Proof. Choose a countable basis \(\{G_i\}_{i\ge1}\) consisting of relatively compact open sets. A diagonal subsequence makes all the limits \[\ell_i=\lim_{j\to\infty}\frac{\log\mu_j(G_i)}{2S_j} \in[-\infty,0]\] exist. Define \(V(x)=\inf_{i:x\in G_i}\ell_i\). If \(V(x)<a\), one of these basis neighborhoods has \(\ell_i<a\), and \(V<a\) throughout that neighborhood. Thus \(V\) is upper semicontinuous.

For any finite \(a>\sup_F V\), every point of \(F\) has a basis neighborhood with exponent less than \(a\). A finite subcover and \(\mu_j(F)\le\sum_i\mu_j(G_i)\) give the compact upper bound by \(a\). Letting \(a\downarrow\sup_F V\) proves the first inequality, also when this supremum is \(-\infty\). If \(x\in G\), a basis neighborhood \(G_i\) with \(x\in G_i\subset G\) gives a lower bound by \(\ell_i\ge V(x)\), proving the second inequality.

For the weighted upper bound, take \(a>\sup_F(V+f)\) and choose the neighborhood at each point small enough that its limiting measure exponent plus the supremum of \(f\) on it is less than \(a\). This is possible by continuity of \(f\) and the definition of \(V\), including at points where \(V=-\infty\). A finite cover and uniform convergence of \(f_j\) on its compact closure give the claimed upper bound. For the lower bound at a point \(x\in G\) with \(V(x)>-\infty\), choose \(G_i\) compactly contained in \(G\) so small that \(f\ge f(x)-\varepsilon\) there. The weighted integral over \(G_i\) has lower exponent at least \(\ell_i+f(x)-\varepsilon\). Let \(\varepsilon\downarrow0\) and then take the supremum over \(x\). ◻

Proof. Suppress the index and write \(v=e^T w\). The conjugated operator is \(P_++P_-\), where \[P_+=\Delta+|\nabla T|^2+K^2p, \qquad P_-=-2\nabla T\cdot\nabla-\Delta T.\] On compactly supported functions, \(P_+\) is symmetric and \(P_-\) is skew-symmetric. Integration by parts gives \[ \begin{split} 2\langle P_+v,P_-v\rangle ={}&4\int T''(\nabla v,\nabla v)\\ &+\int\bigl(4T''(\nabla T,\nabla T)-\Delta^2T +2K^2\nabla p\cdot\nabla T\bigr)v^2. \end{split} \tag{4}\] For clarity, the contribution from \(\Delta\) is \(4\int T''(\nabla v,\nabla v)-\int\Delta^2T\,v^2\). Multiplication by a real function \(a\) contributes \(2\int\nabla a\cdot\nabla T\,v^2\); applying this with \(a=|\nabla T|^2+K^2p\) gives the remaining terms in Equation (4).

If Equation (3) failed on every sufficiently small neighborhood, a diagonal choice would give a subsequence and real smooth \(v_j\) with supports shrinking to \(x_0\), such that \[\|v_j\|_2=1, \qquad \|(P_++P_-)v_j\|_2=o(\sqrt{S_j}D_j).\] All constants below are uniform on one fixed small neighborhood of \(x_0\). There, \(|\nabla T|\le CD\), \(|T''|+|\Delta^2T|\le CS\), and the last hypothesis in Equation (2) controls the potential term in Equation (4). Set \(X=\|P_+v\|_2\), \(Y=\|P_-v\|_2\), and \(G=\|\nabla v\|_2\). Expanding the squared norm and using the identity gives \[X^2+Y^2\le CS(G^2+D^2).\] Testing \(P_+\) against \(v\) gives \(G^2\le CD^2+X\), since \(K=o(D)\). Combining the two inequalities first yields \(X\le C(S+\sqrt S D)\) and then \[G\le CD, \qquad X+Y\le C\sqrt S D.\] Here \(D\ge S\to\infty\) absorbs the term \(S\).

Uniformly on the shrinking supports, \[\nabla T_j/D_j\to a=-b-d\nabla\chi(x_0)\ne0, \qquad T_j''/S_j\to H=-\chi''(x_0).\] The preceding bounds give \[\frac{\sqrt{S_j}}{D_j}\le\frac1{\sqrt{S_j}}\longrightarrow0.\] Dividing the identity obtained by testing \(P_+\) against \(v_j\) by \(D_j^2\) therefore shows \[\int|\nabla v_j/D_j|^2\longrightarrow |a|^2.\] Likewise, \[\left\|\frac{\nabla T_j}{D_j}\cdot \frac{\nabla v_j}{D_j}\right\|_2 \le \frac{\|P_-v_j\|_2+\|\Delta T_j\,v_j\|_2}{2D_j^2} \longrightarrow0.\] Replacing \(\nabla T_j/D_j\) by its uniform limit consequently gives \(\|a\cdot\nabla v_j/D_j\|_2\to0\).

Let \(e\) be a unit vector perpendicular to \(a\). In dimension two, the energy perpendicular to \(a\) has only this one direction. Decomposing \(\nabla v_j/D_j\) in the orthonormal basis \(\{a/|a|,e\}\), the parallel component tends to zero in \(L^2\), whereas the squared norm of the perpendicular component tends to \(|a|^2\). Hence the first term in Equation (4), divided by \(S_jD_j^2\), tends to \(4|a|^2H(e,e)\). The other main Hessian term tends to \(4H(a,a)\). The two remaining terms tend to zero: their absolute values after this division are bounded by \[\frac{C}{D_j^2} \quad\hbox{and}\quad \frac{CK_j^2\|\nabla p_j\|_\infty}{S_jD_j},\] respectively. We have thus proved \[\begin{aligned} \frac{2\langle P_+v_j,P_-v_j\rangle}{S_jD_j^2} &\longrightarrow 4\bigl(|a|^2H(e,e)+H(a,a)\bigr)\\ &=4|a|^2\operatorname{tr}H>0. \end{aligned}\] The squared norms of \(P_+v_j\) and \(P_-v_j\) are nonnegative, so this contradicts the assumed smallness of \(\|(P_++P_-)v_j\|_2^2\). ◻

Proof. Suppose first that an upper test at \(0\) has zero gradient and negative Laplacian; translate the point to \(0\) if necessary. A quadratic Taylor polynomial with its Hessian increased by a sufficiently small positive multiple of the identity gives a quadratic upper test \(h\) with \[\begin{gathered} W(0)=h(0),\qquad \nabla h(0)=0,\qquad \operatorname{tr}h''<0,\\ W(x)<h(x)\quad(0<|x|\le r) \end{gathered}\] for some small \(r>0\). Choose a unit eigenvector \(e\) of \(h''\) with eigenvalue \(\alpha<0\). For small \(t>0\), maximize \(W(x)-h(x)+t e\cdot x\) on \(\overline{\mathbb D_r}\). Upper semicontinuity and the strict boundary gap ensure that a maximizer \(x_t\) is interior. Its value is at least the value \(0\) at the origin, and \(W-h\le0\), so \(e\cdot x_t\ge0\). At \(x_t\) an upper test is \(h(x)-t e\cdot x\), shifted by its contact constant. Its gradient has component \[e\cdot\nabla(h-t e\cdot x)(x_t) =\alpha\, e\cdot x_t-t<0,\] while its Laplacian is still negative. This contradicts the hypothesis. All smooth upper tests at finite points therefore have nonnegative Laplacian.

We obtain harmonic comparison directly. Let \(\overline{\mathbb D_r(a)}\subset\Omega\), let \(h\) be harmonic in \(\mathbb D_r(a)\) and continuous on its closure, and suppose \(W\le h\) on the boundary. If \(W-h\) were positive somewhere inside, then for sufficiently small \(\varepsilon>0\) the function \[h_\varepsilon(x)=h(x)+\varepsilon(r^2-|x-a|^2)\] would still lie below \(W\) at some interior point. A positive maximum of \(W-h_\varepsilon\) is attained at an interior finite point. Adding this maximum to \(h_\varepsilon\) supplies an upper test with Laplacian \(-4\varepsilon<0\), again a contradiction. Thus \(W\) satisfies harmonic comparison. For upper semicontinuous functions this is the usual characterization of subharmonicity, including the identically \(-\infty\) case; see [10]. In particular, local integrability and the submean property of a nontrivial subharmonic function are consequences here, not assumptions used in the argument. ◻

Proof of Theorem 3. Fix such a profile \(V\) and pass to a further subsequence so that \(S_j/D_j\to d\) and \(B_j/D_j\to b\) as in Lemma 4. The profile bounds remain valid on this subsequence. Suppose a smooth \(\chi\) touches \(V\) from above at a finite point \(x_0\), and suppose \[\Delta\chi(x_0)<0, \qquad b+d\nabla\chi(x_0)\ne0.\] By adding a small positive multiple of \(|x-x_0|^4\), we make the touching strict on a sufficiently small closed disk without altering the gradient or Hessian at \(x_0\). Take this disk inside the neighborhood in Lemma 4.

Choose fixed radii \(0<r_0<r_1<r_2<r_3\) inside that neighborhood. Let \(\zeta\) be a smooth cutoff equal to one on \(\mathbb D_{r_1}(x_0)\) and supported in \(\mathbb D_{r_2}(x_0)\). Its derivatives are supported in the annulus between these two disks. Set \[G=\mathbb D_{r_0/2}(x_0),\qquad F=\{x:r_0\le |x-x_0|\le r_3\}.\] We may choose the radii so that the strict touching holds on the whole closed outer disk. Then \(F\) is a compact annulus, \(\sup_F(V-\chi)<0\), and \(\sup_G(V-\chi)=0\).

For completeness, the cutoff error can be controlled with a prefactor independent of the ratio \(D_j/S_j\). Choose a smooth \(\theta\) supported in the interior of \(F\) and equal to one near the derivatives of \(\zeta\). Testing the equation for \(U_j\) against \(\theta^2e^{2T_j}U_j\) and integrating by parts bounds \[\begin{aligned} \int\theta^2e^{2T_j}|\nabla U_j|^2 &\le C\int_F e^{2T_j} \bigl(|\nabla\theta|^2+ \theta^2|\nabla T_j|^2+K_j^2|p_j|\theta^2\bigr)|U_j|^2\\ &\le CD_j^2\int_F e^{2T_j}|U_j|^2. \end{aligned}\] The first inequality absorbs half of the gradient integral by Cauchy–Schwarz; the second uses \(K_j=o(D_j)\) and \(D_j\to\infty\). Since \[(\Delta+K_j^2p_j)(\zeta U_j) =(\Delta\zeta)U_j+2\nabla\zeta\cdot\nabla U_j,\] Lemma 4 gives \[c\sqrt{S_j}D_j\|e^{T_j}U_j\|_{L^2(G)} \le CD_j\|e^{T_j}U_j\|_{L^2(F)}.\] Cancel \(D_j\) before comparing exponential rates. Squaring and multiplying by the common normalization constant \(c_j\) yields \[S_j\int_G e^{-2S_j\chi}\,d\mu_j \le C\int_F e^{-2S_j\chi}\,d\mu_j.\] The weighted bounds of Lemma 2, together with \(\log S_j/S_j\to0\), now give the contradiction \[0=\sup_G(V-\chi)\le\sup_F(V-\chi)<0.\] In particular, no bound on \(\log D_j/S_j\) was required.

We have established the upper-test inequality whenever \(b+d\nabla\chi(x_0)\ne0\). If \(d=0\), then \(|b|=1\) and there is no exceptional gradient; harmonic comparison follows as in Lemma 5. If \(d>0\), put \(W(x)=V(x)+(b/d)\cdot x\). An upper test \(\psi\) for \(W\) corresponds to the upper test \(\chi=\psi-(b/d)\cdot x\) for \(V\), and \(b+d\nabla\chi=d\nabla\psi\). Thus \(W\) satisfies the hypothesis of Lemma 5. It is subharmonic, and subtracting the affine function shows that \(V\) is subharmonic too. ◻

Proof. The positive spectral gap gives \(k\ge k_1>0\). Interpret the log ratio in Equation (5) as \(+\infty\) when the local norm is zero. If the assertion failed, we could choose eigenfunctions \(u_j\), normalized by \(\|u_j\|_{L^2(M)}=1\), such that their log ratios \(R_j\) obey \(R_j\ge j^2 k_j\). This also covers a zero local norm by repeating that eigenfunction. Set \(S_j=j k_j\). Then \(S_j\to\infty\) and \(S_j/k_j\to\infty\), whether or not \(k_j\) tends to infinity.

Apply Lemma 2 on \(M\) to the probability measures \(|u_j|^2\,d\operatorname{vol}_g\). Compactness of \(M\) and the compact upper bound with \(F=M\) give \(0\le\sup_M V\le0\). Upper semicontinuity ensures that this maximum is attained.

In a relatively compact isothermal chart, \(g=p(x)(dx_1^2+dx_2^2)\) with \(p>0\) smooth, and the coordinate functions satisfy \(\Delta U_j+k_j^2pU_j=0\). The positive volume factor \(p\) is bounded above and below on this chart. Multiplying \(|U_j|^2\,dx\) by a fixed positive constant if necessary makes its mass at most one; the resulting logarithmic profile is the restriction of \(V\). Theorem 3 applies with \(B_j=0\), \(K_j=k_j\), and \(D_j=S_j\), since \[\frac{k_j}{S_j}=\frac1j\longrightarrow0, \qquad \frac{k_j^2\|\nabla p\|_\infty}{S_j^2} =\frac{\|\nabla p\|_\infty}{j^2}\longrightarrow0.\] Thus \(V\) is subharmonic in every such chart. The strong maximum principle makes the set where \(V=0\) open; it is also closed by upper semicontinuity and nonempty by the preceding maximum argument. Connectedness gives \(V\equiv0\) on \(M\).

On the other hand, the normalized mass of the given disk satisfies \[\frac{1}{2S_j}\log\int_G |u_j|^2\,d\operatorname{vol}_g =-\frac{R_j}{S_j}\le-j,\] with the same conclusion when the integral is zero. This contradicts the open lower bound in Equation (1), whose right-hand side is \(\sup_GV=0\). The argument proves both positivity of the local norm and Equation (5). It uses only local isothermal charts and therefore imposes no orientability assumption. ◻

Proof. Upper semicontinuity supplies a point \(x_*\in\overline{\mathbb D_1}\) with \(V(x_*)\ge-1\). The disk \(\mathbb D_{10}(x_*)\) contains \(\mathbb D_8\) and is contained in \(\mathbb D_R\). In particular, \(V\) is nontrivial and locally integrable. Since \(V\le0\), the submean inequality gives \[\int_{\mathbb D_8}\abs{V} \le -\int_{\mathbb D_{10}(x_*)}V \le100\pi.\] Its distributional Laplacian is a nonnegative Radon measure. Choose a fixed nonnegative smooth cutoff \(\psi\), supported in \(\mathbb D_8\) and equal to one on \(\mathbb D_5\). If \(\eta=(2\pi)^{-1}\Delta V\), then \[\eta(\mathbb D_5) \le\int\psi\,d\eta =\frac1{2\pi}\int V\Delta\psi \le C\norm{V}_{L^1(\mathbb D_8)}.\] Let \(\nu\) be the restriction of \(\eta\) to \(\mathbb D_4\), and set \(P(x)=\int_{\mathbb D_4}\log\abs{x-z}\,d\nu(z)\). The elementary bound \[\sup_{z\in\mathbb D_4}\int_{\mathbb D_3}\bigl|\log\abs{x-z}\bigr|\,dx<\infty\] shows that \(\norm{P}_{L^1(\mathbb D_3)}\le C\nu(\mathbb D_4)\). The logarithmic fundamental solution identity \(\Delta\log\abs{x-z}=2\pi\delta_z\) implies \(\Delta(V-P)=0\) on \(\mathbb D_3\). Thus \(V-P\) agrees almost everywhere with a harmonic function \(h\). Interior harmonic estimates, applied to its bounded \(L^1(\mathbb D_3)\) norm, bound \(\nabla h\) and \(\nabla^2h\) on \(\mathbb D_2\). Finally, the logarithmic potential of a nonnegative measure has its usual subharmonic representative. Both \(V\) and \(h+P\) are recovered pointwise as limits of their disk averages. Their almost-everywhere equality therefore gives the asserted pointwise representation. ◻

Proof. All disks \(\overline{\mathbb D_{R\rho}(y)}\) lie in \(\mathbb D_2\). The vector \(c_y\) is finite because its kernel is truncated away from its singularity. Put \(W_y(x)=V(x)-c_y\cdot x\), and lower bound its supremum on the inner disk by its disk average there. For the harmonic term, this average is its value at \(y\), after the affine part is accounted for. Taylor’s theorem bounds the corresponding outer deviation by \(C\rho^2\).

For a far source \(\abs{z-y}>2R\rho\), the logarithm is harmonic on the outer disk and its Hessian is bounded there by \(C\abs{z-y}^{-2}\). Its inner disk average is \(\log\abs{y-z}\), and the retained gradient in \(c_y\) removes its first-order term. Its contribution to the outer deviation is therefore at most \(C\rho^2/\abs{z-y}^2\).

For a near source \(\abs{z-y}\le2R\rho\), rescale by \(\rho\). Writing \(t=(z-y)/\rho\), the disk-average formula \[\frac1\pi\int_{\mathbb D_1}\log\abs{\xi-t}\,d\xi = \begin{cases} (\abs{t}^2-1)/2,&\abs{t}\le1,\\ \log\abs{t},&\abs{t}\ge1 \end{cases}\] has lower bound \(-1/2\). On the larger disk the corresponding logarithm is at most \(\log(3R)\), since \(\abs t\le2R\). The common term \(\log\rho\) cancels between these two quantities. Thus the difference between the outer upper bound and the inner average is at most \(\log(3R)+1/2\), uniformly in the source position. This calculation permits an atom at \(y\), or anywhere else in the inner disk: the logarithm remains integrable in the averaging variable. At an outer point where a potential is \(-\infty\), an upper bound causes no difficulty. Because the measure is nonnegative, the kernel bounds may be integrated. On the near region, \[\min\left\{1,\frac{\rho^2}{\abs{z-y}^2}\right\} \ge\frac1{4R^2}.\] Combining the near and far bounds proves (8).

For completeness, fix any \(z\in\mathbb D_4\). The number of grid centers within distance \(r\) of \(z\) is at most \(C(1+r^2/\rho^2)\). There are \(\rho^{-2}\) centers in total. The centers at distance at most \(\rho\) contribute \(C\rho^2\) to the averaged minimum kernel. In each dyadic annulus \(2^\ell\rho<\abs{z-y}\le2^{\ell+1}\rho\), the number of centers is at most \(C2^{2\ell}\), whereas the kernel is at most \(2^{-2\ell}\). All distances are bounded by five, so only \(C(1+\abs{\log\rho})\) annuli occur. Hence, uniformly in \(z\), \[\mathop{\mathrm{avg}}_y\min\left\{1,\frac{\rho^2}{\abs{z-y}^2}\right\} \le C\rho^2(1+\abs{\log\rho}).\] The same annuli, for the truncated gradient kernel, give \[\rho^2\sum_{\substack{y\\\abs{z-y}>2R\rho}} \frac1{\abs{z-y}} \le C.\] Indeed an annulus of radius comparable to \(2^\ell\rho\) contributes at most \(C2^\ell\rho\), and these radii sum to a fixed constant. Integrating the two bounds against the bounded measure \(\nu\), and using the bound for \(\nabla h\), proves both averaged estimates. ◻

Proof. First choose \(A>10R\), using only the uniform estimates of Lemma 9, so large that, with \(\rho=A^{-1}\), \[C\rho^2(1+\abs{\log\rho})<\frac\rho4.\] Choose \(C_1\) strictly larger than the uniform bound for \(\mathop{\mathrm{avg}}_y\abs{c_y}\) in that lemma. We prove that some finite \(C_0\) works for these fixed choices.

Suppose otherwise. There is a sequence of failures of the conclusion for which the hypotheses hold with \(C_0=j\). Use subscripts \(j\) for the associated quantities and rescale each parent square to the unit square centered at zero: \[U_j(X)=u_j(x_{Q_j}+s_jX), \qquad p_j(X)=p(x_{Q_j}+s_jX).\] The rescaled equation is \(\Delta U_j+K_j^2p_jU_j=0\), and \(\norm{p_j}_\infty\le C\), \(\norm{\nabla p_j}_\infty\le Cs_j\). With \(B_j=K_jb_j\) and \(D_j=S_j+\abs{B_j}\), the failure hypotheses imply \[S_j\longrightarrow\infty,\qquad \frac{K_j}{D_j}\longrightarrow0,\qquad \frac{K_j^2\norm{\nabla p_j}_\infty}{S_jD_j} \longrightarrow0.\] No restriction on \(\abs{B_j}/S_j\) has been imposed.

Normalize the tilted measures on \(\mathbb D_R\) by setting \[d\mu_j(X)= \frac{e^{-2B_j\cdot X}\abs{U_j(X)}^2\,dX} {\displaystyle\int_{\mathbb D_R}e^{-2B_j\cdot X}\abs{U_j(X)}^2\,dX}.\] Lemma 2 and Theorem 3 give, after passage to a subsequence, a subharmonic profile \(V\), possibly identically \(-\infty\), on \(\mathbb D_R\). It satisfies \(V\le0\), while \[\mu_j(\mathbb D_1)=e^{-2E(Q_j,b_j)}\ge e^{-2S_j}.\] The compact upper bound in (1), applied to \(\overline{\mathbb D_1}\), therefore gives \(\sup_{\overline{\mathbb D_1}}V\ge-1\). In particular, \(V\) is nontrivial and the two preceding lemmas apply.

For each of the fixed, finitely many child centers \(y\), take the vector \(c_y\) furnished by Lemma 9 for this limiting profile. On the corresponding child \(q_j\), set \[b_{q_j}=b_j+\frac{S_j}{K_j}c_y.\] In parent coordinates the additional weight is \(e^{-2S_jc_y\cdot X}\). To see the cancellation of recentering constants, write \[-K_jb_{q_j}\cdot(X-y) =-B_j\cdot X-S_jc_y\cdot X +(B_j+S_jc_y)\cdot y.\] The last term, as well as the measure normalization and the coordinate Jacobian, cancels in the ratio defining \(E(q_j,b_{q_j})\). Consequently this excess is one half the logarithm of \[\frac{\displaystyle\int_{\mathbb D_{R\rho}(y)} e^{-2S_jc_y\cdot X}\,d\mu_j(X)} {\displaystyle\int_{\mathbb D_\rho(y)} e^{-2S_jc_y\cdot X}\,d\mu_j(X)}.\] Apply the weighted compact upper bound to the numerator, using \(\overline{\mathbb D_{R\rho}(y)}\), and the weighted open lower bound to the denominator. This yields \[\limsup_j\frac{E(q_j,b_{q_j})}{S_j} \le \sup_{\overline{\mathbb D_{R\rho}(y)}}(V(X)-c_y\cdot X) -\sup_{\mathbb D_\rho(y)}(V(X)-c_y\cdot X).\] The inner supremum is finite: a nontrivial subharmonic function is locally integrable, so it cannot equal \(-\infty\) throughout an open disk, and it is bounded above on compact subsets. Both suprema in the displayed difference are therefore finite.

There are only \(A^2\) children, and \(A\) has already been fixed. Averaging the last inequality and invoking Lemma 9 gives \[\limsup_j\mathop{\mathrm{avg}}_{q_j\prec Q_j} \frac{E(q_j,b_{q_j})}{S_j}<\frac\rho4.\] Since \(K_{q_j}=\rho K_j\), it follows, for all sufficiently large \(j\), that \[\mathop{\mathrm{avg}}_{q_j\prec Q_j}m(q_j,b_{q_j}) \le\frac12\frac{S_j}{K_j}.\] At the same time, \[\mathop{\mathrm{avg}}_{q_j\prec Q_j}\abs{b_{q_j}-b_j} =\frac{S_j}{K_j}\mathop{\mathrm{avg}}_y\abs{c_y} \le C_1\frac{S_j}{K_j}.\] Thus these choices satisfy both required inequalities, contrary to the defining failure of the sequence. This proves the proposition. In particular, \(A\) and \(C_1\) have been fixed before \(C_0\), and no positive lower bound for \(K\) was used. ◻

Proof. Write \(K=K_Q\), \(s=s(Q)\), and set \[S=E(Q,b)+C_0+Ks,\qquad B=Kb.\] Thus \(S\ge C_0\), \(E(Q,b)\le S\), and \(S\ge Ks\). Equation (7) gives \[N(Q)\le m(Q,b)+(R+1)\abs{b} \le (R+1)\bigl(m(Q,b)+\abs{b}\bigr).\] Consequently \(S+\abs{B}\ge KN(Q)/(R+1)\); also \(S+\abs{B}\ge K\abs{b}\). Choose, for example, \(L=2C_0(R+1)\). Either of the alternatives in the lemma then implies \[\frac{K}{S+\abs{B}}\le \frac1{C_0}.\] Using \(S\ge Ks\), we obtain the other required smallness condition, \[\frac{K^2s}{S(S+\abs{B})} \le \frac{K}{S+\abs{B}}\le \frac1{C_0}.\] All the hypotheses of Proposition 10 hold. Its conclusions, with \(S/K=m(Q,b)+f_Q\), are precisely (11).

Along a path, \(s(Q_i)=A^{-i}s(I)\) and \(K_{Q_i}=A^{d-i}K_{Q_d}\). Since \(K_{Q_d}\ge a_0\), \[\sum_{i=0}^d \frac{C_0}{K_{Q_i}} \le \frac{C_0}{a_0}\sum_{\ell=0}^d A^{-\ell}, \qquad \sum_{i=0}^d s(Q_i) =s(I)\sum_{i=0}^d A^{-i}.\] The geometric-series bound gives (12). ◻

Proof. There are only finitely many nodes. At each node from which descent continues, choose one of the assignments furnished by Lemma 11. These choices and the stopping rule determine all random variables on the finite probability space of terminal paths. Thus only finite conditional-expectation identities are needed. In expressions with a fixed index, set \(m_i=0\) after stopping; those auxiliary values do not enter the stopped sums.

Let \(d\) be the depth below \(I\), put \(m_0=m(I,b_I)\), and set \[F_I=\frac{A}{A-1}\left(\frac{C_0}{a_0}+s(I)\right), \qquad M=\mathbb E\sum_{i=0}^{T}m_i.\] The event \(\{i<T\}\) is determined before the next child is selected. Applying (11) on this event and summing over \(0\le i<d\) gives \[M-m_0 =\sum_{i=0}^{d-1}\mathbb E\bigl[\mathbf 1_{\{i<T\}}m_{i+1}\bigr] \le \frac12\mathbb E\sum_{i=0}^{T-1}(m_i+f_{Q_i}) \le \frac12 M+\frac12 F_I.\] All the \(m_i\) are nonnegative by Definition 7. It follows that \(M\le 2m_0+F_I\). The tilt estimate in (11) similarly yields \[\mathbb E\sum_{i=0}^{T-1}\abs{b_{i+1}-b_i} \le C_1\mathbb E\sum_{i=0}^{T-1}(m_i+f_{Q_i}) \le 2C_1m_0+2C_1F_I.\] Their sum is at most \[2(1+C_1)m_0+(1+2C_1)F_I.\] Since \(s(I)\le1\), this is bounded by the right-hand side of (13) with a constant depending only on the fixed data. This calculation includes \(T=0\), and it includes both \(m_T\) and the incoming increment when the last square violates a stopping condition. ◻

Proof. If no such pair of thresholds existed, then for each positive integer \(j\) there would be a square \(Q_j\) and a tilt \(b_j\) with \[\abs{b_j}\ge j,\qquad \frac{m(Q_j,b_j)}{\abs{b_j}}\le\frac1j,\qquad N(Q_j)\le L.\] Rescale about the center of \(Q_j\) by its side \(s_j=s(Q_j)\). In the rescaled coordinates, \(U_j\) solves \(\Delta U_j+K_j^2p_jU_j=0\) on \(\mathbb D_R\), where \(K_j=K_{Q_j}\), \(\norm{p_j}_\infty\le C\), and \(\norm{\nabla p_j}_\infty\le Cs_j\). Set \[S_j=K_j\abs{b_j},\qquad B_j=K_jb_j,\qquad D_j=S_j+\abs{B_j}=2S_j.\] The hypotheses of Theorem 3 follow from \[S_j\ge a_0j\longrightarrow\infty,\qquad \frac{K_j}{D_j}=\frac1{2\abs{b_j}}\longrightarrow0,\qquad \frac{K_j^2\norm{\nabla p_j}_\infty}{S_jD_j} \le\frac{Cs_j}{2\abs{b_j}^2}\longrightarrow0.\] Normalize the tilted measures \[d\mu_j(x)=c_j e^{-2B_j\cdot x}\abs{U_j(x)}^2\,dx\] to have mass one on \(\mathbb D_R\). By Lemma 2 and Theorem 3, a subsequence has a subharmonic profile \(V\le0\), with the identically minus infinity case initially allowed. The definitions give \[\frac{\log\mu_j(\mathbb D_1)}{2S_j} =-\frac{E(Q_j,b_j)}{S_j} =-\frac{m(Q_j,b_j)}{\abs{b_j}}\longrightarrow0.\] The upper bound in (1) on the compact set \(\overline{\mathbb D_1}\) therefore implies \(\sup_{\overline{\mathbb D_1}}V=0\). In particular the profile is nontrivial and attains its maximum at an interior point of \(\mathbb D_R\). The strong maximum principle yields \(V=0\) on \(\mathbb D_R\).

Pass to a further subsequence so that \(e_j=b_j/\abs{b_j}\) converges to a unit vector \(e\). Undoing the tilt expresses the untilted growth as \[\frac{E(Q_j,0)}{S_j} = \frac{1}{2S_j}\log \int_{\mathbb D_R}e^{2S_je_j\cdot x}\,d\mu_j(x) - \frac{1}{2S_j}\log \int_{\mathbb D_1}e^{2S_je_j\cdot x}\,d\mu_j(x).\] For any fixed \(1<r<R\), bound the first integral from below by its restriction to \(\mathbb D_r\). The weighted form of Lemma 2, applied from below on the open disk \(\mathbb D_r\) and from above on the compact disk \(\overline{\mathbb D_1}\), gives \[\liminf_{j\to\infty}\frac{E(Q_j,0)}{S_j} \ge \sup_{\mathbb D_r} e\cdot x -\sup_{\overline{\mathbb D_1}} e\cdot x =r-1.\] Letting \(r\uparrow R\) gives a lower bound of \(R-1\), without any claim about mass at the boundary of \(\mathbb D_R\). On the other hand, \[0\le\frac{E(Q_j,0)}{S_j} =\frac{N(Q_j)}{\abs{b_j}} \le\frac{L}{j}\longrightarrow0,\] a contradiction. ◻

Proof. Let \(B_*,\eta_*\) be the thresholds from Lemma 13. Choose \(\gamma\) so that \[0<\gamma<\frac12,\qquad \frac{\gamma}{1-\gamma}\le\eta_*.\] Let \(C_{\mathrm{st}}\) be the constant from Lemma 12. Choose \[0<\sigma\le\min\left\{\frac{\gamma}{2}, \frac{\gamma}{8C_{\mathrm{st}}}\right\}, \qquad P\ge\max\left\{1,\frac{B_*}{1-\gamma}, \frac{2L}{1-\gamma}, \frac{8C_{\mathrm{st}}}{\gamma}\right\}.\] Write \(B_0=\abs{b_I}\). Starting at \(I\), descend using Lemma 11 while \[\abs{b_i-b_I}<\gamma B_0,\qquad m_i<\gamma B_0,\] and stop at the first failure of either inequality or at the terminal level, whichever comes first. These inequalities hold initially by (15). Whenever they hold, we have \[\abs{b_i}>(1-\gamma)B_0\ge B_*, \qquad \abs{b_i}>L, \qquad \frac{m_i}{\abs{b_i}} <\frac{\gamma}{1-\gamma}\le\eta_*.\] Thus every required step is available and Lemma 13 gives \(N(Q_i)>L\) at each such square.

Let \(T\) be this stopping time. Include in the failure event a violation first produced at the terminal level. If a failure occurs, then either \(m_T\ge\gamma B_0\), or the triangle inequality gives \(\sum_{i=0}^{T-1}\abs{b_{i+1}-b_i}\ge\gamma B_0\). Consequently Lemma 12 and Markov’s inequality give \[\begin{split} \mathbb P(\text{failure}) &\le \frac{1}{\gamma B_0} \mathbb E\left[\sum_{i=0}^{T}m_i+ \sum_{i=0}^{T-1}\abs{b_{i+1}-b_i}\right]\\ &\le\frac{C_{\mathrm{st}}}{\gamma} \left(\sigma+\frac1P\right) \le\frac14. \end{split}\] On the complementary event, descent reaches the terminal level and the two strict inequalities hold at every visited square. All of those squares have \(N>L\). This event has probability at least \(3/4\), which proves the asserted lower bound of \(1/2\). The same reasoning includes a subtree of depth zero. ◻