lambda^2\frac{\log1/r_0}{\log1/r_0+C}\Bigr)+C\Bigr]\label{eq:variance32of32Z} \end{align}\tag{50}\] For some \(C>0\), which is uniform in \(z,w,n\) and \(z_i\). Observe also that \[\begin{align} \label{eq:rho32over32sigma942} \frac{\rho}{\sigma_z^2} = \lambda + o(1)\in\Bigl[\lambda\frac{\log1/r_0}{\log1/r_0+C}-C,\lambda\frac{\log1/r_0}{\log1/r_0-C}+C\Bigr], \end{align}\tag{51}\] again for some \(C>0\), which is uniform in \(z,w,n\) and \(z_i\). Hence, conditionally given \[\begin{align} \label{eq:mesoscopic32case3240C14132for32z} X'_{s_z}\in\Bigl[(\alpha_0-\gamma(1+\lambda))\log(1/r_0)-C,(\alpha_0-\gamma(1+\lambda))\log(1/r_0)+C\Bigr], \end{align}\tag{52}\] by 49 , the condition for \(Y'_{s_w}\) can be equivalently reformulated in terms of \(Z\) as \[\begin{align} Z\in(1-\lambda(1 + o(1)))\biggl[(\alpha_0-\gamma\Bigl(1+\lambda(1+o(1))\Bigr)\log(1/r_0),(\alpha_0-\gamma\Bigl(1+\lambda(1+o(1))\Bigr)\log(1/r_0)\biggr], \end{align}\] for an \(o(1)\)-term, which is of the same form as in 50 and 51 . In particular the \(o(1)\)-term is uniform in \(z_i\in\partial_{R(t)}\) and uniform in the choice of \(R(t)\). As \((X'_{s_z},Z)\) is a Gaussian vector with independent entries, we get by the estimates before \[\begin{align} \mathbb{P}^{**}\Bigl[{G_t(z)};{G_t(w)}\Bigr]\le r_0^{\frac{(\gamma(1+\lambda)-\alpha_0)^2}{2}}r_0^{\frac{(1-\lambda)^2(\gamma(1+\lambda)-\alpha_0)^2}{2(1-\lambda^2)}+\epsilon(t)}, \end{align}\] where the \(\epsilon(t)\) is uniform in \(z,w,n\) and \(z_i\) and \(\epsilon(t)\to0\) when \(t\to0\). Using 45 , 46 and \(e^{-n}=r_0^{\lambda}\), we conclude \[\begin{align} \int_{\Omega^t\cap B_{R(t)}(z_i)}&\sum_{n= n(t)}^{N_t}\int_{A_n(z)}\mathbb{P}^{**}\Bigl[{G_t(z)};{G_t(w)}\Bigr]e^{n\gamma^2}dwdz\\ &\le \int_{\Omega^t\cap B_{R(t)}(z_i)}\sum_{n= n(t)}^{N_t}\int_{A_n(z)}\mathbb{P}^{**}\Bigl[G_t(z)\cap G_t(w)\Bigr]e^{n\gamma^2}dwdz\\ &\le\int_{\Omega^t\cap B_{R(t)}(z_i)}\sum_{n= n(t)}^{N_t}\int_{A_n(z)}r_0^{\frac{(\gamma(1+\lambda)-\alpha_0)^2}{2}}r_0^{\frac{(1-\lambda)^2(\gamma(1+\lambda)-\alpha_0)^2}{2(1-\lambda^2)}+\epsilon(t)}e^{n\gamma^2}dwdz\\ &\le C_\partial \sum_{n\ge n(t)}\int_{\Omega^t\cap B_{R(t)}(z_i)}r_0^{\frac{(\gamma(1+\lambda)-\alpha_0)^2}{2}+\frac{(1-\lambda)^2(\gamma(1+\lambda)-\alpha_0)^2}{2(1-\lambda^2)}+\epsilon(t)}e^{n\gamma^2}e^{-2n(1-\mathsf{x})}r_0^{2\mathsf{x}}dz\\ &\le (C_\partial)^2 \sum_{n\ge n(t)}r_0^{\frac{(\gamma(1+\lambda)-\alpha_0)^2}{2}+\frac{(1-\lambda)^2(\gamma(1+\lambda)-\alpha_0)^2}{2(1-\lambda^2)}+\epsilon(t)}e^{n\gamma^2}e^{-2n(1-\mathsf{x})}r_0^{4\mathsf{x}} R(t)^{2(1-\mathsf{x})}\\ &=r_0^{4\mathsf{x}+\epsilon(t)}R(t)^{2(1-\mathsf{x})}\sum_{n\ge n(t)}r_0^{f(\lambda_n)}, \end{align}\] where the \(o(1)\)-term is uniform in \(z,w\) and \(n\) and \[\begin{align} f(\lambda):=\frac{(\gamma(1+\lambda)-\alpha_0)^2}{2}+\frac{(1-\lambda)^2(\gamma(1+\lambda)-\alpha_0)^2}{2(1-\lambda^2)}+\lambda(2(1-\mathsf{x})-\gamma^2). \end{align}\] We estimate \(f(\lambda)\) naively: \[\begin{align} f(\lambda)&=\frac{(\gamma(1+\lambda)-\alpha_0)^2}{1+\lambda}+\lambda(2(1-\mathsf{x})-\gamma^2)\\ &=(\gamma-\alpha_0)^2+\lambda\gamma^2-\frac{\lambda}{1+\lambda}\alpha_0^2+\lambda(2(1-\mathsf{x})-\gamma^2)\\ &=(\gamma-\alpha_0)^2+2\lambda(1-\mathsf{x})-\frac{\lambda}{1+\lambda}\alpha_0^2\\ &\ge(\gamma-\alpha_0)^2+\lambda\Bigl(2(1-\mathsf{x})-{\frac{\alpha_0^2}{2}}\Bigr). \end{align}\] Plugging that into our estimate of the mesoscopic part and using \(\lambda_n=n/(\log1/r_0)\) results in \[\begin{align} \int_{\Omega^t\cap B_{R(t)}(z_i)}\sum_{n=n(t)}^{N_t}\int_{A_n(z)}&\mathbb{P}^{**}\Bigl[{G_t(z)};{G_t(w)}\Bigr]e^{n\gamma^2}dwdz\\&\le t^{4\mathsf{x}\kappa(\alpha_0)+\epsilon(t)} R(t)^{2 (1-\mathsf{x})}\sum_{n\ge n(t)}t^{\kappa(\alpha_0)f(\lambda_n)}\\ &\le t^{\kappa(\alpha_0)(4\mathsf{x}+(\gamma-\alpha_0)^2)+\epsilon(t)} R(t)^{2 (1-\mathsf{x})} \sum_{n\ge n(t)}t^{\kappa(\alpha_0)\lambda_n(2(1-\mathsf{x})-\alpha_0^2/2)}\\ &=t^{2\Delta+\epsilon(t)} R(t)^{2 (1-\mathsf{x})} \sum_{n\ge n(t)}e^ {-n [ (2(1-\mathsf{x}) - \alpha_0^2/2]}\\ &=t^{2\Delta+\epsilon(t)} R(t)^{4(1-\mathsf{x}) - \alpha_0^2/2} . \end{align}\] This concludes the first part of the argument.
Microscopic case. When \(B_{r_z}(z)\) and \(B_{r_w}(w)\) may overlap, the joint covariance structure of the circle-average processes \(X'_{s_z}\) and \(Y'_{s_w}\) becomes pointless. We therefore discard the constraint at \(w\) and impose only a single good event for \(z\). Heuristically, as \(z\) and \(w\) approach each other faster than the radius shrinks, the configuration behaves like a single ball with a double logarithmic insertion at (nearly) the same center. This substantially increases the cost of enforcing the circle average to be no more than \(\alpha_0\log (1/r_0)\) (in fact what will be useful here more specifically will be condition \((C2)\)). This can be used to balance the Girsanov factor \(\vert z-w\vert^{-\gamma^2}\) which otherwise leads to a blowup on the diagonal when \(\gamma\in[\sqrt{2},2)\).

Let us start commenting on the event \((C1)\cap(C2)\) for some \(z\in \Omega^t\) and \(w\in A_n(z)\) such that \(|z-w|<100t^{\kappa(\alpha_0)}\), i.e. \(n>N_t\). As before, we can rephrase this in terms of the Brownian motion \(X'_s:=h'_{e^{-s}}(z)\). Since \(|z-w|\asymp e^{-n}\), this means that \(s\mapsto h_{e^{-s}}(z)\) is a Brownian motion with two additional drifts. The first one is given by \(\gamma s\) and comes from the \(\gamma-\log\)-singularity at \(z\). The second one is coming from the \(\gamma-\log\)-singularity at \(w\), and is thus given by \(\gamma s\) so long as \(s <n\), whereas it is roughly zero when \(s>n\).

Our strategy will thus be to estimate from above \(G_t(z) \cap G_t(w)\) by \((C1)(z)\) and \((C2)(z)\) where we only enforce the condition \((C2)(z)\) at a scale corresponding to the separation between \(z\) and \(w\).

In fact, we will distinguish an additional “intermediate regime”, where \(r_0 /2 \le |z-w| \le 100 r_0\). In this case, essentially \(X'_s\) has drift \(2\gamma\) until roughly \(s=s_z\), so it will in fact suffice to only use \((C1)(z)\). Reasoning as in the mesoscopic case (see 52 with \(\lambda=1\)), but using only the condition \((C1)(z)\), we get \[\begin{align} \label{eq:microscopic32case3240C14132estimate} \mathbb{P}^{**}\Bigl[{G_t(z)};{G_t(w)}\Bigr]\le Cr_0^{\frac{(2\gamma-\alpha_0)^2}{2}}, \end{align}\tag{53}\] where as before \(r_0=t^{\kappa(\alpha_0)}\) and \(C>0\) is uniform in \(z,w\) and \(z_i\). Thus, we estimate the contribution of the part \(r_0/2\le\vert z-w\vert\le 100r_0\) by \[\begin{align} \int_{\Omega^t\cap B_{R(t)}(z_0)}&\int_{r_0/2\le\vert z-w\vert\le100r_0}\mathbb{E}^{**}\Bigl[\boldsymbol{1}_{G_t(z)}\boldsymbol{1}_{G_t(w)}\Bigr]\vert z-w\vert^{-\gamma^2}dwdz \nonumber \\ &\le C\int_{\Omega^t\cap B_{R(t)}(z_0)}\int_{r_0/2\le\vert z-w\vert\le100r_0}r_0^{\frac{(2\gamma-\alpha_0)^2}{2}}r_0^{-\gamma^2}dwdz \nonumber \\ &\le C\int_{\Omega^t\cap B_{R(t)}(z_0)}r_0^{\frac{(2\gamma-\alpha_0)^2}{2}+2-\gamma^2}dz\\ &\le C r_0^{\frac{(2\gamma-\alpha_0)^2}{2}+2-\gamma^2+2\mathsf{x}}. \label{eq:intermediate} \end{align}\tag{54}\] We want to show that this gives a larger exponent than \(2\Delta/\kappa(\alpha_0)\). To prove this we rely on the following two elementary inequalities: \[\label{eq:controlalpha0} \begin{cases} \frac{\alpha_0^2}{2} < 2 (1- \sqrt{\mathsf{x}})^2\\ 2 (1- \sqrt{\mathsf{x}})^2 < 2(1-\mathsf{x}). \end{cases}\tag{55}\] The second inequality is easier to check: we cancel the factors of two and take the square root of both sides (since they are positive) and we need to check that \(1- \sqrt{x} < \sqrt{1-x}\), or, equivalently, \(1 < \sqrt{x} + \sqrt{1-x}\), which is straightforward to check after squaring both sides.

For the first of the inequalities we observe that since \(\alpha_0 = Q - \sqrt{Q^2 - 4 (1-\mathsf{x})}\) from 18 , \(\alpha_0\) is maximized at \(\gamma=2\) by monotonicity (as it is monotone decreasing in \(Q\)). Hence \(\alpha_0\le 2-\sqrt{4-4(1-x)}=2(1-\sqrt{x})\). Using that, we get \[\begin{align} \frac{(2\gamma-\alpha_0)^2}{2}+2-\gamma^2+2\mathsf{x}> 4\mathsf{x}+(\alpha_0-\gamma)^2 = 2\Delta/\kappa(\alpha_0). \end{align}\] Therefore, the contribution of the "intermediate regime" \(r_0/2\le\vert z-w\vert\le 100r_0\) is negligible: that is, the right hand side of 54 is negligible compared to \(t^{2\Delta}\).

Thus, it remains to treat the case \(\vert z-w\vert\le r_0/2\). Fix \(z,w\) such that \(|z-w| \le r_0/2\), \(w \in A_n(z)\). Then if \(s \le n-1\), we have by 47 \[\begin{align} \label{eq:microscopic32part} \Big| h_{e^{-s}}(z)- h'_{e^{-s}}(z)-2\gamma s\Big| \le C , \end{align}\tag{56}\] for some explicit constant \(C>0\). In other words, for points \(y\) at distance \(e^{-s}\) from \(z\), \(G(z,y) = s\), and \(|G(w,y) - s| \le C\). In particular, applying this to \(s = s_z \le n -1\), the condition \((C1)(z)\) implies \[\begin{align} \label{eq:microscopic32part32C134} (C1)''&\qquad X'_{s_z}\in[(\alpha_0-2\gamma+o(1))s_z-C,(\alpha_0-2\gamma+o(1))s_z+C], \end{align}\tag{57}\] where the \(o(1)\)-terms vanishes when \(t\to0\) and are uniform in \(z,w,n\).

We now take \(s = n-1\) in 56 and in condition \((C2)\) from 36 to obtain a condition on the Brownian motion \(X'_s = h'_{e^{-s}}(z)\). However, it is more natural to phrase that condition in terms of the parameter \(q\ge 0\) defining Condition \((C2)\) in 36 , i.e., \((1+q) s_z = s\).

\[\begin{align} \label{eq:microscopic32part32C234} (C2)''\qquad X'_{s} & \le (\alpha_0 - \beta) \log (1/r_0) + s ( \beta - 2\gamma) + C \nonumber \\ & = (\alpha_0-2\gamma+o(1))s_z + (\beta-2\gamma+o(1))qs_z+C, \end{align}\tag{58}\] where \(C>0\) is uniform in \(z,w,n\) and \(t\) and the \(o(1)\)-terms vanishes when \(t\to0\) and is also uniform in \(z,w,n\). So conditional on 57 , 58 implies \[\begin{align} X'_{(1+q)s_z}-X'_{s_z}\le(\beta-2\gamma+o(1))qs_z+C, \end{align}\] where the \(o(1)\)-terms vanishes when \(t\to0\) and is uniform in \(z,w,n\). Now, the left hand side is just an increment of a standard Brownian motion over an interval of duration \(qs_z\). As \(s_z=(1+o(1))\log (1/r_0)\), we have \[\begin{align} \mathbb{P}^{**}[(C2)\vert(C1)]\le \exp\Bigl(-\frac{(\beta-2\gamma+o(1))^2q^2(\log1/r_0)^2+o(1)}{2q\log1/r_0}\Bigr)=r_0^{\frac{(\beta-2\gamma)^2q}{2}+o(1)}, \end{align}\] where all \(o(1)\)-terms are uniform in \(z,w,n\). As the probability of \((C1)''\) can be estimated as in 53 , we get in total \[\begin{align} \mathbb{P}^{**}\Bigl[(C1)\cap (C2)\Bigr] &\le r_0^{\frac{(2\gamma-\alpha_0)^2}{2}+o(1)}r_0^{\frac{(2\gamma-\beta)^2q}{2}+o(1)}. \end{align}\] Now we are able to compute the contribution of the microscopic part. Using \[\vert A_n(z)\vert\le \vert B_{e^{-n}(z)}\vert\le \pi s_z^{2(1+q)} \le C r_0^{2 (1+q)},\] (recall that \(q = q_n\) depends implicitly on \(n\), since \(s_z (1+q) = s= n-1\)), we get: \[\begin{align} &\displaystyle\iint_{z,w \in \Omega^t:|z-w|< r_0/2}\mathbb{P}^{**}\Bigl[{G_t(z)};{G_t(w)}\Bigr]\vert z-w\vert^{-\gamma^2}dzdw\\ &\le C\int_{\Omega^t}\sum_{n=\log1/r_0-O(1)}^{\infty}\mathbb{P}^{**}\Bigl[(C1)(z)\cap(C2)(z)\Bigr]\vert A_n(z)\vert r_0^{-(1+q)\gamma^2}dz\\ &\le\int_{\Omega^t}\sum_{n=0}^{\infty}r_0^{\frac{(2\gamma-\alpha_0)^2}{2}+o(1)}r_0^{\frac{(2\gamma-\beta)^2q}{2}+o(1)}r_0^{(2-\gamma^2)(1+q)}dz\\ &\le \sum_{n=0}^{\infty}r_0^{2\mathsf{x}+\frac{(2\gamma-\alpha_0)^2}{2}+\frac{(2\gamma-\beta)^2q}{2}+(2-\gamma^2)(1+q)+o(1)}=\sum_{n=0}^{\infty}r_0^{f(q_n)+o(1)}, \end{align}\] where all \(o(1)\)-terms are uniform in \(z,w,n\) and \[\begin{align} f(q):=2\mathsf{x}+\frac{(2\gamma-\alpha_0)^2}{2}+\frac{(2\gamma-\beta)^2q}{2}+(2-\gamma^2)(1+q). \end{align}\] One can easily check that \(\alpha_0^2/2<2-2\mathsf{x}\). To do so note that \(\alpha_0\) is maximized at \(\gamma=2\) which gives \(2(1-\sqrt{\mathsf{x}})\) and then use \(2(1-\sqrt{\mathsf{x}})^2<2-2\mathsf{x}\). Choose \(\beta>\gamma\) so close to \(\gamma\) and \(\varepsilon>0\) so small such that \[\begin{align} 2-\gamma^2+\frac{(2\gamma-\beta)^2}{2}> 2-\frac{\gamma^2}{2}-\varepsilon,\qquad 2-\frac{\gamma^2}{2}-2\varepsilon>0,\qquad 2-3\varepsilon&\ge2\mathsf{x}+\frac{\alpha_0^2}{2}. \end{align}\] Then, we get \[\begin{align} 2\mathsf{x}+&\frac{(2\gamma-\alpha_0)^2}{2}+\frac{(2\gamma-\beta)^2q}{2}+(2-\gamma^2)(1+q)\\ &=2\mathsf{x}+\frac{(2\gamma-\alpha_0)^2}{2}-\frac{(2\gamma-\beta)^2}{2}+\Bigl(2-\gamma^2+\frac{(2\gamma-\beta)^2}{2}\Bigr)(q+1)\\ &\ge2\mathsf{x}+\frac{(2\gamma-\alpha_0)^2}{2}-\frac{(2\gamma-\beta)^2}{2}+2-\frac{\gamma^2}{2}-2\varepsilon+(1+q)\varepsilon\\ &\ge4\mathsf{x}+\gamma^2-2\gamma\alpha_0+\varepsilon+\alpha_0^2+(1+q)\varepsilon\\ &=4\mathsf{x}+(\gamma-\alpha_0)^2+\varepsilon+(1+q)\varepsilon. \end{align}\] As \(r_0^{(1+q_n)\varepsilon}=e^{-\varepsilon n(1+o(1))}\), where the \(o(1)\)-term is uniform in \(z,w\) and \(n\), we get for \(t>0\) small enough \[\begin{align} \sum_{n=0}^{\infty}r_0^{f(q_n)+o(1)}\le\sum_{n=0}^{\infty}r_0^{4\mathsf{x}+(\gamma-\alpha_0)^2+\varepsilon}e^{-\varepsilon n(1+o(1))}=t^{2\Delta+\varepsilon}\sum_{n=0}^{\infty}e^{-\varepsilon n(1+o(1))}\le t^{2\Delta+\varepsilon/2}, \end{align}\] which is negligible compared to the contribution coming from the mesoscopic part and thus finishes the proof. ◻

Proof. Let us start showing that \(G_1\) has probability tending to \(1\). By Lemma 9 we get uniformly on all balls \[\begin{align} \mathbb{P}\Bigl[\displaystyle\sup_{z\in B_{2R(t)}(z_i)}\vert h^0(z)-h^0(z_i)\vert\ge \lambda\Bigr]\le ce^{-c'\lambda^2}, \end{align}\] for \(\lambda>1\) and constants \(c,c'>0\). Using that, we conclude \[\begin{align} &\mathbb{P}[G^c]=\mathbb{P}\Bigl[\displaystyle\max_{1\le i \le N(t)}\displaystyle\sup_{z\in B_{2R(t)}(z_i)}\vert h^0(z)-h^0(z_i)\vert> \sqrt{\log (1/t)}\Bigr]\\ &\le\displaystyle\sum_{i=1}^{N(t)}\mathbb{P}\Bigl[\displaystyle\sup_{z\in B_{2R(t)}(z_i)}\vert h^0(z)-h^0(z_i)\vert>\sqrt{\log (1/t)}\Bigr]\\ &\le cN(t)e^{-c'\log (1/t)}=N(t)t^{c'}\rightarrow0, \end{align}\] as \(t \to 0\), since \(N(t)\) diverges to infinity slower than any polynomial in \(t\). Hence \(\mathbb{P}( G_1) \to 1\).

Next, we need to control the circle averages \(h^0_{4R(t)}(z_i)\). Using that \(h^0_{4R(t)}(z_i)=h_{4R(t)}(z_i)\) is a centered Gaussian with variance \(-\log(4R(t))\), we get \[\begin{align} \mathbb{P}[G_2^c]&=\mathbb{P}\Bigl[\displaystyle\max_{1\le i \le N(t)}\vert h^0(z)_{4R(t)}(\omega)\vert>10 \log(N(t))\Bigr]\\ &\le\sum_{i=1}^{N(t)}\mathbb{P}\Bigl[\vert h^0_{4R(t)}(z_i)\vert>10\log(N(t))\Bigr]\\ &\le CN(t)\exp\Bigl(-\frac{100\log(N(t))^2}{2\log(\frac{1}{4R(t)})}\Bigr)\\ &\le CN(t)\exp\Bigl(-50\log(N(t))\Bigr)\\ &=CN(t)^{-49}\to 0,\qquad\textrm{ as }t\to0. \end{align}\] Thus, \(\mathbb{P}(G_2) \to 1\) and the lemma is proved. ◻

Proof. Let us now have a look at what the good event means for \({\mathbf{K} }_i^{h_i}(t)\). On the event \(G_1\cap G_2\) and in \(B_{2R(t)}(z_i)\), we have \(h=h^0+h_i\), where \(\vert h^0\vert\le\vert h^0-h^0(z_i)\vert+\vert h^0(z_i)\vert\le \sqrt{\log(1/t)}+10\log(N(t))\). That means, the harmonic part shifts the Liouville area measure at most by a factor \(e^{\gamma(\sqrt{\log1/t}+10\log(N(t)))}=t^{o(1)}\), i.e. for all \(A\subseteq B_{2R(t)}(z_i)\) \[\begin{align} \mu_\gamma(A)=\lim_{\varepsilon\to0}\int_Ae^{\gamma h_\varepsilon(z)}\varepsilon^{\gamma^2/2}dz\le e^{\gamma(\sqrt{\log1/t}+10\log(N(t)))}\mu^{h_i}_\gamma(A)\le t^{o(1)}\mu^{h_i}_\gamma(A), \end{align}\] Similarly, we get \(\mu_\gamma(A)\ge t^{o(1)}\mu^{h_i}_\gamma(A)\). By a similar argument, we can deduce that, on \(G\), we have \[\begin{align} \nu(t)=t^{o(1)}\nu^{h_i}(t), \end{align}\] where \(\nu^{h_i}\) denotes the time change given by [eq:quantum32clock] with respect to the field \(h_i\). By Lemma 7, we get that \({\mathbf{P} }_z[\sigma^i_{B_{2R(t)}(z_i)}<t]\) vanishes FTP in \(t\), as \(R(t)\) vanishes STP. Thus, we get \[\begin{align} \sigma_{\Omega}=t^{o(1)}\sigma^i_\Omega, \end{align}\] as \(t\to0\). Therefore, (still on the event \(G\)) \[\begin{align} {\mathbf{K} }_{\Omega, z_i}(t)&=\int_{\Omega^t\cap B_{R(t)}(z_i)} {\mathbf{P} }_z[\sigma_\Omega<t]\boldsymbol{1}_{G_t}\mu_\gamma(dz)\\ &=\int_{\Omega^t\cap B_{R(t)}(z_i)} {\mathbf{P} }_z[\sigma^i_\Omega t^{o(1)}<t]\boldsymbol{1}_{G_t}\mu_\gamma(dz) - \int_{\Omega^t\cap B_{R(t)}(z_i)} {\mathbf{P} }_z[\sigma^i_\Omega <t, \sigma_{\Omega}>t]\boldsymbol{1}_{G_t}\mu_\gamma(dz) \\ &\ge\int_{\Omega^t\cap B_{R(t)}(z_i)} {\mathbf{P} }_z[\sigma^i_\Omega t^{o(1)}<t]\boldsymbol{1}_{G_t}\mu^{h_i}_\gamma(dz)t^{o(1)} -\int_{\Omega^t\cap B_{R(t)}(z_i)} {\mathbf{P} }_z[\sigma_{R(t)}<t ]\mu_\gamma(dz)\\ &=t^{o(1)} {\mathbf{K} }^{h_i}_i(t^{1+o(1))}) - X_i, \end{align}\] where \[X_i := \int_{\Omega^t\cap B_{R(t)}(z_i)} {\mathbf{P} }_z[\nu ( \tau_R(t)) <t ]\mu_\gamma(dz).\] Taking expectations, we see that \[\begin{align} \mathbb{E}(X_i\mathbf{1}_{G}) & = \int_{\Omega^t\cap B_{R(t)}(z_i)} \mathbb{E}^*_z \Bigl[ \mathbf{1}_G {\mathbf{P} }_z [ \sigma_{R(t)}< t ]\Bigr] dz. \end{align}\] Arguing as in Corollary 10 (see Lemma 7), we see that on the event \(G\), \[\begin{align} {\mathbf{P} }_z [ \sigma_{R(t)}< t ] & \le P_z [ \hat{\sigma}_1 < t R(t)^{-(2- \gamma^2/2)} e^{\gamma \sqrt{\log (1/t)}} N(t)^{10 \gamma} ]\\ & = P_z [ \hat{\sigma}_1 < t^{1+ o(1)}] \end{align}\] so that, using Lemma 8, \[\begin{align} \mathbb{E}^*_z \Bigl[ \mathbf{1}_{G} {\mathbf{P} }_z [ \nu ( \tau_{R(t)} )< t ]\Bigr] & \le \mathbb{P}^*_z\otimes P_z [ \hat{\sigma}_1 < t^{1+ o(1)}] \le t^{K} \end{align}\] for any \(K>0\) and for all \(t\) sufficiently small, uniformly in \(z \in B_{R(t)} (z_i)\) and \(z_i \in \partial_{R(t)}\). Therefore, \(\mathbb{E}( X_i \mathbf{1}_G) \le t^K\) for all \(t\) sufficiently small. ◻

Proof of Theorem 3. The upper bound on \({\mathbf{K} }_{\Omega}(t)\) follows from the first moment computation (Proposition 6) and Markov’s inequality. We therefore concentrate on the lower bound. Fix \(\varepsilon>0\). Note that by Lemma 15, \[\begin{align} \mathbb{P}( {\mathbf{K} }_{\Omega} (t) \le t^{\Delta + \varepsilon} ) & \le \mathbb{P}( {\mathbf{K} }_{\Omega,z_i} (t) \le t^{\Delta + \varepsilon} , \text{ for all } 1\le i \le N(t))\\ & \le \mathbb{P}(G^c) + \mathbb{P}[t^{o(1)} {\mathbf{K} }^{h_i}_i(t^{1+ o(1)}) - X_i \le t^{\Delta + \varepsilon} \text{ for all } 1\le i \le N(t) ; G] . \end{align}\] Note that by Markov’s inequality, \(\mathbb{P}( X_i \ge t^{\Delta +\varepsilon} ;G) \le t^K\) for some \(K>0\), so that by a union bound, and using the fact that \(N(t) t^K \to 0\), \[\mathbb{P}[ \max_{1\le i \le N(t)} X_i > t^{\Delta + \varepsilon} ; G] \to 0.\] Hence \[\begin{align} \mathbb{P}( {\mathbf{K} }_{\Omega} (t) \le t^{\Delta + \varepsilon} ) & \le o(1) + \mathbb{P}[ \max_{1\le i \le N(t)} {\mathbf{K} }^{h_i}_i(t^{1+ o(1)}) \le 2 t^{\Delta + \varepsilon+ o(1)} ] \end{align}\] It therefore suffices to show: for all \(\varepsilon>0\), \[\begin{align} \label{eq:goal95concl95heat} \mathbb{P}[ \max_{1\le i \le N(t)} {\mathbf{K} }^{h_i}_i(t) \le t^{\Delta + \varepsilon} ] \to 0. \end{align}\tag{61}\] As all \({\mathbf{K} }_i^{h_i}\) are independent, we have \[\begin{align} \mathbb{P}[ \max_{1\le i \le N(t)} {\mathbf{K} }^{h_i}_i(t) \le t^{\Delta + \varepsilon} ] & = \mathbb{P}[ {\mathbf{K} }^{h_1}_1(t) \le t^{\Delta + \varepsilon}]^{N(t)} . \label{eq:heat95concl95prod} \end{align}\tag{62}\] On the other hand, by Lemma 15 again, for each \(1\le i \le N(t)\), \[\begin{align} \mathbb{P}[ {\mathbf{K} }^{h_i}_i(t) \le t^{\Delta + \varepsilon}] &\le \mathbb{P}(G^c) + \mathbb{P}( {\mathbf{K} }_{\Omega, z_i}(t^{1+ o(1)}) \le t^{\Delta + \varepsilon+ o(1)} )\\ & \le N(t)^{-49} + 1-t^{\epsilon(t)}R(t)^{\alpha_0^2/2} \end{align}\] by 59 and Lemma 14. Plugging back into 62 , and using that \(1- x \le e^{-x}\) for any \(x \in \mathbb{R}\), we get \[\begin{align} \mathbb{P}[ \max_{1\le i \le N(t)} {\mathbf{K} }^{h_i}_i(t) \le t^{\Delta + \varepsilon} ] & \le [1- R(t)^{\alpha_0^2/2} t^{\epsilon(t)}+ N(t)^{-49} ]^{N(t)}\\ & \le \exp\left( - N(t) t^{\epsilon(t)} R(t)^{\alpha_0^2/2} + N(t)^{-48} \right) \to 0. \end{align}\] In the last line, to conclude that the right hand side tends to zero, we used the fact that \[\label{eq:R95N} R(t)^{\alpha_0^2/2} t^{\epsilon(t)} N(t) \to \infty.\tag{63}\] Let us justify why this holds. Indeed, given 55 , we have \[R(t)^{\alpha_0^2/2} N(t) t^{\epsilon(t)} \ge R(t)^{-\eta}\cdot R(t)^{2 (1-\mathsf{x}) } N(t) t^{\epsilon(t)}=R(t)^{-\eta/2}t^{\varepsilon(t)} \to \infty\] for \(\eta = 2(1- \mathsf{x}) - \alpha_0^2/2 >0\), and for the conclusion we used the fact that, in Assumption 3, \(N(t) \ge R(t)^{-2 ( 1- \mathsf{x}) + \zeta}\) with \(\zeta = \eta/2\). (Note that this implicitly fixes the value of \(\zeta\) to be used in the definition of \(\partial_{R(t)}\) in Assumption 3, as well as the choice of \(R(t)\), taken so that \(R(t)^{-\eta/2} \gg t^{ \epsilon(t)}\); this is possible as \(\epsilon(t)\) does not depend on \(R(t)\).) This completes the proof of 63 . Altogether, the proof of Theorem 3 is therefore complete. ◻

Proof. The idea is simply to exploit the scale-invariance of the quantum cone, which we first recall. To do so it is useful to indicate the explicit dependence of the area measure \(\mu = \mu_{\tilde{h}}\) with respect to a specific realisation of the quantum cone, \(\tilde{h}\). The scale-invariance of the quantum cone (Theorem 7.18 in [11]) states that the quantum surface described by \((\tilde{h} +C, \mathbb{C}, 0, \infty)\) is identical in law to that described by \((\tilde{h}, \mathbb{C}, 0, \infty)\), for any \(C \in \mathbb{R}\). Thus \(\tilde{h} + C\) is another quantum cone (in an embedding which is not the usual “unit circle embedding”, but embeddings are irrelevant as all quantities of interest here are LQG invariants). Note that \[\mu_{\tilde{h}} (A) = e^{-\gamma C} \mu_{\tilde{h}+C}(A)\] for any Borel set \(A\). Note also that if \((B_t, t \ge 0)\) is a fixed two-dimensional standard Brownian motion, then the quantum clocks with respect to \(\tilde{h}\) and \(\tilde{h}+C\) satisfy a similar relation: \[\nu_{\tilde{h}+C, B} (s) = e^{\gamma C} \nu_{\tilde{h}, B}(s); \quad s \ge 0.\] Thus the corrsponding Liouville Brownian motions \(Z = Z^{\tilde{h},B}(s)\) associated to \(\tilde{h}\) and Brownian path \(B\) satisfy: \[Z^{\tilde{h}, B}(t) = Z^{\tilde{h}+C, B}(t e^{\gamma C}), \text{ for all } t\ge 0.\] Let \(t>0\), and let us fix \(C = \log(1/ t) / \gamma\), so \(e^{\gamma C} = 1/t\). It follows that (using continuity of the heat kernel with respect to the space variable and the dominated convergence theorem) \[\begin{align} {\mathbf{p} }^{\tilde{h}}_t(x,x) &= \lim_{\varepsilon\to 0} \frac{ {\mathbf{P} }_x (Z^{\tilde{h},B}(t) \in B_x(\varepsilon))}{\mu_{\tilde{h}} ( B_x(\varepsilon))}\\ & = \lim_{\varepsilon\to 0} \frac{ {\mathbf{P} }_x (Z^{\tilde{h}+C,B}(1) \in B_x(\varepsilon))}{t\mu_{\tilde{h}+C} ( B_x(\varepsilon))}\\ & = \frac{1}{t} {\mathbf{p} }^{\tilde{h}+C}_1(x,x), \end{align}\] as desired. ◻

Proof. This follows directly from Proposition 17, the local convergence of the pointed quantum surface \((h+C, x)\) as \(C \to \infty\) to a \(\gamma-\)quantum cone as \(C\to \infty\) (in the sense of total variation, say – see Theorem 7.18 in [11]), and the fact that \(\mathbf{p}_t(x,x)\) is measurable with respect to the local environment near \(x\) when \(t \to 0\). Details are left to the reader; alternatively, we provide a complete argument (which yields quantitative bounds on the error) in Lemma 22. ◻

Proof. The proof is similar in both cases (of \(\tilde{\mathbb{P}}\) and \(\mathbb{P}^*_x\)) and we do them in parallel. We will first treat the case \(x = 0\) and then explain how to adapt the argument to the case of \(\mathbb{P}^*_x\) with general \(x \in B_0(1/2)\).

As the heat kernel is an LQG invariant (i.e., invariant under the LQG change of coordinates) we are free to use whichever parametrisation of the surface we wish; it will be convenient to switch between the whole plane model and the strip model, for the calculations below. As mentioned above, we will do so without indicating this in the notations.

Let \(n\ge 1\). We recall that by the general theory of Dirichlet forms, \(t\mapsto \mathbf{p}_t(x,x)\) is monotone decreasing in \(t>0\), and for each fixed \(t>0\), \(y \in \mathbb{C}\mapsto \mathbf{p}_t(x,y)\) is maximised at \(y = x\). Thus \[\mathbf{p}_t (x,x) \le \frac{2}{t}\int_{t/2}^t \mathbf{p}_u (x,x) du.\] Thus raising to the power \(n\ge 1\) and bridge decomposition 66 , as well as Fubini’s theorem, we obtain \[\begin{align} & \mathbb{E}^* [ \mathbf{p}_t(x,x)^n]\\ \le & \frac{2^n}{t^n} \mathbb{E}^*\left[\int_0^\infty \ldots \int_0^\infty \mathbf{1}_{u_1 \in [t/2,t]} \ldots \mathbf{1}_{u_n \in [t/2, t]} \mathbf{p}_{u_1}(x,x) \ldots \mathbf{p}_{u_n}(x,x) du_1 \ldots du_n\right]\\ \le & \frac{2^n}{t^n} \int_0^\infty \ldots \int_0^\infty \mathbb{E}^*\otimes {\mathbf{E} }_{x\overset{u_1}{\rightarrow}x,\ldots x\overset{u_n}{\rightarrow}x} \left[ \mathbf{1}_{\nu^1 (u_1) \in [t/2,t] } \ldots \mathbf{1}_{\nu^n (u_n) \in [t/2,t]} \right] \frac{1}{(2\pi)^n u_1 \ldots u_n} du_1 \ldots du_n \end{align}\] where \({\mathbf{E} }_{x\overset{u_1}{\rightarrow}x,\ldots x\overset{u_n}{\rightarrow}x}\) is the law of \(n\) independent ordinary Brownian bridges from \(x\) to \(x\), of respective durations \(u_1, \ldots u_n\); and \(\nu^1 (\cdot), \ldots, \nu^n(\cdot)\) are the quantum clocks accumulated by each of these bridges in the common environment \(\mathbb{P}^*_x\).

Applying Hölder’s inequality, we get \[\label{eq:HKintegral} \mathbb{E}^*_x [ \mathbf{p}_t(x,x)^n]\le \frac{1}{\pi^n t^n} \left( \int_0^\infty \frac{1}{u} \mathbb{P}^*_x\otimes {\mathbf{P} }_{x\overset{u}{\rightarrow}x}\bigl[\nu(u)\in[t/2,t]\bigr]^{\frac{1}{n}} du \right)^n.\tag{69}\]

In the case of the quantum cone, we recall that by scaling (Proposition 17) it suffices to take \(t = 1\). The bound 69 also holds with \(\tilde{\mathbb{P}}\) replacing \(\mathbb{P}^*_x\).

Thus it suffices to control the integral on the right hand side of 69 (and its analogue for the quantum cone in the case \(t=1\)), which we break into \(u \in (0,t)\), \(u\in (t,1)\), and \(u \in (1, \infty)\). When \(u \in (t,1)\) we simply bound the probability by 1 so the integral is at most \(\log (1/t)\).

So let us assume \(u>1\), and observe that, by monotonicity of \(\nu\), \[\begin{align} \mathbb{P}^*_x\otimes {\mathbf{P} }_{x\overset{u}{\rightarrow}x}\bigl[\nu(u)\in[t/2,t]\bigr] & \le \mathbb{P}^*_x\otimes {\mathbf{P} }_{x\overset{u}{\rightarrow}x}\bigl[\nu(u)\le t \bigr]\\ & \le \mathbb{P}^*_x\otimes {\mathbf{P} }_{x\overset{u}{\rightarrow}x}\bigl[\nu(u/2) \le t\bigr]. \end{align}\] Now, the law of a Brownian bridge of duration \(u\), restricted to \([0,u/2]\), is absolutely continuous with respect to that of an ordinary Brownian motion over the same interval of time, with Radon–Nikodym derivative bounded by a universal constant \(C>0\). Thus for any \(q>0\), using also the monotonicity of \(\nu\) and Markov’s inequality, \[\begin{align} \mathbb{P}^*_x\otimes {\mathbf{P} }_{x\overset{u}{\rightarrow}x}\bigl[\nu(u)\in[t/2,t]\bigr] & \le C \mathbb{P}^*_x\otimes {\mathbf{P} }_x\bigl[\nu(u/2)\le t\bigr] \nonumber \\ & \le C t^{q} \mathbb{E}^*_x \otimes {\mathbf{E} }_x [ \nu (u/2)^{-q} ]. \label{eq:HKbound1} \end{align}\tag{70}\]

We will thus need to control the negative moments of \(\nu (u)\). We defer this to slightly later, explaining first how we take care of the contribution to the integral coming from \(u<t\). In that case, \[\begin{align} \mathbb{P}^*_x\otimes {\mathbf{P} }_{x\overset{u}{\rightarrow}x}\bigl[t/2\le\nu(u) \le t \bigr]& \le\mathbb{P}^*_x\otimes {\mathbf{P} }_{x\overset{u}{\to}x}[\nu(u)\ge t/2]\\ & \le \mathbb{P}^*_x\otimes {\mathbf{P} }_{x\overset{u}{\to}x}[\nu(u/2)\ge t/4] + \mathbb{P}^*_x\otimes {\mathbf{P} }_{x\overset{u}{\to}x}[\nu(u/2, u)\ge t/4] \end{align}\] where \(\nu (u/2, u)\) is the quantum clock accumulated by the Brownian bridge during the interval \([u/2, u]\) (i.e., \(\nu (u/2, u) = \nu(u) - \nu(u/2)\)). By reversibility of the Brownian bridge, and by the same absolute continuity of Brownian bridge with respect to ordinary Brownian motion which was already used in 70 , we get \[\begin{align} \mathbb{P}^*_x\otimes {\mathbf{P} }_{x\overset{u}{\rightarrow}x}\bigl[t/2\le\nu(u) \le t \bigr] & \le 2C \mathbb{P}^*_x\otimes {\mathbf{P} }_{x}[\nu(u/2)\ge t/4] \\ &\le2C \big(\mathbb{P}^*_x\otimes {\mathbf{P} }_{x}[\nu(u/2\wedge\tau_1)\ge t/4]+ {\mathbf{P} }_{x}[\tau_1< u/2]\big), \end{align}\] where we recall that \(\tau_1 = \inf\{ t>0: |X_t-x | > 1\}\) is the first time \(X\) (the Brownian motion, here) leaves a ball of radius 1 around its starting point. Note that \({\mathbf{P} }_{x}[\tau_1< u/2] \to 0\) as \(u \to 0\) FTP (in fact, exponentially fast in \((-1/u)\), by the explicit form of the Gaussian density function), so it suffices to concentrate on the first term. By Markov’s inequality, we again have for all \(q>0\), \[\begin{align} \label{eq:HKbound2} \mathbb{P}^*_x\otimes {\mathbf{P} }_{x}[\nu(u/2\wedge\tau_1)\ge t/4] & \le (t/4)^{-q} \mathbb{E}^*_x \otimes {\mathbf{E} }_x [ \nu (u/2 \wedge \tau_1)^q]. \end{align}\tag{71}\]

Altogether, we get, for any choice of \(q,q'>0\): \[\begin{align} \label{eq:total95integral95moments} \int_0^\infty\frac{1}{u} \mathbb{P}^*_x\otimes {\mathbf{P} }_{x\overset{u}{\rightarrow}x}\bigl[\nu(u)\in[t/2,t]\bigr]^{\frac{1}{n}} du & \le \log (1/t) + C \nonumber \\ & \quad + C_{q,n}t^{q/n} \int_1^\infty \frac{1}{u} \mathbb{E}^*_x \otimes {\mathbf{E} }_x [ \nu(u/2)^{-q}]^{1/n} du \nonumber \\ & \quad + t^{- q'/n}\int_0^t \frac{1}{u} \mathbb{E}^*_x \otimes {\mathbf{E} }_x [ \nu(u/2 \wedge \tau_1)^{q'}]^{1/n} du \end{align}\tag{72}\]

We now state the relevant controls over the moments needed to complete this proof, starting with the control at “large” scales (\(u>1\)):

The interest and potential difficulty here lies in the fact that we want a result valid for all \(u>0\), i.e., including at large scales. The function \(\xi^*\) can be viewed as the \(\gamma\)-quantum cone multifractal spectrum (or that of a whole plane GFF with \(\gamma\)-log singularity at all scales – including large scales, unlike 67 where the logarithmic singularity is only present at scales less than unity).

Thus, applying Lemma 20 with \(q >0\) small enough that \(\xi^*(-q)>0\) (which is possible when \(\gamma<2\)), and plugging into 72 we see that the integral in the second line is bounded by a constant (say \(C\)), so this entire term is bounded by \(Ct^{1/n} \le C\).

The control at small scales (\(u<1\)) is provided by the following lemma.

Again, the proof is deferred to the appendix. Consequently, the integral in the third line of 72 is convergent and thus (crudely) bounded for any \(q'>0\), by some constant \(C_{q',\gamma,n}\). Altogether, this shows that the integral in 72 is bounded by \(C( 1+ \log (1/t) + t^{ - q'/n}) \le C t^{ - q'/n}\) for \(t\) small enough. Thus, combining our estimates and recalling 69 , we see that \[\begin{align} \mathbb{E}^*_x [ \mathbf{p}_t(x,x)^n] & \le \frac{C}{t^n} t^{ - q'} \end{align}\] for \(t\) sufficiently small. Since \(q'>0\) is arbitrary, this concludes the proof of Proposition 19. ◻

Proof. We note that, by construction of the \(\gamma\)-quantum cone, the fields \(\tilde{h}\) and \(h^*_x\) can be made to agree exactly on the ball of radius 1/2, which we denote by \(B\) in this proof. We assume that \(h\) and \(\tilde{h}\) are coupled in this way (for instance, start by sampling \(h\) under \(\mathbb{P}^*_x\), then sample \(\tilde{h}\) according to the conditional law of \(\tilde{h}\) given \(\tilde{h}|_{B}\)). Similar to the in-out decomposition, note that \[\label{eq:inoutball1} \mathbf{p}^*_t(x,x) = \mathbf{p}_t^B (x,x) + \mathbf{p}_t^{ \mathbb{C}\setminus B} (x,x),\tag{75}\] where the first term consists of the heat kernel restricted to trajectories staying entirely in \(B\), while the second denotes the heat kernel restricted to trajectories leaving \(B\) before time \(t\) (e.g., \(\mathbf{p}_t^{ \mathbb{C}\setminus B} (x,x) = \mathbf{p}_t (x,x) {\mathbf{P} }^\gamma_{ x \overset{t}{\to} x} ( \sigma_B< t )\)). For notational ease we have removed the star from the two heat kernels on the right hand side.

Likewise, we may apply the same decomposition to \(\tilde{\mathbf{p} }_t(x,x)\), so that \[\label{eq:inoutball2} \tilde{\mathbf{p}}_t(x,x) = \tilde{\mathbf{p}}_t^B (x,x) + \tilde{\mathbf{p}}_t^{ \mathbb{C}\setminus B} (x,x),\tag{76}\] Since \(\tilde{h}\) and \(h^*_x\) agree in \(B\), we have \[\tilde{\mathbf{p}}_t^B (x,x) = \tilde{\mathbf{p}}_t^B (x,x)\] so that \[|\tilde{\mathbf{p}}_t(x,x) - \mathbf{p}^*_t(x,x)| \le \tilde{\mathbf{p}}_t^{ \mathbb{C}\setminus B} (x,x) + {\mathbf{p}}_t^{ \mathbb{C}\setminus B} (x,x).\] We will bound both terms separately, showing that they converge to 0 in \(L^1(\mathbb{P})\), FTP. The argument is the same for both terms, so let us start with the second one, say. Note that, by reversibility of the Liouville Brownian bridge (which is to say, by reversibility of Liouville Brownian motion with respect to \(L^2 (\mu_h)\)), letting \(\sigma_B\) the first exit time of a given trajectory (which will either be Liouville Brownian motion or bridge below – note that in the latter case we cannot express the Liouville bridge as a time-change of an ordinary Brownian bridge, so cannot write \(\sigma_B = \nu (\tau_B)\) as we would for a Liouville Brownian motion): \[\begin{align} {\mathbf{p} }_t^{\mathbb{C}\setminus B} (x,x) & = {\mathbf{P} }^\gamma_{x \overset{t}{\to} x} ( \sigma_B < t) {\mathbf{p} }_t(x,x) \\ & = 2 {\mathbf{P} }^\gamma_{x \overset{t}{\to} x} ( \sigma_B < t/2) {\mathbf{p} }_t(x,x) \\ & = 2 {\mathbf{E} }^\gamma_x \left[ 1_{\{\sigma_B< t/2\}} {\mathbf{p} }_{t/2} ( \mathbf{Z}_{t/2} , x)\right] \end{align}\] where in the last line we used the explicit construction of the law of Liouville Brownian bridge on \([0,t]\) (or rather, its restriction to \([0, t/2]\) as having the density \({\mathbf{p} }_{t/2} (\cdot, x)/ {\mathbf{p} }_t(x,x)\) with respect to the law of Liouville Brownian motion \((\mathbf{Z}_t, t \ge 0)\).

Now, by the general theory of Dirichlet forms, \({\mathbf{p} }_s(\cdot, x)\) is maximised on the diagonal for any given \(s>0\); thus \[\begin{align} \label{eq:bridge95far} {\mathbf{p} }_t^{\mathbb{C}\setminus B} (x,x) \le 2 {\mathbf{P} }_x^\gamma ( \sigma_B < t/2) {\mathbf{p} }_{t/2} (x,x) . \end{align}\tag{77}\] Let us take expectation with respect to the field \(h^*_x\), and use Cauchy-Schwarz: \[\begin{align} \mathbb{E}^*_x [ {\mathbf{p} }_t^{\mathbb{C}\setminus B} (x,x) ] & \le 2 \mathbb{E}^*_x [ {\mathbf{P} }_x^\gamma ( \sigma_B < t/2)^2]^{1/2} \mathbb{E}^*_x[ {\mathbf{p} }_{t/2} (x,x)^2]^{1/2} . \end{align}\] Concerning the first expectation, \({\mathbf{P} }_x^\gamma ( \sigma_B < t/2) \le 1\) so \({\mathbf{P} }_x^\gamma ( \sigma_B < t/2)^2 \le {\mathbf{P} }_x^\gamma ( \sigma_B < t/2)\), hence by Fubini’s theorem \[\begin{align} \mathbb{E}^*_x [ {\mathbf{p} }_t^{\mathbb{C}\setminus B} (x,x) ] & \le 2 \left[\mathbb{P}^*_x \otimes {\mathbf{P} }^\gamma_x ( \sigma_{B}< t/2) \right]^{1/2} \mathbb{E}^*_x[ {\mathbf{p} }_{t/2} (x,x)^2]^{1/2}\\ & = 2\left[\mathbb{P}^*_x \otimes {\mathbf{P} }_x (\nu(\tau_B) < t/2) \right]^{1/2} \mathbb{E}^*_x[ {\mathbf{p} }_{t/2} (x,x)^2]^{1/2}. \end{align}\] The first term tends to zero FTP by Lemma 8. The second tends to zero no faster than \(t^{-1 + o(1)}\) by Proposition 19. Altogether, this shows that \(\mathbb{E}^*_x [ {\mathbf{p} }_t^{\mathbb{C}\setminus B} (x,x) ] \to 0\), FTP. The same reasoning applies to \(\tilde{\mathbb{E}}[ {\mathbf{p} }_t^{\mathbb{C}\setminus B} (x,x) ]\) (note that Lemma 8 applies also here to show that \(\tilde{\mathbb{P}}\otimes {\mathbf{P} }_x (\nu(\tau_B) < t/2)\), since this estimate concerns only the behaviour of the field inside \(B\), where \(\tilde{h}\) and \(h^*_x\) agree). This concludes the proof of Lemma 22. ◻

Proof. The estimate ?? follows directly from Lemma 22 together with the scale invariance of the heat kernel on the quantum cone (with \(\alpha = \gamma\)) proved in Proposition 17 (and the fact that the expectation of \(\tilde{\mathbf{p} }_1(0,0)\) is finite by Proposition 19). As for ?? , this follows by comparing with 73 . ◻

Proof of Theorem 4. We have already seen in Corollary 18 that \(t {\mathbf{p} }_t^{\Omega}(x,x) \to X\) in law as \(t \to 0\), where \(X = \tilde{\mathbf{p} }_1(0,0)\). In fact, from Remark 23 and Proposition 17 there is a coupling such that \(\mathbb{E}[ | t {\mathbf{p} }_t^{\Omega}(x,x) - X| ] \to 0\), FTP.

From Lemma 24 we furthermore have \(\mathbb{E}[X] = c_\gamma\), and from Proposition 19 we have that \(\mathbb{E}[X^n] < \infty\) for all \(n\ge 0\). ◻

Proof of upper bound of ?? . We start by applying Fubini’s theorem and Girsanov’s lemma, so that \[\begin{align} \mathbb{E}\left[ \int_\Omega {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega} (x,x) \mu_h(dx) \right] & = \int_\Omega \mathbb{E}^*_x\left[ {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega} (x,x) \right] dx. \end{align}\] Arguing as in 77 (but with \(\Omega\) instead of \(B\)), we have \[{\mathbf{p} }_t^{\mathbb{C}\setminus \Omega} (x,x) \le {\mathbf{p} }_{t/2}(x,x) {\mathbf{P} }^\gamma_x ( \sigma_\Omega< t/2)\] Therefore, taking expectation \(\mathbb{E}^*_x\) and applying Hölder’s inequality, we have for some arbitrary \(n\ge 1\) and some constant \(c_n\) depending only on \(n\ge 1\) and \(\gamma\), \[\begin{align} \mathbb{E}\left[ \int_\Omega {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega} (x,x) \mu_h(dx) \right] & \le \int_\Omega \mathbb{E}^*_x [ {\mathbf{p} }_{t/2} (x,x)^{n+1}]^{\tfrac{1}{n+1}} \left(\mathbb{P}^*_x \otimes {\mathbf{P} }_x^\gamma [ \sigma_\Omega < t/2]\right)^{\tfrac{n}{n+1}} dx \\ & \le c_nt^{ -1 +o(1)} \int_\Omega \left(\mathbb{P}^*_x \otimes {\mathbf{P} }_x^\gamma [ \sigma_\Omega < t/2]\right)^{\tfrac{n}{n+1}} dx\\ & \le c_n t^{-1+ o(1)} \left(\int_\Omega \mathbb{P}^*_x \otimes {\mathbf{P} }_x^\gamma [ \sigma_\Omega < t/2] dx\right)^{\tfrac{n}{n+1}} \end{align}\] where in the second line we used Proposition 19, and in the last line we used Jensen’s inequality, since \(x\mapsto x^{n/(n+1)}\) is concave. The integral in the last line is nothing but \(\mathbb{E}[ {\mathbf{K} }_\Omega ( t/2)]\). Therefore, applying Proposition 6, we get \[\begin{align} \mathbb{E}\left[ \int_\Omega {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega} (x,x) \mu_h(dx) \right] \le c_n t^{-1+ o(1)} (t^{\Delta + o(1)})^{\tfrac{n}{n+1}}. \end{align}\] Since \(n\ge 1\) can be chosen arbitrarily large, this proves \[\mathbb{E}\left[ \int_\Omega {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega} (x,x) \mu_h(dx) \right] \le t^{ -1 + \Delta + o(1)},\] which is the desired upper bound. ◻

Proof of lower bound of ?? . As in the proof of Proposition 6, we get a lower bound on the expectation by restricting to points \(x\) at the right distance to the boundary, and to an event \({\mathcal{E} }= {\mathcal{E} }_x\) concerning the behaviour of the GFF near \(x\) similar to those already appearing in the analysis of the heat content, and which ought to provide the dominant contribution. However, the reason that the lower bound is harder is that there is no independence between the \({\mathbf{p} }_t(x,x)\) and the above event \({\mathcal{E} }_x\) which determines the behaviour of the GFF near \(x\); on the contrary, these events may be a priori very correlated. Since we want a lower bound, we cannot (as in the above upper bound) rely on Hölder’s inequality. Thus, an argument is needed, to show that the conditional expectation of \({\mathbf{p} }_t(x,x)\) remains of order \((1/t)\), even when we condition on the event \({\mathcal{E} }_x\).

Let \(\delta>0\), and introduce the following region (our notation here differs slightly from that used in the proof of Proposition 6): \[\begin{align} \Omega^*_{\delta}&:=\Bigl\{x\in\Omega\Big\vert t^{\kappa_0(1+\delta)}\le d(x,\partial^*\Omega)\le 2t^{\kappa_0(1+\delta)}\Bigr\}, \end{align}\] where we recall that \(\partial^*\Omega\) is a connected component of the boundary of \(\Omega\) of positive diameter, and we write \(\kappa_0=\kappa(\alpha_0)\).

We let \(r = t^{\kappa_0(1+\delta)}\) and note that \(\text{dist} (z, \partial^* \Omega) \asymp r\) whenever \(z \in \Omega^*_{\delta}\). We choose a larger scale \(r' = t^{\kappa_0 (1- \delta)}\), so that \(r' /r = t^{ -2 \kappa_0 \delta} \to \infty\) as \(t \to 0\). (More generally the proof will depend on a number of choices of scales which have not been optimised, but are sufficient for the proof). In particular, for \(x \in \Omega_\delta\), the ball \(B_r(x) \subset \Omega\) but a substantial portion of \(B_{r'}(x)\) lies outside \(\Omega\). We will condition the field on its circle average distance \(r'\), as follows: \[\begin{align} {\mathcal{E} }_{x} &:=\{ | h_{r'}(x) - \alpha_0\log(1/r') | \le 1 \} \end{align}\]

Then, obviously, \[\begin{align} \int_\Omega\mathbb{E}^*_x[ {\mathbf{p} }_t^{\mathbb{C}\setminus\Omega}(x,x)]dx&\ge\int_{\Omega^*_{\delta}}\mathbb{E}^*_x[ {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega}(x,x)\boldsymbol{1}_{ {\mathcal{E} }_{x}}]dx =\int_{\Omega^*_{\delta}}\mathbb{E}^*_x[ {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega}(x,x)\vert {\mathcal{E} }_{x}]\mathbb{P}^*_x[ {\mathcal{E} }_{x}]dx. \end{align}\] As in 15 , we have \[\begin{align} \mathbb{P}^*_x[ {\mathcal{E} }_{x}] & = \exp \left( - \tfrac{(\gamma - \alpha_0)^2}{2} \log (1/r') (1+ o(1))\right) \nonumber \\ & = t^{\kappa_0(1-\delta)(\gamma-\alpha_0)^2/2+o(1)}, \label{eq:P42E95x} \end{align}\tag{78}\] so it remains to estimate \(\mathbb{E}^*[ {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega}(x,x)\vert {\mathcal{E} }_{x}]\) for \(x \in \Omega^*_{\delta}\), and more precisely we claim that it suffices to show \[\label{goal:Econd95pt} \mathbb{E}^*[ {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega}(x,x)\vert {\mathcal{E} }_{x}] \ge t^{-1 + o(1)},\tag{79}\] uniformly over \(x \in \Omega^*_{\delta}\). Indeed, if this holds, then \[\begin{align} \int_\Omega\mathbb{E}^*_x[ {\mathbf{p} }_t^{\mathbb{C}\setminus\Omega}(x,x)]dx & \ge t^{\kappa_0 (\gamma - \alpha_0)^2/2 -1 } | \Omega^*_\delta| \times t^{ \delta[ \kappa_0 (\gamma - \alpha_0)^2/2] + o(1)}\\ & \ge t^{2\mathsf{x}\kappa_0 + \kappa_0 ( \gamma - \alpha_0)^2/2 -1 + o(1)} \times t^{ \delta[ \kappa_0 (\gamma - \alpha_0)^2/2] + o(1)}, \end{align}\] where we used our assumption ?? to lower bound \(|\Omega^*_\delta|\). Recalling from 19 that \(\kappa_0 (2\mathsf{x}+ ( \gamma - \alpha_0)^2/2) = \Delta\), we get \[\int_\Omega\mathbb{E}^*_x[ {\mathbf{p} }_t^{\mathbb{C}\setminus\Omega}(x,x)]dx\ge t^{\Delta -1} \times t^{ \delta[ \kappa_0 (\gamma - \alpha_0)^2/2] + o(1)},\] which proves the lower bound of ?? since \(\delta\) is arbitrary.

We thus focus on 79 . Even though \({\mathbf{p} }_t^{\mathbb{C}\setminus \Omega}(x,x)\) is far from independent from the event \({\mathcal{E} }_{x}\), it is nevertheless reasonable to hope for such a lower bound in view of the a.s. uniform (over the entire domain) lower bound on the on-diagonal heat kernel \(\mathbf{p}_t(x,x)\) of the form \(t^{-1+ o(1)}\) obtained in [41]. Of course, the latter is an a.s. bound, rather than a bound in expectation, so it cannot be directly used, but this suggests it is indeed reasonable to hope for such a conditional bound.

To see 79 we first change the field \(h^*_x\) to the \(\gamma\)-quantum cone. Arguing as in Lemma 22, the error we make in doing so tends to zero FTP: \[\mathbb{E}^*_x[ {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega} (x,x) | {\mathcal{E} }_x ] = \tilde{\mathbb{E}}_x[ {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega} (x,x) | {\mathcal{E} }_x ] + o_{\text{FTP}}(1),\] where \(\tilde{\mathbb{E}}_x\) denotes the law of a \(\gamma\)-quantum cone translated so \(x\) is at the origin. (Briefly, the only difference with the setting of Lemma 22 is that here we are conditioning on \({\mathcal{E} }_x\). Recall that Lemma 22 relies on coupling \(h^*_x\) and the translated quantum cone in such a way that they agree exactly in a ball of radius at least \(1/2\) around \(x\), hence the events \({\mathcal{E} }_x\) are the same for both fields and thus the argument from that Lemma applies verbatim.) It thus suffices to prove \[\label{eq:goal95cond95pt95cone} \tilde{\mathbb{E}}_x [ {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega}(x,x) | {\mathcal{E} }_x] \ge c t^{-1}\tag{80}\] for \(t\) small enough, uniformly in \(x \in \Omega^*_\delta\). Let us apply a scaling to the quantum cone, as in Proposition 17, i.e., consider the “unit circle embedding” \(\bar h_x\) of \(\tilde{h}_x + C\), where \(\tilde{h}_x\) has the law \(\tilde{\mathbb{P}}_x\), of the translated \(\gamma\)-quantum cone and \(C = (1/\gamma) \log (1/t)\). In other words, we apply the LQG change of coordinate: \[\bar h_x(\cdot) = \tilde{h}_x ( r_C\cdot) + C+ Q \log r_C,\] where \(r_C = e^{ - s_C}\) is defined by \[s_C = \inf\{ s \in \mathbb{R}: \tilde{X}_s - (Q- \gamma) s = -C\}\] and where \(\tilde{X}_s\) describes the radial part of the \(\gamma\)-quantum cone, as in 68 (i.e., the circle average at distance \(\rho\) from the origin of \(\tilde{h}\) is \(\tilde{X}_{\log (1/ \rho)} + \gamma \log (1 / \rho)\)).

As explained in Proposition 17, we have \(\tilde{\mathbf{p} }_t(x,x) = (1/t) \bar {\mathbf{p} }_1(x,x)\), where \(\tilde{\mathbf{p} }\) and \(\bar {\mathbf{p} }\) denote the heat kernels associated respectively to \(\tilde{h}_x\) and to \(\bar h_x\). Furthermore, \[{\mathbf{p} }_t^{\mathbb{C}\setminus \Omega}(x,x) = \frac{1}{t} \bar {\mathbf{p} }^{\mathbb{C}\setminus r_C^{-1} \Omega}_1(x,x)\] where \(r_C^{-1} \Omega\) denotes, with a slight abuse of notation, the image of \(\Omega\) under \(z \mapsto (z-x)r_C^{-1} + x\). Furthermore, when we phrase the event \({\mathcal{E} }_x\) in terms of \(\bar h_x\) instead of in terms of \(\tilde{h}_x\), \[\bar {\mathcal{E} }_x = \Bigl\{ \Big|(\bar h_x)_{r'/r_C}(x) - \alpha_0 \log (1/r') + C + Q\log r_C\Big| \le 1 \Bigr\},\] where \((\bar h_x)_{\rho}(x)\) denotes the circle average of \(\bar h_x\) at distance \(\rho\) from \(x\). Thus, using the invariance of the heat kernel under LQG coordinate changes, \[\label{eq:rescaled95hk} \tilde{\mathbb{E}}_x [ {\mathbf{p} }_t^{\mathbb{C}\setminus \Omega} (x,x) | {\mathcal{E} }_x ] = \frac{1}{t} \bar \mathbb{E}_x [ {\mathbf{p} }_1^{ \mathbb{C}\setminus r_C^{-1} \Omega } (x,x) | \bar {\mathcal{E} }_x].\tag{81}\]

We claim that:

• (Large scales.) The effect of the conditioning becomes negligible as \(t\to 0\), i.e., the conditional law of \(\bar h_x\) given \({\mathcal{E} }_x\) converges (say in the sense of total variation locally on any compact) to that of the unconditional \(\gamma\) quantum-cone \(\tilde{\mathbb{P}}_x\): \[\label{eq:simpl1} \bar \mathbb{P}_x ( \cdot | \bar {\mathcal{E} }_x) \Rightarrow \tilde{\mathbb{P}}_x\tag{82}\]

• (Small scales.) The requirement that trajectories leave the scaled domain \(r_C^{-1} \Omega\) in the computation of the heat kernel at time 1 also becomes irrelevant, since this domain shrinks to the origin \(x\); in other words, \[\label{eq:simpl2} {\mathbf{p} }_1^{\mathbb{C}\setminus r_C^{-1} \Omega} (x,x) \to {\mathbf{p} }_1^{\mathbb{C}} (x,x)\tag{83}\]

Assuming these two facts, one sees that, using 81 and Fatou’s lemma, \[\begin{align} \nopagebreak \liminf_{t\to 0} t \tilde{\mathbb{E}}_x [ {\mathbf{p} }_t^{ \mathbb{C}\setminus \Omega} (x,x) | {\mathcal{E} }_x] &\ge \liminf_{t\to 0} \bar \mathbb{E}_x[ {\mathbf{p} }_1^{ \mathbb{C}\setminus r_C^{-1} \Omega} (x,x) | \bar {\mathcal{E} }_x] \nonumber \nopagebreak \\ & \ge \tilde{\mathbb{E}}_x[ {\mathbf{p} }^{\mathbb{C}}_1(x,x) ] = c_\gamma \label{eq:Fatou} \end{align}\tag{84}\] where in the last line we used ?? . So, this proves 80 (in fact, a more precise version of 80 in which there is no term of the form \(t^{o(1)}\) and the constant in front is explicit, and is given by \(c_\gamma\)).

The proof of these two facts 82 and 83 is encapsulated by the following lemma.

Proof. To prove this lemma it is preferrable to change coordinates to the cylinder \({\mathcal{S} }\), as in 68 .

In these coordinates, the conditioning \({\mathcal{E} }_x\) for \(h_x\) is simply that \[\tilde{\mathcal{E} }_x = \Bigl\{\Big|\tilde{X}_{\log (1/r') } - (\alpha_0 - Q) \log (1/r') \Big| \le 1 \Bigr\}.\] By the Markov property, the conditional behaviour of \(\tilde{X}_s - (Q -\gamma ) s\) for \(s \ge \log (1/r')\), given \(\tilde{\mathcal{E} }_x\), is that of a Brownian motion with drift \(- (Q-\gamma)<0\), starting at time \(\log (1/r')\) at a position which is within 1 of the value of \((\alpha_0 -Q) \log (1/r')\). The key observation is that, given \(\tilde{\mathcal{E} }_x\), \[\label{eq:sC} s_C - \log (1/r') \to +\infty\tag{85}\] in probability as \(t\to 0\) (which, in the original coordinate system of \(\mathbb{C}\), boils down to the claimed fact that \(r'/r_C \to \infty\)). To see 85 , note that \[\label{eq:C95r39} C> ( Q - \alpha_0) \log (1/r')\tag{86}\] and more precisely \[C - ( Q - \alpha_0) \log (1/r') = \frac{\delta}{ \gamma} \log (1/t) \to \infty\] since \(r' = t^{\kappa_0 (1- \delta)}\) and \(\kappa_0 = 1/ ( \gamma( Q - \alpha_0))\), by definition of \(\kappa(\alpha)\) in ?? . (Visually, the meaning of 86 is that the horizontal line \(y = - C\) lies much below the line \(y = (\alpha_0 - Q) \log (1/r')\), which is the value of \(\tilde{X}\) at time \(\log (1/r')\) given \(\tilde{\mathcal{E} }_x\).) Therefore, applying the Markov property, the difference \(s_C - \log (1/r')\) (which is the first hitting time of some large negative value by a Brownian motion with negative drift) is to first order governed by the law of large numbers: \[\label{eq:LLN} \frac{ s_C - \log (1/r')}{\log (1/t)} \to \frac{\delta}{\gamma( Q - \gamma)}\tag{87}\] in probability, which certainly proves 85 . Applying the Markov property of \(\tilde{X}\) (under \(\tilde{\mathbb{P}}_x ( \cdot | \tilde{\mathcal{E} }_x)\)) at time \(\log (1/r')\), and using the exact same proof as that of Theorem 7.11 in [11] we deduce that the law of \((\tilde{X}_{s_C + s} + C + ( \gamma - Q) s)_{s\in \mathbb{R}}\), given \(\tilde{\mathcal{E} }_x\), converges (in the sense of total variation, on any interval of the form \([s_0, \infty) \subset \mathbb{R}\) – crucially we do not consider just compact intervals) to that of the unconditional law of \((\tilde{X}_s + (\gamma - Q) s)_{s\in \mathbb{R}}\). After returning to the original coordinate system of \(\mathbb{C}\), this shows that the conditional law of \(\bar h_x\), given \(\bar {\mathcal{E} }_x\), converges to the unconditional law of \(\tilde{\mathbb{P}}_x\).

To conclude, it remains to prove that \(r/ r_C \to 0\) in probability as \(t \to 0\). However, this follows at once from 87 , the fact that \(\log (1/r) - \log (1/r') = 2\kappa_0 \delta \log (1/t)\), and the inequality \[\begin{align} \label{eq:kappa32and32Q} 2\kappa_0 > \frac{1}{\gamma(Q - \gamma)}. \end{align}\tag{88}\] Indeed, recall that \(\kappa_0=(\gamma(Q-\alpha_0))^{-1}\), so 88 is equivalent to \(2(Q-\gamma)>Q-\alpha_0\), which can be reformulated as \[\begin{align} 2(Q-\gamma)>\sqrt{Q^2-4(1-\mathsf{x})}. \end{align}\] Since \(4(1-\mathsf{x})\le4\), it suffices to show (after squaring and rearranging) that \[\begin{align} 3Q^2-8Q\gamma+4\gamma^2+4>0. \end{align}\] Using \(8\gamma Q=16+4\gamma^2\), this is equivalent to \(3Q-12>0\), which is true as \(Q>2\). This shows that \(r/ r_C \to 0\) in probability.

This implies that the distance from \(x\) to the rescaled component of the boundary \(\partial^* \Omega\), \(\partial^* (r_C^{-1} \Omega)\), tends to 0 as \(t\to0\), and so Liouville Brownian motion (and thus also Liouville Brownian bridge by absolute continuity), is guaranteed to leave \(r_C^{-1} \Omega\) after any strictly positive amount of time with probability tending to 1 as \(t\to 0\), since \(\partial^* \Omega\) is connected and has positive diameter (so the diameter even tends to infinity after rescaling by \(r_C^{-1}\), though it is not necessary to exploit this fact). Thus \[\bar {\mathbf{p} }_1^{ \mathbb{C}\setminus r_C^{-1} \Omega} (x,x) \to \bar {\mathbf{p} }_1^{\mathbb{C}} ( x,x)\] in law, as desired. This finishes the proof of Lemma 26. ◻

As explained in 84 , this proves 80 , and thus, 79 . In turn, this proves ?? and thus concludes the proof of Proposition 25. ◻

Proof of Lemma 20. Our first observation is that it suffices to consider the case \(x = 0\). In the case of the quantum cone this is directly assumed. In the case of the whole-plane GFF, recall that the whole plane GFF is invariant under all Möbius transforms up to additive constant (see, e.g., Lemma 6.26 in [11]). This implies that \(h^*_x = h^*_0 - \Omega\) in law, where \(\Omega\) is a Gaussian random variable with fixed variance (corresponding to the average value of \(h^*_0\) on the circle of radius 1 around \(-x\)). The two random variables on the right hand side of this identity in law are not independent, but \(\mathbb{E}[(e^{\gamma \Omega})^p]< \infty\) for all \(p>0\), so by Hölder’s inequality it suffices to prove the result with \(h^*_0\) in place of \(h^*_x\), as announced.

The key observation is that both fields \(h^*_0\) and \(\tilde{h}\) can be coupled to a third field, which we will denote by \(\hat{h}\), in such a way that \[\label{eq:domination} h^*_0 \ge \hat{h},\quad \tilde{h} \ge \hat{h}\tag{89}\] (in the sense that the difference of the distributions is a nonnegative function), and where the field \(\hat{h}\) can be simply written as \[\label{eq:32hat32h} \hat{h} (\cdot) = h + \gamma \log (\tfrac{1}{|z|}); \quad \quad z \in \mathbb{C}\setminus \{0\}.\tag{90}\] This is essentially obvious in the case of \(h^*_0\), and amounts to the observation that \[h^*_0(z) = h (z) + \gamma\log G(0, z)\ge h + \gamma \log ( 1/ |z|), \quad \quad z \in \mathbb{C}\setminus \{0\}.\]There is in fact equality in the unit disc \(B_0(1)\), and outside the unit by definition of \(G\) (see 12 ) we have \(G(0, \cdot) = 0\) while \(\log (1/|z|) \le 0\).

In the case of \(\tilde{h}\) this requires a bit more arguments and boils down to the fact that if \((X_t, t \ge 0)\) is a Brownian motion conditioned so that \(X_t - (Q- \gamma ) t \ge 0\) for all \(t \ge 0\), then \(X_t\) stochastically dominates a Brownian motion. This can be seen either from an explicit computation of the generator, which has a positive drift (see, e.g., the discussion after Definition 7.7 in [11] where the generator is computed) or, more conceptually, can be seen from the fact that the conditioned process is a certain Doob \(h\)-transform, where the associated harmonic function, which is here very explicit as it is just an exponential function, is monotone increasing (see e.g. [42] for related arguments).

It thus suffices to prove the result with \(\hat{h}\) instead of \(h^*_x\) or \(\tilde{h}\), and we will denote the corresponding measure by \(\hat{\mu}\) and quantum clock by \(\hat{\nu}(u), u \ge 0\).

Now, note that by conformal invariance of the whole plane GFF, \((h(zR)_{z\in \mathbb{C}})\) is a full plane GFF, with a different additive normalisation. Thus \(h(zR) - h_R(0)\) has the same law as \(h\) (where \(h_r(x)\) denote the circle average of \(h\) at radius \(r\) around \(x\)), and hence \[\label{eq:scaling32hat32h} \hat{h}(zR) {=} \hat{h}'(z) + \gamma\log (\frac{1}{R})+ h_R(0)\tag{91}\] where \(\hat{h}'\) has the same law as \(\hat{h}\). Fix a Brownian motion \((B_t, t \ge 0)\) and for \(t\ge 0\), let \(B'_t = R B_{t/R^2}\). Then \((B'_t)_{t\ge 0}\) is also a standard Brownian motion. Let \(\hat{\nu} (u)\) denote the quantum clock of \(B\) with respect to \(\hat{h}\) and let \(\hat{\nu}'(u)\) denote the quantum clock of \(B'\) with respect to \(\hat{h}'\). Fix \(u>0\). Then writing down explicitly the definition of the quantum clock, and noting that \[\hat{h}_\varepsilon(zR) = ( \hat{h}')_{\varepsilon/R} (z) + \gamma \log (1/R) + h_R(0),\] we see that \[\label{eq:scaling32clock} \hat{\nu} (u) = e^{\gamma h_R(0) + \gamma^2 \log (1/R)} \hat{\nu}'( u/R^2) R^{2 + \gamma^2/2}.\tag{92}\] Thus if we suppose \(u = R^2\) we find \[\hat{\nu} (u) = e^{ \gamma h_R(0)} \hat{\nu}'(1) R^{ 2- \gamma^2/2}.\] Raising to the power \(-q\) and taking expectations on both sides, \[\begin{align} \mathbb{E}[ \hat{\nu} (u)^{-q} ] & = R^{-q( 2- \gamma^2/2)} \mathbb{E}[ e^{ - q \gamma h_R(0)} \hat{\nu}'(1)^{-q} ]. \end{align}\] Unlike in the exactly scale invariant case discussed e.g. in [11], the two random variables inside the expectation in the right hand side above are not independent. Nevertheless, we can use Hölder’s inequality: for every \(m>1\), letting \(m'\) be its Hölder conjugate (so \(1/m + 1/m' = 1\)) we have \[\mathbb{E}[ \hat{\nu} (u)^{-q} ] \le u^{ - q (1 - \gamma^2/4)} \mathbb{E}[ e^{ - q \gamma m h_R(0)}]^{1/m} \mathbb{E}[ \hat{\nu}'(1)^{ -q m'}]^{1/m'}\] The first expectation can be computed exactly (and is equal to \(e^{ \gamma^2 m^2 q^2 \mathrm{Var}(h_R(0)) /2}\)). Since GMC admits negative moments of all order (Theorem 3.42 in [11]), the second expectation is some fixed number \(C = C_{q,m, \gamma}\) which depends only on \(q, m\) and of course \(\gamma\). Since \(\mathrm{Var}(h_R(0)) = \log R + O(1) = (1/2) \log u + O(1)\), we see that for any \(m >1\), \[\begin{align} \mathbb{E}[ \hat{\nu} (u)^{-q} ] & \le C u^{ - q (1 - \gamma^2/4) } (u^{1/2})^{q^2 m^2 \gamma^2/2}. \end{align}\] By choosing \(m>1\) as close to 1 as needed, we see that the exponent of \(u\) above can be taken to be as close to \[- q( 1- \gamma^2/4) + q^2 \gamma^2/4 = \xi^*(-q)/2\] as desired. This concludes the proof of Lemma 20. ◻

Proof. Let \(\phi:\Omega\to\Omega'\) be a conformal isomorphism and let \((B_s)_{s\ge0}\) be a planar Brownian motion starting at some \(z\in\Omega\). Then the Liouville Brownian motion associated to \((\Omega,h)\) is given by \(\mathbf{Z}_t = B_{\nu^{-1}(t)}\), where \(\nu\) is the quantum clock, \[\label{bcodkhvq} \nu(t) = \lim_{\varepsilon\to 0} \varepsilon^{\gamma^2/2}\int_0^t e^{\gamma h_\varepsilon(B_s) } \mathrm{d} s .\tag{93}\] By Theorem 1.3 in [15] the process \((\mathbf{Z}_t, t \ge0)\) satisfies the change of coordinate formulae and so is itself intrinsic to LQG: \(\phi(\mathbf{Z}_t)\) is a Liouville Brownian motion in \(\Omega'\) associated to the field \(h\circ \phi^{-1} + Q \log | ( \phi^{-1})'|\), where \(Q = \tfrac{2}{\gamma} + \tfrac{\gamma}{2}\). This is precisely \(h'\) since \((\Omega, h)\) and \((\Omega', h')\) are equivalent as quantum surfaces.

Furthermore, \(\phi(\mathbf{Z}_t)=: \mathbf{Z}'_t\) starts at \(\mathbf{Z}'_0 = \phi(z)\). Thus, if \(\tau_\Omega\) is the exit time by \(B\) of the domain \(\Omega\), and \(\sigma_\Omega\) (resp. \(\sigma'_{\Omega'}\)) is the exit time of \(\Omega\) (resp. \(\Omega'\)) by \(\mathbf{Z}\) (resp. \(\mathbf{Z}'\)) \(\sigma_\Omega = \nu (\tau_\Omega)\) and \(\sigma_\Omega = \sigma'_{\Omega'}\). Thus we find \(\mathbf{P}_z ( \sigma_\Omega < t) = \mathbf{P}_{\phi(z)} ( \sigma'_{\Omega'} < t)\). It follows that the heat content on \(\Omega\) associated to \(h\) is equal to that on \(\Omega'\) associated to \(h'\), as desired. ◻

Proof. We have to estimate the outer and inner \(\varepsilon\)-neighborhood for \(\varepsilon>0\) small.
Inner neighborhood. Without the rectangles, the segment from \((0,1)\) to \((1,1)\) would contribute an \(\varepsilon\)-neighborhood of area exactly \(\varepsilon\). We now quantify the additional area contributed by each attached rectangle, distinguishing two regimes:
Consider the area of the \(\varepsilon-\)(inner) neighbourhood of \(\tilde{R}_n\). Suppose first that \(2\varepsilon\le a_n\) (as shown in the first two rectangles in Figure 3). Compared to the flat case of \(\partial S\), the amount that is added comes from the neighbourhood of the vertical sides of \(R_n\) (their \(\varepsilon\)-neighbourhood are distinct since \(2\varepsilon\le a_n\)) and some rounding around the corners. Hence \[|\partial ( R_n \setminus \ell_n)_\varepsilon| =2\varepsilon b_n-2c\varepsilon^2,\] where we recall that \(b_n\) is the rectangle’s height and the term with \(c\in(0,1)\) accounts for the rounded corners near the two lower corners.

Now suppose instead that \(2\varepsilon> a_n\). Then the \(\varepsilon\)-neighborhood fills the rectangle completely (see the third rectangle in Figure 3), so the added area is \[|\partial ( R_n \setminus \ell_n)_\varepsilon| =a_nb_n-\tilde{c}a_n^2,\] where \(\tilde{c}\in(0,2)\) can be computed explicitly and again comes from the rounded corners.

Note that \(2\varepsilon\le a_n\) holds precisely when \(n\le\lfloor \varepsilon^{-1/3}\rfloor\). Thus, the total area of the inner \(\varepsilon\)-neighbourhood of this part of \(\partial \Omega\) can be obtained by summing the contribution of all rectangles: \[\begin{align} I_\varepsilon&=\varepsilon+\sum_{n=1}^{\varepsilon^{-1/3+o(1)}}(2\varepsilon b_n-2\varepsilon^2 c)+\sum_{n=\epsilon^{-1/3+o(1)}}^{\infty}a_n(b_n-a_n\tilde{c})\\ &=\varepsilon+\sum_{n=1}^{\varepsilon^{-1/3+o(1)}}\Bigl(\frac{2\varepsilon}{\log(n)}-2\varepsilon^2 c\Bigr)+\sum_{n=\varepsilon^{-1/3+o(1)}}^{\infty}\frac{2}{n^3\log(n)}-\frac{4\tilde{c}}{n^6}\\ &=\varepsilon+\varepsilon^{1-1/3+o(1)}+(\varepsilon^{-1/3})^{-(2-1)+o(1)}=\varepsilon^{2/3+o(1)}, \end{align}\] where \(o(1)\) vanishes when \(\varepsilon\to0\). In conclusion, the inner Minkowski dimension is \(2-\lim_{\varepsilon\to0}\log(I_\varepsilon)/\log(\varepsilon^{-1})=2-2/3=4/3\).
Outer neighborhood. Similar considerations apply, estimating the contribution of each rectangle. Now we must also account for interactions between neighboring rectangles, depending on whether they are sufficiently separated, corresponding to whether \(2\varepsilon\le c_n\) or not.

If \(2\varepsilon\le c_n\), the gap between \(R_n\) and \(R_{n+1}\) is large enough that their outer \(\varepsilon\)-neighborhoods do not overlap. The \(n\)-th rectangle then contributes an additional area \(2\varepsilon b_n-2c\varepsilon^2\), with the negative term arising from the rounded corners (for some \(c\in(0,1)\)).

If \(2\varepsilon>c_n\), the \(\varepsilon\)-neighborhoods overlap in the gap. In this case, the extra area placed between the two rectangles is at most, say, \(2b_n c_n \le C n^{-2}/\log (n)\).

Combining these two regimes, we obtain the stated decomposition of the total contribution: \[\begin{align} O_\varepsilon&=\varepsilon+\sum_{n=1}^{\varepsilon^{-1/2+o(1)}}(\frac{2\varepsilon}{\log(n)}-2\varepsilon^2 c)+C \sum_{n=\varepsilon^{-1/2+o(1)}}^{\infty}\frac{n^{-2+o(1)}}{\log(n)}\\ &=\varepsilon+\varepsilon^{1-1/2+o(1)}+\varepsilon^{-1(-1/2)+o(1)}\\ &=\varepsilon^{1/2+o(1)} \end{align}\] so the outer Minkowski dimension is \(2-\lim_{\varepsilon\to0}\log(O_\varepsilon)/\log(\varepsilon)=2-1/2=3/2\). This completes the proof of Lemma 29. ◻