18.42.12 Gaussian transitions at curvature boundaries77CA2588A10E

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. $\square$