18.42 Localization on curvesB3DCF0B463CE

Proof. We compare $B_ m$ with the model kernels of section 18.41 on a shrinking disk, which gives the upper bound. For the lower bound, we cut off model polynomials concentrated near $x$ and correct them to global sections by an $L^2$ estimate. Put $z_ m=z(x_ m)$, $\rho _ m=|z_ m|$, $R_ m=(q_ m/(rm))^{1/(2r)}$ and

Then $R_ m\to 0$ and $\rho _ m/R_ m\to 1$, and $T_ m>0$ for all sufficiently large $m$. Positive constants denoted by $M$, with subscripts when needed, may change between estimates. They are independent of $m$ and of the argument of $z_ m$; they may depend on the fixed geometric data, $r$, $\tau $, and fixed auxiliary cutoffs or radii. We write $q=q_ m$ when no confusion is possible. The model densities $B_{m,r}^{(q)}$ are those of section 18.41.

Step 1: the kernel on a shrinking disk. Fix a real $A>1$ and set $D_ m=\{ |z|<A R_ m\} $. On $D_ m$,

The first limit follows from $q_ mR_ m=r^{-1/(2r)}(q_ m^{2r+1}/m)^{1/(2r)}\to 0$. Hence, on $D_ m$, the norm defined by $h^ m$ and $\nu $ and the radial norm with weight $\mathrm{e}^{-m|z|^{2r}}\, \mathrm{d}A$ differ by a factor $1+o(1)$, uniformly over all holomorphic functions on $D_ m$. The same comparison holds for the corresponding pointwise weights at $z_ m$.

For integers $k\geq 0$, let $I_{m,k}$ and $J_{m,k}$ be the squared norms of $z^ k$ on $D_ m$ and on the whole plane, respectively, with this radial weight. Polar integration gives

Here $b,U>0$. By Parseval's identity on circles and Tonelli's theorem, the normalized monomials form a complete orthonormal system in the space of holomorphic functions on $D_ m$ with finite radial norm: the squared norm of a Taylor series is the sum of the squared norms of its monomials, so its partial sums converge in that norm. Therefore the density of the subspace of functions vanishing to order at least $q$ is

Point evaluation is continuous by the local submean inequality, so this sum is also the supremum of $\mathrm{e}^{-T_ m}|f(z_ m)|^2$ over the unit ball of that subspace.

Choose fixed real numbers $1<c<d<A^{2r}$ and put $k_ m=\lfloor rcT_ m\rfloor $. For large $m$, $k_ m\geq q$ and $U_ m\geq dT_ m$. Uniformly for $0\leq k\leq k_ m+1$,

for some fixed $\eta >0$. Indeed, let $G_ b$ be a random variable with Gamma density $t^{b-1}\mathrm{e}^{-t}/\Gamma (b)$ on $(0,\infty )$. A change of variable gives $\mathbb E\mathrm{e}^{sG_ b}=(1-s)^{-b}$ for $0<s<1$. With $b=(k+1)/r\leq cT_ m+O_ r(1)$, Markov's inequality yields

Since $d>c$, choose $s>0$ small enough that $sd+c\log (1-s)>0$. The right side is at most $M\mathrm{e}^{-\eta T_ m}$, as required.

The Cauchy–Schwarz inequality for the positive radial measure gives $I_{m,k+1}^2\leq I_{m,k}I_{m,k+2}$, so $I_{m,k+1}/I_{m,k}$ increases with $k$. By (18.42.3) and the asymptotics of Gamma ratios,

Choose $\kappa \in (c^{-1/r},1)$. For all large $m$, the ratio of consecutive terms $\rho _ m^{2k}/I_{m,k}$ is at most $\kappa $ for every $k\geq k_ m$. Consequently,

For the last equality, replace $I_{m,k_ m}$ by $J_{m,k_ m}$ using (18.42.3). The resulting term is one of the terms of the series for $B_{m,r}^{(k_ m)}(z_ m)$. In (18.41.7) for this kernel, each residue threshold divided by $T_ m$ tends to $c>1$, so Chebyshev's inequality makes all the Poisson tails tend to zero, and the comparison error is $O(T_ m^{-1/2})$. Hence $B_{m,r}^{(k_ m)}(z_ m)=o(S_ m)$.

For the terms with $q\leq k\leq k_ m$, we use (18.42.3). Since $I_{m,k}\leq J_{m,k}$ for every $k$, we obtain

By proposition 18.41.13,

The global norm of a selected section dominates its norm on $D_ m$. Restriction to $D_ m$ and the preceding comparisons of norms therefore give $B_ m(x_ m)\leq (1+o(1))B_ m^ D(z_ m)$, also when $V_ m=0$. This proves (18.42.2).

Step 2: polynomials with almost all their norm near the point. From now on we assume the curvature hypotheses of the last assertion. Fix a real $c>1$ and put $K_ m=\lfloor cq_ m\rfloor $. Define

For large $m$ these are well defined, $\| p_ m\| _{m,r}=1$, and

Here the first equality holds because $p_ m$ is the normalized evaluation vector at $z_ m$ of the span of $z^{q_ m},\ldots ,z^{K_ m}$. The omitted kernel $B_{m,r}^{(K_ m+1)}(z_ m)$ is $o(S_ m)$ by (18.41.7), since its Poisson thresholds divided by $T_ m$ tend to $c>1$; the second equality then follows from proposition 18.41.13.

Choose $\delta \in (0,1/2)$ so that the coordinate disk $|z|<\delta $ satisfies $\phi (z)\geq \beta |z|^{2r}$ for a fixed $\beta \in (1/2,1)$, and such that $a$ is bounded above there. For a sufficiently large fixed $A>1$,

To see this, write $p_ m(z)=\sum _{k=q_ m}^{K_ m}b_{m,k}z^ k$, so that $\sum _ k|b_{m,k}|^2J_{m,k}=1$. The lower bound for $\phi $, the orthogonality of monomials on circles and polar integration bound the left side by a fixed constant times

The exponential moment of $G_ b$ at $s=1/2$ gives, for $b,U>0$,

Choose $A$ so large that $\beta A^{2r}/2>c\log (2/\beta )+1$. Since $(k+1)/r\leq (cq_ m+1)/r$, the last factor in the sum is $O(\mathrm{e}^{-q_ m/r})$, uniformly in $k$. This proves (18.42.6). On $|z|<A R_ m$, the norm defined by $h^ m$ and $\nu $ and the radial model norm differ by a factor $1+o(1)$, as shown in Step 1, and the model norm of $p_ m$ on this disk is at most $1$.

Fix a smooth cutoff function $\chi $ with $0\leq \chi \leq 1$, compactly supported in $|z|<\delta $ and equal to one near zero, and extend $\chi p_ m e^{\otimes m}$ by zero to $C$. The preceding estimates give

The numbers $\delta ,\beta ,A$ and the cutoff $\chi $ are fixed before $m$ tends to infinity.

Step 3: a global correction with the same vanishing condition. Let $f_ m=\bar\partial (\chi p_ m e^{\otimes m})$. Its support is contained in a fixed annulus on which $\phi $ has a positive lower bound. The Cauchy–Schwarz inequality and the identity $\sum _ k|b_{m,k}|^2J_{m,k}=1$ give, for $|z|<\delta <1$,

Here we used that $\Gamma (b)$ has a positive lower bound for $b\geq 1/r$. Since $q_ m\log m=o(m)$, the factor $\mathrm{e}^{-m\phi }$ on the annulus dominates this growth of the coefficients.

Regard $\nu $ as a Kähler form on the curve; it defines pointwise norms of forms and the induced metric $h_{K_ C^{-1}}$ on the anticanonical bundle $K_ C^{-1}$. Choose a smooth cutoff function $\eta $ with $0\leq \eta \leq 1$, supported in the coordinate chart and equal to one near $x$; shrink its support if necessary. For a sufficiently small fixed $t_0>0$, the global smooth functions

extended by zero outside the chart, are nonpositive. Their complex Hessians are nonnegative where $\eta =1$, and elsewhere they are bounded below independently of $t$, because $\eta =1$ near $x$.

Choose a fixed smooth real function $\psi $ that equals a sufficiently large positive multiple of $|z|^2$ near $x$ and is cut off inside the chart, so that

on a neighborhood of $x$ where $\eta =1$. Here $\mathrm{i}\Theta $ denotes the Chern curvature form, without the factor $1/(2\pi )$, so that the weight $\mathrm{e}^{-\psi }$ contributes $\mathrm{i}\partial \bar\partial \psi $. For all large $m$ and all $0<t\leq t_0$, equip $E_ m=L^ m\otimes K_ C^{-1}$ with the smooth metric

This metric satisfies

Indeed, near $x$ the two terms involving $h$ and $\ell _ t$ are nonnegative, and the other two together are at least $2\nu $. On the complement of a fixed such neighborhood, compactness and strict positivity give a lower bound $c_0m\nu $ for $m\mathrm{i}\Theta (L,h)$, with $c_0>0$. Since $q_ m=o(m)$, this term absorbs all remaining negative terms, which are bounded by $M(q_ m+1)\nu $.

On the support of $f_ m$, the functions $\ell _ t$ are bounded below uniformly in $t$. The polynomial estimate, the bounded derivatives of $\chi $, and $q_ m\log m=o(m)$ therefore give a constant $a_0>0$ such that, for all large $m$,

On a curve, $K_ C=\Lambda ^{1,0}$, so an $L^ m$-valued $(0,1)$-form is the same as an $E_ m$-valued $(1,1)$-form, and the induced metrics give it the same pointwise norm; the anticanonical factor in $E_ m$, whose curvature enters (18.42.8), accounts for this change of bidegree. In bidegree $(1,1)$ on a curve, (18.42.8) makes the curvature operator at least the identity. Since the curve is compact and every $(1,1)$-form on it is $\bar\partial $-closed, the $L^2$ existence theorem for complete Kähler manifolds [Theorem 5.1, DemAM11] applies. It gives an $L^ m$-valued weak solution $u_{m,t}$ with

Now fix $m$ and let $t\downarrow 0$. Since $\ell _ t\leq 0$ and $\psi $ is bounded, the solutions $u_{m,t}$ are bounded in the original $L^2$ norm, and a subsequence converges weakly to some $u_ m$. Continuity of $\bar\partial $ in the sense of distributions gives $\bar\partial u_ m=f_ m$. For fixed $t'>0$ and $t<t'$, we have $\ell _ t\leq \ell _{t'}$, so the same bound holds in the smooth weighted space with weight $\mathrm{e}^{-\psi -q_ m\ell _{t'}}$, whose norm is equivalent to the original one for this fixed $m$ and $t'$. Weak lower semicontinuity, followed by monotone convergence as $t'\downarrow 0$, gives

In particular, $\| u_ m\| _ m^2\leq M\mathrm{e}^{-a_0m}$. Near $x$ the right side $f_ m$ vanishes, so the Cauchy–Riemann equation, in the sense of distributions, makes $u_ m$ holomorphic in the frame $e^{\otimes m}$. The smooth factors in the weight are bounded above and below near $x$ for each fixed $m$, so the coefficient of $u_ m$ is square integrable against $|z|^{-2q_ m}$. By lemma 5.37.1 in dimension one, this forces vanishing order at least $q_ m$ at $x$.

Since $p_ m$ also vanishes to order at least $q_ m$ at zero, it follows that $s_ m=\chi p_ m e^{\otimes m}-u_ m$ is a global holomorphic section in $V_ m$. By (18.42.7), $\| s_ m\| _ m\leq 1+o(1)$.

Step 4: evaluation at the moving point. For large $m$, the disk $D_ m$ lies in the region where $\chi =1$ and $f_ m=0$. Apply the local comparison of norms and the extremal property of $B_ m^ D$ to the holomorphic function representing $u_ m$ on $D_ m$, which vanishes to order at least $q_ m$. By (18.42.4),

The pointwise factor $\mathrm{e}^{-m\phi (z_ m)}$ differs from $\mathrm{e}^{-T_ m}$ by a factor $1+o(1)$. Since $\Phi (\tau )>0$, (18.42.5) and the triangle inequality give

In particular, these sections are nonzero for large $m$. The extremal characterization of $B_ m$ and $\| s_ m\| _ m\leq 1+o(1)$ yield the matching lower bound, which proves the limit. $\square$

Proof. Fix $k\geq 3$ and $0<\alpha <1$. Let $A^{k,\alpha }$ be the real Banach space of functions that are holomorphic on $\mathbb D$, of class $C^{k,\alpha }$ on its closure, and real at zero. For $h\in A^{k,\alpha }$ near zero set $f(u)=u(1+h(u))$. For small $h$, the map $f$ is univalent, with $f(0)=0$ and $f'(0)>0$. On $|u|=1$ define

where $P_-$ is the projection onto the strictly negative Fourier modes and $[\cdot ]_0$ is the constant Fourier coefficient.

The Cauchy projections are bounded on these Hölder spaces. This follows from the Hilbert kernel on the circle: subtract the value at the evaluation point, split the integral at twice the length of the increment, and use the Hölder bound for differences on the inner part and the bound for the differentiated kernel on the outer part. This gives the $C^{0,\alpha }$ bound, and commuting angular derivatives with the projections gives the $C^{k,\alpha }$ bound. The superposition operator $h\mapsto (Q_ R)_\zeta \circ f$ involves no derivatives of $h$ and is smooth. Thus $T$ is a smooth map into the product of the space of negative-mode functions with $\mathbb R$. At $(R,h)=(0,0)$ we have $T=0$ and

This is a bounded real-linear isomorphism, so the implicit function theorem in Banach spaces gives a unique normalized branch $f_ R$ of solutions of $T(R,h)=0$, depending smoothly on $R$ in the fixed space, with $\| f_ R-\mathrm{id}\| =O(R)$. The branches obtained for larger $k$ agree with this one by uniqueness in the smaller space. For each fixed order, this gives the spatial estimates and the expansions in $R$ needed below, on a parameter interval that depends on the order.

The equation $T=0$ prescribes only the real part of $[K]_0$; we show that the imaginary part vanishes as well. Since $P_-K=0$, the function $K$ extends holomorphically to $\mathbb D$, and $K(0)=[K]_0$ is purely imaginary. On the other hand, Stokes' formula gives

On the circle the integrand equals $(r+K)f_ R'/f_ R$, whose residue at zero is $r+K(0)$. Hence $K(0)$ is real as well, and therefore zero. Consequently

with $H_ R$ holomorphic on $\Omega _ R=f_ R(\mathbb D)$, and $S_ R=(Q_ R)_\zeta $ on the boundary of $\Omega _ R$. Let

Its complex gradient agrees with that of $Q_ R$ on the boundary, so the difference $Q_ R-U_ R$ is constant there; choose the constant $C_ R$ to make it zero. Then the difference and its gradient vanish on the boundary. Since the tangential second derivative of the difference also vanishes there, its inward normal second derivative equals $\Delta Q_ R>0$. Hence $Q_ R-U_ R>0$ in an inner collar of uniform width. At $R=0$,

By uniform convergence, this gap keeps a positive minimum on the remaining compact subsets of the interior for small $R$, and the logarithmic term controls a neighborhood of zero.

Set $U_ R=Q_ R$ outside $\Omega _ R$. Since values and gradients match across the boundary, the Laplacian of the patched function carries no measure on the boundary. Its curvature is $r\delta _0$ inside $\Omega _ R$ and $\mathrm{dd}^{\mathrm{c}}Q_ R\geq 0$ outside, so it is an admissible minorant. If $v$ is any competitor, then $v-U_ R$ is subharmonic on $\Omega _ R\setminus \{ 0\} $, bounded above near zero and nonpositive on the boundary. Removability of the singularity at zero and the maximum principle give $v\leq U_ R$ in $\Omega _ R$, and outside $\Omega _ R$ this follows from $v\leq Q_ R$. This identifies the envelope and shows that there are no further noncontact components. The radial graph property and the estimates for the harmonic part follow from the normalized map and its dependence on the parameter. We henceforth write $F_ R=f_ R$. $\square$

Proof. Take $b_ R=|z|^2/R^2$ on a disk containing $D_ t$, and cut it off before the radius $3R$, so that its negative curvature is supported outside $D_ t$. This function is uniformly bounded and its Hessian is $O(R^{-2})$. On the support of its negative curvature, the curvature $m\rho $ is comparable to $q/R^2$; since $q\to \infty $, it absorbs the negative part. Inside $D_ t$, the function $b_ R$ contributes the curvature $c/R^2$. Elsewhere in a fixed coordinate disk, the original curvature is bounded below by a fixed multiple of $m|z|^{2r-2}$, and off that disk it is at least $cm$. Since $mR^2\to \infty $, all these bounds dominate $c/R^2$ and the bounded Ricci term of $\nu $.

We use the $L^2$ estimate for $\bar\partial $ with a pointwise lower bound for the curvature of a singular metric, in the form described in section 25.16. As in the proof of theorem 18.42.1, we regard $L^ m$-valued $(0,1)$-forms as $(1,1)$-forms with values in $L^ m\otimes K_ C^{-1}$ and sections of $L^ m$ as $(1,0)$-forms with values in the same bundle; this identification preserves the pointwise norms and brings in the Ricci term of $\nu $. The estimate then reads as follows: if $H$ is a singular metric on $L^ m$ and the curvature of $L^ m\otimes K_ C^{-1}$ is at least $\lambda \nu $ for a continuous function $\lambda >0$, then

On the support of $\alpha $ we have $\lambda ^{-1}=O(R^2/q)$, $\phi _ t=\phi $, and $b_ R$ is bounded. Elsewhere $\phi _ t\leq \phi $ gives $\mathrm{e}^{-m\phi }\leq C\mathrm{e}^{-\psi _ m}$. These comparisons yield (18.42.14). Choosing the solution of minimal norm in the singular weighted Hilbert space makes the correction linear in its source.

The estimate requires a continuous lower bound $\lambda $. We take $\lambda $ equal to $c/R^2$ inside $D_ t$ and to $cq/R^2$ on the exterior annulus, interpolating across a thin exterior collar of $\Gamma _ t$. For the cutoff sources used below, whose supports are separated from $\Gamma _ t$, the collar can be chosen to miss the source. For an exterior source whose support accumulates on $\Gamma _ t$, we let the collar shrink to $\Gamma _ t$; since $\Gamma _ t$ has zero area, dominated convergence in the integrals of $\lambda ^{-1}|\alpha |^2$ and weak compactness in the fixed weighted Hilbert space give the same estimate. The singular metric is handled by approximation, which gives the same estimate for a weak solution. Where the source vanishes, the Cauchy–Riemann equation in the sense of distributions makes the solution holomorphic. Near $x$, the weight is $|z|^{-2q}$ times a smooth positive factor. A holomorphic term $z^ j$ is square integrable against it precisely when $j\geq q$, so the solution has the required vanishing order. $\square$

Proof. We pass to logarithmic coordinates adapted to $\Gamma _ t$, in which the norm of a section of order $q$ becomes Gaussian in the distance to the boundary. Restriction to a thin ring gives the upper bound, and localized trial sections corrected by lemma 18.42.13 give the lower bound. Finally we compare with the full kernel to obtain saturation. Throughout the proof, $c$ and $C$ denote positive constants independent of $m$, which may change from line to line.

Step 1: the effective potential. Define on the unit disk

where the logarithm uses the branch near one. Then $A_ R=1/2+O(R)$ and $F_ R=w+O(R)$ in every fixed norm used below. The effective potential $\Psi _ R=(Q_ R-U_ R)\circ F_ R$ is positive inside the disk, and its value and gradient vanish on the boundary. For $w=\mathrm{e}^{v+\mathrm{i}\theta }$ it has the expansion

and the exact quadratic coefficient is $2(Q_ R)_{\zeta \bar\zeta }(F_ R(\mathrm{e}^{\mathrm{i}\theta }))|F_ R'(\mathrm{e}^{\mathrm{i}\theta })|^2$.

The maps need not extend holomorphically across the circle. To obtain charts that also cover a neighborhood outside the circle, we truncate their Taylor series at the degree $K=\lfloor q^{1/16}\rfloor $ and denote the truncations by $F_ K,A_ K$. The Taylor coefficients of a function bounded in a sufficiently high fixed Hölder norm decay faster than a large fixed power of their index, so the truncation error is $O(q^{-10})$, together with the fixed number of derivatives needed here. Fix a small $c_0>0$. On $|w|\leq 1+c_0/K$ the derivatives of the truncations remain bounded and $F_ K-w,A_ K-1/2=O(R)$, because the factor $(1+c_0/K)^ K\leq \mathrm{e}^{c_0}$ controls the sums. In particular, $F_ K$ is univalent there for small $R$. Put $\delta =c_0/K$ and

The errors in the value and the first derivative of $\Psi _ K$ on the circle are $O(q^{-10})$, and uniformly for $|v|\leq \delta $,

Inside the disk, the effective potential has uniform positive quadratic growth near the circle, a positive lower bound on compact annuli farther inside, and logarithmic growth at zero.

In the coordinate $z=RF_ K(w)$, a local holomorphic section of order at least $q$ at $x$ is uniquely represented, in the original frame, as

with $f$ holomorphic near zero. The volume in this chart is $R^2(1+O(R))\, \mathrm{d}A_ w$. Put $\eta =\sqrt{rq}\, v$. On each band $|\eta |\leq N$ with $N$ fixed,

Step 2: the upper bound. Let the observation point have coordinate $\eta _0$ in a fixed bounded interval. Restricting the norm to the band $|\eta |\leq N$ bounds the kernel from above by the kernel of the restricted space. For $f(w)=\sum _{j\geq 0}b_ jw^ j$, the limiting diagonal norms of the monomials on this band are

After normalization by $rq/(\pi R^2)$, the limiting upper kernel is

The passage from sums to integrals uses Riemann sums with mesh $1/\sqrt{rq}$. To control their tails, bound $I_ N(s)$ from below by its integral over a fixed subinterval of $[-N,N]$ lying beyond every observation point $\eta _0$; this gives an exponential bound uniform in the mesh. Thus we may first pass to the Riemann integral and then let $N\to \infty $. Since $I_\infty (s)=\sqrt{\pi /2}\, \mathrm{e}^{s^2/2}$, the expression (18.42.20) tends to

Step 3: the lower bound. Fix $T>0$, and use (18.42.19) with $f$ in the span of the monomials $w^ j$, $0\leq j\leq T\sqrt{rq}$. For $f=\sum _ jb_ jw^ j$, put $\| s\| _{\mathrm{trial}}^2=\sum _ j|b_ j|^2n_ j$, with $n_ j$ given by (18.42.23) below. Let $\chi $ be a radial cutoff function in the $w$-plane, supported in $|w|\leq 1+\delta $ and equal to one on $|w|\leq 1+\delta /2$. The annulus on which $\chi $ is not constant lies outside $D_ t$. Equations (18.42.18) and (18.42.19) bound the squared norm of the source, as a quadratic form on the entire span, by

the linear term $2jv$ in the exponent is absorbed because $\sqrt q\, \delta \to \infty $. Lemma 18.42.13 gives a correction $g_ s$, depending linearly on $s$, with $\bar\partial g_ s=\bar\partial (\chi s)$ and

The constants here depend on $m$ only through $R$ and $q$. The factor $R^2$ in (18.42.14) cancels the factor $R^{-2}$ coming from the derivative of the cutoff; this cancellation is needed even when $q$ grows more slowly than every power of $m$.

On this span, the Gram form of the global norm of $\chi s$ converges, in the sense of relative operator norm, to the diagonal form

Indeed, the ratio of the weights is uniformly close to one on each fixed band in $\eta $. Outside such a band, the radial bound for the weight and Parseval's identity on circles bound arbitrary combinations by diagonal tails $\mathrm{e}^{-c\eta ^2+2T|\eta |}$. We first pass to the limit on a fixed band and then enlarge it. The positive lower bound of the effective potential on compact annuli and its logarithmic singularity at zero control the remaining part of the disk. This argument controls the whole quadratic form, without a factor depending on the dimension of the span.

To evaluate the corrections we need a pointwise bound. Let $u$ be a section holomorphic on a disk of radius $R/\sqrt q$ centered at a point $z$ with $|z|$ comparable to $R$. Then a local submean estimate gives

To prove it, multiply $u$ by the exponential of a holomorphic polynomial that removes the harmonic part of the linear and quadratic Taylor terms of $m\phi $ at $z$. After rescaling the disk to unit size, the remaining quadratic coefficient is bounded and the third-order error is $O(q^{-1/2})$, which proves the estimate. It applies to the corrections, because their sources lie on the cutoff annulus, away from the observation points. Together with (18.42.22), it shows that the kernel of the corrected global trial sections differs by $o(q/R^2)$ from the local trial kernel. After normalization, the lower limit is the integral in (18.42.21) with upper limit $T$ instead of infinity, and letting $T\to \infty $ gives the matching lower bound.

Step 4: return to the original coordinate. Let $p$ be the nearest boundary point and $\mathrm{e}^{\mathrm{i}\theta }$ its preimage under $F_ R$. Then

Moreover, $m\rho (z)/a(z)=(rq/R^2)(1+o(1))$ in the band $|d_ t(z)|\leq MR/\sqrt q$. These identities convert $2\eta _0$ into the argument in (18.42.16) and give the normalization of the density there. This proves (18.42.16).

Step 5: the full density and saturation. Let $B_ m^{\rm full}$ be the density of the full space $H^0(C,L^ m)$. At points with $|z|\asymp R$, it satisfies

uniformly in an $o(R)$-neighborhood of $\Gamma _ t$. We give a direct localization argument, which also yields the saturation statements of the theorem. Remove the harmonic Taylor terms of $m\phi $ at the observation point and rescale by $(m\rho (z))^{-1/2}$. On every fixed rescaled disk, the remaining weight tends to $|u|^2$, where $u$ is the rescaled coordinate, and the volume density tends to the constant $a(z)$. The Taylor estimate is uniform, because the third derivatives of $\phi $ at radius $R$ are $O(R^{2r-3})$, so the normalized third-order error is $O(q^{-1/2})$. Restriction to such disks bounds the evaluation norm from above, and the Gaussian norm of the constant function on them tends to $\pi $ as their radius grows. This proves the upper bound in (18.42.25).

For the lower bound, take the holomorphic exponential canceling the harmonic Taylor terms, and cut it off on a disk of radius $\varepsilon R$ centered at the observation point, where $\varepsilon \to 0$ and $\sqrt q\, \varepsilon \to \infty $. Its local squared norm is $(\pi a(z)/(m\rho (z)))(1+o(1))$ times its squared weighted value at the center, which is one. The quadratic lower bound for the weight bounds the source on the cutoff annulus by an exponential factor $\mathrm{e}^{-cq\varepsilon ^2}$. Add the same bounded function of scale $R$ to the original metric and use the pointwise $\bar\partial $ estimate. The relative squared norm of the correction is bounded by $C(q\varepsilon ^2)^{-1}\mathrm{e}^{-cq\varepsilon ^2}=o(1)$, after decreasing $c$ if necessary, and its value at the center is $o(1)$ by (18.42.24). This proves the matching lower bound.

For saturation inside $D_ t$, use the polynomial chart above and put $u=1-|w|>0$. Restrict the norm in (18.42.19) to the exterior half-band $0\leq v\leq c/\sqrt q$. On this half-band every monomial $w^ j$, $j\geq 0$, has norm at least $cR^2/\sqrt q$. Parseval's identity on circles and the Cauchy–Schwarz inequality give

If $u\sqrt q\to \infty $, this bound divided by $q/R^2$ tends to zero. The error $O(q^{-10})$ of the chart changes neither the hypothesis nor the conclusion.

For saturation outside $D_ t$, cut off the localized Gaussian section at $z$ on a disk of radius

This disk lies entirely outside $D_ t$, and again $\varepsilon \to 0$ and $\sqrt q\, \varepsilon \to \infty $. Apply lemma 18.42.13 with the singular obstacle metric, which equals the original metric on the support of the source. The correction has negligible norm and value, and it vanishes to order $q$ at $x$. Thus a member of $V_{m,q}$ asymptotically attains the full Gaussian density, and (18.42.25) gives the reverse inequality. This proves saturation to one. $\square$

Proof. By lemma 18.42.10, $c_ R-1=O(R)$ in every fixed norm required here. The observation points in (18.42.28) satisfy

Since the slope of the boundary graph tends to zero, the argument of $\Phi $ in theorem 18.42.15 is $\tau +o(1)$ whenever $\sqrt q R\to 0$, and this condition is equivalent to $q=o(m^{1/(r+1)})$. Moreover, $a(z)=1+o(1)$ and $\rho (z)=r^2|z|^{2r-2}(1+o(1))$ at these points, so the two normalizations agree. Thus theorem 18.42.15 proves the law for all data in this range.

The same reasoning shows that (18.42.29) is sufficient. For necessity, suppose that (18.42.29) fails. On a subsequence, choose an angle realizing the maximal boundary displacement and observe at $|z|=R$ with $\tau =0$. A nonzero finite subsequential limit of the scaled displacement gives a nonzero Gaussian argument, and an infinite one gives saturation to zero or one. Radial and normal distances have relative error tending to zero, because the graph has vanishing slope. In either case the limit differs from $1/2$.

It remains to construct the compact examples. On $\mathbb P^1$ choose a radial metric on $\mathcal O(d)$ whose local potential $f(u)$, $u=|w|^2$, equals $u^ r$ near zero, such that $g(u)=uf'(u)$ is strictly increasing on $(0,\infty )$ and tends to $d$, and such that $g(u)=d-\epsilon /u$ for large $u$, with a constant $\epsilon >0$. Such a smooth interpolation exists for a sufficiently large integer $d$: prescribe a positive function $g'$, equal to $r^2u^{r-1}$ near zero and to $\epsilon /u^2$ near infinity, with total integral $d$. Then $f=d\log u+\mathrm{constant}+\epsilon /u$ near infinity, so the potential in the frame at infinity is smooth with strictly positive curvature. Choose a smooth radial volume form equal to $\mathrm{d}A_ w$ near zero.

For sufficiently small $t$, the hole $D_ t$ lies in the region where $f(u)=u^ r$ and equals the disk $|w|<R$: rotational invariance and uniqueness make it a disk, and its curvature mass is $rR^{2r}=t$. Hence the transition and saturation statements of theorem 18.42.15 apply to this compact example.

Now use instead the coordinate $z$ defined by $w=z+cz^2$, with a constant $c>0$. In this coordinate the same compact metric satisfies $\phi (z)=|z|^{2r}+O(|z|^{2r+1})$, the volume form still has $a(0)=1$, and vanishing orders are unchanged. At the uncorrected observation point $z=R$ with $\tau =0$,

and the normalizing factors computed in the two coordinates have ratio tending to one. If $q$ is not $o(m^{1/(r+1)})$, there is a subsequence on which $\sqrt q R$ tends to a positive finite number or to infinity. Theorem 18.42.15, together with its saturation statement when the argument diverges, then gives a limit strictly larger than $1/2$. This proves that the range is sharp. $\square$