Lemma 23.1.253F868B74CEBD

Proof. We first derive an exact formula for $Q[W]$ on $[0,b]$, then bound its negative part separately on $[0,a^{-1}]$ and on $[a^{-1},b]$, integrate these bounds, and finally examine the ends of the radial line.

Step 1: the Schur complement on $[0,b]$. Write $F=F_\eta $ for the heat potential. On $[0,b]$ we have $q\geq 1$, hence $\vartheta (q)=1$ and $W=F(t(x),w)$. Set

The heat equation gives $F_ t=h-\eta $, and the chain rule gives

Substituting these in $Q[W]$, using $h_ t=h((\log k_ t)_{w\bar w}+|\partial _{\bar w}\log k_ t|^2)$, and completing the square, we obtain the exact identity

By (23.1.14), the part $\sigma ^2th[1+t(\log k_ t)_{w\bar w}]$ is nonnegative.

Step 2: the initial interval $[0,a^{-1}]$. Here

Moreover $q_0-1/2\leq q\leq q_0$ and $t\leq \varepsilon \mathrm{e}^{q_0}$, and the latter tends to zero faster than any power of $\eta $. In particular,

At each point where $\chi (ax)>0$, apply lemma 23.1.19 with $\kappa \chi (ax)^2$ in place of $\kappa $; its constant is uniform in $x$. The positive square in (23.1.27) then absorbs the negative term $-\kappa \chi ^2tk_ t$ coming from $\sigma _ xtF_ t$. The remaining part of $\sigma _ xtF_ t$, which involves $\chi '$, is at least $-\eta $, because $F_ t\geq -\eta $. The other negative terms in (23.1.27) are $O(\eta )$, and so is the exponential error of lemma 23.1.19, since $\sigma ^2t\leq \eta ^2$. Thus $d_\eta \leq C\eta $ on this interval. At points where $\chi (ax)=0$, the smooth nonnegative function $\chi $ has a minimum, so $\chi '(ax)=0$ and directly

since $\mu \leq \mu (a^{-1})<1-\eta $ for small $\eta $.

Step 3: the interval $[a^{-1},b]$. Here the formulas simplify to

For $t\leq 1$, lemma 23.1.19 and (23.1.27) give

Indeed, $-\eta \sigma ^2t=-2\kappa q^2t$ and $\sigma _ xtF_ t=-\kappa tk_ t+\kappa t$; these are the two terms for which the scale was chosen. When $t\geq 1$, we insert the exponential derivative estimates (23.1.16) directly in the definition of $Q$. The denominator stays above $1-\eta $, and the chain rule gives

The perturbing derivatives are bounded by constants times $\eta \mathrm{e}^{-t/4}$, $\eta \sigma t \mathrm{e}^{-t/4}$ and $\eta [(|\sigma _ x|+\sigma ^2)t+\sigma ^2t^2]\mathrm{e}^{-t/4}$. Since $q,t\geq 1$ here, each is bounded by a constant times the right side of (23.1.30). Since $1+W_{w\bar w}\geq 1-\eta $, subtracting the two quotients gives

The square of the mixed derivative is at most $C\kappa \eta q^2t^2\mathrm{e}^{-t/2}$, so it satisfies the same bound. Since $Q[0]>0$, this proves (23.1.30).

Step 4: integration of the negative part. The constant $2\eta $ and the initial interval contribute $O(\eta b)=o(1)$, by (23.1.23). In the small-time region, $q\geq \sqrt L$ and $t=\eta ^{-1}\mathrm{e}^{-q^2}$. Therefore

For the exponential term the integral is at most

The interval $[\sqrt L,2\sqrt L]$ has length $O(\sqrt L)$. For $q\geq 2\sqrt L$ and sufficiently small $\eta $, $\eta ^2\mathrm{e}^{q^2}/q^2\geq \mathrm{e}^{q^2/4}$. The integrand on this remaining interval is bounded by $\exp (-\mathrm{e}^{q^2/4}/(C\kappa ))$, which is integrable on $[1,+\infty )$. The prefactor is $O(\sqrt\eta )$, and the initial interval contributes $O(\sqrt{\eta L})$, so the full expression is $o(1)$. In the large-time region, (23.1.28) gives

This proves the integral assertion on $[0,b]$.

Step 5: the radial slope. The negative part of $W_ x$ is uniformly small as well. For $t\leq 1$, we have $F_ t\geq -\eta $ and $0\leq \sigma \leq 2aq$, so

The last inequality follows because $t\sqrt{L+\log (1/t)}$ is increasing on $(0,1]$ when $L>1$. For $t\geq 1$ and $q\geq 1$, use $|W_ x|\leq C\sqrt\eta \, qt \mathrm{e}^{-t/4}$ and $q\leq \sqrt L$.

Step 6: the ends. It remains to consider the two ends, including the cutoff region. If $b\leq x\leq x_*$, then $0\leq q\leq 1$ and $1/(\mathrm{e}\eta )\leq t\leq 1/\eta $. The derivatives of the cutoffs and scale have at most polynomial growth in $\eta ^{-1}$. Thus (23.1.16) implies

But $Q[0]=\mu (1-\mu )\geq \mathrm{e}^{-x_*}/4$ on this interval, and (23.1.23) shows that the error is $o(\mathrm{e}^{-x_*})$. Hence $Q[W]>0$ there. For $x\geq x_*$, $W=0$ and positivity is immediate. For $x\leq 0$, $W_ x=0$ and

since $\mu \leq 1/2$. On $[-2,-1]$, the bounds $1/9\leq \mu \leq 3/10$ and $1-\eta \geq 9/10$ give $Q[W]\geq 2/27$. Finally, $Q[W]$ is continuous, with positive denominator, so the maximum of its negative part over the compact fiber $E$ is continuous in $x$. This proves the assertions about support and continuity, and completes the proof. $\square$