Sharp nodal length on smooth surfaces

Proof. Choose a countable basis \{G_i\}_{i\ge1} consisting of relatively compact open sets. A diagonal subsequence makes all the limits

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)-\eps there. The weighted integral over G_i has lower exponent at least \ell_i+f(x)-\eps. Let \eps\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

On compactly supported functions, P_+ is symmetric and P_- is skew-symmetric. Integration by parts gives

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 (2.4).

If Equation (2.3) 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

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 (2.2) controls the potential term in Equation (2.4). Set X=\|P_+v\|_2, Y=\|P_-v\|_2, and G=\|\nabla v\|_2. Expanding the squared norm and using the identity gives

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

Here D\ge S\to\infty absorbs the term S.

Uniformly on the shrinking supports,

The preceding bounds give

Dividing the identity obtained by testing P_+ against v_j by D_j^2 therefore shows

Likewise,

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 (2.4), 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

respectively. We have thus proved

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

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{\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

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{\D_r(a)}\subset\Omega, let h be harmonic in \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 \eps>0 the function

would still lie below W at some interior point. A positive maximum of W-h_\eps is attained at an interior finite point. Adding this maximum to h_\eps supplies an upper test with Laplacian -4\eps<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, Chapter 2]. 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 2.2. 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 2.3. The profile bounds remain valid on this subsequence. Suppose a smooth \chi touches V from above at a finite point x_0, and suppose

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 2.3.

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 \D_{r_1}(x_0) and supported in \D_{r_2}(x_0). Its derivatives are supported in the annulus between these two disks. Set

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

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

Lemma 2.3 gives

Cancel D_j before comparing exponential rates. Squaring and multiplying by the common normalization constant c_j yields

The weighted bounds of Lemma 2.1, together with \log S_j/S_j\to0, now give the contradiction

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 2.4. 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 2.4. 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 (2.5) 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 2.1 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 2.2 applies with B_j=0, K_j=k_j, and D_j=S_j, since

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

with the same conclusion when the integral is zero. This contradicts the open lower bound in Equation (2.1), whose right-hand side is \sup_GV=0. The argument proves both positivity of the local norm and Equation (2.5). It uses only local isothermal charts and therefore imposes no orientability assumption.□

Proof. Upper semicontinuity supplies a point x_*\in\overline{\D_1} with V(x_*)\ge-1. The disk \D_{10}(x_*) contains \D_8 and is contained in \D_R. In particular, V is nontrivial and locally integrable. Since V\le0, the submean inequality gives

Its distributional Laplacian is a nonnegative Radon measure. Choose a fixed nonnegative smooth cutoff \psi, supported in \D_8 and equal to one on \D_5. If \eta=(2\pi)^{-1}\Delta V, then

Let \nu be the restriction of \eta to \D_4, and set P(x)=\int_{\D_4}\log\abs{x-z}\,d\nu(z). The elementary bound

shows that \norm{P}_{L^1(\D_3)}\le C\nu(\D_4). The logarithmic fundamental solution identity \Delta\log\abs{x-z}=2\pi\delta_z implies \Delta(V-P)=0 on \D_3. Thus V-P agrees almost everywhere with a harmonic function h. Interior harmonic estimates, applied to its bounded L^1(\D_3) norm, bound \nabla h and \nabla^2h on \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{\D_{R\rho}(y)} lie in \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

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,

Combining the near and far bounds proves (3.3).

For completeness, fix any z\in\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,

The same annuli, for the truncated gradient kernel, give

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 3.3, so large that, with \rho=A^{-1},

Choose C_1 strictly larger than the uniform bound for \av_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:

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

No restriction on \abs{B_j}/S_j has been imposed.

Normalize the tilted measures on \D_R by setting

Lemma 2.1 and Theorem 2.2 give, after passage to a subsequence, a subharmonic profile V, possibly identically -\infty, on \D_R. It satisfies V\le0, while

The compact upper bound in (2.1), applied to \overline{\D_1}, therefore gives \sup_{\overline{\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 3.3 for this limiting profile. On the corresponding child q_j, set

In parent coordinates the additional weight is e^{-2S_jc_y\cdot X}. To see the cancellation of recentering constants, write

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

Apply the weighted compact upper bound to the numerator, using \overline{\D_{R\rho}(y)}, and the weighted open lower bound to the denominator. This yields

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 3.3 gives

Since K_{q_j}=\rho K_j, it follows, for all sufficiently large j, that

At the same time,

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

Thus S\ge C_0, E(Q,b)\le S, and S\ge Ks. Equation (3.2) gives

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

Using S\ge Ks, we obtain the other required smallness condition,

All the hypotheses of Proposition 3.4 hold. Its conclusions, with S/K=m(Q,b)+f_Q, are precisely (4.1).

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,

The geometric-series bound gives (4.2).□

Proof. There are only finitely many nodes. At each node from which descent continues, choose one of the assignments furnished by Lemma 4.1. 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

The event \{i<T\} is determined before the next child is selected. Applying (4.1) on this event and summing over 0\le i<d gives

All the m_i are nonnegative by Definition 3.1. It follows that M\le 2m_0+F_I. The tilt estimate in (4.1) similarly yields

Their sum is at most

Since s(I)\le1, this is bounded by the right-hand side of (4.3) 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

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 \D_R, where K_j=K_{Q_j}, \norm{p_j}_\infty\le C, and \norm{\nabla p_j}_\infty\le Cs_j. Set

The hypotheses of Theorem 2.2 follow from

Normalize the tilted measures

to have mass one on \D_R. By Lemma 2.1 and Theorem 2.2, a subsequence has a subharmonic profile V\le0, with the identically minus infinity case initially allowed. The definitions give

The upper bound in (2.1) on the compact set \overline{\D_1} therefore implies \sup_{\overline{\D_1}}V=0. In particular the profile is nontrivial and attains its maximum at an interior point of \D_R. The strong maximum principle yields V=0 on \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

For any fixed 1<r<R, bound the first integral from below by its restriction to \D_r. The weighted form of Lemma 2.1, applied from below on the open disk \D_r and from above on the compact disk \overline{\D_1}, gives

Letting r\uparrow R gives a lower bound of R-1, without any claim about mass at the boundary of \D_R. On the other hand,

a contradiction.□

Proof. Let B_*,\eta_* be the thresholds from Lemma 4.3. Choose \gamma so that

Let C_{\mathrm{st}} be the constant from Lemma 4.2. Choose

Write B_0=\abs{b_I}. Starting at I, descend using Lemma 4.1 while

and stop at the first failure of either inequality or at the terminal level, whichever comes first. These inequalities hold initially by (4.5). Whenever they hold, we have

Thus every required step is available and Lemma 4.3 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 4.2 and Markov’s inequality give

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.□

Proof. Immediately before an entry the path is inactive. An inactive square cannot have N>H, since that would start an excursion, so N(Q^p)\le H. Apply Proposition 3.4 at Q^p with b=0 and S=HK_{Q^p}+C_0. Indeed, E(Q^p,0)\le HK_{Q^p}\le S, and

since s(Q^p)\le1. Thus (3.4) holds when H is sufficiently large. All terms in the averages in (3.5) are nonnegative. There are A^2 children, so the tilt b_Q assigned to the particular child Q satisfies

Combining these inequalities with (3.2) and K_{Q^p}\ge a_0 proves the upper bound in (5.1).□

Proof. The contradiction argument below extracts a nonconstant profile, finds a nearby square with small affine excess relative to a nonzero slope, and transfers that square to Proposition 4.4.

We may require H to exceed any fixed threshold needed above. Suppose there is no triple (H,\delta,J) with the asserted uniform property. Choose H_j\to\infty, all above these fixed thresholds and at least 2, and apply this negation with \delta_j=1/H_j and J_j=\lceil H_j\rceil. For each j there are an eigenfunction, a finite tree, and an entry Q_j for which (5.2) fails with these parameters.

There must be more than J_j levels strictly below Q_j. Otherwise the finer-entry count would be at most J_j everywhere, and Y_{Q_j}=Q_j would satisfy (5.2). In particular, every fixed number of further levels is eventually available.

Rescale about the center x_j of the parent Q_j^p, using its side s_j=s(Q_j^p) as unit length. Write k_j for the eigenfrequency of u_j, and put

The rescaled equation is \Delta U_j+K_j^2p_jU_j=0. Its coefficients satisfy \|p_j\|_\infty\le C and \|\nabla p_j\|_\infty\le Cs_j. Moreover,

Normalize |U_j|^2\,d\xi to have mass one on \D_R. Lemma 2.1 and Theorem 2.2, with zero tilt, give a subharmonic profile V on \D_R satisfying

For the second assertion, the parent inequality gives E(Q_j^p,0)/S_j\le1, and the compact upper bound in (2.1) applies to the closed unit disk. Thus V is not identically -\infty. Its supremum on every nonempty open disk in \D_R is therefore finite.

There are only A^2 possible positions of the entry inside its parent. After a subsequence its rescaled center is a fixed point y_e. The entry inequality and the compact upper and open lower bounds in (2.1) imply

The finiteness of the second supremum justifies the subtraction of these bounds. Since A>10R, both disks are contained in \D_2. Consequently V is nonconstant on \D_2.

The representation in Lemma 3.2 implies V\in W^{1,1}(\D_2): the gradient of the logarithmic kernel is locally integrable in two dimensions, and the representing measure has finite mass. There is a Lebesgue point z\in\D_2 of \nabla V for which c=\nabla V(z)\ne0. Indeed, otherwise the weak gradient would vanish almost everywhere on the connected disk, forcing V to be constant almost everywhere; its subharmonic representative, which is determined by local averages, would then be constant everywhere.

Extend the fine subdivision grids in the rescaled parent coordinates to the whole plane. For r=A^{-m}, choose a grid square with center y_m such that |y_m-z|\le r. These grids have edges -\tfrac12+nA^{-m} in each coordinate and are the same for every j. We claim that

To prove this, let W(x)=V(x)-c\cdot x, and let a_r be its average on B_r=\D_{(2R+2)r}(z). For large m these disks are contained in \D_2. Poincare’s inequality and the Lebesgue-point property give

If x\in\overline{\D_{Rr}(y_m)}, the disk \D_{(R+1)r}(x) lies in B_r. The submean inequality for the subharmonic function W therefore gives, uniformly in such x,

On the other hand,

This proves (5.3). Subharmonicity is essential here: it turns the average control into a uniform upper bound.

Now fix m\ge2 sufficiently large that Rr\le1/4 and the quotient of the left side of (5.3) by r|c| is strictly less than \sigma. In the original coordinates let I_j be the square with center x_j+s_jy_m and side s_jr, and assign it the tilt b_{I_j}=H_jc. The additional weight in parent coordinates is e^{-2S_jc\cdot\xi}, up to a constant canceled in the norm ratio. The weighted bounds of Lemma 2.1 therefore yield

The factor is exactly 1/(r|c|), since K_{I_j}=rK_j and S_j=H_jK_j. Because c\ne0, we also have |H_jc|\ge P for all large j. Thus (4.5) holds at I_j.

We verify that Proposition 4.4 can be used on the whole subtree of I_j down to the original bottom. The chosen m is fixed before j tends to infinity, so that level is eventually available. All squares of a given extended-grid level have the same physical scale as the original tree, and hence have K\ge a_0. The center y_m stays in \D_3, and Rr\le1/4. The disks required for I_j and all its descendants are consequently contained in \D_{5s_j}(x_j): even the root margin has rescaled radius 2Rr, and |y_m|+2Rr<3+1/2<5. Since Q_j^p\subset Q_0, we have |x_j-x_{Q_0}|\le s_0/\sqrt2 and s_j\le s_0, so this disk lies in \D_{6s_0}(x_{Q_0})\subset\D_{2Rs_0}(x_{Q_0}), within the fixed coordinate patch. This also covers the larger disk margins required when regarding I_j as the root of a subtree.

Proposition 4.4 now gives a measurable subset of I_j of area at least |I_j|/2 on which every visited square from I_j through the bottom has N>L. The extended grids align with the original root: a root edge differs from the corresponding parent edge by an integer multiple of s_j, and the extended parent grid has rescaled edges -1/2+nA^{-m}. Thus a fine grid square is either contained in Q_0 or has interior disjoint from it. In the latter case its points belong to no entry of the original tree. In the former case its descendants are exactly those of the original tree. From the level of I_j onward there can be at most one entry along any of these paths, since two starts require an intervening value N\le L. Before that level there are at most m relevant levels below the tested entry. Thus the number of strictly finer entries is at most m+1 on this subset.

Finally, the subset lies in 20A Q_j: in parent units that dilated square has half-side 10, whereas y_m\in\D_3 and the entry center lies in the unit parent square. Its area relative to the entry is at least

Both this positive number and m+1 are fixed independently of j. They eventually satisfy the tested requirements \delta_j=1/H_j and J_j=\lceil H_j\rceil, contradicting the choice of Q_j. This proves the proposition.□

Proof. Ignore grid boundaries and the boundaries of the finitely many dilates DQ with Q\in\mathcal F; these form a null set. Define

Squares of a fixed rank n have disjoint interiors. At each rank, process the squares in decreasing order of side length, retaining a square if its D-dilate has interior disjoint from the previously retained dilates. Let \mathcal S be the union of these selected families. Every discarded square Q has its dilate meeting that of a selected square P with s(P)\ge s(Q); hence Q\subset(2D+1)P. Covering the original disjoint squares at that rank by these larger dilates and then summing over ranks gives

The selected witness sets at a single rank are disjoint up to null sets. It remains to control overlap between different ranks. For a point y outside the ignored boundaries, put

If a witness point y lies inside its square Q, every larger square containing y is an ancestor of Q, so the witness assumption makes n_{\mathrm{tot}}(y) differ from n(Q) by at most J. An exterior witness point can also disagree with the ancestry of Q at larger scales. For Q\in\mathcal S and y\in Y_Q, let

count the mismatches between membership of y in a larger square and that square’s being an ancestor of Q. If b is the number of larger squares in \mathcal F containing y, then

At most one square of side s(Q) contains y, and the witness assumption bounds the number of smaller squares containing it by J. Decomposing n_{\mathrm{tot}}(y) into these three size ranges gives

Thus a uniform bound for d_Q(y) restricts the ranks that can contribute at y to a fixed interval. We next show that the total witness area weighted by d_Q is controlled:

where B depends only on A,D.

Fix F\in\mathcal F and a finer side length t=A^{-\ell}s(F), \ell\ge1. The grid is laminar: each square Q of side t is either contained in F or interior-disjoint from it. If it can contribute a mismatch, a point of DQ lies on the opposite side of \partial F from Q. It follows that the whole square Q lies in the collar

where \dist_\infty is distance for the maximum norm on \R^2. Figure 1 shows the two possible directions of this mismatch near one face of F. This collar has area at most

All squares of side t, across all ranks, have disjoint interiors. Since |Y_Q|\le D^2|Q|, their total contribution for this fixed F and t is at most D^2 times that collar area. Summing the geometric series over \ell\ge1 and then summing over F proves (5.7). For example, one may take

Choose an integer M\ge1 with M\ge2B/\alpha, and set Z_Q=\{y\in Y_Q:d_Q(y)\le M\}. Remove also the null sets where the witness assumption fails. Markov’s inequality and (5.7) give

For y\in Z_Q, Equation (5.6) now gives

Only 2(M+J+1)+1 ranks can therefore contribute at y. At a given rank the selected witness sets are disjoint, so this is also a bound on the total overlap of the Z_Q.

For D\ge1, the inclusions Q\subset Q_0 imply DQ\subset DQ_0. Integrating the overlap bound over this common region yields

This proves (5.5).□

Proof. Fix the threshold H supplied by Proposition 5.3. A terminal path on which no excursion is active has terminal growth N\le H. Every other terminal path has exactly one final active entry. For an entry Q, let \Omega_Q\subset Q be the union of terminal squares on which that entry’s excursion is still active. These sets are disjoint as Q varies.

Starting separately at each entry Q, set b_Q=0 and perform the descent of Lemma 4.1, stopping at the first square with N\le L or at the bottom, whichever comes first. Every continuing square has N>L, so the descent is available regardless of its current tilt. The low-growth stopping square, if any, is included as the last produced value; no further step is needed there. These separate choices of tilts do not affect excursion membership, which is defined entirely by the untilted quantity N.

Along a random descendant path from Q, denote the stopping time by T and the successive values by m_i,b_i. Since b_0=0, (3.2) gives

The uniform distribution on paths agrees with normalized planar area on Q. Lemma 4.2 and Lemma 5.2 therefore imply

Here N_{\mathrm{term}} is constant on each terminal square, with value N of that square. The first inequality merely restricts a nonnegative integral to the paths still active at the bottom; it does not condition on that event or divide by its probability.

Sum over all entries and use (5.5), then include the inactive terminal paths. This gives

Equal terminal areas turn this integral bound into the first inequality of (5.8). The second follows because every terminal square has K_q=ks_{\mathrm{term}} and E(q,0)=K_qN(q).□

Proof. Let x_I be the center of I, and put

The disk of physical radius 180s is available because R=1000. Choose \eps so small that (60\eps)^2\norm{p}_\infty<\eps_0 uniformly over the coordinate regions in use. Then F satisfies Theorem 6.1.

An interior H^2 estimate for \Delta F=-qF, followed by the two-dimensional Sobolev embedding, gives

Only the uniform bound on q is needed here. On the smaller disk,

Consequently

The same L^2 version of the local length bound is recorded directly in [12, Theorem 3.1.1].

In physical coordinates the two norm radii are 165s and 15s. Their common change-of-variables factor cancels, and nesting gives

All these norms are nonzero by Proposition 2.5. The length disk becomes \D_s(x_I) and contains the closed square I, whose corner distance is s/\sqrt2. Hausdorff length scales by 60s. Theorem 6.1 therefore yields (6.1).□

Proof of Theorem 1.1. Choose finitely many smooth isothermal neighborhoods and root squares Q_0 whose interiors cover M. Make their sides s_0\le1 small enough that each closed disk of radius 2Rs_0 about a root center is contained in its coordinate neighborhood. The conformal factors and their first derivatives have uniform bounds on the finitely many regions used below, and the conformal factors are bounded away from zero there. Proposition 2.5 gives a uniform initial bound N(Q_0)\le C_{\mathrm{init}}.

Choose the subdivision constants from Proposition 3.4, and then the small \eps of Lemma 6.2. Write k_{\min}=\sqrt{\lambda_1(M,g)}>0. A single permissible lower scale bound for all roots is

Now fix the descent and persistence constants with this a_0. This order of choices makes every constant depend only on the fixed metric and cover.

For a given eigenfunction, subdivide each root to the first level at which ks\le\eps, where s is that level’s common side. If this level is below the root, its parent has kAs>\eps, so ks>\eps/A. If it is the root, then ks\ge k_{\min}s_0. In both cases, every visited square satisfies K_Q\ge a_0. Theorem 5.5 therefore gives

There are (s_0/s)^2 terminal squares. Applying Lemma 6.2 and summing their lengths yields

Subadditivity suffices even if a nodal arc lies on a grid edge.

On each convex coordinate square, the inverse coordinate map is Lipschitz into (M,d_g) with constant at most \sqrt{\sup p}: the coordinate line segment is an admissible path for the Riemannian distance. Hence the preceding Euclidean Hausdorff-length bound gives the same bound for \mathcal H_g^1 up to a fixed factor. Sum over the finite root cover to obtain \mathcal H_g^1(Z_u)\le C(M,g)k. The coordinate choices and all constants were independent of the eigenfunction and of its eigenspace. Since k=\sqrt\lambda, this proves the theorem.□