23.1.50 Heat concentration at large depth92674EE5B61B

Proof. For the upper bound use $k_ r\geq 0$ on $(0,1]$ and (23.1.16). The nonnegative Fourier coefficients imply $k_ r(w)\leq k_ r(0)$, and Gaussian summation gives $k_ r(0)\leq 2/r+C r^{-1}\mathrm e^{-1/(2r)}$ for $r\leq 1$. Its integrable error proves the lower bound. On the indicated annulus the nonzero-lattice terms and their derivatives are exponentially small relative to the zero-lattice term, proving the last two identities. For the first, split the defining integral at a fixed small $r_0$. On $[t,r_0]$ the nonzero-lattice contributions have bounded integral, and the difference between $(2/r)\mathrm e^{-|w|^2/r}$ and $2/r$ has integral at most $2|w|^2/t\leq 8$. The remaining integral is uniformly bounded. $\square$

Proof. Fix $\kappa $ as in lemma 23.1.19, also with $\kappa <1/100$. Choose a small fixed $c>0$ later and set

Use the cutoffs $\chi $ and $\vartheta $ from the earlier heat-time construction. Then $Z_\eta =F_\eta (\varepsilon ,\cdot )$ for $x\leq 0$, $Z_\eta =0$ for $x\geq x_*$, and

The larger value of $q_0$ changes the integrated positivity defect; we estimate it at this scale. Write $r_-=\max (-r,0)$ for the negative part of a real number $r$.

On $[0,b]$ we have $\vartheta =1$. Write $F=F_\eta $ and put $\sigma =(\log t)_ x$, $h=\eta k_ t$ and $\zeta =\sigma t\partial _{\bar w}\log k_ t/\mu $. The identity (23.1.27) applies to $Z_\eta $ on this interval. The contribution involving $h[1+t(\log k_ t)_{w\bar w}]$ is nonnegative. On $[0,a^{-1}]$,

This upper bound for $t$ is smaller than every power of $\eta $. Thus $\kappa q\chi 't\leq \eta $ and $\sigma ^2t\leq \eta ^2$. Use (23.1.20) with $\kappa \chi ^2$ where $\chi >0$, and $F_ t\geq -\eta $ for the $\chi '$ term. It follows that $[Q[Z_\eta ]]_-\leq C\eta $. Where $\chi =0$ also $\chi '=0$, and $Q[Z_\eta ]=\mu -\mu ^2/(1-\eta +h)>0$, since $x\leq a^{-1}=o(1)$.

On $[a^{-1},b]$, $\sigma =2aq$, $\sigma _ x=-\kappa /\eta $, $\mathrm dx=-\mathrm dq/a$. For $t\leq 1$, (23.1.27) and (23.1.20) give

For $t\geq 1$, (23.1.16) gives instead $[Q[Z_\eta ]]_-\leq C\kappa q^2t^2\mathrm e^{-t/4}$. Put $L=\log (1/\eta )$. Integrating the $q^2t$ term uses $\int _{\sqrt L}^\infty q^2\mathrm e^{-q^2}\, \mathrm dq\leq C\sqrt L\eta $. For the exponential term split at $2\sqrt L$: the initial interval has length $O(\sqrt L)$, and on the rest $\eta ^2\mathrm e^{q^2}/q^2\geq \mathrm e^{q^2/4}$ for small $\eta $. Its remaining integral is uniformly bounded even with upper endpoint infinity. For large time use $\mathrm dx=\sqrt{\eta /(2\kappa )}\, \mathrm dt/(tq)$, $q\leq \sqrt L$, and the integrability of $t\mathrm e^{-t/4}$. Including the constant term and initial interval yields

All constants here are independent of $q_0,c,\eta $ with fixed $\kappa $. The bound is small after choosing $c$, although it need not tend to zero.

Also $\sup [(Z_\eta )_ x]_-=O(\sqrt{\eta L})$: for $t\leq 1$ use $F_ t\geq -\eta $ and $t\sqrt{L+\log (1/t)}\leq \sqrt L$; for $t\geq 1$ use (23.1.16). On $[b,x_*]$, $0\leq q\leq 1$ and $(\mathrm e\eta )^{-1}\leq t\leq \eta ^{-1}$, so differentiated cutoffs and (23.1.16) give

Choose $c$ so that $\ell <1/(10\mathrm e)$. Since $Q[0]=\mu (1-\mu )\geq \mathrm e^{-x_*}/4$ here, this implies $Q[Z_\eta ]>0$ on the cutoff interval. For $x\leq 0$ use $Q[Z_\eta ]\geq \mu (1-\mu /(1-\eta ))>0$, and for $x\geq x_*$ use $Z_\eta =0$. On $[-2,-1]$ the same formula gives $Q[Z_\eta ]\geq 2/27$.

The continuous defect $d_\eta (x)=\max _ w[Q[Z_\eta ]]_-$ is therefore supported in $[0,b]$. Choose a smooth nonnegative $r_\eta $, supported in $(-1/2,b+1/2)$, with $r_\eta >d_\eta $ on $[0,b]$ and $m_\eta :=\int r_\eta \leq \int d_\eta +\eta $. Uniform approximation and a positive bump give this choice for each fixed $\eta $; no uniform derivative bound is needed. Fix a smooth probability bump $\psi $ supported in $(-2,-1)$ and put

The support order implies $-m_\eta \leq g_\eta '\leq 0$; $g_\eta =0$ at the negative end and is constant at the positive end. Furthermore $g_\eta \leq 0$ and $|g_\eta |\leq m_\eta (b+3)=O_ c(\eta ^{-1})$. Shrink the fixed $c$ so that (23.1.53) ensures

for all sufficiently small $\eta $. Now $Q[V_\eta ]=Q[Z_\eta ]+r_\eta -m_\eta \psi >0$ everywhere: use the majorant on $[0,b]$, the margin $2/27$ on $(-2,-1)$, and the established positivity elsewhere. The vertical coefficient is above $1-\eta $. For the horizontal coefficient, derivatives vanish for $x\leq -2$; on $[-2,0]$ use $\mu \geq \mu (-2)>m_\eta $; on $[0,\infty )$ use $\mu \geq 1/2$ and $(Z_\eta )_ x\geq -o(1)$. Thus (23.1.8) gives a smooth globally strictly positive potential.

On $x<-3$, $V_\eta =F_\eta (\varepsilon ,\cdot )$ has fiber mean zero, so $0\leq \sup _ XV_\eta \leq C\eta $ by lemma 23.1.51 and $g_\eta \leq 0$. Set $u_\eta =V_\eta -\sup _ XV_\eta $. Where the cutoff is nonzero $t\geq \varepsilon $, giving

Multiplication by a cutoff between zero and one preserves this lower bound for $Z_\eta $, including where $F_\eta $ is positive.

Mass and depth on the concentration region. Let $K_\eta $ consist of all radial angular directions and

using the small lift near zero in $E$. This lies in the linear transport region, with $\vartheta =1$, $x\asymp _ c\eta ^{-1}\to \infty $, $\sigma =2aq\asymp _ c\eta ^{-2}$, and superexponentially small $t$. Lemma 23.1.51 gives, uniformly,

In (23.1.27) the first $h$ contribution is nonnegative, $\sigma _ xtF_ t=-\kappa t(k_ t-1)\geq -3\kappa $, and the completed square tends to one. Its other negative terms tend to zero. Thus $Q[Z_\eta ]\geq 1/4$ for small $\eta $. On the core $\psi =0$ and $g_\eta ''=r_\eta \geq 0$, so $Q[V_\eta ]\geq 1/4$. Also

Indeed $(Z_\eta )_ x=\eta \sigma (tk_ t-t)$ and $\mu +g_\eta '>0$ here; the vertical inequality follows from $1-\eta +\eta k_ t$.

In (23.1.43), $\Delta _{V_\eta }=(1+(V_\eta )_{w\bar w})Q[V_\eta ]$. The fiber annulus has area $3t/2$. Integrating the preceding lower bounds and using $\sigma \, \mathrm dx=-2q\, \mathrm dq$ yields

The annular value estimate also gives

As $\eta q_0^2=c\eta ^{-2}+\eta \log (1/\eta )$, this is bounded above and below by fixed positive multiples of $\eta ^{-2}$ on the core. Together with (23.1.54), this proves the asserted mass and depth bounds. All these potentials are bounded: truncations beyond their individual depth equal them everywhere, so the defining extended product equals (23.1.43). $\square$

Proof. On $K_\eta $, monotonicity of $W$ gives

Every sufficiently large $T$ equals $B\eta ^{-2}$ for some sufficiently small $\eta $, proving the necessary bound. Also $K_\eta \subset \{ -u_\eta >A\eta ^{-2}/2\} $, so uniform deep tails force this lower bound to tend to zero. $\square$

Proof. First select the correction with additional local control. Put

Then $t(x_{\rm pre})=\eta ^3$, $x_{\rm pre}\to \infty $, and $b-x_{\rm pre}=O(\sqrt{\eta L})\to 0$. For a fixed large $K$ the preceding defect estimates give $d_\eta \leq (K/2)\eta $ on $[0,x_{\rm pre}]$: on the linear portion $q^2t\leq 4L\eta ^3$, and the exponential error is $O(\exp (-1/(C\kappa \eta ^2L)))=o(\eta )$. On the initial portion use the previously proved $C\eta $ bound. The integral of $d_\eta $ on $[x_{\rm pre},b]$ tends to zero by the same integrated estimates and this interval's length.

Choose a fixed left cutoff, zero for $x\leq -1/2$ and one for $x\geq -1/4$, and a translated right cutoff, one for $x\leq x_{\rm pre}$ and zero for $x\geq x_{\rm pre}+1/4$. Let $r^0_\eta $ be $K\eta $ times their product. It strictly exceeds $d_\eta $ on $[0,x_{\rm pre}]$. The continuous function $e_\eta =\max (d_\eta -r^0_\eta ,0)$ is supported in $[x_{\rm pre},b]$. Add a smooth nonnegative majorant $s_\eta >e_\eta $ on that interval, supported in $(x_{\rm pre}-1/4,b+1/2)$, with $\int s_\eta \leq \int e_\eta +\eta $. Set $r_\eta =r^0_\eta +s_\eta $. Then

After shrinking the fixed $c$, this is bounded by a fixed $m_0$ within all the positivity tolerances above. The conclusions of theorem 23.1.52 therefore hold for this choice, with fixed $A,B,D$. On every fixed compact interval $I\Subset (0,\infty )$, for sufficiently small $\eta $ we have $r_\eta =K\eta $ and exactly

Here $b_0,b_1$ depend only on the fixed left cutoff. Globally $0\leq -g_\eta (x)\leq m_0(x+2)$ for $x\geq 0$.

Let $G(w)=-\int _0^\infty (k_ r(w)-1)\, \mathrm dr$ for $w\ne 0$ on $E$. Gaussian integration gives the uniform bound

Indeed the nearest Gaussian has integral bounded by $C+2\log ^+(1/\operatorname {dist}_ E(w,0)^2)$ on $(0,1]$; the other lattice terms have bounded integral and large times decay. The same bound applies after the cutoff. On a fixed $I\Subset (0,\infty )$, $\vartheta =1$ for small $\eta $ and $t_\eta (x)\leq \exp (-c/(2\eta ^3))$. Every fixed $x$ derivative of $t_\eta $ has this bound times a polynomial in $\eta ^{-1}$. On compact fiber sets away from zero,

To verify derivatives, subtract $G$ from $G_ t=F_\eta (t,\cdot )/\eta $: $G_ t-G=\int _0^ t(k_ r-1)\, \mathrm dr=-t+O(\exp (-d^2/(Ct)))$ when $d$ stays positive. Differentiated Gaussian errors have only polynomial factors in $t^{-1}$, and repeated $x$ differentiation preserves their decay. Together with (23.1.57) and $0\leq \sup V_\eta \leq C\eta $, this gives uniform $C^\ell $ bounds for $u_\eta $ on each such fixed compact product set, for all sufficiently small $\eta $. The threshold may depend on the set and derivative order.

Fix $a_ j=2^{-j-2}$, $j\geq 1$, so $\sum a_ j=1/4$. Choose scales $\eta _ j\downarrow 0$ recursively and write $u_ j=u_{\eta _ j}$, $K_ j=K_{\eta _ j}$. Require the minimum $x$ on $K_ j$ to exceed every earlier $x_{*,i}$ by two. Choose a relatively compact product neighborhood $O_ j$ of $K_ j$, with positive $x$ interval and fiber closure away from zero, on which the earlier $u_ i$ are constant. The $K_ j$ are pairwise disjoint. With $E_ j=\sum _{i<j}a_ i\| u_ i\| _\infty $, impose

For each earlier $O_ i$, require the preceding uniform derivative bounds to hold for $u_ j$ through order $j$. All are finite constraints at stage $j$, achievable by making $\eta _ j$ smaller. They ensure that the future series $\sum _{k>i}a_ ku_ k$ converges in $C^\infty $ on compact subsets of $O_ i$: for every fixed derivative order the sufficiently late terms have a uniform bound there, and $\sum a_ k<\infty $.

These constraints also give $\sum _{k>j}a_ k\eta _ k\leq a_ j\eta _ j$. On $K_ j$ the coordinate and distance estimates are

Using (23.1.58), the global slope bound on $g_ k$, and the supremum constants, the future depth is at most

The last term is at most a fixed multiple of $a_ j\eta _ j^{-2}$ since $\eta _ j\leq a_ j$. Including the earlier and current terms, there is a fixed $C_*>0$, independent of the scales, such that

Fix this constant before the scale choice. Since $\sqrt T W(T)$ is unbounded, at each stage choose arbitrarily large $T_ j$ such that

and so small a scale that all finite preceding constraints hold. Such values occur arbitrarily far out because $W$ is finite on compact intervals. Thus the recursive choice is possible.

The smooth nonpositive partial sums $v_ N=\sum _{j\leq N}a_ ju_ j$ decrease, and

By theorem 5.25.2, $\int _ X(-v_ N)\, \mathrm dV\leq C\sum _ j a_ j<+\infty $. Thus corollary 5.25.3 gives a $\omega $-plurisubharmonic limit $v$ with $\omega +\mathrm{dd}^{\mathrm{c}}v\geq 3\omega /4$. Its supremum is finite and nonpositive.

On $O_ j$ the earlier terms are constant and the future series converges smoothly. Thus $v$ is smooth there and $\omega +\mathrm{dd}^{\mathrm{c}}v\geq a_ j(\omega +\mathrm{dd}^{\mathrm{c}}u_ j)$ pointwise. The determinant inequality for positive Hermitian matrices gives on $K_ j$

On these sets the extended product is the smooth product: on a compact neighborhood of $K_ j$ all sufficiently deep canonical truncations equal $v$. Set $U=v-\sup _ Xv$. Since $\sup v\leq 0$, $-U\leq -v$, and its local currents are unchanged. Monotonicity of $W$ and (23.1.59) give

The disjoint $K_ j$ prove infinite total mass for this single potential. They lie in its smooth locus, so no mass is assigned to poles. $\square$