$\begin{split}\frac{d}{dr}\left(\frac{1}{r^{6}}\int_{\\{t=-r^{2}\\}}(u-p_{2})\zeta\,Q\,G\right)&=\frac{1}{r^{4}}\int_{\\{t=-1\\}}\big{\\{}\text{\small\bf
H}\big{(}(u-p_{2})\zeta\big{)}_{r}\,Q-\big{(}(u-p_{2})\zeta\big{)}_{r}\,\text{\small\bf
H}Q\big{\\}}\,G\\\ &\geq\frac{1}{r^{4}}\int_{\\{t=-1\\}}\text{\small\bf
H}\big{(}(u-p_{2})\zeta\big{)}_{r}\,Q\,G\\\ &=\frac{1}{r^{7}}\int_{\\{x\in
B_{1/2},\ t=-r^{2}\\}}r^{2}\chi_{\\{u=0\\}}\,Q\,G+O(e^{-1/r}).\end{split}$

Note that $Q(x)\leq C_{1}|x_{n}||x|^{2}$ with
$C_{1}:=C_{n}\|Q\|_{L^{2}(\mathscr{C}_{1})}$ (see (9.1)). Also, since
$\lambda^{2nd}=3$, it follows by Proposition 6.6 that

$\\{u=0\\}\cap\\{|x|<1/2,t=-r^{2}\\}\subset\\{|x_{n}|\leq
C_{2}(|x^{\prime}|+r)^{2}\\},$

where $C_{2}$ depends only on $n$ and $\|u\|_{L^{\infty}}$. Hence,

$\begin{split}\frac{1}{r^{7}}\int_{\\{x\in B_{1/2},\
t=-r^{2}\\}}r^{2}\chi_{\\{u=0\\}}\,Q\,G\geq-\frac{C_{1}}{r^{5}}\int_{\\{x\in
B_{1/2},\ t=-r^{2}\\}}\chi_{\\{|x_{n}|\leq
C_{2}(|x^{\prime}|+r)^{2}\\}}|x_{n}||x|^{2}G\geq-C,\end{split}$

where the constant $C$ depends only on $n$ and $\|u\|_{L^{\infty}}$, and the
lemma follows. ∎

As a consequence, we find:

######  Corollary 9.4.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), and
assume that $(0,0)\in\Sigma_{n-1}^{=3}$. Then the limit

$\lim_{r\downarrow 0}\frac{(u-p_{2})_{r}}{r^{3}}\qquad\mbox{in }\,W^{1,2}_{\rm
loc}\big{(}\mathbb{R}^{n}\times(-\infty,0]\big{)}$

exists in the weak topology, and it is of the form (9.1).

###### Proof.

Given any sequence $r_{k}\downarrow 0$, by Proposition 6.7 and Corollary 8.5
there exists subsequence such that

$r_{k_{\ell}}^{-3}(u-p_{2})_{r_{k_{\ell}}}\rightharpoonup Q\qquad\mbox{in
}\,W^{1,2}_{\rm loc}\big{(}\mathbb{R}^{n}\times(-\infty,0]\big{)},$

where $Q$ is some 3-homogeneous solution of (6.17). Now, assume that we have
two limits along different sequences:

$\big{(}r_{k}^{(1)}\big{)}^{-3}w_{r_{k}^{(1)}}\rightarrow
Q^{(1)}\qquad\mbox{and}\qquad\big{(}r_{k}^{(1)}\big{)}^{-3}w_{r_{k}^{(2)}}\rightarrow
Q^{(2)}.$ (9.3)

Then, by Lemma 9.3, for fixed $j\in\\{1,2\\}$ we have that

$\left(\frac{1}{r^{6}}\int_{\\{t=-1\\}}r^{-3}\big{(}(u-p_{2})\zeta\big{)}Q^{(j)}G\right)+C\|Q\|_{L^{2}(\mathscr{C}_{1})}r$

is nondecreasing in $r$. Hence, using (9.3), and letting $r_{k}^{(j)}\to 0$,
we deduce that

$\int_{\\{t=-1\\}}Q^{(1)}Q^{(j)}G=\int_{\\{t=-1\\}}Q^{(2)}Q^{(j)}G\quad\text{for
$j=1,2$}\qquad\Rightarrow\qquad\int_{\\{t=-1\\}}(Q^{(1)}-Q^{(2)})^{2}G=0.$

This implies that $Q^{(1)}\equiv Q^{(2)}$ on $\\{t=-1\\}$ and hence, by
homogeneity, in all of $\mathbb{R}^{n}\times(-\infty,0]$. ∎

We now investigate the structure of the second blow-ups at at “most points” of
$\Sigma_{n-1}^{=3}$. For this, we need a new dimension reduction lemma.

######  Lemma 9.5.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), and let
$(0,0)\in\Sigma_{n-1}^{=3}$. Assume there exists a sequence of singular points
$(x_{k},t_{k})\in\Sigma_{n-1}^{=3}$, with $|x_{k}|\leq r_{k}$ for some
$r_{k}\downarrow 0$, and $t_{k}\leq 0$. Assume also that
$\tilde{w}_{r_{k}}\rightarrow q$, where $w:=u-p_{2}$, and that
$y_{k}:=\frac{x_{k}}{r_{k}}\to y_{\infty}\neq 0$, and let $q^{even}$ denote
the even symmetrisation of $q$ with respect to the hyperplane $\\{p_{2}=0\\}$.

Then $y_{\infty}\in\\{p_{2}=0\\}$, and $q^{even}(x^{\prime},x_{n})$ is
translation invariant in the direction of $y_{\infty}$.

###### Proof.

Without loss of generality we assume $p_{2}=\frac{1}{2}x_{n}^{2}$. Since
$(0,0)\in\Sigma\subset\\{u=0\\}$, it follows from Lemma 8.3 applied at
$(x_{k},t_{k})$ that

$(0,0)\not\in B_{r}(x_{k})\times[t_{k}+Cr^{2},1)\qquad\forall\,r\in(0,1/2),$

where $C$ is independent of $k$. As a consequence, choosing $r=r_{k}$ and
recalling that $|x_{k}|\leq r_{k}$, we get

$-Cr_{k}^{2}\leq-C|x_{k}|^{2}\leq t_{k}\leq 0.$ (9.4)

Let $s_{k}:=\frac{t_{k}}{r_{k}^{2}}$. Up to taking a subsequence we may assume
that $(y_{k},s_{k})\to(y_{\infty},s_{\infty})$, where $-C\leq s_{\infty}\leq
0$. Also, by Lemma 7.5 we obtain that $y_{\infty}\in\\{p_{2}=0\\}$.

Next, define
$P_{k}:=r_{k}^{2}\big{(}p_{2}(y_{k}+\cdot)-p_{2,x_{k},t_{k}}\big{)}$ (note
that this is a harmonic polynomial of degree two), and set

$\begin{split}w_{r_{k}}(y_{k}+\cdot,s_{k}+\cdot)+P_{k}&=u(x_{k}+r_{k}\cdot,t_{k}+r_{k}^{2}\,\cdot\,)-r_{k}^{2}p_{2}(y_{k}+\cdot)+r_{k}^{2}\big{(}p_{2}(y_{k}+\cdot)-p_{2,x_{k},t_{k}}\big{)}\\\
&=u(x_{k}+r_{k}\cdot,t_{k}+r_{k}^{2}\,\cdot\,)-p_{2,x_{k},t_{k}}(r_{k}\,\cdot\,)=:\bar{w}_{k}\end{split}$

Since $(x_{k},t_{k})\in\Sigma_{n-1}^{=3}$ we have $\phi(0^{+}\bar{w}_{k})=3$
and therefore, by Lemma 5.6,

$c\bigg{(}\frac{\varrho^{\prime}}{\varrho}\bigg{)}^{\negmedspace
6}\leq\frac{H(\varrho^{\prime},\bar{w}_{k}\zeta(r_{k}\cdot))}{H(\varrho,\bar{w}_{k}\zeta(r_{k}\cdot))}\leq
C_{\delta}\bigg{(}\frac{\varrho^{\prime}}{\varrho}\bigg{)}^{\negmedspace
6+\delta}\qquad\negmedspace\negmedspace\mbox{and}\qquad
c\bigg{(}\frac{\varrho^{\prime}}{\varrho}\bigg{)}^{\negmedspace
6}\leq\frac{H(\varrho^{\prime},w_{r_{k}}\zeta(r_{k}\cdot))}{H(\varrho,w_{r_{k}}\zeta(r_{k}\cdot))}\leq
C_{\delta}\bigg{(}\frac{\varrho^{\prime}}{\varrho}\bigg{)}^{\negmedspace
6+\delta},$ (9.5)

whenever $0<\varrho<\varrho^{\prime}\ll r_{k}^{-1}$. Also, by Lemma 6.3, for
all $0<\varrho\ll r_{k}^{-1}$ we have

$H(\varrho,\bar{w}_{k}\zeta(r_{k}\cdot))\asymp{\mathchoice{\kern
4.30554pt\vbox{\hrule height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern
4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\mathscr{C}_{\varrho}}(\bar{w}_{k})^{2}\qquad\mbox{and}\qquad
H(\varrho,w_{r_{k}}\zeta(r_{k}\cdot))\asymp{\mathchoice{\kern
4.30554pt\vbox{\hrule height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern
4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\mathscr{C}_{\varrho}}(w_{r_{k}})^{2},$
(9.6)

where $X\asymp Y$ here means $X\leq CY$ and $Y\leq CX$ with $C$ depending only
on $n$ and $\|u\|_{L^{\infty}}$.

Now, recall $0\geq s_{k}\to s_{\infty}>-\infty$ and $y_{k}\to
y_{\infty}\in\overline{B}_{1}\cap\\{p_{2}=0\\}$, define
$a_{k}:=H(1,w_{r_{k}}\zeta(r_{k}\cdot))^{1/2}$ and
$b_{k}:=\|P_{k}\|_{L^{2}(B_{1})}$, and let us show that $b_{k}\leq Ca_{k}$ as
$k\to\infty$. Indeed, using (9.5)-(9.6) with $\varrho=1$ and
$\varrho^{\prime}=R\gg 1$ (we want $R^{2}$ to be larger than, say,
$-2s_{\infty}$), we get

$\|\bar{w}_{k}\|_{L^{2}(\mathscr{C}_{R})}\leq\|w_{r_{k}}(y_{k}+\cdot,s_{k}+\cdot)+P_{k}\|_{L^{2}(\mathscr{C}_{R})}+\leq\|w_{r_{k}}\|_{L^{2}(\mathscr{C}_{2R})}+C(R)\|P_{k}\|_{L^{2}(B_{1})}\leq
C(R)(a_{k}+b_{k}).$

We now claim that $b_{k}\leq Ca_{k}.$ Indeed, if by contradiction $b_{k}\gg
a_{k}$ as $k\to\infty$ then, up to subsequence (note that any sequence of
quadratic polynomials bounded in $L^{2}$ is pre-compact),

$\lim_{k}\frac{\tilde{w}_{k}}{a_{k}+b_{k}}=\lim_{k}\frac{P_{k}}{b_{k}}=:\tilde{P}=\,[\mbox{second
order harmonic polynomial}]\qquad\text{in $L^{2}_{\rm
loc}\big{(}\mathbb{R}^{n}\times(-\infty,0]\big{)}$. }$

On the other hand, it follows by (9.5) and (9.6) that, for
$0<\varrho<\varrho^{\prime}\ll r_{k}^{-1}$,

$c\bigg{(}\frac{\varrho^{\prime}}{\varrho}\bigg{)}^{\negmedspace
6}\leq\frac{{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\mathscr{C}_{\varrho^{\prime}}}\left(\frac{\bar{w}_{k}}{a_{k}+b_{k}}\right)^{2}}{{\mathchoice{\kern
4.30554pt\vbox{\hrule height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern
4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\mathscr{C}_{\varrho}}\left(\frac{\bar{w}_{k}}{a_{k}+b_{k}}\right)^{2}}.$
(9.7)

Taking the limit this yields

$c\bigg{(}\frac{\varrho^{\prime}}{\varrho}\bigg{)}^{\negmedspace
6}\leq\frac{{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\mathscr{C}_{\varrho^{\prime}}}(\tilde{P})^{2}}{{\mathchoice{\kern
4.30554pt\vbox{\hrule height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern
4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\mathscr{C}_{\varrho}}(\tilde{P})^{2}}\qquad\forall\,0<\varrho<\varrho^{\prime},$

a contradiction since $\tilde{P}$ is quadratic (and hence cannot satisfy a
cubic growth).

This proves that $b_{k}\leq Ca_{k}$, hence

$\left\|\frac{P_{k}(\cdot-y_{k})}{a_{k}}\right\|_{L^{2}(B_{1})}\leq
C,\quad\mbox{ and therefore }\quad\frac{P_{k}(\cdot-
y_{k})}{a_{k}}=:\bar{P}_{k}\rightarrow\bar{P}_{\infty}$

as $k\to\infty$ (up to a subsequence). Now, a simple computation —see the
proof of (3.11) in [FS17, Lemma 3.3]— shows that

$\bar{P}_{\infty}\mbox{ is odd with respect to $\\{p_{2}=0\\}$}.$ (9.8)

Therefore, since $\frac{w_{r_{k}}}{a_{k}}=\tilde{w}_{r_{k}}\to q$ in
$L^{2}_{\rm loc}(\mathbb{R}^{n}\times(-\infty,0])$, using again (9.7) (now
with $a_{k}$ instead of $a_{k}+b_{k}$ in the denominator) we obtain

$c\bigg{(}\frac{\varrho^{\prime}}{\varrho}\bigg{)}^{\negmedspace
6}\leq\frac{{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{(y_{k},s_{k})+\mathscr{C}_{\varrho^{\prime}}}(\tilde{w}_{r_{k}}+\bar{P}_{k})^{2}}{{\mathchoice{\kern
4.30554pt\vbox{\hrule height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern
4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{(y_{k},s_{k})+\mathscr{C}_{\varrho}}(\tilde{w}_{r_{k}}+\bar{P}_{k})^{2}}\qquad\Rightarrow\qquad
c\bigg{(}\frac{\varrho^{\prime}}{\varrho}\bigg{)}^{\negmedspace
6}\leq\frac{{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{(y_{\infty},s_{\infty})+\mathscr{C}_{\varrho^{\prime}}}(q+\bar{P}_{\infty})^{2}}{{\mathchoice{\kern
4.30554pt\vbox{\hrule height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern
4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{(y_{\infty},s_{\infty})+\mathscr{C}_{\varrho}}(q+\bar{P}_{\infty})^{2}}.$

In particular, choosing $\varrho^{\prime}=1$ and taking even parts, (9.8)
implies that

$\left({\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{(y_{\infty},s_{\infty})+\mathscr{C}_{\varrho}}(q+\bar{P}_{\infty})^{2}\right)^{1/2}\leq
C\varrho^{3}\quad\mbox{and}\quad\left({\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{(y_{\infty},s_{\infty})+\mathscr{C}_{\varrho}}(q^{even})^{2}\right)^{1/2}\leq
C\varrho^{3}\qquad\forall\,\varrho\in(0,1).$ (9.9)

Note now that, thanks to Proposition 6.7, $q$ (and therefore also $q^{even}$)
is a $3$-homogeneous solution of the parabolic Signorini problem in
$\mathbb{R}^{n}\times(-\infty,0]$ with obstacle zero on $\\{p_{2}=0\\}$. Then,
(9.9) shows that $(y_{\infty},s_{\infty})$ is a singular point for $q^{even}$
of homogeneity at least three. In addition, since $q^{even}$ is 3-homogeenous
we have

$3\leq\phi\big{(}0^{+},q^{even}(y_{\infty}+\,\cdot\,,s_{\infty}+\,\cdot\,)\big{)}\leq\phi\big{(}\infty,q^{even}(y_{\infty}+\,\cdot\,,s_{\infty}+\,\cdot\,)\big{)}=\phi(\infty,q)=3.$

This implies that
$r\mapsto\phi\big{(}r,q^{even}(y_{\infty}+\,\cdot\,,s_{\infty}+\,\cdot\,)\big{)}$
is constantly equal to 3, therefore
$q^{even}(y_{\infty}+\,\cdot\,,s_{\infty}+\,\cdot\,)$ is also 3-homogeneous.
In particular

$q^{even}(y_{\infty}+\,\cdot\,,s_{\infty}+\,\cdot\,)\equiv\lim_{R\to\infty}R^{-3}q^{even}(y_{\infty}+R\,\cdot\,,s_{\infty}+R^{2}\,\cdot\,)\equiv
q.$

This implies that $q^{even}$ is translation invariant in the direction
$y_{\infty}$,777A possible way to show this, is the following. By homogeneity
of $q^{even}(y_{\infty}+\,\cdot\,,s_{\infty}+\,\cdot\,)$ and $q^{even}$, it
holds  $q^{even}(\delta
y_{\infty}+x,\delta^{2}s_{\infty}+t)=\delta^{3}q^{even}(y_{\infty}+(x/\delta),s_{\infty}+(t/\delta^{2}))=\delta^{3}q^{even}(x/\delta,t/\delta^{2})=q^{even}(x,t)\qquad\forall\,\delta>0.$
Subtracting $q(x,t)$ to both sides, dividing by $\delta$, and sending
$\delta\downarrow 0$ we obtain $y_{\infty}\cdot\nabla q(x,t)\equiv 0$.
concluding the proof of the lemma. ∎

For the sequel, it will be useful to introduce the following:

######  Definition 9.6.

Given $u:B_{1}\times(-1,1)\to[0,\infty)$ a bounded solution of (3.1), we
define $\Sigma^{*}\subset\Sigma_{n-1}^{=3}$ as the set of singular points
$(x_{\circ},t_{\circ})$ such that, in some coordinate system where
$p_{2,x_{\circ},t_{\circ}}=\frac{1}{2}(x_{n})^{2}$, we have that

$p_{3,x_{\circ},t_{\circ}}^{*}:=\lim_{k}\frac{u(x_{\circ}+r\,\cdot\,,t_{\circ}+r^{2}\,\cdot\,)-r^{2}\,p_{2,x_{\circ},t_{\circ}}}{r^{3}}$

can be written as

$a_{x_{\circ},t_{\circ}}|x_{n}|\big{(}x_{n}^{2}+6t\big{)}+p_{3,x_{\circ},t_{\circ}},$

for some $a_{x_{\circ},t_{\circ}}\geq 0$ and $p_{3,x_{\circ},t_{\circ}}$ an
odd 3-homogeneous caloric polynomial.

The motivation of the previous definition is given by the following lemma,
which follows from Lemma 9.5 and Proposition 7.6.

######  Lemma 9.7.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1). Then

$\dim_{\mathcal{H}}\big{(}\pi_{x}(\Sigma_{n-1}^{=3}\setminus\Sigma^{*})\big{)}\leq
n-2.$

###### Proof.

We begin by observing the following: Let $Q=Q(x,t)$ be an 3-homogenous
solution of the parabolic Signorini problem, which is even with respect to
$\\{x_{n}=0\\}$ and it is invariant with respect to translations parallel to
$\\{x_{n}=0\\}$. Then $Q$ depends on the variables $x_{n}$ and $t$, and hence
it must be of the form

$Q(x,t)=a|x_{n}|\bigg{(}\frac{x_{n}^{2}}{6}+t\bigg{)},$ (9.10)

for some $a\geq 0$. We now show that this implies the following:

Claim. Let $(x_{\circ},t_{\circ})\in\Sigma_{n-1}^{=3}\setminus\Sigma^{*}$, and
$(x_{k}^{(j)},t_{k}^{(j)})$ a sequence of singular points in
$\Sigma_{n-1}^{=3}$, with $t_{k}^{(j)}\leq t_{\circ}$, such that
$y_{k}^{(j)}:=\frac{x_{k}^{(j)}-x_{\circ}}{r_{k}}\to y_{\infty}^{(j)}\neq 0$.
Then $y_{\infty}^{(j)}\in\\{x_{n}=0\\}$, and the set
$\big{\\{}y_{\infty}^{(j)}\big{\\}}_{j}$ has rank at most $n-2$.

Indeed, applying Corollary 9.4 and Lemma 9.5 to the function
$u(x_{\circ}+\,\cdot\,,t_{\circ}+\,\cdot\,)$, we deduce that

$\frac{u(x_{\circ}+r_{k}\,\cdot\,,t_{\circ}+r_{k}^{2}\,\cdot\,)-r_{k}^{2}\,p_{2,x_{\circ},t_{\circ}}}{r_{k}^{3}}\to
q$

and $y_{\infty}^{(j)}\in\\{x_{n}=0\\}$. Then, if the rank of
$\big{\\{}y_{\infty}^{(j)}\big{\\}}_{j}$ was $n-1$, Lemma 9.5 would imply that
$q^{even}$ is of the form (9.10). Recalling Definition 9.6, this would mean
that $(x_{\circ},t_{\circ})\in\Sigma^{*}$, a contradiction.

Thanks to the Claim, we can apply Proposition 7.6 to deduce the desired
dimensional bound on $\pi_{x}(\Sigma_{n-1}^{=3}\setminus\Sigma^{*})$ (more
precisely, since the control that we have now is for “points from the past”
instead of on “points in the future”, we need to apply Proposition 7.6 to the
set $E:=\\{(x,-t)\,:\,(x,t)\in\Sigma_{n-1}^{=3}\setminus\Sigma^{*}\\}$). ∎

We now show that, for any $(x_{\circ},t_{\circ})\in\Sigma^{*}$, the
coefficient $a_{x_{\circ},t_{\circ}}$ must be strictly positive.

######  Lemma 9.8.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), and
$\varrho\in(0,1)$. There exists $c_{\varrho}>0$ such that the following holds:
For any $(x_{\circ},t_{\circ})\in\Sigma^{*}\cap
B_{1-\varrho}\times(-1+\varrho^{2},1)$, let $a_{x_{\circ},t_{\circ}}$ be the
coefficient of the non-caloric part of $p_{3,x_{\circ},t_{\circ}}^{*}$, as in
Definition 9.6. Then $a_{x_{\circ},t_{\circ}}\geq c_{\varrho}$.

###### Proof.

Let $(x_{\circ},t_{\circ})\in\Sigma^{*}\cap
B_{1-\varrho}\times(-1+\varrho^{2},1)$. By Lemma 8.4, there exist positive
constants $c$ and $r_{\circ}$ such that

${\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{B_{r}}\partial_{t}u(x_{\circ}+\,\cdot\,,t_{\circ}-r^{2})\geq
cr\qquad\mbox{ in }B_{r},\quad\forall\,r\in(0,r_{\circ}).$

Equivalently, defining
$w:=u(x_{\circ}+\,\cdot\,,t_{\circ}+\cdot)-p_{2,x_{\circ},t_{\circ}}$ and
noting that $\partial_{t}w=\partial_{t}u\geq 0$, we have

${\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{B_{1}\times\\{-1\\}}\partial_{t}(r^{-3}w_{r})\geq
c>0\qquad\,\quad\forall\,r\in(0,r_{\circ}).$

Taking the limit as $r\to 0$, we deduce that

${\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{B_{1}\times\\{-1\\}}\partial_{t}\big{(}a_{x_{\circ},t_{\circ}}|x_{n}|\big{(}x_{n}^{2}+6t\big{)}+p_{3,x_{\circ},t_{\circ}}\big{)}\geq
c>0.$

Since $p_{3,x_{\circ},t_{\circ}}$ is odd, so is
$\partial_{t}p_{3,x_{\circ},t_{\circ}}$ and therefore ${\mathchoice{\kern
4.30554pt\vbox{\hrule height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern
4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{B_{1}\times\\{-1\\}}\partial_{t}p_{3,x_{\circ},t_{\circ}}=0$.
The lemma follows. ∎

##  10\. A useful monotone quantity

Let us define a smoothed version of $H$ given by

$\tilde{H}(\varrho,w):=\int_{1}^{2}H(\varrho\theta,w)d\theta$ (10.1)

The goal of this section is to establish the following monotonicity formula.
This will be crucial in the next sections in order to establish higher order
regularity at (most) singular points in $\Sigma^{*}$.

######  Proposition 10.1.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), let
$(0,0)\in\Sigma^{*}$, and assume that $p_{2}=\frac{1}{2}x_{n}^{2}$ and
$p_{3}^{*}=\overline{a}|x_{n}|(x_{n}^{2}+6t)+p_{3}$,.

Given $C_{\circ}>0$ and $M\in(6,7)$, there exist $\varepsilon_{\circ}>0$ and
$R_{\circ}\geq 1$ such that the following holds: Assume that there exists
$r\in(0,\varepsilon_{\circ})$ such that

$\|u-p_{2}-p_{3}^{*}\|_{L^{\infty}(B_{R_{\circ}\rho}\times(-2\varrho^{2},\varrho^{2}/2))}\leq\varepsilon_{\circ}\varrho^{3}\quad\mbox{
for all }\varrho\in(0,r).$ (10.2)

Denote $w:=(u-p_{2}-Q)\zeta$ with $Q=x_{n}Q_{2}+\bar{a}|x_{n}|(x_{n}^{2}+6t)$,
where $\bar{a}\in[0,C_{\circ}]$ and $Q_{2}$ is any 2-homogeneous polynomial
satisfying $\text{\small\bf H}(x_{n}Q_{2})=0$ and
$\|Q_{2}\|_{L^{2}(\mathscr{C}_{1})}\leq C_{\circ}$. Then

$\mathfrak{h}^{\frac{5}{2}}(\varrho,w):=\max_{R\in[1,1/\varrho]}R^{-M}\varrho^{-5}\tilde{H}(R\varrho,w)$
(10.3)

is monotone nondecreasing for $\varrho\in(0,r)$.

The idea behind the quantity $\mathfrak{h}^{\frac{5}{2}}(\varrho,w)$ is the
following. Ideally, we would like to have that $\varrho^{-6}H(\varrho,w)$ is
monotone at every point. Unfortunately this is false at points in $\Sigma^{*}$
because $p_{3}^{*}$ is not caloric. This means that we must look for a
monotone quantity of lower order, and this is why we look at a quantity of the
form $\varrho^{-5}H(\varrho,w)$. Our monotone quantity
$\mathfrak{h}^{\frac{5}{2}}$ is basically a modified version of this, in which
we first consider an averaged version of $H$, given by (10.1), and then define
$\mathfrak{h}^{\frac{5}{2}}$ as in (10.3) in order to control the growth
between different scales (see (10.5)). It is thanks to these (small but
crucial) modifications that we can prove that the quantity is monotone.

Before proving the monotonicity formula, we need the following:

######  Lemma 10.2.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1). Given
positive constants $c_{\circ}$, $r_{\circ}$, $C_{\circ}$, and $R_{\circ}$,
there exists $\delta>0$ such that the following holds: Let $Q_{2}$ be any
2-homogeneous polynomial such that $\text{\small\bf H}(x_{n}Q_{2})=0$ and
$\|Q_{2}\|_{L^{2}(\mathscr{C}_{1})}\leq C_{\circ}$. Assume that

$\frac{1}{r^{3}}\bigg{(}u(r\,\cdot\,,r^{2}\,\cdot\,)-r^{2}\frac{x_{n}^{2}}{2}\bigg{)}\leq
c_{\circ}|x_{n}|t+C_{\circ}|x_{n}|^{3}+x_{n}Q_{2}(x^{\prime})+\delta\quad\mbox{inside
}\big{\\{}|x|\leq 2R_{\circ},-4\leq t\leq-1\big{\\}}.$

for all $r\in(0,r_{\circ})$. Then

$u(r\,\cdot\,,r^{2}\,\cdot\,)\leq Cr^{4}\quad\mbox{ on }\
\\{x_{n}=0\\}\cap\\{|x|\leq R_{\circ},t=-1\\},$

for all $r\in(0,r_{\circ})$.

###### Proof.

Fix $z\in\\{x_{n}=0\\}\cap\\{|x|\leq R_{\circ},\,t=-1\\}$, and define

$\phi(x,t):=-nx_{n}^{2}+|x^{\prime}-z^{\prime}|^{2}-2(1+t)+x_{n}Q_{2}(x^{\prime})$

and

$v(x,t):=\frac{1}{r^{3}}\bigg{(}u(rx,r^{2}t)-r^{2}\frac{x_{n}^{2}}{2}\bigg{)}.$

Also, consider the set $U:=\\{|x^{\prime}-z^{\prime}|\leq
1,\,|x_{n}|\leq\varrho,\,-2\leq t\leq-1\\}$, with $\varrho>0$ to be fixed. We
choose first $\varrho>0$ and then $\delta>0$, both conveniently small, so that
for all $x\in\partial_{par}U$ we have

$\begin{split}v(x)&\leq
c_{\circ}|x_{n}|t+C_{\circ}|x_{n}|^{3}+x_{n}Q_{2}(x)+\delta\\\
&\leq\begin{cases}-c_{\circ}\varrho+C_{\circ}\varrho^{3}+x_{n}Q_{2}(x)+\delta\leq-2n\varrho^{2}+x_{n}Q_{2}(x)\leq\phi(x)\quad&\mbox{for
}|x_{n}|=\varrho\\\
-c_{\circ}|x_{n}|+C_{\circ}\varrho^{3}+x_{n}Q_{2}(x)+\delta\leq-2n\varrho^{2}+x_{n}Q_{2}(x)+1\leq\phi(x)&\mbox{for
}|x_{n}|\leq\varrho,|x^{\prime}-z^{\prime}|=1\\\
C_{\circ}\varrho^{3}+x_{n}Q_{2}(x)+\delta\leq-2n\varrho^{2}+x_{n}Q_{2}(x)+2\leq\phi(x)&\mbox{for
}|x_{n}|\leq\varrho,t=-2.\end{cases}\end{split}$

Now, consider the function

$\psi(x):=\frac{1}{2r}x_{n}^{2}+\phi(x)+Cr,$

where $C$ is a large enough constant, chosen so that $\psi$ is nonnegative in
$U$ (note that $\frac{1}{2r}x_{n}^{2}+Cr\geq\sqrt{C}|x_{n}|$, so such a
constant $C$ exists independently of $r$). Since $\text{\small\bf H}\phi=0$,
we have $\text{\small\bf H}\psi\leq\frac{1}{r}$ in $U$. Also, by the argument
above, $\frac{1}{r^{3}}u(r\,\cdot\,,r^{2}\,\cdot\,)\leq\psi$ on
$\partial_{par}U$. Therefore, since $\psi\geq 0$ in $U$ and $\text{\small\bf
H}\big{(}\frac{1}{r^{3}}u(r\,\cdot\,,r^{2}\,\cdot\,)\big{)}\geq\frac{1}{r}$
whenever $\frac{1}{r^{3}}u(r\,\cdot\,,r^{2}\,\cdot\,)>0$, it follows from the
comparison principle that

$\frac{1}{r^{3}}u(r\,\cdot\,,r^{2}\,\cdot\,)\leq\psi\quad\mbox{in }U.$

In particular, choosing $x=z$ and $t=-1$ we get (recall that $z_{n}=0$)

$\frac{1}{r^{3}}u(rz,-r^{2})\leq\psi(z,-1)=Cr,$

and the lemma follows. ∎

We can now prove Proposition 10.1.

###### Proof of Proposition 10.1.

Note that $\mathfrak{h}^{\frac{5}{2}}(\cdot,w)$ is a Lipschitz function, so it
suffices to prove that its derivative is nonnegative at every
differentiability point. Thus, we compute
$\frac{d}{d\varrho}\big{|}_{\varrho=\varrho_{\circ}}\mathfrak{h}^{\frac{5}{2}}(\varrho,w)$
and we note that two possible alternatives arise:

(a)
$\max_{R\in[1,1/\varrho_{\circ}]}R^{-M}\tilde{H}(R\varrho_{\circ},w)>\tilde{H}(\varrho_{\circ},w)$;

(b)
$\max_{R\in[1,1/\varrho_{\circ}]}R^{-M}\tilde{H}(R\varrho_{\circ},w)=\tilde{H}(\varrho_{\circ},w)$.

If $(a)$ holds, consider $R_{\circ}\in(1,1/\varrho_{\circ}]$ such that
$\max_{R\in[1,1/\varrho_{\circ}]}R^{-M}\tilde{H}(R\varrho_{\circ},w)=R_{\circ}^{-M}\tilde{H}(R_{\circ}\varrho_{\circ},w)$.
By continuity, for $s\in(0,1)$ sufficiently close to $1$ we have

$\varrho_{\circ}^{5}\mathfrak{h}^{\frac{5}{2}}(\varrho_{\circ},w)=(R_{\circ})^{-M}\tilde{H}(R_{\circ}\varrho_{\circ},w)=\max_{R^{\prime}\in[s,1/\varrho_{\circ}]}(R^{\prime})^{-M}\tilde{H}(R^{\prime}\varrho_{\circ},w)\\\
=s^{-M}\max_{R\in[1,1/(s\varrho_{\circ})]}R^{-M}\tilde{H}(Rs\varrho_{\circ},w)=s^{-M}(s\varrho_{\circ})^{5}\mathfrak{h}^{\frac{5}{2}}(s\varrho_{\circ},w).$

Hence

$\mathfrak{h}^{\frac{5}{2}}(s\varrho_{\circ},w)=s^{M-5}\mathfrak{h}^{\frac{5}{2}}(\varrho_{\circ},w)\qquad\text{for
$s\in(0,1)$ sufficiently close to $1$,}$

and therefore
$\frac{d}{d\varrho}\big{|}_{\varrho=\varrho_{\circ}}\mathfrak{h}^{\frac{5}{2}}(\varrho,w)=\frac{M-5}{\varrho_{\circ}}\mathfrak{h}^{\frac{5}{2}}(\varrho_{\circ},w)\geq
0$. This proves the result whenever $(a)$ holds.

Instead, when (b) holds, we note that
$\mathfrak{h}^{\frac{5}{2}}(\varrho,w)\geq\varrho^{-5}\tilde{H}(\varrho,w)$
with equality for $\varrho=\varrho_{\circ}$. Therefore

$\frac{\mathfrak{h}^{\frac{5}{2}}(\varrho,w)-\mathfrak{h}^{\frac{5}{2}}(\varrho_{\circ},w)}{\varrho-\varrho_{\circ}}\geq\frac{\varrho^{-5}\tilde{H}(\varrho,w)-\varrho_{\circ}^{-5}\tilde{H}(\varrho_{\circ},w)}{\varrho-\varrho_{\circ}}\qquad\Rightarrow\qquad\frac{d}{d\varrho}\bigg{|}_{\varrho=\varrho_{\circ}}\mathfrak{h}^{\frac{5}{2}}(\cdot,w)=\frac{d}{d\varrho}\bigg{|}_{\varrho=\varrho_{\circ}}\big{(}\varrho^{-5}\tilde{H}(\rho,w)\big{)}.$

In addition, if (b) holds then we have the extra growth information

$\tilde{H}(R\varrho_{\circ},w)\leq\tilde{H}(\varrho_{\circ},w)R^{M}\quad\mbox{for
all }R\in[1,1/\varrho_{\circ}],$ (10.4)

and in particular, taking $R=1/\varrho_{\circ}$, we get

$\varrho_{\circ}^{M}\tilde{H}(1,w)\leq\tilde{H}(\varrho_{\circ},w).$ (10.5)

Therefore, to conclude the proof of the lemma, it is enough to show that
$\frac{d}{d\varrho}\big{|}_{\varrho=\varrho_{\circ}}\big{(}\varrho^{-5}\tilde{H}(\varrho,w)\big{)}\geq
0$ at a every scale $\varrho_{\circ}$ where (10.4) holds. We divide the proof
of this fact in four steps.

$\bullet$ Step 1. Denote

$v=(u-p_{2})\zeta,\qquad
h=(x_{n}Q_{2}+\overline{a}\Psi)\zeta,\qquad\Psi(x,t):=|x_{n}|(x_{n}^{2}+6t),$

where $\zeta$ is the usual spatial cut-off. Recalling that $w=v-h$ we first
show that, at any small enough scale $\varrho_{\circ}$ at which (10.4) holds,
we have

$\frac{d}{d\varrho}\bigg{|}_{\varrho=\varrho_{\circ}}\big{(}\varrho^{-5}\tilde{H}(\varrho,w)\big{)}\geq\frac{1}{\varrho_{\circ}^{6}}\bigg{(}\frac{\tilde{H}(\varrho_{\circ},v-h)}{2}+4\int_{-2\varrho_{\circ}^{2}}^{-\varrho_{\circ}^{2}}\biggl{[}\int_{\mathbb{R}^{n}}\big{\\{}v\text{\small\bf
H}v-(v-h)\text{\small\bf H}(v-h)\big{\\}}G\biggr{]}\,dt\bigg{)}$ (10.6)

Observe that, by (10.1), we have

$\frac{d}{d\varrho}\tilde{H}(\varrho,w):=\int_{1}^{2}\frac{dH}{d\varrho}(\varrho\theta,w)\,\theta\,d\theta,$
(10.7)

so we need to compute $\frac{d}{d\varrho}H(\varrho\theta,w)$. Using the
identity from Lemma 4.2

$\langle h,Zv\rangle_{\varrho}=\langle
Zh,v\rangle_{\varrho}+2\varrho^{2}\big{(}\langle h,\text{\small\bf
H}v\rangle_{\varrho}-\langle\text{\small\bf H}h,v\rangle_{\varrho}\big{)},$

we obtain

$\begin{split}\frac{d}{d\varrho}H(\varrho,v-h)&=\frac{2}{\varrho}\langle
v-h,Z(v-h)\rangle_{\varrho}\\\ &=\frac{2}{\varrho}\big{(}\langle
v,Zv\rangle_{\varrho}-\langle h,Zv\rangle_{\varrho}-\langle
v,Zh\rangle_{\varrho}+\langle h,Zh\rangle_{\varrho}\big{)}\\\
&=\frac{2}{\varrho}\big{(}\langle v,Zv\rangle_{\varrho}-2\langle
v,Zh\rangle_{\varrho}-2\varrho^{2}\big{(}\langle h,\text{\small\bf
H}v\rangle_{\varrho}-\langle\text{\small\bf
H}h,v\rangle_{\varrho}\big{)}+\langle h,Zh\rangle_{\varrho}\big{)}\\\
&\geq\frac{2}{\varrho}\big{(}3\langle v,v\rangle_{\varrho}+2\varrho^{2}\langle
v,\text{\small\bf H}v\rangle_{\varrho}-6\langle
v,h\rangle_{\varrho}-2\varrho^{2}\big{(}\langle h,\text{\small\bf
H}v\rangle_{\varrho}-\langle\text{\small\bf
H}h,v\rangle_{\varrho}\big{)}+3\langle
h,h\rangle_{\varrho}\big{)}-Ce^{-1/\varrho}\\\
&=\frac{2}{\varrho}\big{(}3H(\varrho,v-h)+2\varrho^{2}\big{(}\langle
v,\text{\small\bf H}v\rangle_{\varrho}-\langle h,\text{\small\bf
H}v\rangle_{\varrho}+\langle\text{\small\bf
H}h,v\rangle_{\varrho}\big{)}\big{)}-Ce^{-1/\varrho}\\\
&\geq\frac{2}{\varrho}\big{(}3H(\varrho,v-h)+2\varrho^{2}\big{(}2\langle
v,\text{\small\bf H}v\rangle_{\varrho}-\langle v-h,\text{\small\bf
H}(v-h)\rangle_{\varrho}-2\langle h,\text{\small\bf
H}v\rangle_{\varrho}\big{)}\big{)}-Ce^{-1/\varrho}.\end{split}$

Here, we used that $Zh=3h$, that $\langle v,Zv\rangle_{\varrho}\geq 3\langle
v,v\rangle_{\varrho}+2\varrho^{2}\langle v,\text{\small\bf
H}v\rangle_{\varrho}-Ce^{-1/\varrho}$ (this follows from (4.3), Proposition
5.4, and the fact that the frequency at $(0,0)$ is $3$), and that
$\big{|}\langle h,\text{\small\bf H}h\rangle_{\varrho}\big{|}\leq
Ce^{-1/\varrho}$. Therefore

$\begin{split}\frac{d}{d\varrho}\big{(}\varrho^{-5}H(\varrho,w)\big{)}&=\frac{1}{\varrho^{6}}\big{(}\varrho{\textstyle\frac{d}{d\varrho}}H(\varrho,w)-5H(\varrho,w)\big{)}\\\
&\geq\frac{1}{\varrho^{6}}\big{(}H(\varrho,v-h)+4\varrho^{2}\big{(}2\langle
v,\text{\small\bf H}v\rangle_{\varrho}-\langle v-h,\text{\small\bf
H}(v-h)\rangle_{\varrho}-2\langle h,\text{\small\bf
H}v\rangle_{\varrho}\big{)}\big{)}-Ce^{-1/\varrho}.\end{split}$

Now recall that $v=(u-p_{2})\zeta$ and hence

$(u-p_{2})\text{\small\bf H}(u-p_{2})\geq 0\qquad\Rightarrow\qquad\langle
v,\text{\small\bf H}v\rangle_{\varrho}\geq-Ce^{-1/\varrho}.$

On the other hand, since $\text{\small\bf H}(u-p_{2})=-\chi_{\\{u=0\\}}$,
$h=(x_{n}Q_{2}+\bar{a}\psi)\zeta$, and $|x_{n}Q_{2}+\bar{a}\psi|\leq
C|x_{n}||x|^{2}$, we obtain

$\big{|}\langle h,\text{\small\bf
H}v\rangle_{\varrho}\big{|}\leq\int_{\\{t=-\varrho^{2}\\}}|h|\,\chi_{\\{u=0\\}}G+Ce^{-1/\varrho}\leq
C\int_{\\{t=-\varrho^{2}\\}}C|x_{n}|\,|x|^{2}\chi_{\\{u=0\\}}G+Ce^{-1/\varrho}.$

Also, using that (as a consequence of Proposition 6.6)

$\\{u=0\\}\cap\\{|x|<1/2,t=-\varrho^{2}\\}\subset\\{|x_{n}|\leq
C(|x^{\prime}|+\varrho)^{2}\\},$ (10.8)

we obtain

$\big{|}\langle h,\text{\small\bf
H}v\rangle_{\varrho}\big{|}\leq\int_{\\{t=-\varrho^{2}\\}}C(\varrho+|x|)^{4}\chi_{\\{|x_{n}|\leq
C(|x^{\prime}|+\varrho)^{2}\\}}G+Ce^{-1/\varrho}\leq C\varrho^{5}.$ (10.9)

Therefore, we have shown that

$\frac{d}{d\varrho}\big{(}\varrho^{-5}H(\varrho,w)\big{)}\geq\frac{1}{\varrho^{6}}\bigg{(}H(\varrho,v-h)+4\varrho^{2}\big{(}\langle
v,\text{\small\bf H}v\rangle_{\varrho}-\langle v-h,\text{\small\bf
H}(v-h)\rangle_{\varrho}\big{)}-C\varrho^{7}\bigg{)}-Ce^{-1/\varrho}$

Applying this inequality with $H(\varrho\theta,w)$ in place of $H(\varrho,w)$,
and integrating with respect to $\theta\,d\theta$ over $[1,2]$, we obtain

$\frac{d}{d\varrho}\big{(}\varrho^{-5}\tilde{H}(\varrho,w)\big{)}\geq\frac{1}{\varrho^{6}}\bigg{(}\tilde{H}(\varrho,v-h)+4\int_{-2\varrho^{2}}^{-\varrho^{2}}\bigg{[}\int_{\mathbb{R}^{n}}\big{\\{}v\text{\small\bf
H}v-(v-h)\text{\small\bf
H}(v-h)\big{\\}}G\bigg{]}\,dt-C\varrho^{7}\bigg{)}-Ce^{-1/\varrho}.$

Since $\tilde{H}(\varrho_{\circ},v-h)\geq
c\varrho_{\circ}^{M}\gg\varrho_{\circ}^{7}\gg Ce^{-1/\varrho_{\circ}}$ (recall
(10.5)), (10.6) follows.

$\bullet$ Step 2. Let

$w_{\varrho_{\circ}}:=(v-h)(\varrho_{\circ}\,\cdot\,,\varrho_{\circ}^{2}\,\cdot\,)=\big{(}(u-p_{2}-p_{3}-\bar{a}\Psi)\zeta\big{)}(\varrho_{\circ}\,\cdot\,,\varrho_{\circ}^{2}\,\cdot\,).$

We now show that for any $\varepsilon>0$ there exists $\delta>0$ such that, if
$\varrho_{\circ}$ is sufficiently small,

$\int_{-2}^{-1}\bigg{(}\int_{\mathbb{R}^{n}}|w_{\varrho_{\circ}}\text{\small\bf
H}w_{\varrho_{\circ}}|\min\\{G,\delta\\}\bigg{)}\,dt\leq\varepsilon\tilde{H}(\varrho_{\circ},w_{\varrho_{\circ}}).$
(10.10)

This will be used in order to bound the integral in (10.6) outside a large
ball, as we will see in Step 4 below.

We first note that, inside $B_{1/2}\times(-1,0)$, it holds

$v\text{\small\bf H}v=(u-p_{2})\text{\small\bf H}(u-p_{2})\geq 0,\quad
h\text{\small\bf H}h=0,\quad\text{and}\quad-v\text{\small\bf
H}h=-u\text{\small\bf H}h\geq 0.$

Therefore, using (10.8) similarly to the argument in the previous step, we get

$\begin{split}(u-p_{2}-x_{n}Q_{2}-\bar{a}\Psi)\text{\small\bf
H}(u-p_{2}-x_{n}Q_{2}-\bar{a}\Psi)&\geq-(x_{n}Q_{2}+\bar{a}\Psi)\text{\small\bf
H}(u-p_{2})\\\ &\geq-C(\varrho+|x|)^{4}\chi_{\\{|x_{n}|\leq
C(|x^{\prime}|+\varrho)^{2}\\}}\quad\mbox{in
}B_{1/2}\times(-\varrho^{2},0).\end{split}$

Hence, after scaling and taking into account the error introduced by the cut-
off $\zeta$, for all $\varrho\in(0,1/2)$ it holds

$w_{\varrho_{\circ}}\text{\small\bf
H}w_{\varrho_{\circ}}\geq-C\varrho_{\circ}^{2}\min\big{\\{}(\varrho+\varrho_{\circ}|x|)^{4}\chi_{\\{\varrho_{\circ}|x_{n}|\leq
C(\varrho+\varrho_{\circ}|x^{\prime}|)^{2}\\}},1\big{\\}}\quad\mbox{ in
}\mathbb{R}^{n}\times(-\varrho^{2}/\varrho_{\circ}^{2},0).$

Equivalently, setting $\varrho=R\varrho_{\circ}$, for all
$R\in\big{(}0,\frac{1}{2\varrho_{\circ}}\big{)}$ we have

$w_{\varrho_{\circ}}\text{\small\bf
H}w_{\varrho_{\circ}}\geq-C\varrho_{\circ}^{6}\min\big{\\{}(R+|x|)^{4}\chi_{\\{|x_{n}|\leq
C\varrho_{\circ}(R+|x^{\prime}|)^{2}\\}},\varrho_{\circ}^{-4}\big{\\}}\quad\mbox{
in }\mathbb{R}^{n}\times(-R^{2},0).$

Integrating this bound for $R\in(1,2)$, we obtain

$\int_{-4}^{-1}\int_{\mathbb{R}^{n}}\big{(}w_{\varrho_{\circ}}\text{\small\bf
H}w_{\varrho_{\circ}}\big{)}_{-}G\leq\int_{\mathbb{R}^{n}}C\varrho_{\circ}^{6}(4+|x|)^{4}\chi_{\\{|x_{n}|\leq
C\varrho_{\circ}(4+|x|)^{2}\\}}G\leq C\varrho_{\circ}^{7}.$ (10.11)

Now, note that the following identity holds for any function
$\phi=\phi(x,t)\in C^{1}_{t}C^{1,1}_{x}$ with compact support and for any
smooth $\xi=\xi(x,t)$:

$2\int_{\mathbb{R}^{n}\times\\{t\\}}\phi\,\text{\small\bf
H}\phi\,\xi+2\int_{\mathbb{R}^{n}\times\\{t\\}}|\nabla\phi|^{2}\xi+\frac{d}{dt}\int_{\mathbb{R}^{n}\times\\{t\\}}\phi^{2}\xi=\int_{\mathbb{R}^{n}\times\\{t\\}}\phi^{2}(\Delta+\partial_{t})\xi.$
(10.12)

In particular, choosing $\phi=w_{\varrho_{\circ}}$ and $\xi=G$, we get

$2\int_{\mathbb{R}^{n}\times\\{t\\}}w_{\varrho_{\circ}}\,\text{\small\bf
H}w_{\varrho_{\circ}}\,G+2\int_{\mathbb{R}^{n}\times\\{t\\}}|\nabla
w_{\varrho_{\circ}}|^{2}G+\frac{d}{dt}\int_{\mathbb{R}^{n}\times\\{t\\}}w_{\varrho_{\circ}}^{2}G=0.$
(10.13)

Integrating between $t=-\theta$ and $t=-1$, we obtain

$2\int_{-\theta}^{-1}\biggl{(}\int_{\mathbb{R}^{n}}w_{\varrho_{\circ}}\text{\small\bf
H}w_{\varrho_{\circ}}\,G\biggr{)}\,dt+2\int_{-\theta}^{-1}\biggl{(}\int_{\mathbb{R}^{n}}|\nabla
w_{\varrho_{\circ}}|^{2}\,G\biggr{)}\,dt+\sup_{t\in(-\theta,-1)}\int_{\mathbb{R}^{n}\times\\{t\\}}w_{\varrho_{\circ}}^{2}G\\\
\leq\int_{\\{t=-\theta\\}}w_{\varrho_{\circ}}^{2}G\leq\int_{-4}^{-3}\int_{\mathbb{R}^{n}}w_{\varrho_{\circ}}^{2}G,$
(10.14)

where $\theta\in(3,4)$ is chosen (thanks to Fubini) so that
$\int_{\\{t=-\theta\\}}w_{\varrho_{\circ}}^{2}G\leq\int_{-4}^{-3}\int_{B_{2R}}w_{\varrho_{\circ}}^{2}$.
Hence, since
$\int_{-4}^{-3}\int_{\mathbb{R}^{n}}w_{\varrho_{\circ}}^{2}G\leq\tilde{H}(2,w_{\varrho_{\circ}})\leq
C\tilde{H}(1,w_{\varrho_{\circ}})$ (by the definition of $\tilde{H}$ and using
(10.4) with $R=2$), recalling (10.11) we deduce

$2\int_{-\theta}^{-1}\bigg{(}\int_{\mathbb{R}^{n}}|\nabla
w_{\varrho_{\circ}}|^{2}\,G\bigg{)}\,dt+\sup_{t\in(-\theta,-1)}\int_{\mathbb{R}^{n}\times\\{t\\}}w_{\varrho_{\circ}}^{2}G\leq
C\tilde{H}(1,w_{\varrho_{\circ}})+C\varrho_{\circ}^{7}\leq
C\tilde{H}(1,w_{\varrho_{\circ}}),$

where the last inequality follows from (10.5) (indeed, since $M<7$,
$\tilde{H}(1,w_{\varrho_{\circ}})\gg\varrho_{\circ}^{7}$ for
$\varrho_{\circ}>0$ small).

In particular, again by Fubini, there exists $\tilde{\theta}\in(2,3)$ such
that

$\int_{\mathbb{R}^{n}\times\\{-\tilde{\theta}\\}}|\nabla
w_{\varrho_{\circ}}|^{2}\,G+\int_{\mathbb{R}^{n}\times\\{-\tilde{\theta}\\}}w_{\varrho_{\circ}}^{2}G\leq
C\tilde{H}(1,w_{\varrho_{\circ}}).$

We now claim that this implies, thanks to the log-Sobolev inequality, that

$\int_{\mathbb{R}^{n}\times\\{\tilde{\theta}\\}}w_{\varrho_{\circ}}^{2}\min\\{G,\delta\\}\leq\varepsilon(\delta)\tilde{H}(1,w_{\varrho_{\circ}}),$
(10.15)

where $\varepsilon(\delta)\to 0$ as $\delta\to 0$.

Indeed, the function
$f:=\frac{w_{\varrho_{\circ}}(\tilde{\theta}^{1/2}\,\cdot\,,\tilde{\theta}\,\cdot)}{C\tilde{H}(1,w_{\varrho_{\circ}})^{1/2}}$
satisfies

$\int_{\mathbb{R}^{n}}|\nabla f|^{2}\,dm\leq
1\quad\mbox{and}\quad\int_{\mathbb{R}^{n}}|f|^{2}\,dm\leq 1,$

where $dm=G(\,\cdot\,,-1)dx$ is the Gaussian measure. Hence, given
$\lambda>1$, we can apply the Gaussian log-Sobolev inequality to
$F:=\max\\{|f|,1\\}$ (cp. Lemma 6.4) to get

$\log\lambda\int_{\mathbb{R}^{n}}f^{2}\chi_{\\{f^{2}>\lambda\\}}dm\leq\int_{\mathbb{R}^{n}}F^{2}\log
F^{2}dm\leq\int_{\mathbb{R}^{n}}|\nabla
F|^{2}dm+\biggl{(}\int_{\mathbb{R}^{n}}F^{2}dm\biggr{)}\,\log\biggl{(}\int_{\mathbb{R}^{n}}F^{2}dm\biggr{)}\\\
\leq\int_{\mathbb{R}^{n}}|\nabla
f|^{2}dm+\biggl{(}1+\int_{\mathbb{R}^{n}}f^{2}dm\biggr{)}\,\log\biggl{(}1+\int_{\mathbb{R}^{n}}f^{2}dm\biggr{)}\leq
1+2\log 2<3.$

Thus, for any $R>0$ we have

$\int_{\mathbb{R}^{n}\setminus B_{R}}f^{2}dm\leq\int_{\mathbb{R}^{n}\setminus
B_{R}}f^{2}\chi_{\\{f^{2}\leq\lambda\\}}dm+\int_{\mathbb{R}^{n}\setminus
B_{R}}f^{2}\chi_{\\{f^{2}>\lambda\\}}dm\leq\lambda\,\int_{\mathbb{R}^{n}\setminus
B_{R}}dm+\frac{3}{\log\lambda}.$

Choosing $\lambda=\lambda_{\varepsilon}$ large enough so that
$3/\log\lambda_{\varepsilon}<\varepsilon/4$, and then $R_{\varepsilon}>0$ so
that $\lambda_{\varepsilon}\,\int_{\mathbb{R}^{n}\setminus
B_{R_{\varepsilon}}}dm<\varepsilon/4$, we deduce that

$\int_{\mathbb{R}^{n}\setminus
B_{R_{\varepsilon}}}f^{2}dm<\frac{\varepsilon}{2}.$

Therefore, (10.15) follows by taking $\delta$ small enough.

Now, since $(\Delta+\partial_{t})\min\\{G,\delta\\}\leq 0$, we can use again
(10.12) with $\phi=w_{\varrho_{\circ}}$ and $\xi=\min\\{G,\delta\\}$ to get

$2\int_{\mathbb{R}^{n}\times\\{t\\}}w_{\varrho_{\circ}}\,\text{\small\bf
H}w_{\varrho_{\circ}}\,\min\\{G,\delta\\}+2\int_{\mathbb{R}^{n}\times\\{t\\}}|\nabla
w_{\varrho_{\circ}}|^{2}\min\\{G,\delta\\}+\frac{d}{dt}\int_{\mathbb{R}^{n}\times\\{t\\}}w_{\varrho_{\circ}}^{2}\min\\{G,\delta\\}\leq
0.$ (10.16)

Integrating between $t=-\tilde{\theta}$ and $t=-1$, exactly as before (but now
taking advantage of (10.12)) we obtain

$\begin{split}2\int_{-\tilde{\theta}}^{-1}\biggl{(}\int_{\mathbb{R}^{n}}|w_{\varrho_{\circ}}\text{\small\bf
H}w_{\varrho_{\circ}}|\,\min\\{G,\delta\\}\biggr{)}\,dt\leq\varepsilon(\delta)\tilde{H}(1,w_{\varrho_{\circ}})+C\varrho_{\circ}^{7}\leq
2\varepsilon(\delta)\tilde{H}(1,w_{\varrho_{\circ}}),\end{split}$

if $\varrho_{\circ}$ is sufficiently small. Since $\tilde{\theta}\in(2,3)$,
this concludes the proof of (10.10).

$\bullet$ Step 3. Let $u_{\varrho}:=u(\varrho\,\cdot\,,\varrho^{2}\,\cdot\,)$.
We now show that, for any fixed $R_{\circ}>0$, there exists $r>0$ such that

$0\leq-\int_{-2}^{-1}\biggl{(}\int_{B_{R_{\circ}}}u_{\varrho}\text{\small\bf
H}(\varrho^{3}\Psi)\biggr{)}\,dt\leq
C\varrho^{7}\qquad\forall\,\varrho\in(0,r).$ (10.17)

Indeed, thanks to Lemma 9.8 (recall that $(p_{3}^{*})^{even}$ denotes the even
symmetrization with respect to $\\{p_{2}=0\\}=\\{x_{n}=0\\}$), it holds

$(p_{3}^{*})^{even}=\bar{a}(t|x_{n}|+{\textstyle\frac{1}{6}}|x_{n}|^{3})\qquad\text{with
}\bar{a}\geq c_{\circ}>0.$

Therefore (10.2) implies that

$\frac{1}{\varrho^{3}}\bigg{(}u(\varrho\,\cdot\,,\varrho^{2}\,\cdot\,)-r^{2}\frac{x_{n}^{2}}{2}\bigg{)}\leq
c_{\circ}|x_{n}|t+C|x_{n}|^{3}+p_{3}^{odd}(x)+\varepsilon_{\circ}\qquad\mbox{in
}\\{|x|\leq 2R_{\circ},-4\leq t\leq-1\\},$

and so Lemma 10.2 implies

$0\leq u(\varrho\,\cdot\,,\varrho^{2}\,\cdot\,)\leq C\varrho^{4}\quad\mbox{on
}\\{x_{n}=0\\}\cap B_{R_{\circ}}.$

Recalling that $\Psi(x,t)=t|x_{n}|+|x_{n}|^{3}/6$ (in particular, its heat
operator is concentrated on $\\{x_{n}=0\\}$), (10.17) follows easily.

$\bullet$ Step 4. We finally conclude the proof of the lemma by combining
(10.6), (10.10), and (10.17).

Indeed, recall that we want to show that
$\frac{d}{d\varrho}|_{\varrho=\varrho_{\circ}}\big{(}\varrho^{-5}\tilde{H}(\varrho,w)\big{)}\geq
0$ at every scale $\varrho_{\circ}$ where (10.4)-(10.5) holds. In view of
(10.6), it is enough to prove that

$\int_{-2\varrho_{\circ}^{2}}^{-\varrho_{\circ}^{2}}\int_{\mathbb{R}^{n}}\big{\\{}v\text{\small\bf
H}v-(v-h)\text{\small\bf
H}(v-h)\big{\\}}G\geq-\frac{1}{10}\tilde{H}(\varrho_{\circ},v-h).$ (10.18)

We split the integral

$\int_{-2\varrho_{\circ}^{2}}^{-\varrho_{\circ}^{2}}\int_{\mathbb{R}^{n}}\big{\\{}v\text{\small\bf
H}v-(v-h)\text{\small\bf
H}(v-h)\big{\\}}G=\int_{-2}^{-1}\int_{\mathbb{R}^{n}}\big{\\{}v_{\varrho_{\circ}}\text{\small\bf
H}v_{\varrho_{\circ}}-w_{\varrho_{\circ}}\text{\small\bf
H}w_{\varrho_{\circ}}\big{\\}}G$

into two pieces, according to the partition
$\mathbb{R}^{n}\times(-2,-1)=B_{R_{\circ}}\times(-2,-1)\cup(\mathbb{R}^{n}\setminus
B_{R_{\circ}})\times(-2,-1)$.

Note that, using Lemma 5.2, (10.10), and the bound $G\leq Ce^{-cR^{2}}$ in
$(B_{2R}\setminus B_{R})\times(-2,-1)$, we get

$\begin{split}\int_{-2}^{-1}\int_{(\mathbb{R}^{n}\setminus
B_{R_{\circ}})}\big{\\{}v_{\varrho_{\circ}}\text{\small\bf
H}v_{\varrho_{\circ}}-w_{\varrho_{\circ}}\text{\small\bf
H}w_{\varrho_{\circ}}\big{\\}}G&\geq-\int_{-2}^{-1}\int_{(\mathbb{R}^{n}\setminus
B_{R_{\circ}})}|w_{\varrho_{\circ}}\text{\small\bf
H}w_{\varrho_{\circ}}|G-Ce^{-c\varrho_{\circ}}\geq-\frac{1}{20}\tilde{H}(\varrho_{\circ},v-h),\end{split}$

provided that $R_{\circ}$ is taken sufficiently large. (Here we used that,
given $\delta_{\circ}>0$, we have $\min\\{G,\delta_{\circ}\\}=G$ in
$\mathbb{R}^{n}\setminus B_{R_{\circ}}$ if $R_{\circ}$ is sufficiently large.)

On the other hand, to estimate the integral in $B_{R_{\circ}}\times(-2,-1)$ we
use that $w_{\varrho}=v_{\varrho}+\varrho^{3}(x_{n}Q_{2}+\bar{a}\Psi)$ inside
the integration domain, that $\text{\small\bf H}(x_{n}Q_{2})=0$, and
$\Psi\text{\small\bf H}\Psi\equiv 0$. This implies that

$\int_{-2}^{-1}\int_{B_{R_{\circ}}}\big{\\{}v_{\varrho_{\circ}}\text{\small\bf
H}v_{\varrho_{\circ}}-w_{\varrho_{\circ}}\text{\small\bf
H}w_{\varrho_{\circ}}\big{\\}}G=\int_{-2}^{-1}\int_{B_{R_{\circ}}}\big{\\{}v_{\varrho_{\circ}}\text{\small\bf
H}(\varrho_{\circ}^{3}\bar{a}\Psi)+(\varrho_{\circ}^{3}(x_{n}Q_{2}+\bar{a}\Psi))\text{\small\bf
H}v_{\varrho_{\circ}}\big{\\}}G$

Now, as in (10.9) we have the bound

$\bigg{|}\int_{-2}^{-1}\int_{B_{R_{\circ}}}(\varrho_{\circ}^{3}(x_{n}Q_{2}+\bar{a}\Psi))\text{\small\bf
H}v_{\varrho_{\circ}}G\bigg{|}\leq C\varrho_{\circ}^{7}.$

Also, using (10.17),

$0\leq-\int_{-2}^{-1}\int_{B_{R_{\circ}}}v_{\varrho_{\circ}}\text{\small\bf
H}(\varrho_{\circ}^{3}\Psi)G=-\int_{-2}^{-1}\int_{B_{R_{\circ}}}u_{\varrho_{\circ}}\text{\small\bf
H}(\varrho_{\circ}^{3}\Psi)\leq C\varrho_{\circ}^{7},$

thus, recalling (10.5),

$\int_{-2}^{-1}\int_{B_{R_{\circ}}}\big{\\{}v_{\varrho_{\circ}}\text{\small\bf
H}v_{\varrho_{\circ}}-w_{\varrho_{\circ}}\text{\small\bf
H}w_{\varrho_{\circ}}\big{\\}}G\geq-C\varrho_{\circ}^{7}\gg-\frac{c}{20}\varrho_{\circ}^{M}\geq-\frac{1}{20}H(\varrho_{\circ},v-h).$

This concludes the proof of (10.18) and therefore the proof of the
proposition. ∎

##  11\. The set $\Sigma^{*}$: $\varepsilon$-flatness vs accelerated decay

To continue our analysis, we now make a finite partition of $\Sigma^{*}$ as
follows: given $\delta>0$, we consider

$\Sigma^{*}=\bigcup_{\ell\in\mathbb{N}}\Sigma^{*}_{\delta,\ell},\qquad\text{with}\qquad\Sigma^{*}_{\delta,\ell}:=\bigg{\\{}(x_{\circ},t_{\circ})\in\Sigma^{*}\
:H(1,p^{*}_{3,x_{\circ},t_{\circ}})\in\frac{\delta}{2}[\ell,\ell+1)\bigg{\\}}.$
(11.1)

Since $1/C_{\varrho}\leq H(1,p^{*}_{3,x_{\circ},t_{\circ}})\leq C_{\varrho}$
for all
$(x_{\circ},t_{\circ})\in\Sigma^{*}\cap\big{(}B_{1-\varrho}\times(-1+\varrho,1)\big{)}$
(by Corollary 8.5), for any given $\delta>0$ the number of sets in the
previous partition restricted to compact subsets of $B_{1}\times(-1,1)$ is
bounded (their number is roughly $C/\delta$).

As we will see, the advantage of considering this partition is the following
key property:

######  Lemma 11.1.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1). For any
$\varepsilon>0$, $R_{\circ}>1$, and $\varrho\in(0,1)$, there exists
$\bar{\delta}>0$ such that the following holds provided
$\delta\leq\bar{\delta}$:
Given $\ell\in\mathbb{N}$ and
$(x_{\circ},t_{\circ})\in\Sigma_{\delta,\ell}^{*}\cap(B_{1-\varrho}\times(-1+\varrho,1))$,
there exists $r_{\circ}=r_{\circ}(x_{\circ},t_{\circ})>0$ such that

$\big{\|}u(x_{1}+\cdot,t_{1}+\cdot)-p_{2,x_{1},t_{1}}-p_{3,x_{1},t_{1}}^{*}\big{\|}_{L^{\infty}(B_{R_{\circ}r}\times(-2r^{2},-r^{2}/2))}\leq\varepsilon
r^{3}\qquad\forall\,r\in(0,r_{\circ})$

for all $(x_{1},t_{1})\in\Sigma^{*}_{\delta,\ell}$ such that
$|x_{1}-x_{\circ}|\leq r_{\circ}$.

###### Proof.

Let $\bar{\varepsilon}>0$ be a small constant to be fixed later, and let
$(x_{1},t_{1})$ be as in the statement. Define

$\boldsymbol{v}^{r}:=\frac{u(x_{1}+r\cdot,t_{1}+r^{2}\cdot)-r^{2}p_{2,x_{1},t_{1}}}{r^{3}}.$

We divide the proof in two steps.

$\bullet$ Step 1. We show that

$\|\boldsymbol{v}^{r}-p^{*}_{3,x_{1},t_{1}}\|^{2}_{L^{2}(B_{2R_{\circ}}\times(-4,-1/2))}\leq
C(R_{\circ})\bar{\varepsilon}\qquad\forall\,r\in(0,r_{\circ}).$ (11.2)

Indeed, by definition of $\Sigma_{\delta,\ell}^{*}$, since
$(x_{\circ},t_{\circ}),(x_{1},t_{1})\in\Sigma_{\delta,\ell}^{*}$ it holds

$\bigg{|}\int_{\\{t=-1\\}}(p_{3,x_{1},t_{1}}^{*})^{2}G-\int_{\\{t=-1\\}}(p_{3,x_{\circ},t_{\circ}}^{*})^{2}G\bigg{|}\leq\delta.$
(11.3)

Also, by Lemma 9.3, the quantity

$\mathcal{I}_{x_{1},t_{1}}(r,Q):=\frac{1}{r^{6}}\int_{\\{t=-r^{2}\\}}\big{(}u(x_{1}+\,\cdot\,,t_{1}+\,\cdot\,)-p_{2,x_{1},t_{1}}\big{)}\zeta\,Q\,G$

satisfies

$\frac{d}{dr}\mathcal{I}_{x_{1},t_{1}}(r,Q)\geq-C\|Q\|_{L^{2}(\mathscr{C}_{1})}$

for any $(x_{1},t_{1})\in\Sigma^{*}_{\delta,\ell}\cap
B_{1-\varrho}\times(-1+\varrho,1)$ and for any given 3-homogenous solution $Q$
of (6.17).

Hence, on the one hand we have

$\begin{split}\int_{\\{t=-1\\}}\boldsymbol{v}^{r}\,p^{*}_{3,x_{1},t_{1}}G=\mathcal{I}_{x_{1},t_{1}}(r,p^{*}_{3,x_{1},t_{1}})\geq\mathcal{I}_{x_{1},t_{1}}(0^{+},p^{*}_{3,x_{1},t_{1}})-Cr=\int_{\\{t=-1\\}}(p_{3,x_{1},t_{1}})^{2}G-Cr.\end{split}$
(11.4)

On the other hand, given $\varepsilon^{\prime}>0$, we can fix $s_{\circ}>0$
—depending on $(x_{\circ},t_{\circ})$ and $\varepsilon^{\prime}$— such that

$\int_{\\{t=-1\\}}(p_{3,x_{\circ},t_{\circ}}^{*})^{2}G+\varepsilon^{\prime}\geq
s_{\circ}^{-6}H\big{(}s_{\circ},(u(x_{\circ}+\,\cdot\,,t_{\circ}+\,\cdot\,)-p_{2,x_{\circ},t_{\circ}})\zeta\big{)}.$

Note that, since

$u(x_{1}+\,\cdot,t_{1}+\cdot\big{)}-p_{2,x_{1},t_{1}}\longrightarrow
u(x_{\circ}+\,\cdot,t_{\circ}+\cdot\big{)}-p_{2,x_{\circ},t_{\circ}}\quad\mbox{as
}(x_{1},t_{1})\to(x_{\circ},t_{\circ}),$

then for $(x_{1},t_{1})$ sufficiently close to $(x_{\circ},t_{\circ})$ and for
$r\in(0,s_{\circ})$ we will have

$\begin{split}\int_{\\{t=-1\\}}(p_{3,x_{\circ},t_{\circ}}^{*})^{2}G+2\varepsilon^{\prime}&\geq
s_{\circ}^{-6}H\big{(}s_{\circ},(u(x_{\circ}+\,\cdot\,,t_{\circ}+\,\cdot\,)-p_{2,x_{\circ},t_{\circ}})\zeta\big{)}+\varepsilon^{\prime}\\\
&=s_{\circ}^{-6}H\big{(}1,(u(x_{\circ}+s_{\circ}\,\cdot\,,t_{\circ}+s_{\circ}^{2}\,\cdot\,)-s_{\circ}^{2}p_{2,x_{\circ},t_{\circ}})\zeta\big{)}+\varepsilon^{\prime}\\\
&\geq
s_{\circ}^{-6}H\big{(}1,(u(x_{1}+s_{\circ}\,\cdot\,,t_{1}+s_{\circ}^{2}\,\cdot\,)-s_{\circ}^{2}p_{2,x_{1},t_{1}})\zeta\big{)}\\\
&=s_{\circ}^{-6}H\big{(}s_{\circ},(u(x_{1}+\cdot,t_{1}+\cdot)-p_{2,x_{1},t_{1}})\zeta\big{)}\\\
&\geq
r^{-6}H\big{(}r,(u(x_{1}+\cdot,t_{1}+\cdot)-p_{2,x_{1},t_{1}})\zeta\big{)}+O(e^{-c/s_{\circ}})\\\
&=\int_{\\{t=-1\\}}\big{(}\boldsymbol{v}^{r}\big{)}^{2}G+O(e^{-c/s_{\circ}}).\end{split}$

Combining this estimate with (11.4) and (11.3), and choosing first $s_{\circ}$
small enough and then $(x_{1},t_{1})$ sufficiently close to
$(x_{\circ},t_{\circ})$, we obtain

$\begin{split}2\int_{\\{t=-1\\}}\boldsymbol{v}^{r}\,p^{*}_{3,x_{1},t_{1}}G\geq\int_{\\{t=-1\\}}\big{(}(\boldsymbol{v}^{r})^{2}+(p^{*}_{3,x_{1},t_{1}})^{2}\big{)}G-\bar{\varepsilon}\qquad\forall\,r\in(0,s_{\circ}).\end{split}$

In other words,

$\int_{\\{t=-1\\}}(\boldsymbol{v}^{r}-p^{*}_{3,x_{1},t_{1}})^{2}G\leq\bar{\varepsilon}\qquad\forall\,r\in(0,s_{\circ}),$

and (11.2) follows.

$\bullet$ Step 2. We upgrade the $L^{2}$ bound from (11.2) to an $L^{\infty}$
bound.

For this, recall that $(p^{*}_{3,x_{1},t_{1}})^{2}$ is always divisible by
$p_{2,x_{1},t_{1}}$,therefore

$\mathcal{P}_{1}:=p_{2,x_{1},t_{1}p_{3}}+p^{*}_{3,x_{1},t_{1}}+{\textstyle\frac{(p^{*}_{3,x_{1},t_{1}})^{2}}{4p_{2,x_{1},t_{1}}}}\geq
0$

satisfies

$\text{\small\bf H}\mathcal{P}_{1}\leq 1+\text{\small\bf
H}\left({\textstyle\frac{(p^{*}_{3,x_{1},t_{1}})^{2}}{4p_{2,x_{1},t_{1}}}}\right)\leq
1+C(|x|^{2}+t).$

Thus, since $\text{\small\bf H}u=1$ in $\\{u>0\\}$,

$\text{\small\bf
H}\big{(}u(x_{1}+r\cdot,t_{1}+r^{2}\cdot)-\mathcal{P}_{1}(r\cdot,r^{2}\cdot)\big{)}_{+}\geq-
Cr^{4}\qquad\mbox{in }B_{2R_{\circ}r}\times(-3r^{2},-r^{2}/2).$

On the other hand since $\text{\small\bf H}\mathcal{P}_{1}=1+O(|x|^{2}+t)$
inside $\\{\mathcal{P}_{1}>0\\}$, while $\text{\small\bf H}u\leq 1$ and $u\geq
0$, we have

$\text{\small\bf
H}\big{(}u(x_{1}+r\cdot,t_{1}+r^{2}\cdot)-\mathcal{P}_{1}(r\cdot,r^{2}\cdot)\big{)}_{-}\geq-
Cr^{4}\qquad\mbox{in }B_{2R_{\circ}r}\times(-3r^{2},-r^{2}/2).$

This proves that

$\text{\small\bf
H}\big{|}u(x_{1}+r\cdot,t_{1}+r^{2}\cdot)-\mathcal{P}_{1}(r\cdot,r^{2}\cdot)\big{|}\geq-
Cr^{4}\qquad\mbox{in }B_{2R_{\circ}r}\times(-3r^{2},-r^{2}/2).$

Also, (11.2) implies that

$\big{\|}u(x_{1}+r\,\cdot\,,t_{1}+r^{2}\,\cdot\,)-\mathcal{P}_{1}(r\,\cdot\,,r^{2}\,\cdot\,)\big{\|}_{L^{2}(B_{2R_{\circ}r}\times(-3r^{2},-r^{2}/2))}\leq\bar{\varepsilon}r^{3}+Cr^{4}.$

Therefore, it follows from the one-sided Harnack inequality for the heat
equation that

$\big{\|}u(x_{1}+r\,\cdot\,,t_{1}+r^{2}\,\cdot\,)-\mathcal{P}_{1}(r\,\cdot\,,r^{2}\,\cdot\,)\big{\|}_{L^{\infty}(B_{R_{\circ}r}\times(-2r^{2},-r^{2}/2))}\leq
C\bar{\varepsilon}r^{3}+Cr^{4}.$

This implies the conclusion of the lemma by taking $\bar{\varepsilon}$ and $r$
sufficienlty small. ∎

We have the following:

######  Corollary 11.2.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1). Given
$\varrho\in(0,1)$, $C_{\circ}$, and $M\in(6,7)$, $\bar{\delta}>0$ such that
the following holds provided $\delta\leq\bar{\delta}$:
Let $\mathfrak{h}^{\frac{5}{2}}$ be as in (10.3), and recall (11.1). For any
given $\ell\in\mathbb{N}$ and
$(x_{\circ},t_{\circ})\in\Sigma_{\delta,\ell}^{*}\cap\big{(}B_{1-\varrho}\times(-1+\varrho,1)\big{)}$,
there exists $r_{\circ}=r_{\circ}(x_{\circ},t_{\circ})>0$ such that

$\mathfrak{h}^{\frac{5}{2}}\big{(}r,(u(x_{1}+\cdot,t_{1}+\cdot)-p_{2,x_{1},t_{1}}-Q_{x_{1},t_{1}})\zeta\big{)}\qquad\text{is
monotone nondecreasing for $r\in(0,r_{\circ})$}$

for all $(x_{1},t_{1})\in\Sigma^{*}_{\delta,\ell}$ with $|x_{1}-x_{\circ}|\leq
r_{\circ}$. Here $Q_{x_{1},t_{1}}=x_{n}Q_{2}+\bar{a}|x_{n}|(x_{n}^{2}+6t)$ in
coordinates such that $p_{2,x_{1},t_{1}}=\frac{1}{2}x_{n}^{2}$, where
$\bar{a}\in[0,C_{\circ}]$ and $Q_{2}$ is any 2-homogeneous polynomial
satisfying $\text{\small\bf H}(x_{n}Q_{2})=0$ and
$\|Q_{2}\|_{L^{2}(\mathscr{C}_{1})}\leq C_{\circ}$.

###### Proof.

It follows from Proposition 10.1 and Lemma 11.1. ∎

The next proposition will be crucial in our argument. It provides us with the
following powerful decay property: if $\Sigma^{*}_{\delta,\ell}$ is not
$\varepsilon$-flat at some small scale $r$, then we have an accelerated decay
for $u$ at such scale. This result will be at the core of the argument in the
next section, where we will prove that we have an accelerated decay at
$\Sigma^{*}$, outside a set of Hausdorff dimension at most $n-2$.

Recall that, given a function $q$, we denote by $q^{even}$ the even
symmetrization of $q$ with respect to the hyperplane $\\{p_{2}=0\\}$.

######  Proposition 11.3.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a solution of (3.1), let
$\varrho>0$, and let $\delta>0$ be given by Corollary 11.2. Let
$(x_{\circ},t_{\circ})\in\Sigma^{*}_{\delta,\ell}\cap\big{(}B_{1-\varrho}\times(-1+\varrho,1)\big{)}$
for some $\ell\in\mathbb{N}$, and choose coordinates so that
$p_{2,x_{\circ},t_{\circ}}=\frac{1}{2}x_{n}^{2}$ and
$(p_{3,x_{\circ},t_{\circ}}^{*})^{\rm even}=a_{x_{\circ},t_{\circ}}\Psi$,
where

$\Psi(x,t):=|x_{n}|(x_{n}^{2}+6t).$

Let $\hat{\alpha}:=1/3$, and for $r\in(0,\varrho]$ define

$A(r):=\min_{a\in\mathbb{R}}\big{\|}(u(x_{\circ}+\,\cdot,t_{\circ}+\,\cdot\,)-p_{2,x_{\circ},t_{\circ}}-p_{3,x_{\circ},t_{\circ}}^{*}-a\Psi)_{r}\big{\|}_{L^{2}(\mathscr{C}_{1})}\qquad\text{and}\qquad\Theta(r):=\max_{s\in[r,\varrho]}(r/s)^{3+\hat{\alpha}}A(s)\,.$

Then the following two properties hold.

(1) Fix $\varepsilon>0$. For any $\varepsilon_{1}$ there exists
$r_{\varepsilon_{1}}>0$ such that, for every $r\in(0,r_{\varepsilon_{1}})$,
the following holds:
Consider the “$\varepsilon$-flatness property”

$\pi_{x}(\Sigma^{*}_{\delta,\ell}\cap\\{t\leq t_{\circ}\\})\cap
B_{r}(x_{\circ})\subset L+\varepsilon B_{r}\qquad\text{for some linear space
$L$ with ${\rm dim}(L)\leq n-2$.}$ (11.5)

If (11.5) fails, then

$A(r)\leq\varepsilon_{1}\,\Theta(r).$

(2) For any $\varepsilon_{2}>0$ there exists
$\lambda_{\varepsilon_{2}}\in(0,1)$ such that, for all $r\in(0,\varrho)$ and
$\lambda\in(0,\lambda_{\varepsilon_{2}}]$,

$\Theta(\lambda r)\leq\lambda^{3-\varepsilon_{2}}\,\Theta(r).$

###### Proof.

Assume for notational simplicity that $(x_{\circ},t_{\circ})=(0,0)$ and
$\varrho=1$. We begin by proving (1).

$\bullet$ Proof of (1). Set

$\overline{a}_{r}:={\rm
arg\,min}_{a\in\mathbb{R}}\big{\|}(u-p_{2}-p_{3}^{*}-a\Psi)_{r}\big{\|}^{2}_{L^{2}(\mathscr{C}_{1})},\qquad
w^{r}:=\frac{(u-p_{2}-p_{3}^{*}-\overline{a}_{r}\Psi)_{r}}{\Theta(r)}.$

Note that, by construction, for any $s>0$ we have $\Theta(2s)\leq
2^{3+\hat{\alpha}}\Theta(s)$. Hence

$\big{\|}(u-p_{2}-p_{3}^{*}-\overline{a}_{2^{j}r}\Psi)_{2^{j}r}\big{\|}_{L^{2}(\mathscr{C}_{1})}=A(2^{j}r)\leq\Theta(2^{j}r)\leq(2^{j})^{3+\hat{\alpha}}\Theta(r)\qquad\forall\,j\geq
1.$

Rescaling and using the triangle inequality between two consecutive dyadic
scales, this gives

$\begin{split}c\big{|}\overline{a}_{2^{j+1}r}-\overline{a}_{2^{j}r}\big{|}2^{3j}&=\big{\|}(\overline{a}_{2^{j+1}r}\Psi-\overline{a}_{2^{j}r}\Psi)_{2^{j}r}\big{\|}_{L^{2}(\mathscr{C}_{1})}\\\
&\leq\big{\|}\big{(}u-p_{2}-p_{3}^{*}-\overline{a}_{2^{j+1}r}\Psi\big{)}_{2^{j}r}\big{\|}_{L^{2}(\mathscr{C}_{2})}+\big{\|}\big{(}u-p_{2}-p_{3}^{*}-\overline{a}_{2^{j}r}\Psi\big{)}_{2^{j}r}\big{\|}_{L^{2}(\mathscr{C}_{1})}\\\
&=2^{\frac{n+2}{2}}A(2^{j+1}r)+A(2^{j}r)\leq
C2^{(3+\hat{\alpha})j}\Theta(r).\end{split}$

Hence, summing the geometric series and using again the triangle inequality,
we obtain $|\overline{a}_{2^{j}r}-\overline{a}_{r}|\leq
C2^{j\alpha\circ}\Theta(r)$, which implies the growth control

$\big{\|}w^{r}(R\,\cdot\,,R^{2}\,\cdot\,)\big{\|}_{L^{2}(\mathscr{C}_{1})}\leq
CR^{3+\hat{\alpha}}\qquad\forall R\in(1,1/r).$ (11.6)

Now, assume by contradiction that there exist a constant $\varepsilon_{1}>0$
and a sequence $r_{k}\downarrow 0$ for which (11.5) fails at the scales
$r=r_{k}$ but $A(r_{r})\geq\varepsilon_{1}\,\Theta(r_{k})$, or equivalently

$\|w^{r_{k}}\|_{L^{2}(\mathscr{C}_{1})}>\varepsilon_{1},$ (11.7)

while the negation of (11.5) gives the existence of
$x_{k}^{(1)},x_{k}^{(2)},\dots,x_{k}^{(n-1)}\in\pi_{x}(\Sigma^{*}\cap\\{t\leq
0\\})\cap B_{r_{k}}$ such that
$y_{k}^{(i)}:=x_{k}^{(i)}/r_{k}\in\overline{B_{1}}$ satisfy

${\rm dist}\big{(}\\{y_{k}^{(i)}\\}_{1\leq i\leq
n-1},L\big{)}\geq\varepsilon\,\quad\ \forall\,L\subset\mathbb{R}^{n}\mbox{
linear space with }{\rm dim}(L)=n-2.$

The outline of the proof will go as follows.

In Step 1, we show that $w^{r_{k}}$ converges (up to a subsequence) strongly
in $L^{2}_{\rm loc}(\mathbb{R}^{n}\times(-\infty,0])$ to a some function
$w^{0}\not\equiv 0$ which satisfying the growth condition

$\|w^{0}(R\,\cdot\,,R^{2}\,\cdot\,)\|_{L^{2}(\mathscr{C}_{1})}\leq
R^{3+\hat{\alpha}},\qquad\forall R\in(1,\infty)$ (11.8)

and solves

$\begin{cases}\text{\small\bf H}w^{0}=0\quad&\mbox{ in }\\{x_{n}\neq 0\\}\\\
w^{0}=0&\mbox{ on }\\{x_{n}=0\\}.\end{cases}$ (11.9)

This will imply that $w^{0}$ is a caloric polynomial degree 3 in each side of
$\\{x_{n}=0\\}$.

In Step 2, we use Proposition 10.1 to show that

$\mbox{ $w^{0}$ is homogeneous of degree $3$}.$ (11.10)

In Step 3, we use the monotonicity formulae from Lemma 9.3 to show that

$w^{0}$ is even in the variable $x_{n}$, (11.11)

so that, up to a rotation fixing $\boldsymbol{e}_{n}$, $w^{0}$ must be of the
form

$w^{0}(x,t)\equiv
a\Psi+\sum_{i=1}^{n-1}a_{i}\,|x_{n}|(-3x_{i}^{2}+x_{n}^{2}).$ (11.12)

In addition we show that, thanks to the choice of $\bar{a}_{r}$, $w^{0}$
satisfies

$\int_{\mathscr{C}_{1}}w^{0}\Psi=0$ (11.13)

and hence $a=0$.

Finally, in Step 4 we exploit again the monotonicity formula from Proposition
10.1, but now at points $(x_{k}^{(i)},t_{k}^{(i)})$, in order to reach a
contradiction with (11.12)-(11.13) to complete the proof.

We now develop Steps 1-4.

$\bullet$ Step 1. We prove first the compactness of the functions $w^{r_{k}}$
in $C^{0}_{\rm loc}(\mathbb{R}^{n}\times(-\infty,0])$. To prove this we show
that, for any given $R>0$, there exists $k_{\circ}$ such that, for $k\geq
k_{\circ}$, we have

$\int_{B_{R}\times(-R^{2},0)}\bigl{(}|\nabla
w^{r_{k}}|^{2}+|\partial_{t}w|^{2}\bigr{)}\leq C_{R}.$ (11.14)

This will give compactness $w^{r_{k}}$ in $L^{2}_{\rm loc}$, and we will later
upgrade this convergence to a locally uniform one.

Since $\Theta(r_{k})\geq cr_{k}^{3+\hat{\alpha}}$ (thanks to (11.7)), arguing
similarly to (10.11) we get

$w^{r_{k}}\text{\small\bf
H}w^{r_{k}}\geq-C(r_{k})^{-2\hat{\alpha}}R^{4}\chi_{\\{|x_{n}|\leq
Cr_{k}R^{2}\\}}\quad\mbox{ in }\mathscr{C}_{r}.$ (11.15)

Using (10.12) with $\phi=w^{r_{k}}$ and with $\xi$ replaced by
$\xi_{R}(x):=\xi(x/R)$, where $\xi\in C^{\infty}_{c}(B_{2})$ is a nonnegative
cut-off function satisfying $\xi\equiv 1$ in $B_{1}$, we obtain

$\int_{B_{2R}\times\\{t\\}}\big{(}|w^{r_{k}}\text{\small\bf
H}w^{r_{k}}|+|\nabla
w^{r_{k}}|^{2}\big{)}\xi_{R}+\frac{d}{dt}\int_{B_{2R}\times\\{t\\}}(w^{r_{k}})^{2}\xi_{R}\leq\frac{C_{n}}{R^{2}}\int_{B_{2R}\times\\{t\\}}(w^{r_{k}})^{2}+Cr_{k}^{1-2\hat{\alpha}}R^{n+5}.$
(11.16)

for $t\in(-R^{2},0)$ and $R\in(0,1/2r_{k})$. Integrating (11.16) in time and
using (11.6), this yields

$\frac{1}{R^{2}}\int_{\mathscr{C}_{R}}|w^{r_{k}}|^{2}+\int_{\mathscr{C}_{R}}|\nabla
w^{r_{k}}|^{2}\leq C(R).$ (11.17)

Also, since $\partial_{tt}(u-p_{2}-p_{3,r}^{*})=\partial_{tt}u\geq-C$ (see
Proposition 3.4), it follows by scaling that $\partial_{tt}w_{r_{k}}\geq-
Cr_{k}^{1-\hat{\alpha}}$.

We now use the following elementary bound: for $f:(0,R)\to\mathbb{R}$,

${\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{0}^{R}R^{2}|f^{\prime}|^{2}\leq
C\bigg{(}{{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{0}^{R}|f|^{2}}+\sup_{(0,R)}R^{4}(f^{\prime\prime})^{2}_{-}\bigg{)},$

Applying this with $f(\,\cdot\,)=u(x,\,\cdot\,)$ for each fixed $x\in B_{R}$,
and then integrating in the $x$ variable, it follows by (11.17) that

$\int_{\mathscr{C}_{R}}|\partial_{t}w^{r_{k}}|^{2}dx\,dt\leq C(R)+o_{k}(1),$

where $o_{k}(1)\to 0$ as $k\to\infty$. This gives (11.14), and thus
$w^{r_{k}}\rightarrow w^{0}$ in $L^{2}_{\rm
loc}(\mathbb{R}^{n}\times(-\infty,0])$. Notice also that $w^{0}$ is nonzero by
(11.7). Moreover, reasoning exactly as in Step 2 in the proof of Lemma 11.1,
such $L^{2}_{\rm loc}$ convergence implies $C^{0}_{\rm loc}$ convergence.

The growth condition (11.8) follows from (11.6), so it remains to show that
$w^{0}$ is satisfies (11.9).

First, using Lemma 10.2,

$w^{r_{k}}|_{\\{x_{n}=0\\}}\rightarrow 0\quad\mbox{ in }C^{0}_{\rm
loc}(\\{x_{n}=0\\}).$

On the other hand, since $\text{\small\bf H}w^{r_{k}}=0$ in
$\mathbb{R}^{n}\setminus(\\{u_{r_{k}}=0\\}\cup\\{x_{n}=0\\})$ and the
Hausdorff distance of $\\{u_{r_{k}}=0\\}$ from $\\{x_{n}=0\\}$ converges
locally to zero, we obtain that $\text{\small\bf H}w^{r_{k}}\to 0$ in every
compact subset of $\\{x_{n}\neq 0\\}$. It follows that $w^{0}$ solves (11.9)
(in the weak sense).

Finally, by the growth condition (11.8) and the Liouville theorem for the heat
equation in a half-space, we deduce that $w^{0}$ is a caloric polynomial of
degree 3 on each side of $\\{x_{n}=0\\}$. Furthermore, since
$w^{0}|_{\\{x_{n}=0\\}}=0$, both such polynomials are divisible by $x_{n}$.

$\bullet$ Step 2. We show that $w^{0}$ must be homogeneous of degree three,
i.e., it cannot have linear or quadratic terms. This will follow by using
that, by Proposition 10.1, we have that

$\rho\mapsto\mathfrak{h}^{\frac{5}{2}}\big{(}\rho,(u-p_{2}-p_{3}^{*}-\overline{a}_{r_{k}}\Psi)\zeta\big{)}\quad\mbox{is
monotone increasing}.$ (11.18)

Indeed, rescaling (11.18), we have

$\rho\mapsto\max_{R\in[1,1/(r_{k}\rho)]}R^{-M}\rho^{-5}\tilde{H}\big{(}R\rho,w^{r_{k}}\zeta(\,\cdot\,/r_{k})\big{)}\quad\mbox{is
monotone increasing}.$ (11.19)

Now, taking $2(3+\hat{\alpha})<M<7$ (recall that $\hat{\alpha}=1/3$), thanks
to the growth control (11.6) we have

$\max_{R\geq
1}R^{-M}\tilde{H}\big{(}R,w^{r_{k}}\zeta(\,\cdot\,/r_{k})\big{)}\leq C_{1},$

with $C_{1}$ independent of $k$. Hence, by monotonicity, for all
$\rho\in(0,1)$ we have

$\rho^{-5}\tilde{H}\big{(}\rho,w^{r_{k}}\zeta(\,\cdot\,/r_{k})\big{)}\leq\max_{R\in[1,1/(r_{k}\rho)]}R^{-M}\rho^{-5}\tilde{H}\big{(}R\rho,w^{r_{k}}\zeta(\,\cdot\,/r_{k})\big{)}\leq
C_{1},$

and letting $k\to\infty$ we get

$\tilde{H}\big{(}\rho,w^{0}\big{)}\leq C_{1}\rho^{5}.$ (11.20)

Since we already know that $w^{0}$ must be a degree three polynomial when
restricted to each side of $\\{x_{n}=0\\}$, using (11.20) we conclude that it
must be 3-homogenous.

$\bullet$ Step 3. We now show that $w^{0}$ must be of the form

$w^{0}(x,t)\equiv\sum_{i=1}^{n-1}a_{i}\,|x_{n}|(-3x_{i}^{2}+x_{n}^{2}).$
(11.21)

We first show that $w^{0}$ satisfies (11.11). Indeed, by Lemma 9.3, we have

$\frac{1}{r^{3}}\int_{\\{t=-r^{2}\\}}(u-p_{2})\zeta\,P\,G\geq\int_{\\{t=1\\}}p_{3}^{*}\,P\,G-C(P)r$

for all $P$ odd 3-homogeneous caloric polynomial. Therefore, using that
$\Psi^{odd}=0$, we have

$\int_{\\{t=-1\\}}w^{r_{k}}\zeta(r_{k}\,\cdot\,)\,P\,G=\int_{\\{t=-r_{k}^{2}\\}}\frac{u-p_{2}-p_{3}^{*}-\overline{a}_{r}\Psi}{\Theta(r_{k})}\,\zeta\,P\,G-Cr_{k}^{4}/\Theta(r_{k}).$

Since $\Theta(r_{k})\geq cr_{k}^{3+\hat{\alpha}}$, it follows that
$r_{k}^{4}/\Theta(r_{k})\to 0$ as $k\to\infty$, and therefore

$\int_{\\{t=-1\\}}w^{0}\,P\,G\geq 0,\quad\mbox{for every 3-homogeneous odd
caloric polynomial $P$}.$

This implies that $w^{0}$ must be even, and therefore must be of the form
(11.12).

In addition, by definition of $\overline{a}_{r}$ and $w^{r}$ we obtain, for
all $r>0$, the following orthogonality condition holds:

$\int_{\mathscr{C}_{1}}w_{r}\,\Psi=0.$

Taking the limit as $r=r_{k}\downarrow 0$ we obtain (11.13), and (11.21)
follows.

$\bullet$ Step 4. Let $\boldsymbol{R}_{k}^{(i)}\in SO(n)$ denote the rotation
mapping $\\{p_{2,x_{k}^{(i)},t_{k}^{(i)}}=0\\}$ to
$\\{x_{n}=0\\}=\\{p_{2}=0\\}$ such that $\boldsymbol{R}_{k}^{(i)}-{\rm Id}$
has minimal Hilbert-Schmidt norm. Since
$p_{2,x_{k}^{(i)},t_{k}^{(i)}}-p_{2}(x_{k}^{(i)}+r_{k}\cdot,t_{k}^{(i)}+r^{2}\cdot)=O(r_{k}^{3})$
(this follows, for instance, using Corollary 8.5 and the triangle inequality),
we have that $|\boldsymbol{R}_{k}^{(i)}-{\rm Id}|\leq Cr_{k}$. Also, since
$(x_{k}^{(i)},t_{k}^{(i)})\in\Sigma_{n-1}$, we have
$p_{2,x_{k}^{(i)},t_{k}^{(i)}}=p_{2}\circ\boldsymbol{R}_{k}^{(i)}$.
Furthermore, thanks to Proposition 10.1, we know that

$\mathfrak{h}^{\frac{5}{2}}\big{(}r,(u(x_{k}^{(i)}+\cdot,t_{k}^{(i)}+\cdot)-(p_{2}-p_{3}^{*}-\overline{a}_{r_{k}}\Psi)\circ\boldsymbol{R}_{k})\zeta\big{)}\quad\mbox{is
monotone increasing}.$ (11.22)

Note that, since $(x_{k}^{(i)},t_{k}^{(i)})\in\Sigma_{n-1}$, we have
$p_{2,x_{k}^{(i)},t_{k}^{(i)}}=p_{2}\circ\boldsymbol{R}_{k}^{(i)}$.

Define now

$D_{k}^{(i)}:=\big{(}r_{k}^{2}p_{2}+r_{k}^{3}(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)\big{)}(x_{k}^{(i)}+\,\cdot\,,t_{k}^{(i)}+\,\cdot\,)-\big{(}r_{k}^{2}p_{2}+r_{k}^{3}(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)\big{)}\circ\boldsymbol{R}_{k}^{(i)}.$

Then

$\begin{split}W_{k}^{(i)}(x):&=u(x_{k}^{(i)}+r_{k}\,\cdot\,,t_{k}^{(i)}+r_{k}\,\cdot\,)-r_{k}^{2}p_{2,x_{k}^{(i)},t_{k}^{(i)}}-r_{k}^{3}(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)\circ\boldsymbol{R}_{k}^{(i)}\\\
&=(u-p_{2}-p_{3}^{*}-\overline{a}_{r_{k}}\Psi)(x_{k}^{(i)}+r_{k}\,\cdot\,,t_{k}^{(i)}+r_{k}\,\cdot\,)+D_{k}^{(i)}\\\
&=\Theta(r_{k})w^{r_{k}}(y_{k}^{(i)}+\,\cdot\,,s_{k}^{(i)}+\,\cdot\,)+D_{k}^{(i)}.\end{split}$

We want to compute the limit $D_{k}^{(i)}/\Theta(r_{k})$ for $i=1,\ldots,n-1$.
There are two alternatives:

  * (i)

either there exists $i\in\\{1,\ldots,n-1\\}$ such that
$\Theta(r_{k})=o\left(\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}\right)$ as
$k\to\infty$,

  * (ii)

or there exists $c>0$ such that $\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}\leq
c\Theta(r_{k})$ for all $i=1,\ldots,n-1$.

We want to reach a contradiction in both cases.

_\- Case (i)_. Fix $i\in\\{1,\ldots,n-1\\}$ such that
$\Theta(r_{k})=o\left(\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}\right)$. Then,
up to a subsequence we have

$\lim_{k}\frac{W_{k}^{(i)}}{\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}}=\lim_{k}\frac{D_{k}^{(i)}}{\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}}=:\tilde{D}^{(i)}_{\infty}.$

Note that, since $D_{k}^{(i)}$ belongs to a finite dimensional space, the
convergence is strong and
$\|\tilde{D}^{(i)}_{\infty}\|_{L^{2}(\mathscr{C}_{1})}=1$.

We want to understand the structure of $D_{\infty}^{(i)}$. First, since
$p_{2}=\frac{1}{2}x_{n}^{2}$ we have

$\frac{r_{k}^{2}\,p_{2}(x_{k}^{(i)}+\,\cdot\,,t_{k}^{(i)}+\,\cdot\,)-r_{k}^{2}\,p_{2}\circ\boldsymbol{R}_{k}^{(i)}}{\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}}\to
x_{n}(a^{\prime}\cdot x^{\prime})+c_{1}x_{n},$

where $a^{\prime}\in\mathbb{R}^{n-1}$ and $x^{\prime}=(x_{1},\dots,x_{n-1})$.

On the other hand,

$\begin{split}&\frac{r_{k}^{3}\,(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)(x_{k}^{(i)}+\,\cdot\,,t_{k}^{(i)}+\,\cdot\,)-r_{k}^{3}\,(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)\circ\boldsymbol{R}_{k}^{(i)}}{\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}}=\\\
&=\frac{r_{k}^{3}\big{(}(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)(x_{k}^{(i)}+\,\cdot\,,t_{k}^{(i)}+\,\cdot\,)-(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)\big{)}}{\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}}+\frac{r_{k}^{3}\big{(}(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)-(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)\circ\boldsymbol{R}_{k}^{(i)}\big{)}}{\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}}\end{split}$

Hence, since $(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)^{even}$ is a positive
multiple of $|x_{n}|(x_{n}^{2}+6t)$, we get

$\frac{r_{k}^{3}\big{(}(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)(x_{k}^{(i)}+\,\cdot\,,t_{k}^{(i)}+\,\cdot\,)-(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)\big{)}}{\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}}\to
c_{2}{\rm sign}(x_{n})x_{n}^{2}+c_{3}|x_{n}|+\big{[}\mbox{odd quadratic
polynomial}\big{]}.$

Also, recalling that $\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}\geq
r_{k}^{3+\hat{\alpha}}\gg r_{k}^{4}$ and $|\boldsymbol{R}_{k}^{(i)}-{\rm
Id}|\leq Cr_{k}$, we have

$\frac{r_{k}^{3}\big{(}(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)-(p_{3}^{*}+\overline{a}_{r_{k}}\Psi)\circ\boldsymbol{R}_{k}^{(i)}\big{)}}{\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}}\to
0.$

This proves that

$\tilde{D}_{\infty}^{(i)}=c_{2}{\rm
sign}(x_{n})x_{n}^{2}+c_{3}|x_{n}|+\big{[}\mbox{odd quadratic
polynomial}\big{]}.$ (11.23)

On the other hand, using (11.22) and recalling the definition of
$\mathfrak{h}^{\frac{5}{2}}$ in (10.3), we obtain that

$\rho\mapsto\max_{R\in[1,1/(r_{k}\rho)]}R^{-M}\rho^{-5}\tilde{H}\big{(}R\rho,W_{k}^{(i)}\zeta(\,\cdot\,/r_{k})/\|D_{k}\|_{L^{2}(\mathscr{C}_{1})}\big{)}\quad\mbox{is
monotone increasing},$ (11.24)

and by the same argument as in (11.20), we get

$\tilde{H}\big{(}\rho,W_{k}^{(i)}\zeta(\,\cdot\,/r_{k})/\|D_{k}\|_{L^{2}(\mathscr{C}_{1})}\big{)}\leq
C\rho^{5}\qquad\Rightarrow\qquad\tilde{H}(\rho,\tilde{D}_{\infty})\leq
C\rho^{5}.$

Recalling (11.23) we deduce that $\tilde{D}_{\infty}=0$, a contradiction.

_\- Case (ii)_. In this case, for each $i=1,\ldots,n-1$ we have

$\lim_{k}\frac{W_{k}^{(i)}}{\Theta(r_{k})}=w^{r_{k}}(y_{k}^{(i)}+\,\cdot\,,s_{k}^{(i)}+\cdot)+\lim_{k}\frac{D_{k}^{(i)}}{\Theta(r_{k})}$

where $y_{k}^{(i)}:=x_{k}^{(i)}/r_{k}\in\overline{B_{1}}$ and $-C\leq
s_{k}^{(i)}=t_{k}^{(i)}/r_{k}^{2}\leq 0$ (cp. (9.4)).

Up to taking a subsequence we can define
$\tilde{D}^{(i)}_{\infty}:=\lim_{k}\frac{D_{k}^{(i)}}{\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}}$,
so that
$\lim_{k}\frac{D_{k}^{(i)}}{\Theta(r_{k})}=\lim_{k}\frac{\|D_{k}^{(i)}\|_{L^{2}(\mathscr{C}_{1})}}{\Theta(r_{k})}\tilde{D}_{\infty}^{(i)}$
and therefore

$\lim_{k}\frac{W_{k}^{(i)}}{\Theta(r_{k})}=w^{0}(y_{\infty}^{(i)}+\,\cdot\,,s_{\infty}^{(i)}+\,\cdot\,)+c^{(i)}\tilde{D}^{(i)}_{\infty},$
(11.25)

where the points $y_{\infty}^{(i)}\in\overline{B_{1}}\cap\\{x_{n}=0\\}$ are
linearly independent, $-C\leq s_{\infty}^{(i)}\leq 0$, $c^{(i)}\geq 0$, and
$\tilde{D}^{(i)}_{\infty}$ is of the form (11.23) (by the same argument as the
one above, in Case (i)).

Also, thanks to (11.22) and recalling the definition of
$\mathfrak{h}^{\frac{5}{2}}$ in (10.3), we have that

$\rho\mapsto\max_{R\in[1,1/(r_{k}\rho)]}R^{-M}\rho^{-5}\tilde{H}\big{(}R\rho,W_{k}^{(i)}\zeta(\,\cdot\,/r_{k})/\Theta(r_{k})\big{)}\quad\mbox{is
monotone increasing},$ (11.26)

and as in Case (i) we get

$\tilde{H}(\rho,w^{0}(y_{\infty}^{(i)}+\,\cdot\,,s_{\infty}^{(i)}+\,\cdot\,)+c^{(i)}\tilde{D}^{(i)}_{\infty})\leq
C\rho^{5}\qquad\forall\,i=1,\ldots,n-1.$ (11.27)

Recall that $w^{0}$ is of the form (11.21), while each
$\tilde{D}^{(i)}_{\infty}$ is of the form (11.23). Hence, since the points
$y_{\infty}^{(i)}$ are $n-1$ linearly independent in $\\{x_{n}=0\\}$, one can
easily check that the only possibility is $w^{0}\equiv 0$ and $c^{i}=0$ for
all $i$. However, recalling (11.14), taking the limit in (11.7) we obtain
$\|w^{0}\|_{L^{2}(\mathscr{C}_{1})}\geq\varepsilon_{1}$, a contradiction.

This completes the proof of part (1) of the proposition.

$\bullet$ Proof of (2). This follows from a modified (and simpler) version of
the argument given for part (1). We will prove that, for any given
$\varepsilon_{2}>0$, there exists $C_{\circ}\geq 1$ such that we have

$A(\lambda r)\leq
C_{\circ}\lambda^{3-\varepsilon_{2}/2}\,\Theta(r)\qquad\forall\,r,\lambda\in(0,1).$
(11.28)

First notice that (11.28) implies the conclusion. Indeed, by the definition of
$\Theta$ and (11.28),

$\begin{split}\Theta(\lambda r)&=\max_{s\in[\lambda r,1]}(\lambda
r/s)^{3+\hat{\alpha}}A(s)=\max\Big{\\{}\lambda^{3+\hat{\alpha}}\Theta(r),\max_{s\in[\lambda
r,r]}(\lambda r/s)^{3+\hat{\alpha}}A(s)\Big{\\}}\\\
&\leq\max\left\\{\lambda^{3+\hat{\alpha}}\Theta(r),C_{\circ}\lambda^{3-\varepsilon_{2}/2}\,\Theta(r)\right\\}=C_{\circ}\lambda^{3-\varepsilon_{2}/2}\,\Theta(r)\leq\lambda^{3-\varepsilon_{2}}\,\Theta(r),\end{split}$

for $\lambda>0$ sufficiently small.

To prove (11.28) we reason by contradiction and compactness. Assume by
contradiction that we have sequences $r_{k}\in(0,1)$ and $\lambda_{k}\in(0,1)$
such that

$A(\lambda_{k}r_{k})\geq k\lambda_{k}^{3-\varepsilon_{2}/2}\,\Theta(r_{k}),$
(11.29)

and define

$\Omega_{k}(\lambda)=\max_{\lambda^{\prime}\in[\lambda,1]}(\lambda/\lambda^{\prime})^{3-\varepsilon_{2}/2}\Theta(\lambda^{\prime}r_{k}).$

Using (11.29) and the definition of $\Theta$ we have

$\lambda_{k}^{-3+\varepsilon_{2}/2}\frac{\Theta(\lambda_{k}r_{k})}{\Theta(r_{k})}\to\infty\qquad\text{as
}k\to\infty.$ (11.30)

Note that

for any $\theta_{\circ}\in(0,1)$, there exists a constant
$C_{\theta_{\circ}}>0$ such that $\Theta(\theta_{\circ}r)\leq
C_{\theta_{\circ}}\Theta(r)$ for all $r\in(0,1)$. (11.31)

Also, by (11.30) and the definition of $\Omega_{k}$, there exists
$\lambda^{\prime}_{k}\in[\lambda_{k},1]$ such that

$\Omega_{k}(\lambda_{k})=(\lambda_{k}/\lambda^{\prime}_{k})^{3-\varepsilon_{2}/2}\Theta(\lambda^{\prime}_{k}r_{k})=\max_{\lambda^{\prime}\in[\lambda_{k}^{\prime},1]}(\lambda_{k}/\lambda^{\prime})^{3-\varepsilon_{2}/2}\Theta(\lambda^{\prime}r_{k})\geq(\lambda_{k})^{3-\varepsilon_{2}/2}\Theta(r_{k}).$

Thanks to these two facts we deduce that $\lambda^{\prime}_{k}\to 0$. Also, by
the maximality of $\lambda_{k}^{\prime}$,

$\Omega_{k}(\tau)=(\tau/\lambda^{\prime}_{k})^{3-\varepsilon_{2}/2}\Theta(\lambda^{\prime}_{k}r_{k})\qquad\forall\,\tau\in[\lambda_{k},\lambda_{k}^{\prime}],$

and by the definitions of $\Theta$ and $\Omega_{k}$ we deduce that

$A(\lambda^{\prime}_{k}r_{k})=\Theta(\lambda^{\prime}_{k}r_{k})=\Omega_{k}(\lambda_{k}^{\prime}).$
(11.32)

We now define

$W^{k}:=\frac{(u-p_{2}-p_{3}^{*}-\overline{a}_{\lambda_{k}r_{k}}\Psi)_{\lambda^{\prime}_{k}r_{k}}}{\Omega_{k}(\lambda^{\prime}_{k})}.$

By (11.32) we have

$\|W^{k}\|_{L^{2}(\mathscr{C}_{1})}\geq 1.$

Also, by construction,

$\big{\|}W^{k}(R\,\cdot\,,R^{2}\,\cdot\,)\big{\|}_{L^{2}(\mathscr{C}_{1})}\leq\begin{cases}CR^{3-\varepsilon_{2}/2}\quad&\mbox{for
}R\in[1,1/\lambda_{k}^{\prime}]\\\
C(1/\lambda_{k}^{\prime})^{3-\varepsilon_{2}/2}(R/\lambda_{k}^{\prime})^{3+\alpha}\quad&\mbox{for
}R\in[1/\lambda_{k}^{\prime},1/(\lambda_{k}^{\prime}r_{k})].\\\ \end{cases}$
(11.33)

Hence, repeating the same reasoning as in Step 1 above, we find that
$W^{k}\to\widetilde{W}$ where

$\|\widetilde{W}\|_{L^{2}(\mathscr{C}_{1})}\geq 1,\qquad\text{\small\bf
H}\widetilde{W}=0\quad\mbox{in }\\{x_{n}\neq
0\\},\qquad\widetilde{W}=0\quad\mbox{on }\\{x_{n}=0\\}.$

Furthermore, thanks to (11.33) and the fact that $\lambda_{k}^{\prime}\to 0$,
$\widetilde{W}$ satisfies the sub-cubic growth

$\big{\|}\widetilde{W}(R\,\cdot\,,R^{2}\,\cdot\,)\big{\|}_{L^{2}(\mathscr{C}_{1})}\leq
CR^{3-\varepsilon_{2}/2}\quad\mbox{for all }R\geq 1.$

This implies that $\widetilde{W}$ is a caloric polynomial of degree 2 on each
side of the hyperplane $\\{x_{n}=0\\}$. On the other hand, using (11.33)
again, we can repeat the argument in Step 2 above to show that
$\tilde{H}\big{(}\rho,\widetilde{W}\big{)}\leq C_{1}\rho^{5}$ for all $\rho>0$
(cp. (11.20)). This implies that $\widetilde{W}\equiv 0$, a contradiction. ∎

##  12\. An estimate of order $3+\beta$ at “most” singular points

In this section we prove that, at “most” points of $\Sigma^{*}$, we have a
decay of order $3+\beta$ as in the following definition.

######  Definition 12.1.

Denote by $\Sigma^{\diamond}$ the set of points
$(x_{\circ},t_{\circ})\in\Sigma^{*}$ such that, for some positive numbers
$\beta$ and $r_{\circ}$ (which may depend on the point), we have

$\big{\|}\big{(}u(x_{\circ}+\,\cdot\,,t_{\circ}+\,\cdot\,)-p_{2,x_{\circ},t_{\circ}}-p_{3,x_{\circ},t_{\circ}}^{*}\big{)}_{r}\big{\|}_{L^{2}(\mathscr{C}_{1})}\leq
r^{3+\beta}\quad\mbox{for all }r\in(0,r_{\circ}).$ (12.1)

We want to show the following:

######  Proposition 12.2.

Assume that $u:B_{1}\times(-1,1)\to[0,\infty)$ is a solution of (3.1). Then,

${\rm
dim}_{\mathcal{H}}\big{(}\pi_{x}(\Sigma^{*}\setminus\Sigma^{\diamond})\big{)}\leq
n-2.$

We will prove Proposition 12.2 by carefully exploiting Proposition 11.3,
combined with some delicate covering arguments.

Let $\delta>0$ be given by Corollary 11.2, and $\ell\in\mathbb{N}$. Also, let
$\varepsilon>0$ to be fixed later. Given
$(x_{\circ},t_{\circ})\in\Sigma^{*}_{\delta,\ell}$, let
$\kappa_{\circ}=\lfloor\log_{2}r_{\circ}\rfloor+1$. For $j\geq
k_{x_{\circ},t_{\circ}}$ we define

$N_{\varepsilon}(x_{\circ},t_{\circ},j):=\\#\big{\\{}i\in\\{\kappa_{\circ},\ldots,j\\}\
:\mbox{ \eqref{eq:Pr} applied to $u(x_{\circ}+\,\cdot\,,\
t_{\circ}+\,\cdot\,)$ does \emph{not} hold at scale $r=2^{-i}$}\big{\\}},$
(12.2)

and

$\omega_{\varepsilon}(x_{\circ},t_{\circ}):=\liminf_{j\to\infty}\frac{N_{\varepsilon}(x_{\circ},t_{\circ},j)}{j}.$
(12.3)

Notice that $\omega_{\varepsilon}$ is nonincreasing in $\varepsilon$.

The goal of this section will be to prove the following two properties:

  1. (i)

if $\omega_{\varepsilon}(x_{\circ},t_{\circ})>0$ for some $\varepsilon>0$,
then $(x_{\circ},t_{\circ})\in\Sigma^{\diamond}$;

  2. (ii)

for any $\ell\in\mathbb{N}$, the $\pi_{x}$-projection of the set
$\\{(x_{\circ},t_{\circ})\in\Sigma^{*}_{\delta,\ell}\,:\,\omega_{\varepsilon}(x_{\circ},t_{\circ})=0\
\forall\,\varepsilon>0\\}$ has Hausdorff dimension at most $n-2$.

The implication (i) will follow by Proposition 11.3(2), thanks to the
definition of $\omega(x_{\circ},t_{\circ})$. It is the content of the
following:

######  Lemma 12.3.

Assume that $\omega_{\varepsilon}(x_{\circ},t_{\circ})>0$ for some
$\varepsilon>0$. Then there exist $r_{\circ},\beta>0$ such that (12.1) holds.

###### Proof.

For notational simplicity we assume that $(x_{\circ},t_{\circ})=(0,0)$. By
assumption there exists $N_{\circ}\in\mathbb{N}$ such that

$\omega_{\varepsilon}(0,0):=\liminf_{j\to\infty}\frac{N_{\varepsilon}(0,0,j)}{j}\geq\frac{2}{N_{\circ}}.$

Hence, there exists $j_{\circ}$ such that

$\frac{N(0,0,j)}{j}\geq\frac{1}{N_{\circ}}\qquad\forall\,j\geq j_{\circ}.$

This means that there exists a set
$\mathcal{I}_{\varepsilon}\subset\\{\kappa_{\circ},\ldots\\}$ such that

$\mbox{ \eqref{eq:Pr} does not hold at scales $r=2^{-i}$ for all
$i\in\mathcal{I}_{\varepsilon}$,
}\qquad\\#\big{\\{}\mathcal{I}_{{\varepsilon}}\cap\\{\kappa_{\circ},\dots,j\\}\big{\\}}\geq\frac{j}{N_{\circ}}\quad\forall\,j\geq
j_{\circ}.$ (12.4)

In particular, given $\varepsilon_{1}>0$, up to taking $j_{\circ}$ larger if
needed (depending on $\varepsilon_{1}$), it follows from Proposition 11.3(1)
that

$A(2^{-i})\leq\varepsilon_{1}\Theta(2^{-i}),\qquad\mbox{whenever
}i\in\mathcal{I}_{{\varepsilon}}.$ (12.5)

Also, given $\varepsilon_{2}>0$, it follows from Proposition 11.3(2) that

$A(\lambda
2^{-i})\leq\lambda^{3-\varepsilon_{2}}\,\Theta(2^{-i}),\quad\mbox{whenever
}i\not\in\mathcal{I}_{{\varepsilon}}\quad\mbox{and}\quad\lambda\in(0,\lambda_{\varepsilon_{2}}).$
(12.6)

We now start to fix the different parameters.

First, we set $\beta:=\frac{1}{8N_{\circ}}$ and choose $\varepsilon_{2}>0$
small enough so that

$(3-\varepsilon_{2})\bigg{(}1-\frac{1}{N_{\circ}}\bigg{)}+\frac{10}{3N_{\circ}}\geq
3+2\beta.$ (12.7)

Then, we fix $M\in\mathbb{N}$ such that $2^{-M}<\lambda_{\varepsilon_{2}}$ and
we note that, by the definition of $A$ (see Proposition 11.3)

$i\in\mathcal{I}_{\varepsilon}\qquad\Rightarrow\qquad A(\lambda
2^{-i})\leq\lambda^{-(n+2)}A(2^{-i})\leq\lambda^{-(n+2)}\varepsilon_{1}\Theta(2^{-i}).$

Hence, choosing $\varepsilon_{1}>0$ small so that
$2^{(n+2)M}\varepsilon_{1}\leq 2^{-\frac{10}{3}M}$, it follows from the bound
above and the definition of $\Theta$ (see Proposition 11.3 and recall that
$\hat{\alpha}=1/3$) that

$i\in\mathcal{I}_{\varepsilon}\qquad\Rightarrow\qquad\Theta(2^{-i-M})\leq
2^{-\frac{10}{3}M}\Theta(2^{-i}).$ (12.8)

Now, to conclude the proof, given $k\in\mathbb{N}$ such that
$j_{k}:=\kappa_{\circ}+(kN_{\circ}+1)M\geq j_{\circ}$, we note that there
exists $m_{k}\in\\{1,\dots,M\\}$ such that

$\\#\big{(}\mathcal{I}_{{\varepsilon}}\cap(M\mathbb{Z}+\kappa_{\circ}+m_{j})\cap\\{\kappa_{\circ},\dots,j_{k}\\}\big{)}\geq
k.$

Hence, if we set
$\tilde{\mathcal{I}}_{k}:=\big{(}\mathcal{I}_{{\varepsilon}}\cap(M\mathbb{Z}+\kappa_{\circ}+m_{k})\cap\\{\kappa_{\circ},\ldots,j_{k}\\}\big{)}$,
it follows from (12.8) that

$i\in\tilde{\mathcal{I}}_{k}\qquad\Rightarrow\qquad\Theta(2^{-i-M})\leq
2^{-4M}\Theta(2^{-i}).$

On the other hand, by Proposition 11.3(2) we have

$i\in\bigl{(}(M\mathbb{Z}+\kappa_{\circ}+m_{k})\cap\\{\kappa_{\circ},\ldots,j\\}\bigr{)}\setminus\tilde{\mathcal{I}}_{k}\qquad\Rightarrow\qquad\Theta(2^{-i-M})\leq
2^{-(3-\varepsilon_{2})M}\Theta(2^{-i}).$

Combining these two informations and recalling that
$\\#\tilde{\mathcal{I}}_{k}\geq k$, it follows from (12.7) that

$\Theta(2^{-(\kappa_{\circ}+m_{k}+kN_{\circ}M)})\leq
2^{-\left(\frac{10}{3}M\,\\#\tilde{\mathcal{I}}_{k}+(3-\varepsilon_{2})[kN_{\circ}M-\\#\tilde{\mathcal{I}}_{k}]\right)}\Theta(2^{-(\kappa_{\circ}+m_{k})})\\\
\leq
2^{-[10/3+(3-\varepsilon_{2})(N_{0}-1)]kM}\Theta(2^{-(\kappa_{\circ}+m_{k})})\leq
2^{-(3+2\beta)kN_{\circ}M}\Theta(2^{-(\kappa_{\circ}+m_{k})}).$

Since $m_{k}\in\\{0,\ldots,M-1\\}$, using (11.31) we get

$\Theta(2^{-(\kappa_{\circ}+M+kN_{\circ}M)})\leq
C_{M}2^{-(3+2\beta)kN_{\circ}M}\Theta(2^{-\kappa_{\circ}})\qquad\forall\,k\in\mathbb{N},$

for some constant $C_{M}$ depending only on $M$. Thus, by (11.31) and the
definition of $\Theta$ we get

$A(r)\leq\Theta(r)\leq C_{N_{0},M,\kappa_{0}}r^{3+2\beta}\quad\mbox{for
}r\in(0,2^{-\kappa_{\circ}}),$

and the lemma follows by choosing $r_{\circ}$ sufficiently small. ∎

We now want to show property (ii). This will follow from the following GMT
results.

######  Lemma 12.4.

Let $m\leq n$ be positive integers. Given $\alpha\in(0,1)$ there exists
$\rho_{\circ}>0$ so that, for any $\rho\leq\rho_{\circ}$, there exist small
constants $\varepsilon_{\alpha,\rho},\omega_{\alpha,\rho}>0$ such that the
following statement holds for $\varepsilon\leq\varepsilon_{\alpha,\rho}$ and
$\omega\leq\omega_{\alpha,\rho}$.

Let $E\subset B_{1/2}(z)\subset\mathbb{R}^{n}$ for some $z\in\mathbb{R}^{n}$.
For every $x\in E$ and $j\geq 1$, define

$\begin{split}N^{\varepsilon,\rho}_{m}(E,x,j)&:=\\#\big{\\{}i\in\\{0,\dots,j\\}\
:\ E\cap B_{\rho^{i}}(x)\subset y+L+B_{\rho^{i}\varepsilon}\\\ &\hskip
56.9055pt\mbox{ for some $y\in\mathbb{R}^{n}$ and $L\subset\mathbb{R}^{n}$
linear subspace with ${\rm dim}(L)\leq m$}\big{\\}}.\end{split}$ (12.9)

Assume that, for some $j\geq 1$, we have

$N^{\varepsilon,\rho}_{m}(E,x,j)\geq(1-\omega)j\quad\mbox{ for all }x\in E.$
(12.10)

Then $E$ can be covered by $\rho^{-(m+\alpha)j}$ balls of radius $\rho^{j}$.

To prove it, we will need the following simple result (see [FRS20, Lemma
7.2]):

######  Lemma 12.5.

Let $B_{r}(x)\subset\mathbb{R}^{n}$ be an open ball, and $L$ be a
$m$-dimensional linear subspace (not necessarily passing through $x$). Let
$\beta_{1}>m$. Then there exists $\hat{\tau}=\hat{\tau}(m,\beta_{1})>0$ such
that the following holds.

Let $F\subset\mathbb{R}^{n}$ satisfy

$F\subset B_{r}(x)\cap\\{y\,:\,{\rm dist}(y,L)\leq\tau r\\},\qquad\text{for
some $0<\tau\leq\hat{\tau}$, $x\in\mathbb{R}^{n}$, $r>0$.}$

Then $F$ be covered with $\gamma^{-\beta_{1}}$ balls of radius $\gamma r$
centered at points of $F$, where $\gamma:=5\tau$.

We now prove Lemma 12.4.

###### Proof of Lemma 12.4.

We will prove the following:

Claim. Let $\ell\geq 1$ and $0\leq k\leq\ell$. If $E\subset B_{1/2}(z)$
satisfies

$N^{\varepsilon,\rho}_{m}(E,x;\ell)\geq k\quad\mbox{ for all }x\in E,$ (12.11)

then $E$ can be covered by $\rho^{-(n+\alpha)(\ell-k)}\rho^{-(m+\alpha)k}$
balls of radius $\rho^{\ell}$.

To prove the Claim, we shall proceed by induction on $\ell\geq 1$. However, we
remark first that the case $k=0$ and $\ell$ arbitrary, we simply use that
$E\subset B_{1/2}$ to deduce that $E$ can be covered by
$C_{n}\rho^{n}\leq\rho^{-(n+\alpha)}$ balls of radius $\rho$ (for $\rho$
small). Hence, in the induction procedure, we can assume that $k\geq 1$.

$\bullet$ The case $\ell=1$. In this case the only option is $\ell=k=1$. Then
the assumption $N^{\varepsilon,\rho}_{m}(E,x,1)\geq 1$ implies $E\cap
B_{1/2}(x)\subset y+L+B_{\varepsilon}$, so Lemma 12.5 with $\tau=\rho/5$ and
$r=1/2$ yields that $E$ can be covered by $\rho^{-(m+\alpha)}$ balls of radius
$\rho$ (provided $\rho$ is small enough so that $\tau\leq\hat{\tau}$, and then
$\varepsilon$ is chosen sufficiently small so that $\varepsilon\leq\tau r$).

$\bullet$ The inductive step. Assume that the claim is true for $\ell-1\geq 1$
and for all $k=1,\ldots\ell-1$. We now prove it for $\ell$, and
$k=1,\ldots,\ell$. There are two cases to consider.

_\- Case 1._ Assume there exists $x\in E$ such that

$E\cap B_{1}(x)\subset x+L+B_{\varepsilon}.$ (12.12)

Then, since $E\subset B_{1/2}(z)$, as before we can apply Lemma 12.5 to deduce
that $E$ can be covered (provided $\rho$ and $\varepsilon$ are chosen small
enough) by $\rho^{-(m+\alpha)}$ balls of radius $\rho/2$. Let us call these
balls $\\{B_{\rho/2}(z_{q})\\}_{q\in\mathcal{I}}$,
$\\#\mathcal{I}\leq\rho^{-(m+\alpha)}$.

Now, for any $q\in\mathcal{I}$ we define $E_{q}:=\frac{1}{\rho}\big{(}E\cap
B_{\rho/2}(z_{q})\big{)}\subset B_{1/2}(z_{q}/\rho)$ and we observe that,
thanks to (12.11),

$N^{\varepsilon,\rho}_{m}(E_{q},x;\ell-1)\geq
k-1\qquad\forall\,q\in\mathcal{I}.$

Therefore, by induction hypothesis, each set $E_{q}$ can be covered by
$\rho^{-(n+\alpha)(\ell-k)}\rho^{-(m+\alpha)(k-1)}$ balls of radius
$\rho^{\ell-1}$. This implies that the union of all these balls multiplied by
a factor $\rho$ covers $E$, and the total number of such balls is
$\rho^{-(n+\alpha)(\ell-k)}\rho^{-(m+\alpha)(k-1)}\\#\mathcal{I}\leq\rho^{-(n+\alpha)(\ell-k)}\rho^{-(m+\alpha)}$,
as desired.

_\- Case 2._ Assume that for none of the points $x\in E$ (12.12) holds. In
particular this implies that $k\leq\ell-1$ and therefore, thanks to (12.11),
for all $x\in E$ we have

$k\leq N^{\varepsilon,\rho}_{m}(E,x;\ell)=\\#\big{\\{}i\in\\{1,\dots,\ell\\}\
:\ E\cap B_{\rho^{i}}(x)\subset x+L+B_{\rho^{i}\varepsilon}\big{\\}}.$ (12.13)

(Note that, in the formula above, $i\geq 1$ instead of $i\geq 0$ as in the
definition of $N^{\varepsilon}$.)

Now, let us cover the ball $B_{1/2}(z)$ by $\rho^{-(n+\alpha)}$ balls
$\\{B_{\rho/2}(z_{q})\\}_{q\in\mathcal{I}}$ of radius $\rho/2$. In particular
these balls cover $E$. As in Case 1 we define
$E_{q}:=\frac{1}{\rho}\big{(}E\cap B_{\rho/2}(z_{q})\big{)}\subset
B_{1/2}(z_{q}/\rho)$, and it follows from (12.13) that

$N^{\varepsilon,\rho}_{m}(E_{q},x;\ell-1)\geq
k\qquad\forall\,q\in\mathcal{I}.$

By induction hypothesis, each set $E_{q}$ can be covered by
$\rho^{-(n+\alpha)(\ell-1-k)}\rho^{-(m+\alpha)k}$ balls of radius
$\rho^{\ell-1}$, and therefore $E$ can be covered by
$\rho^{-(n+\alpha)(\ell-1-k)}\rho^{-(m+\alpha)k}\\#\mathcal{I}\leq\rho^{-(n+\alpha)(\ell-k)}\rho^{-(m+\alpha)}$
balls of radius $\rho^{\ell}$. Thus, the Claim is proved.

Using the Claim, we finally show the result. Indeed, setting $\ell=j$ and
$k=\lfloor(1-\omega)j\rfloor$, we know that for any $\hat{\alpha}>0$ the set
$E$ can be covered by
$\rho^{-(n+\hat{\alpha})(j-\lfloor(1-\omega)j\rfloor)-(m+\hat{\alpha})\lfloor(1-\omega)j\rfloor}\leq\rho^{-(m+\hat{\alpha}+(n-m)\omega)j-1}$
balls of radius $\rho^{j}$. Choosing $\hat{\alpha}<\alpha-(n-m)\omega$, the
result follows by ensuring that $\rho$ is sufficiently small. ∎

######  Lemma 12.6.

Let $E\subset\mathbb{R}^{n}$. Given $\varepsilon,\omega>0$ small and $m\leq
n$, assume that

$\limsup_{j\to\infty}\frac{N^{\varepsilon,1/2}_{m}(E,x,j)}{j}\geq
1-\omega\qquad\forall\,x\in E,$ (12.14)

where $N^{\varepsilon,1/2}_{m}$ is defined as in (12.9). Then

${\rm dim}_{\mathcal{H}}(E)\leq m+\alpha,$

where $\alpha(n,\varepsilon,\omega)\to 0$ as $(\varepsilon,\omega)\to(0,0)$.

###### Proof.

Up to taking countable unions of we may assume that $E\subset B_{1/2}(z)$ for
some $z\in\mathbb{R}^{n}$.

We being by observing that, as a consequence of (12.14) and the definition
$N^{\varepsilon,1/2}_{m}$,

$\limsup_{j\to\infty}\frac{N^{\varepsilon,2^{-M}}_{m}(E,x,j)}{j}\geq
1-M\omega\qquad\forall\,M\geq 1.$ (12.15)

Hence we fix $\rho:=2^{-M}$ with $M$ sufficiently large so that
$2^{-M}\leq\rho_{\circ}$, where $\rho_{\circ}$ is given by Lemma 12.4.

Now, given $k\in\mathbb{N}$ large, we define

$j(x,k):=\min\\{\,j\geq k\ :\
N^{\varepsilon,\rho}_{m}(E,x,j)\geq(1-2M\omega)j\,\\}.$

Thanks to (12.15), we can partition $E$ as

$E=\bigcup_{\ell=k}^{\infty}E_{\ell}\qquad\mbox{ with}\quad E_{\ell}:=\\{x\in
E\ :\ j(x,k)=\ell\\}.$

Hence, given $\alpha>0$, provided $\varepsilon,\omega$ are small enough we can
apply Lemma 12.4 to deduce that $E_{\ell}\subset E\subset B_{1/2}(z)$ can be
covered by $\rho^{-(m+\alpha/2)\ell}$ balls of radius $\rho^{\ell}$.
Therefore, $E$ can be covered by balls of radius $r_{i}=\rho^{\ell_{i}}$ with
$\ell_{i}\geq k$ and $\\#\\{i:\ \ell_{i}=\ell\\}\leq\rho^{-(m+\alpha/2)\ell}$.
This implies that

$\mathcal{H}^{m+\alpha}_{2^{-k}}(E)\leq
C_{m+\alpha}\sum_{i}r_{i}^{m+\alpha}\leq
C_{m+\alpha}\sum_{\ell=k}^{\infty}\rho^{-(m+\alpha/2)\ell}(\rho^{\ell})^{m+\alpha}=C_{m+\alpha}\sum_{\ell=k}^{\infty}(\rho^{\alpha/2})^{\ell},$

and the right hand side can be made arbitrarily small by choosing $k$ large.
This proves that $\mathcal{H}^{m+\alpha}(E)=0$, and therefore ${\rm
dim}_{\mathcal{H}}(E)\leq m+\alpha$, as desired.

∎

We will also need the following modification of Lemma 12.6.

######  Proposition 12.7.

Let $E\subset\mathbb{R}^{n}\times\mathbb{R}$. Given $\varepsilon,\omega>0$
small, $m\leq n$, $(x,t)\in E$, and $j\geq 1$, define

$\begin{split}\widetilde{N}^{\varepsilon,\rho}_{m}(E,x,t,j):=\\#\big{\\{}&i\in
0,1,\dots,j\ :\
\pi_{x}\big{(}E\cap(\overline{B_{\rho^{i}}}(x)\times(-\infty,t])\big{)}\subset
x+L+\overline{B_{\rho^{i}\varepsilon}}\\\ &\hskip 85.35826pt\mbox{ for some
linear subspace $L\subset\mathbb{R}^{n}$ with ${\rm
dim}(L)=m$}\big{\\}}.\end{split}$ (12.16)

Assume that

$\limsup_{j\to\infty}\frac{\widetilde{N}^{\varepsilon,1/2}_{m}(E,x,t,j)}{j}\geq
1-\omega\qquad\forall\,(x,t)\in E.$ (12.17)

Then

${\rm dim}_{\mathcal{H}}\big{(}\pi_{x}(E)\big{)}\leq m+\alpha,$

where $\alpha(n,\varepsilon,\omega)\to 0$ as $(\varepsilon,\omega)\to(0,0)$.

###### Proof.

Fix $\alpha>0$, and let $\rho=2^{-M}$ with $M$ chosen large enough so that
$\rho^{\alpha/4}\leq\frac{1}{4}$ and $\rho\leq\rho_{\circ}$, with
$\rho_{\circ}$ given by Lemma 12.4. Up to taking countable unions, we may
assume that $\pi_{x}(E)\subset\overline{B_{1/4}}(z)\times[-1,1]$. Also, as in
the proof of the previous lemma, (12.17) implies

$\limsup_{j\to\infty}\widetilde{N}^{\varepsilon,\rho}(E,x,j)\geq 1-M\omega.$
(12.18)

Hence, given $k\in\mathbb{N}$ large, we define

$j(x,k):=\min\\{\,j\geq k\
:\widetilde{N}^{\varepsilon,\rho}_{m}(E,x,j)\geq(1-2M\omega)j\,\\},$

and we partition

$E=\bigcup_{\ell=k}^{\infty}E_{\ell}\qquad\mbox{with}\quad E_{\ell}:=\\{x\in
E\ :\ j(x,k)=\ell\\}.$

Since

$\widetilde{N}^{\varepsilon,\rho}_{m}(E_{\ell},x,t,\ell)\geq(1-2N\omega)\ell\qquad\forall\,(x,t)\in
E_{\ell},$

for any $(x,t)\in E_{\ell}$ there exist a subset
$\sigma_{x,t}\subset\\{0,\ldots,\ell\\}$ satisfying

$\\#\sigma_{x,t}\geq(1-2N\omega)\ell\qquad\text{and}\qquad\pi_{x}\big{(}E\cap(\overline{B_{\rho^{i}}}(x)\times(-\infty,t])\big{)}\subset
x+L_{x,i}+\overline{B_{\rho^{i}\varepsilon}}\quad\forall\,i\in\sigma_{x,t},$

where $L_{x,i}\subset\mathbb{R}^{n}$ is some liner subspace with ${\rm
dim}(L)=m$.

Let us further decompose each $E_{\ell}$ as follows:

$E_{\ell}=\bigcup_{\begin{subarray}{c}\ell\in\mathbb{N}\\\
\sigma\subset\\{0,\ldots,\ell\\}\end{subarray}}E_{\ell,\sigma}\qquad\mbox{where}\quad
E_{\ell,\sigma}:=\\{(x,t)\in E_{\ell}\ :\ \sigma_{x,t}=\sigma\\},$

and let us show that

$N^{\varepsilon/\rho,\rho}_{m}\left(\pi_{x}(E_{\ell,\sigma}),x,t,\ell\right)\geq(1-2M\omega)\ell-1\geq(1-3M\omega)\ell\qquad\forall\,(x,t)\in
E_{\ell,\sigma},$ (12.19)

provided that $\ell$ is large enough, where $N^{\varepsilon,r}_{m}$ is defined
as in (12.9) .

Indeed, by the definition of $E_{\ell,\sigma}$ we have

$\pi_{x}\big{(}E_{\ell,\sigma}\cap(\overline{B_{\rho^{i}}}(x)\times(-\infty,t])\big{)}\subset
x+L_{x,t,i}+\overline{B_{\rho^{i}\varepsilon}}\quad\forall\,(x,t)\in
E_{\ell,\sigma},\ \forall\,i\in\sigma.$ (12.20)

The first important observation is that (12.20) holds also for all
$(\bar{x},\bar{t})$ in the closure $\overline{E_{\ell,\sigma}}$ of
$E_{\ell,\sigma}$, provided that we define $L_{\bar{x},\bar{t},i}$ for
$(\bar{x},\bar{t})\in\overline{E_{\ell,\sigma}}\setminus E_{\ell,\sigma}$ as a
limit of hyperplanes $L_{x,t,i}$ for $(x,t)\in E_{\ell,\sigma}$ converging to
$(\bar{x},\bar{t})$.

Now, given $(x,t)\in E_{\ell,\sigma}$ and $i\in\\{1,\ldots,\ell\\}$ such that
$(i-1)\in\sigma$, we choose
$\bar{t}:=\max\pi_{t}\big{(}\overline{B_{\rho^{i}}(x)}\times[-1,1]\cap\overline{E_{\ell,\sigma}}\big{)}$
and let
$(\bar{x},\bar{t})\in\overline{B_{\rho^{i}}(x)}\times[-1,1]\cap\overline{E_{\ell,\sigma}}$
a corresponding point of “maximal time”. Since $(i-1)\in\sigma$, then (12.20)
holds for the point $(\bar{x},\bar{t})$ with $i$ replaced by $i-1$. Thus,
using that $B_{\rho^{i}(x)}\subset B_{\rho^{i-1}}(\bar{x})$, we get

$\ \pi_{x}\big{(}E_{\ell,\sigma}\big{)}\cap
B_{\rho^{i}}(x)\subset\pi_{x}\left(\overline{E_{\ell,\sigma}}\cap(\overline{B_{\rho^{i-1}}}(\bar{x})\times(-\infty,\bar{t}])\right)\subset\bar{x}+L_{\bar{x},t,i}+\overline{B_{\rho^{i-1}\varepsilon}}.$

In other words, for any given $(x,t)\in E_{\ell,\sigma}$ and for all scales
$i\in\\{1,2,\dots,\ell\\}$ such that $(i-1)\in\ \sigma$ (a set of cardinality
at least $(1-3M\omega)\ell$), we have

$\pi_{x}\big{(}E_{\ell,\sigma}\big{)}\cap B_{\rho^{i}}(x)\subset
y+L+\overline{B_{\rho^{i}\varepsilon/\rho}},$

for some $y\in\mathbb{R}^{n}$ and a linear subspace $L\subset\mathbb{R}^{n}$
of dimension $m$. Thus (12.19) follows.

To conclude, choosing $\omega$ and $\varepsilon$ small enough and applying
Lemma 12.4, we deduce that $\pi_{x}(E_{\ell,\sigma})\subset B_{1/2}(z)$ can be
covered by $\rho^{-(m+\alpha/4)\ell}$ balls of radius $\rho^{\ell}$.
Therefore, since the set of all possible choices of
$\sigma\in\\{0,\ldots,\ell\\}$ has $2^{\ell+1}$ elements, we see that each set
$\pi_{x}(E_{\ell})$ can be covered by $2^{\ell+1}\rho^{-(m+\alpha/4)\ell}$
balls of radius $\rho^{\ell}$. Since $\rho$ was chosen so that
$\rho^{\alpha/4}\geq\frac{1}{4}$, this implies that $\pi_{x}(E_{\ell})$ can be
covered by $\rho^{-(m+\alpha/2)\ell}$ balls of radius $\varrho^{\ell}$.

Adding these bounds over $\ell\geq k$ and letting $k\to\infty$, we conclude
that $\pi_{x}(E_{\ell})$ has zero $\mathcal{H}^{m+\alpha}$-measure (cp. proof
of Lemma 12.6). ∎

We can finally prove Proposition 12.2.

###### Proof of Proposition 12.2.

Let $\delta>0$ be given by Corollary 11.2, and split
$\Sigma^{*}=\cup_{\ell}\Sigma^{*}_{\delta,\ell}$. It suffices to show that
${\rm
dim}_{\mathcal{H}}(\Sigma^{*}_{\delta,\ell}\setminus\Sigma^{\diamond})\leq
n-2$.

For any $\varepsilon>0$ and any
$(x_{\circ},t_{\circ})\in\Sigma^{*}_{\delta,\ell}$, let $\omega_{\varepsilon}$
be given by (12.3). By Lemma 12.3, if
$\omega_{\varepsilon}(x_{\circ},t_{\circ})>0$ then
$(x_{\circ},t_{\circ})\in\Sigma^{\diamond}$. On the other hand, Proposition
12.7 applied with $m=n-2$ and $E=\Sigma^{*}_{\delta,\ell}$ implies that ${\rm
dim}_{\mathcal{H}}(\\{(x_{\circ},t_{\circ})\in\Sigma^{*}_{\delta,\ell}:\omega_{\varepsilon}(x_{\circ},t_{\circ})=0\
\text{for all}\,\varepsilon>0\\})\leq n-2$. Thus ${\rm
dim}_{\mathcal{H}}(\Sigma^{*}_{\delta,\ell}\setminus\Sigma^{\diamond})\leq
n-2$. ∎

##  13\. Enhanced decay towards polynomial Ansatz

In this section we will show that, once we have an estimate of order $3+\beta$
for some $\beta>0$ at a singular point, then we can actually prove a
$C^{\infty}$ estimate.

Assume with not loss of generality that $(0,0)\in\Sigma^{\diamond}$ (see
Definition 12.1). Then there exist $\beta\in(0,\frac{1}{8})$ and $C_{\circ}>0$
such that

$\big{\|}(u-p_{2}-p_{3}^{*})_{r}\big{\|}_{L^{2}(\mathscr{C}_{1})}\leq
C_{\circ}r^{3+2\beta}\quad\mbox{for all }r\in(0,1).$ (13.1)

Moreover, we may choose coordinates so that $p_{2}=\frac{1}{2}x_{n}^{2}$ and
$p_{3}^{*}=a|x_{n}|(x_{n}^{2}+6t)+p_{3}$.

In the following result, and throughout this section, we denote

$\Omega^{\beta}_{R}:=\left\\{(x,t)\in\mathbb{R}^{n}\times\mathbb{R}\,:\,|x|<R(-t)^{\frac{1}{2+\beta}},\
t\leq 0\right\\}.$ (13.2)

Recall that $\mathscr{C}_{r}:=B_{r}\times(-r^{2},0)$ denotes a parabolic
cylinder. Also, , to avoid unnecessary parentheses, given $z\in\mathbb{R}$ we
denote $z_{+}^{2}=(z_{+})^{2}$ (and analogously for $z_{-}^{2}$).

######  Theorem 13.1.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a solution of (3.1) and assume
$(0,0)\in\Sigma^{\diamond}$. Then there exists $r_{\circ}>0$, depending only
on $n$, $\beta$, and $C_{\circ}$, such that inside

$V:=\Omega^{\beta}_{1}\cap\mathscr{C}_{r_{\circ}}$

the positivity set $\\{u>0\\}$ has two connected components and
$u=u^{(1)}+u^{(2)}$ in $V$, where $u^{(i)}$ are two solutions of the parabolic
obstacle problem inside $V$ with disjoint supports satisfying

$\frac{u^{(1)}(rx)}{r^{2}}\to\frac{1}{2}(x_{n})_{+}^{2}\quad\text{and}\quad\frac{u^{(2)}(rx)}{r^{2}}\to\frac{1}{2}(x_{n})_{-}^{2}\qquad\text{as
}r\to 0.$ (13.3)

This result will allow us to break the function $u$ —which has a singular
point at the origin— into two separate functions $u_{1}$ and $u_{1}$ that
behave _as if_ the origin was a _regular_ point for each of them. This is
because, with the parabolic scaling, the domain $\Omega^{\beta}_{R}$ near the
origin is almost equivalent to a full cylinder.

To prove Theorem 13.1 we need a barrier argument to show that the contact set
$\\{u=0\\}$ splits $\Omega^{\beta}_{1}\cap\mathscr{C}_{r_{\circ}}$ into two
disconnected pieces. This barrier is constructed in the following:

######  Lemma 13.2.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a solution of (3.1), and assume that
$(0,0)\in\Sigma^{\diamond}$, i.e., $u$ satisfies 13.1. Choose coordinates so
that $p_{2}=\frac{1}{2}x_{n}^{2}$ and
$p_{3}^{*}=a|x_{n}|(x_{n}^{2}+6t)+p_{3}$.

Then there exists $r_{\circ}>0$, depending only on $n$, $\beta$, and
$C_{\circ}$, such that

$\\{x_{n}+p_{3}/x_{n}=0\\}\cap\Omega^{\beta}_{1}\cap\mathscr{C}_{r_{\circ}}\subset\\{u=0\\}.$

(Recall that $p_{3}(x,t)$ is an odd polynomial, hence it is divisible by
$x_{n}$.)

###### Proof.

First of all we note that, by exactly the same argument as the one given in
Step 2 of the proof of Lemma 11.1, the $L^{2}$ bound from (13.1) can be
upgraded to and $L^{\infty}$ (up to enlarging the constant $C_{\circ}$).
Therefore, we may assume that

$\big{\|}u-p_{2}-p_{3}^{*}\big{\|}_{L^{\infty}(\mathscr{C}_{r})}\leq
C_{\circ}r^{3+2\beta}\quad\mbox{for all }r\in(0,1),$ (13.4)

where $\beta\in(0,\frac{1}{8})$.

Now, let $\boldsymbol{P}(x,t):=\frac{1}{2}(x_{n}+p_{3}(x,t)/x_{n})^{2}$ and
fix
$(\bar{x},\bar{t})\in\\{\boldsymbol{P}=0\\}\cap\Omega^{\beta}_{1}\cap\mathscr{C}_{r_{\circ}}$,
with $r_{\circ}$ small to be chosen. We need to prove that
$u(\bar{x},\bar{t})=0$.

First observe that, since $p_{3}^{*}=a|x_{n}|(x_{n}^{2}+6t)+p_{3}$ with $a\geq
c_{1}>0$ (see Lemma 9.8), (13.4) implies that

$\frac{1}{r^{3}}(u-\boldsymbol{P})_{r}\leq
c|x_{n}|t+C|x_{n}|^{3}+C_{\circ}r^{2\beta}\quad\mbox{in }\mathscr{C}_{2}.$
(13.5)

We will use a modification of the proof of Lemma 10.2, using as barrier

$\Psi_{\bar{y},\bar{s}}(x,t):=r^{-3}\boldsymbol{P}_{r}(x,t)+r^{\beta}\big{(}|x^{\prime}-\bar{y}^{\prime}|^{2}-(t-\bar{s})-2n\,r^{-2}\boldsymbol{P}_{r}(x,t)\big{)},$

where $(\bar{y},\bar{s})\in\\{\boldsymbol{P}_{r}=0\\}\cap\mathscr{C}_{1}$ is
any given point satisfying $\bar{s}\leq r^{\beta}$.

Note that $\text{\small\bf H}\boldsymbol{P}_{r}=r^{2}+O(r^{4})$ inside
$\mathscr{C}_{1}$. Hence, for $r$ sufficiently small we have

$\text{\small\bf
H}\Psi_{\bar{y},\bar{s}}\leq\frac{1}{r}+Cr+Cr^{\beta}\big{(}2(n-1)+1-2n(1+O(r^{2}))\big{)}\leq\frac{1}{r}.$

Consider now the domain
$U_{\varrho}:=\left\\{|x^{\prime}-\bar{y}^{\prime}|\leq\frac{1}{10},|x_{n}|\leq\varrho,\frac{1}{10}\leq
t-\bar{s}\leq 0\right\\}$ with $\varrho:=r^{2\beta/3}$. Using (13.5) and the
fact that $r^{-2}\boldsymbol{P}_{r}=\frac{1}{2}x_{n}^{2}+O(r)$ and
$\bar{s}\leq r^{\beta}$, on
$\partial_{par}U_{\varrho}\cap\\{|x_{n}|=\varrho\\}$ we have

$r^{-3}u_{r}-\Psi_{\bar{y},\bar{s}}\leq
2n\,r^{\beta}r^{-2}\boldsymbol{P}_{r}+\frac{c}{2}\bar{s}\varrho+C\varrho^{3}+C_{\circ}r^{2\beta}\\\
\leq
2n\,r^{\beta}(\varrho^{2}+Cr)-\frac{c}{2}r^{\beta}\varrho+C\varrho^{3}+C_{\circ}r^{2\beta}\leq
Cr^{2\beta}-\frac{c}{2}r^{5\beta/3}<0$ (13.6)

for $r$ small enough. Also it is easy to show that also
$r^{-3}u_{r}-\Psi_{\bar{x},\bar{t}}\leq 0$ on the remaining pieces of
$\partial_{par}U_{\varrho}$, thanks to the positive term
$r^{\beta}\big{(}|x^{\prime}-\bar{y}^{\prime}|-(t-\bar{s})\big{)}$ appearing
in the definition of $\Psi_{\bar{x},\bar{t}}$.

Hence, since $r^{-3}\text{\small\bf H}u_{r}=\frac{1}{r}\geq\text{\small\bf
H}\Psi_{\bar{x},\bar{t}}$ inside $\\{u_{r}>0\\}$ and $\Psi_{\bar{x},\bar{t}}$
is nonnegative, it follows from the maximum principle that

$0\leq r^{-3}u_{r}\leq\Psi_{\bar{y},\bar{s}}\quad\mbox{in }U_{\varrho}.$

In particular, evaluating at $(\bar{y},\bar{s})$ we obtain $0\leq
r^{-3}u_{r}(\bar{y},\bar{s})\leq\Psi_{\bar{x},\bar{t}}(\bar{x},\bar{t})=0$.
Since $(\bar{y},\bar{s})\in\\{\boldsymbol{P}_{r}=0\\}\cap\mathscr{C}_{1}$ was
an arbitrary point satisfying $\bar{s}\leq r^{\beta}$, after rescaling we
obtain

$u=0\quad\mbox{on
}\\{\boldsymbol{P}=0\\}\cap\mathscr{C}_{r}\cap\left\\{t\leq-r^{2+\beta}\right\\}$

whenever $r$ is sufficiently small, as desired. ∎

We can now prove Theorem 13.1.

###### Proof of Theorem 13.1.

Using Lemma 13.2 we see that $\\{u>0\\}$ is split into two connected
components inside $\Omega^{\beta}_{1}\cap\mathscr{C}_{r_{\circ}}$ — the number
of connected components cannot be larger since $\partial_{nn}u\geq 0$ in
$B_{r_{\circ}}\times(-r_{\circ}^{2},r_{\circ}^{2})$ thanks to Lemma 3.8. Then
$u=u^{(1)}+u^{(2)}$, where $u^{(1)}$ and $u^{(2)}$ are respectively supported
in $\\{x_{n}+p_{3}(x,t)/x_{n}>0\\}$ and $\\{x_{n}+p_{3}(x,t)/x_{n}<0\\}$
inside $\Omega^{\beta}_{1}\cap\mathscr{C}_{r_{\circ}}$. Finally, (13.3)
follows from the convergence $r^{-2}u(rx)\to\frac{1}{2}x_{n}^{2}$ and the fact
that, after rescaling, the positivity set of $u^{(1)}$ (resp. $u^{(2)}$)
converges to $\\{x_{n}>0\\}$ (resp. $\\{x_{n}<0\\}$). ∎

Once $u$ is split into two separate functions $u^{(1)}$ and $u^{(2)}$ in
$\Omega^{\beta}_{R}\cap\mathscr{C}_{r_{\circ}}$, we now look for a
$C^{\infty}$ expansion for each of these two functions. For this, we need
first to construct a series of ansatz for the Taylor expansion of these
functions.

######  Definition 13.3.

Let $k\geq 3$, and let $(Q_{\ell})_{2\leq\ell\leq k-1}$ be a family of
polynomials such that $Q_{\ell}=Q_{\ell}(x,t)$ is (parabolically) homogeneous
of degree $\ell$ and satisfies $\text{\small\bf H}(x_{n}Q_{\ell})\equiv 0.$ We
define the polynomial $\mathscr{A}_{k}=\mathscr{A}_{k}[Q_{2},\dots,Q_{k-1}]$
as follows.

For the base case $k=3$, we set

$\mathscr{A}_{3}=\mathscr{A}_{3}[Q_{2}](x,t):=x_{n}+Q_{2}(x,t)+x_{n}R_{2}(x,t),$

where $R_{2}$ is the unique 2-homogeneous caloric polynomial satisfying888The
fact that (13.7) has exactly one solution can be shown as follows: write
$\text{\small\bf
H}\bigg{(}\frac{1}{2}(Q_{2})^{2}\bigg{)}=:T_{2,2}(x^{\prime})+x_{n}T_{2,1}(x^{\prime})+x_{n}^{2}T_{2,0},$
where $T_{2}^{\ell}$ is an $\ell$-homogeneous polynomial in the variables
$x^{\prime}$, and
$R_{2}=S_{2,2}(x^{\prime})+x_{n}S_{2,1}(x^{\prime})+x_{n}^{2}S_{x,0},$ where
$S_{2,\ell}$ is a $\ell$-homogeneous polynomial in the variables $x^{\prime}$.
Then (13.7) amounts to
$2S_{2,2}(x^{\prime})+6x_{n}S_{2,1}(x^{\prime})+x_{n}^{2}(12S_{2,0}+\text{\small\bf
H}S_{2,2})=-T_{2,2}(x^{\prime})-x_{n}T_{2,1}(x^{\prime})-x_{n}^{2}T_{2,0},$
which leads to a linear system with triangular structure, for which the unique
solution is given by  $S_{2,2}=-\frac{1}{2}T_{2,2},\qquad
S_{2,1}=-\frac{1}{6}T_{2,1},\qquad
S_{2,0}=-\frac{1}{12}(T_{2,0}+\text{\small\bf H}S_{2,2}).$

$\text{\small\bf H}\bigg{(}\frac{1}{2}(Q_{2})^{2}+x_{n}^{2}R_{2}\bigg{)}=0.$
(13.7)

Note that (13.7) is equivalent to

$\text{\small\bf
H}\bigg{(}\frac{1}{2}\mathscr{A}_{3}^{2}\bigg{)}=1+O\big{(}(|x|+|t|^{1/2})^{3}\big{)}.$

For $k\geq 3$, we inductively define

$\mathscr{A}_{k+1}\big{[}(Q_{\ell})_{2\leq\ell\leq
k}\big{]}:=\mathscr{A}_{k}\big{[}(Q_{\ell})_{2\leq\ell\leq
k-1}\big{]}+Q_{k}+x_{n}R_{k},$

where $R_{k}$ is a $k$-homogeneous caloric polynomial characterized by the
identity

$\text{\small\bf
H}\bigg{(}\frac{1}{2}\mathscr{A}_{k+1}^{2}\bigg{)}=1+O\big{(}(|x|+|t|^{1/2})^{k+1}\big{)}.$
(13.8)

As before, this equation has exactly one solution $R_{k}$. Indeed, since
$\mathscr{A}_{k}=x_{n}+O(|x|^{2}+|t|)$, it follows by induction that

$\begin{split}\text{\small\bf
H}\bigg{(}\frac{1}{2}\mathscr{A}_{k+1}^{2}\bigg{)}=\text{\small\bf
H}\bigg{(}\frac{1}{2}\mathscr{A}_{k}^{2}+x_{n}Q_{k}+Q_{2}Q_{k-1}+x_{n}^{2}R_{k}\bigg{)}+O\big{(}(|x|+|t|^{1/2})^{k+1}\big{)}\end{split}$

Hence, if we define

$\text{\small\bf
H}\bigg{(}\frac{1}{2}\mathscr{A}_{k}^{2}+x_{n}Q_{k}+Q_{2}Q_{k-1}\bigg{)}=:1+\sum_{j=0}^{k}x_{n}^{k-j}T_{k,j}(x^{\prime})+O\big{(}(|x|+|t|^{1/2})^{k+1}\big{)}$

and write

$R_{k}:=\sum_{j=0}^{k}x_{n}^{k-j}S_{k,j}(x^{\prime}),$

then (13.8) is equivalent to

$\sum_{j=0}^{k}(k-j+2)(k-j+1)x_{n}^{k-j}S_{k,j}(x^{\prime})+\sum_{j=0}^{k-2}x_{n}^{k-j}\text{\small\bf
H}S_{k,j+2}=-\sum_{j=0}^{k}x_{n}^{k-j}T_{k,j}(x^{\prime}).$

This leads to a triangular system, whose unique solution is given by

$S_{k,k}=-\frac{1}{2}T_{k,k},\quad S_{k,k-1}=-\frac{1}{6}T_{k,k-1},\quad
S_{k,j}=-\frac{1}{(k-j+2)(k-j+1)}(T_{k,j}+\text{\small\bf
H}S_{k,j+2})\quad\text{for }j\in\\{0,\ldots,k-2\\}.$

The previous definition aims to construct a Taylor expansion which is
compatible with the PDE satisfied by our solution. However, the previous
formulas do not take care of possible translations and rotation. This is the
purpose of the next definition.

######  Definition 13.4.

Let $k\geq 3$, and let $(Q_{\ell})_{2\leq\ell\leq k-1}$ be a family of
parabolically homogeneous polynomials of degree $\ell$ satisfying
$\text{\small\bf H}(x_{n}Q_{\ell})\equiv 0.$ Then, given $\tau\in\mathbb{R}$
and a rotation $\boldsymbol{R}\in SO(n)$, we define

$\mathscr{P}_{k}=\mathscr{P}_{k}\big{[}Q_{2},\dots,Q_{k-1},\tau,\boldsymbol{R}\big{]}(x,t)$

by

$\mathscr{P}_{k}(x,t):=\frac{1}{2}\left(\mathscr{A}_{k}\big{[}Q_{2},\dots,Q_{k-1}\big{]}\right)_{+}^{2}(\boldsymbol{R}(x+\tau\boldsymbol{e}_{n}),t),$

where $\mathscr{A}_{k}$ is given in Definition 13.3.

Our next goal will be to prove an $\varepsilon$-regularity result in the
domain $\Omega^{\beta}_{R}$. Notice that this result is for “one-sided”
solutions, i.e., it will be applied separately to $u^{(1)}$ and $u^{(2)}$.

######  Theorem 13.5.

Given $\alpha\in(0,1)$, $\beta>0$, $k\geq 4$, and $\delta\in(0,1)$, there
exist positive constants $\bar{r}$ and $\varepsilon_{\circ}$, depending only
on $n$, $\beta$, and $k$, such that the following holds.

Let $u$ satisfy $\text{\small\bf H}u=\chi_{\\{u>0\\}}$, $u\geq 0$, and
$\partial_{t}u\geq 0$ inside $\Omega^{\beta}_{1}\cap\mathscr{C}_{1}$. Assume
that $u(0,0)=0$ and

$\bigg{\|}u-\frac{1}{2}(x_{n})_{+}^{2}\bigg{\|}_{L^{\infty}(\Omega^{\beta}_{1}\cap\mathscr{C}_{r_{\circ}})}\leq\varepsilon_{\circ}.$

Then, for some Ansatz
$\mathscr{P}_{k}=\mathscr{P}_{k}\big{[}(Q_{\ell})_{2\leq\ell\leq
k-1},\tau,\boldsymbol{R}\big{]}$ we have

$\|u-\mathscr{P}_{k}\|_{L^{\infty}(B_{r}\times(-r^{2},-\delta r^{2}))}\leq
r^{k+1+\alpha}$

for all $r\in(0,\bar{r})$.

As a corollary of this result, it follows that the free boundary is
$C^{\infty}$ near every _regular_ free boundary point. However, thanks to the
fact that our result looks at the solution only inside $\Omega^{\beta}_{1}$,
thanks to Theorem 13.1 it can be applied as well to the case of singular
points in $\Sigma^{\diamond}$.

Before proving Theorem 13.5, we will need several ingredients. We start with a
compactness result.

######  Lemma 13.6.

Given $n$, $k$, $\beta$, $\alpha$, and $\varepsilon_{\circ}$, there exists a
positive constant $M_{\circ}$ such that the following holds. Assume that
$N_{\circ}\geq M_{\circ}$ and that
$w:\Omega^{\beta}_{N_{\circ}}\cap\mathscr{C}_{M_{\circ}}\to\mathbb{R}$
satisfies

$\bigg{(}\int_{\Omega^{\beta}_{N_{\circ}}\cap\mathscr{C}_{1}}|w_{2^{m}}|^{2}+|\nabla
w_{2^{m}}|^{2}+|\partial_{t}w_{2^{m}}|^{2}\,dx\,dt\bigg{)}^{1/2}\leq(2^{m})^{k+1+\alpha}\quad\mbox{for
}0\leq m\leq M_{\circ},$ (13.9)

where $w_{2^{m}}(x,t):=w(2^{m}x,4^{m}t)$. Assume in addition that

$\begin{cases}|\text{\small\bf H}w|\leq\frac{1}{M_{\circ}}\ &\mbox{ in }\
\Omega^{\beta}_{N_{\circ}}\cap\big{(}B_{M_{\circ}}\times(-M_{\circ},-{\textstyle\frac{1}{M_{\circ}}})\big{)}\cap\big{\\{}x_{n}\geq{\textstyle\frac{1}{M_{\circ}}}\big{\\}}\\\
w=0&\mbox{ in }\
\Omega^{\beta}_{N_{\circ}}\cap\big{(}B_{M_{\circ}}\times(-M_{\circ},-{\textstyle\frac{1}{M_{\circ}}})\big{)}\cap\big{\\{}x_{n}\leq-{\textstyle\frac{1}{M_{\circ}}}\big{\\}}.\end{cases}$
(13.10)

Finally, suppose that

$\bigg{|}\int_{-2}^{-1}\int_{B_{M_{\circ}}}(x_{n})_{+}w\,Q\,G\,dx\,dt\bigg{|}\leq\frac{1}{M_{\circ}}$
(13.11)

for any $Q=Q(x,t)$ parabolically $\ell$-homogeneous polynomial for some
$0\leq\ell\leq k$ which satisfies $\text{\small\bf H}(x_{n}Q)=0$. Then

$\bigg{(}\int_{\Omega^{\beta}_{N_{\circ}}\cap\mathscr{C}_{1}}|w|^{2}\bigg{)}^{1/2}\leq\varepsilon_{\circ}.$

###### Proof.

We split the proof in two steps.

$\bullet$ _Step 1_. We show the following classification result for ancient
solutions in the whole space. Let
$w:\mathbb{R}^{n}\times(-\infty,0)\to\mathbb{R}$ be such that both $|w|^{2}$
and $|\nabla w|^{2}$ are locally integrable and satisfy

$\bigg{(}\int_{\mathscr{C}_{1}}|w_{2^{m}}|^{2}+|\nabla
w_{2^{m}}|^{2}+|\partial_{t}w_{2^{m}}|^{2}\,dx\,dt\bigg{)}^{1/2}\leq(2^{m})^{k+1+\alpha}\quad\mbox{for
all }m\geq 1,$ (13.12)

and assume in addition that

$\text{\small\bf H}w=0\quad\mbox{ in }\\{x_{n}>0\\},\qquad w|_{\\{x_{n}\leq
0\\}}\equiv 0,$ (13.13)

and

$\int_{-2}^{-1}\int_{\mathbb{R}^{n}}(x_{n})_{+}w\,Q\,G\,dx\,dt=0$ (13.14)

for any polynomial $Q=Q(x,t)$ which parabolically $\ell$-homogeneous for some
$0\leq\ell\leq k$ and satisfies $\text{\small\bf H}(x_{n}Q)=0$. Then,

$w\equiv 0.$

To show this we notice that, thanks to (13.12) and (13.13), it follows from
the Liouville Theorem for the heat equation that the odd extension of $w$
across $\\{x_{n}=0\\}$ is a caloric polynomial of degree at most $k+1$
vanishing on $x_{n}=0$. Since (13.14) implies that $w$ must be orthogonal to
all such polynomials (recall that $w$ vanished for $x_{n}\leq 0$), we conclude
that $w\equiv 0$.

$\bullet$ _Step 2_. Since $H^{1}_{\rm loc}(\mathbb{R}^{n}\times(-\infty,0])$
embeds compactly in $L^{2}_{\rm loc}((\mathbb{R}^{n}\times(-\infty,0])$, the
desired result follows immediately by compactness as $M_{\circ}\to\infty$. ∎

We also need the following estimate. Notice that when the integrals below are
taken in $\mathscr{C}_{2}$ and $\mathscr{C}_{1}$, then the estimate is more
standard —and we proved it in Lemma 6.1. Now, we need such estimate in the new
domains $\Omega^{\beta}_{R}$, and thus it becomes a bit more delicate.

######  Lemma 13.7.

Fix $\beta\in(0,1)$, and let $u_{i}:\mathscr{C}_{2}\to\mathbb{R}$, $i=1,2,$ be
two solutions of (6.1) inside $\Omega^{\beta}_{R}\cap\mathscr{C}_{2}$, with
$|\varepsilon_{i}(x,t)|\leq\bar{\varepsilon}<\frac{1}{100}$. Set
$w:=u_{1}-u_{2}$. Then, for any $\theta\in(\frac{1}{2},1)$ there exist
constants $R_{\circ},C>0$, depending only on $n$, $\beta$, and $\theta$, such
that

$\int_{\Omega^{\beta}_{\theta^{2}R}\cap\mathscr{C}_{1}}|\nabla
w|^{2}+|w\,\text{\small\bf H}w|+|\partial_{t}w|^{2}\leq
C\bigg{(}\int_{\Omega^{\beta}_{R}\cap\mathscr{C}_{2}}w^{2}+\bar{\varepsilon}^{2}\bigg{)}\qquad\forall\,R\geq
R_{\circ}.$ (13.15)

###### Proof.

As in the proof of Lemma (6.1) we have

$w\,\text{\small\bf
H}w\geq-2\bar{\varepsilon}|w|\quad\mbox{and}\quad\partial_{t}w\,\text{\small\bf
H}w\geq-2\bar{\varepsilon}|\partial_{t}w|.$

The idea is to follow the proof of Lemma (6.1), but multiplying the spatial
cut-offs $\tilde{\eta}(x)$ and $\eta(x)$ by an extra appropriately chosen
space-time cut-off in order to make sure that we only evaluate the equation
inside its domain $\Omega^{\beta}_{R}\cap\mathscr{C}_{2}$.

Set $\alpha:=\frac{1}{2+\beta}$ and note that $\alpha\in(1/3,1/2)$. The new
cut-off will be of the form

$\xi^{\beta}_{R}=\xi^{\beta}_{R}(x,t):=\xi\left(\frac{|x|}{R(-t)^{\frac{1}{2+\beta}}}\right)\quad\mbox{for
some suitable cut-off $\xi\in C^{1,1}([0,1])$ satisfying
$\xi|_{[0,\theta]}\equiv 1$}.$

More precisely, we choose $\xi\in C^{1,1}([0,1])$ which satisfies the
following properties:

$\xi(s):=\begin{cases}1\quad&\mbox{for }s\in[0,\theta]\\\ \mbox{concave and
decreasing}&\mbox{for
}s\in\big{[}\theta,{\textstyle\frac{\theta+1}{2}}\big{]}\\\
(1-s)^{2+p}&\mbox{for
}s\in\big{[}{\textstyle\frac{\theta+1}{2}},1\big{]},\end{cases}$ (13.16)

where $p\in(2,\infty)$ satisfies $\frac{p}{p+1}=2\alpha$. It is important to
notice that, since $\xi^{\prime}\leq 0$,

$\partial_{t}\xi^{\beta}_{R}(x,t)\leq 0.$ (13.17)

We claim that

$\Delta\xi^{\beta}_{R}+\partial_{t}\xi^{\beta}_{R}\leq C_{\circ},\quad\mbox{
for }(x,t)\in\mathbb{R}^{n}\times(-4,0),$ (13.18)

where $C_{\circ}$ depends only on $n$, $\theta$, $\beta$, but not on $R$.

Indeed, defining $s=s(x,t):=\frac{|x|}{R(-t)^{\alpha}}$, since the inequality
is trivial for $s\leq\theta$ we have

$\Delta\xi^{\beta}_{R}+\partial_{t}\xi^{\beta}_{R}=\frac{1}{R^{2}(-t)^{2\alpha}}\bigg{(}\xi^{\prime\prime}(s)+\frac{(n-1)}{s}\xi^{\prime}(s)\bigg{)}+\xi^{\prime}(s)\frac{\alpha
s}{(-t)}\leq\frac{\xi^{\prime\prime}(s)}{R^{2}(-t)^{2\alpha}}+\frac{\xi^{\prime}(s)}{6(-t)},$

where we used that $\xi^{\prime}\leq 0$ and that $\alpha s\geq 1/6$ for
$s\geq\theta\geq 1/2$. Since $\xi$ is concave and decreasing inside
$\big{[}\theta,\frac{\theta+1}{2}\big{]}$, the right hand side above is
negative for $s\in\big{[}\theta,\frac{\theta+1}{2}\big{]}$. Hence, to prove
(13.18) we only need to bound the previous expression for
$s\in\big{[}\frac{\theta+1}{2},1\big{]}$, where we have $\xi(s)=(1-s)^{2+p}$.
Recalling that $2\alpha=\frac{p}{p+1}$ and writing $\tau:=(1-s)^{p+1}$, for
$R\geq 1$ we obtain

$\begin{split}\frac{1}{R^{2}(-t)^{2\alpha}}\xi^{\prime\prime}(s)+\frac{\xi^{\prime}(s)}{-6t}&=(2+p)\bigg{(}\frac{(1+p)}{R^{2}(-t)^{2\alpha}}(1-s)^{p}+\frac{1}{6t}(1-s)^{p+1}\bigg{)}\\\
&=\frac{(2+p)}{6}\bigg{(}\frac{6(1+p)}{R^{2}}\bigg{(}\frac{\tau}{-t}\bigg{)}^{2\alpha}-\bigg{(}\frac{\tau}{-t}\bigg{)}\bigg{)}\leq
C_{p,\theta}\qquad\forall\,\tau\in(0,1),\end{split}$

where we used that $2\alpha\in(0,1)$. This concludes the proof of (13.18).

We also note that

$32|\nabla\xi^{\beta}_{R}|+\partial_{t}\xi^{\beta}_{R}\leq 0\quad\mbox{ for
}(x,t)\in\mathbb{R}^{n}\times(-4,0).$ (13.19)

Indeed, as before we only need to check the inequality for $s\geq\theta$, and
we have

$32|\nabla\xi_{\beta}|+\partial_{t}\xi_{\beta}=\frac{32}{R(-t)^{\alpha}}|\xi^{\prime}(s)|-\xi^{\prime}(s)\frac{\alpha
s}{t}\leq\frac{32}{R(-t)^{\alpha}}|\xi^{\prime}(s)|+\xi^{\prime}(s)\frac{1}{6(-t)}\leq
0,$

provided that $R\geq R_{\circ}$ is large enough.

We can now prove (13.15). As in the proof of Lemma 6.1, let

$\tilde{\eta}=\tilde{\eta}(x)\in C^{\infty}_{c}(B_{5/3})\quad\mbox{such that
$\tilde{\eta}\equiv 1$ in $\overline{B}_{4/3}$, $0\leq\tilde{\eta}\leq 1$ in
$B_{5/3}$, and $|\nabla\tilde{\eta}|+|D^{2}\tilde{\eta}|\leq 10$.}$ (13.20)

Then

$\begin{split}\frac{d}{dt}\int_{B_{5/3}\times\\{t\\}}w^{2}\tilde{\eta}\,\xi^{\beta}_{R}+2\int_{B_{2}\times\\{t\\}}(w\,\text{\small\bf
H}w+|\nabla
w|^{2})\,\eta\,\tilde{\xi}^{\beta}_{R}&=\int_{B_{2}\times\\{t\\}}w^{2}(\partial_{t}+\Delta)\big{(}\tilde{\eta}\xi^{\beta}_{R}\big{)}\\\
&=\int_{B_{2}\times\\{t\\}}w^{2}\big{(}\tilde{\eta}(\partial_{t}+\Delta)\xi^{\beta}_{R}+2\nabla\tilde{\eta}\cdot\nabla\xi^{\beta}_{R}+\Delta\tilde{\eta}\,\xi^{\beta}_{R}\big{)}.\end{split}$

Note that, thanks to (13.18),
$\tilde{\eta}(\partial_{t}+\Delta)\xi^{\beta}_{R}\leq
C_{\circ}\chi_{B_{5/3}}(x)\chi_{\Omega^{\beta}_{R}}$. Also, the term
$\nabla\tilde{\eta}\cdot\nabla\xi^{\beta}_{R}$ is also bounded because
$\frac{1}{R(-t)^{\alpha}}\sim 1$ on the intersection of the supports of
$\nabla\tilde{\eta}$ and $\nabla\xi^{\beta}_{R}$. Therefore, we obtain

$\frac{d}{dt}\int_{B_{5/3}\times\\{t\\}}w^{2}\tilde{\eta}\,\xi^{\beta}_{R}+2\int_{B_{2}\times\\{t\\}}(w\,\text{\small\bf
H}w+|\nabla w|^{2})\,\tilde{\eta}\,\xi^{\beta}_{R}\leq
C\int_{B_{5/3}\times\\{t\\}}w^{2}\chi_{\Omega^{\beta}_{R}}.$ (13.21)

Integrating (13.21) with respect to $t\in(-T_{*},0)$ where $T_{*}\in(3/2,2)$
satisfies

$\int_{(B_{5/3}\times\\{-T_{*}\\})\cap\Omega^{\beta}_{R}}w^{2}\xi^{\beta}_{R}\leq
10\int_{(B_{5/3}\times(-2,0))\cap\Omega^{\beta}_{R}}w^{2}\xi^{\beta}_{R},$

and recalling that $w\,\text{\small\bf H}w\geq-2\bar{\varepsilon}|w|$ and
that, by construction, $\tilde{\eta}\equiv 1$ in $B_{4/3}$ and
$\xi^{\beta}_{R}\equiv 1$ in $\Omega^{\beta}_{\theta R}$, we obtain

$\begin{split}\int_{-3/2}^{0}\int_{B_{4/3}}\big{(}|w\,\text{\small\bf
H}w|+|\nabla w|^{2}\big{)}\chi_{\Omega^{\beta}_{\theta R}}\leq
C\bigg{(}\int_{(B_{5/3}\times(-2,0))\cap\Omega^{\beta}_{R}}w^{2}+\bar{\varepsilon}|w|\bigg{)}\leq
C\bigg{(}\int_{(B_{5/3}\times(-2,0))\cap\Omega^{\beta}_{R}}w^{2}+\bar{\varepsilon}^{2}\bigg{)}.\end{split}$
(13.22)

This proves the estimates for $|\nabla w|^{2}$ and $|w\,\text{\small\bf H}w|$
in (13.15).

To control $|\partial_{t}w|^{2}$, we choose a spatial cut-off $\eta\in
C^{\infty}_{c}(B_{4/3})$ such that $\eta\equiv 1$ in $\overline{B}_{1}$. Since
$\partial_{t}w\text{\small\bf H}w\geq-2|\partial_{t}w|\bar{\varepsilon}$ we
obtain (using the inequality $4ab\leq\frac{2}{M}a^{2}+2Mb^{2}$ and choosing
$M=8$)

$\begin{split}2\int_{B_{2}\times\\{t\\}}&(\partial_{t}w)^{2}\eta^{2}(\xi^{\beta}_{\theta
R})^{2}+\frac{d}{dt}\int_{B_{2}\times\\{t\\}}|\nabla
w|^{2}\eta^{2}(\xi^{\beta}_{\theta R})^{2}-\int_{B_{2}\times\\{t\\}}|\nabla
w|^{2}\partial_{t}\big{[}\eta^{2}\,(\xi^{\beta}_{\theta R})^{2}\big{]}=\\\
&=2\int_{B_{2}\times\\{t\\}}(\partial_{t}w)^{2}\eta^{2}(\xi^{\beta}_{\theta
R})^{2}+\int_{B_{2}\times\\{t\\}}2\nabla
w\cdot\nabla\partial_{t}w\,\eta^{2}(\xi^{\beta}_{\theta R})^{2}\\\
&=\int_{B_{2}\times\\{t\\}}2(\partial_{t}w-\Delta
w)\,\partial_{t}w\,\eta^{2}(\xi^{\beta}_{\theta R})^{2}-4\nabla
w\cdot\nabla\eta\,\partial_{t}w\,\eta\,(\xi^{\beta}_{\theta
R})^{2}-4\partial_{t}w\,\nabla w\cdot\nabla\xi^{\beta}_{\theta
R}\,\eta^{2}\xi^{\beta}_{\theta R}\\\
&\leq\int_{B_{2}\times\\{t\\}}4\bar{\varepsilon}|\partial_{t}w|\eta^{2}(\xi^{\beta}_{\theta
R})^{2}+\int_{B_{2}\times\\{t\\}}4M|\nabla
w|^{2}\big{(}|\nabla\eta|^{2}(\xi^{\beta}_{\theta
R})^{2}+|\nabla\xi^{\beta}_{\theta
R}|^{2}\eta^{2}\big{)}+\frac{4}{M}(\partial_{t}w)^{2}\eta^{2}(\xi^{\beta}_{\theta
R})^{2}\\\
&\leq\int_{B_{2}\times\\{t\\}}(\partial_{t}w)^{2}\eta^{2}(\xi^{\beta}_{\theta
R})^{2}+C\bar{\varepsilon}^{2}+32\int_{B_{2}\times\\{t\\}}|\nabla
w|^{2}\big{(}|\nabla\eta|^{2}(\xi^{\beta}_{\theta
R})^{2}+|\nabla\xi^{\beta}_{\theta R}|^{2}\eta^{2}\big{)}.\end{split}$

Hence, recalling (13.19), since $\xi^{\beta}_{\theta R}$ is supported in
$\Omega^{\beta}_{\theta R}$, we obtain

$\int_{B_{2}\times\\{t\\}}(\partial_{t}w)^{2}\eta^{2}(\xi^{\beta}_{\theta
R})^{2}+\frac{d}{dt}\int_{B_{2}\times\\{t\\}}|\nabla
w|^{2}\eta^{2}(\xi^{\beta}_{\theta R})^{2}\leq
C\bigg{(}\int_{(B_{4/3}\times\\{t\\})\cap\Omega^{\beta}_{\theta R}}|\nabla
w|^{2}+\bar{\varepsilon}^{2}\bigg{)}.$

Integrating with respect to $t\in(-T_{\diamond},0)$ where
$T_{\diamond}\in(1,3/2)$ is chosen so to satisfy

$\int_{(B_{4/3}\times\\{T_{\diamond}\\})\cap\Omega^{\beta}_{\theta R}}|\nabla
w|^{2}\eta^{2}(\xi^{\beta}_{\theta R})^{2}\leq
10\int_{(B_{4/3}\times(-3/2,1))\cap\Omega^{\beta}_{\theta R}}|\nabla
w|^{2}\eta^{2}(\xi^{\beta}_{\theta R})^{2},$

and recalling that $\eta\equiv 1$ in $B_{1}$ and $\xi^{\beta}_{\theta R}\equiv
1$ in $\Omega^{\beta}_{\theta^{2}R}$, we finally get

$\int_{\Omega^{\beta}_{\theta^{2}R}\cap\mathscr{C}_{1}}|\partial_{t}w|^{2}\leq
C\bigg{(}\int_{\Omega^{\beta}_{R}\cap\mathscr{C}_{2}}|\nabla
w|^{2}+\bar{\varepsilon}^{2}\bigg{)}.$

This completes the proof of the estimate (13.15), and the Lemma follows. ∎

Recall now the set of Ansatz given in Definition 13.4. We note that our
parametrization of possible Ansatz given in Definition 13.4 is non-injective:
for every ${\boldsymbol{S}}\in SO(n)$ that fixes $\\{x_{n}=0\\}$ we have

$\mathscr{P}_{k}\big{[}\tau,\boldsymbol{R}{\boldsymbol{S}^{-1}},{\boldsymbol{S}}_{*}Q_{2},\dots,{\boldsymbol{S}}_{*}Q_{k-1}\big{]}=\mathscr{P}_{k}\big{[}\tau,\boldsymbol{R},Q_{2},\dots,Q_{k-1}\big{]},\quad\mbox{where}\
{\boldsymbol{S}}_{*}P(x,t):=P({\boldsymbol{S}}x,t).$ (13.23)

In Lemma 13.9 below, we will consider the set of Ansatz that are a “small
perturbation” of $\frac{1}{2}(x_{n})_{+}^{2}$, and it will be convenient to
provide an injective and smooth parametrization of such subset.

######  Definition 13.8.

Given $\delta>0$ small, let

$\mathcal{N}_{\delta}:=\big{\\{}\mbox{$\mathscr{P}_{k}\big{[}\tau,\boldsymbol{R},Q_{2},\dots,Q_{k-1}\big{]}$
: $|\tau|\leq\delta$,
\,$|\boldsymbol{R}\boldsymbol{e}_{n}-\boldsymbol{e}_{n}|\leq\delta$,
\,$\max_{\ell}\|Q_{\ell}\|_{L^{2}(Q_{1})}\leq\delta$}\big{\\}}.$

Our goal is to find a smooth bijective parameterization of
$\mathcal{N}_{\delta}$. Keeping (13.23) in mind, given $e\in\mathbb{S}^{n-1}$
with $|e-\boldsymbol{e}_{n}|\leq\delta$ we define

$\boldsymbol{R}_{e}:=\arg\min\big{\\{}\|\boldsymbol{R}-{\rm Id}\|_{HS}\ :\
\boldsymbol{R}\in SO(n),\quad\boldsymbol{R}\boldsymbol{e}_{n}=e\big{\\}}.$

Note that $\boldsymbol{R}_{e}$ is a rotation of angle
$\angle(\boldsymbol{e}_{n},e)$ in the plane generated by $\boldsymbol{e}_{n}$
and $e$ and leaves all the vectors orthogonal to this plane invariant (when
$e=\boldsymbol{e}_{n}$ then $\boldsymbol{R}_{e}={Id}$). So, it makes sense to
consider the map

$\big{(}\tau,e,Q_{2},\dots,Q_{k-1}\big{)}\ \longmapsto\
\mathscr{P}_{k}\big{[}\tau,\boldsymbol{R}_{e},Q_{2},\dots,Q_{k-1}\big{]},$
(13.24)

and we want to show that it is a bijection for $\delta$ small.

For technical convenience, instead of considering $\mathscr{P}_{k}$, it makes
sense to replace it with the polynomial of degree $2k$ that equals
$\mathscr{P}_{k}$ inside $\\{\mathscr{P}_{k}>0\\}$, namely

$\mathscr{P}_{k}(x,t)=\frac{1}{2}\left(\mathscr{A}_{k}\big{[}Q_{2},\dots,Q_{k-1}\big{]}\right)_{+}^{2}(\boldsymbol{R}(x+\tau\boldsymbol{e}_{n}),t)\quad\leadsto\quad\widetilde{\mathscr{P}}_{k}(x,t)=\frac{1}{2}\left(\mathscr{A}_{k}\big{[}Q_{2},\dots,Q_{k-1}\big{]}\right)^{2}(\boldsymbol{R}(x+\tau\boldsymbol{e}_{n}),t).$
(13.25)

In this way, the map (13.24) takes value in the smooth manifold of $2k$
homogeneous polynomials. In the next lemma we compute the differential of this
map at $\tau=0$, $e=\boldsymbol{e}_{n}$, $(Q_{\ell})=0$, and prove that it is
a smooth diffeomorphism for $\delta$ small. Also, we show that
$\widetilde{\mathscr{P}}_{k}$ is completely determined, in a continuous way,
by its coefficients of degree $\leq k$.

######  Lemma 13.9.

Let $\mathscr{P}_{k},\mathcal{N}_{\delta}$ be as in Definitions 13.4 and 13.8.
For $\mathscr{P}_{k}\in\mathcal{N}_{\delta}$, let
$\widetilde{\mathscr{P}}_{k}$ denote the polynomial of degree $2k$ which
equals $\mathscr{P}_{k}$ in $\\{\mathscr{P}_{k}>0\\}$, see (13.25). Then, the
differential of the map

$\big{(}\tau,e,Q_{2},\dots,Q_{k-1}\big{)}\
\mapsto\widetilde{\mathscr{P}}_{k}\big{[}\tau,\boldsymbol{R}_{e},Q_{2},\dots,Q_{k-1}\big{]}$
(13.26)

at the point
$\mathcal{O}:=\big{(}\tau=0,e=\boldsymbol{e}_{n},Q_{2}=0,\dots,Q_{k-1}=0\big{)}$
has the following diagonal structure:

$\begin{split}\frac{d}{d\varepsilon}\bigg{|}_{\varepsilon=0}\widetilde{\mathscr{P}}_{k}\big{[}\varepsilon\tau^{\prime},\boldsymbol{R}_{\boldsymbol{e}_{n}},0,\dots,0\big{]}&=x_{n}\tau^{\prime}\\\
\frac{d}{d\varepsilon}\bigg{|}_{\varepsilon=0}\widetilde{\mathscr{P}}_{k}\big{[}0,\boldsymbol{R}_{\frac{\boldsymbol{e}_{n}+\varepsilon
e^{\prime}}{|\boldsymbol{e}_{n}+\varepsilon
e^{\prime}|}},0,\dots,0\big{]}&=x_{n}(x\cdot e^{\prime})\\\
\frac{d}{d\varepsilon}\bigg{|}_{\varepsilon=0}\widetilde{\mathscr{P}}_{k}\big{[}0,\boldsymbol{R}_{\boldsymbol{e}_{n}},0,\dots,\varepsilon
Q_{\ell}^{\prime},\dots,0\big{]}&=x_{n}Q_{\ell}^{\prime}\qquad\mbox{for
}2\leq\ell\leq k-1\end{split}$ (13.27)

where $\tau^{\prime}\in\mathbb{R}$,
$e^{\prime}\in\mathbb{S}^{n-1}\cap\\{x_{n}=0\\}$, and $Q_{\ell}^{\prime}$ are
homogeneous polynomials of degree $\ell$ satisfying $\text{\small\bf
H}(x_{n}Q_{\ell}^{\prime})=0$. In particular, (13.26) gives a diffeomorphic
parametrization of $\mathcal{N}_{\delta}$.

Moreover, for $0\leq i\leq 2k$, let $\varpi_{i}[\widetilde{\mathscr{P}}_{k}]$
denote the $i$-homogeneous part of $\widetilde{\mathscr{P}}_{k}$. Then the
differential of the map

$\big{(}\tau,e,Q_{2},Q_{3},\dots,Q_{k-1}\big{)}\
\mapsto\left(\varpi_{1}\big{(}\widetilde{\mathscr{P}}_{k}\big{)},\varpi_{2}\big{(}\widetilde{\mathscr{P}}_{k}\big{)},\dots,\varpi_{k}\big{(}\widetilde{\mathscr{P}}_{k}\big{)}\right)$

at $\mathcal{O}$ has maximal rank. In particular there exists $\delta_{k}>0$
small enough such that, inside $\mathcal{N}_{\delta_{k}}$, the whole
polynomial $\widetilde{\mathscr{P}}_{k}$ is determined (in a continuous way)
by its coefficients of terms of degree $\leq k$.

###### Proof.

The result follows from a direct computation, recalling that

$\widetilde{\mathscr{P}}_{k}\big{[}\tau,\boldsymbol{R}_{e},Q_{2},Q_{3},\dots,Q_{k-1}\big{]}(x,t)=\frac{1}{2}\big{(}x_{n}+Q_{2}+x_{n}R_{2}+\dots+Q_{k-1}+x_{n}R_{k-1}\big{)}^{2}(\boldsymbol{R}_{e}x+\tau\boldsymbol{e}_{n},t).$

Also, for $\delta$ small, the map is a diffeomorphism thanks to the implicit
function theorem. ∎

Combining all the previous results we can now prove the following result, from
which we will deduce Theorem 13.5.

######  Proposition 13.10.

Given $\alpha\in(0,1)$, $\beta>0$, and $k\geq 3$, let $\delta=\delta_{k}$ be
given by Lemma 13.9. There exist positive constants $N_{\circ}$ and
$\varepsilon_{\circ}$, depending only on $n$, $\beta$, and $k$, such that the
following holds.

Let $u$ satisfy $\text{\small\bf H}u=\chi_{\\{u>0\\}}$, $u\geq 0$, and
$\partial_{t}u\geq 0$ inside $\Omega^{\beta}_{N_{\circ}}\cap\mathscr{C}_{1}$.
Assume that, for some integer $j\geq N_{\circ}$ and a sequence of Ansatz
$\mathscr{P}_{k}^{(\ell)}$, we have

$\bigg{(}\int_{\Omega^{\beta}_{N_{\circ}}\cap\mathscr{C}_{1}}\left(u-\mathscr{P}_{k}^{(m)}\right)^{2}(2^{-m}x,4^{-m}t)\,dx\,dt\bigg{)}^{1/2}\leq\varepsilon_{\circ}(2^{-m})^{k+\alpha}\quad\mbox{for
}0\leq m\leq j,$ (13.28)

where

$\left\|\mathscr{P}_{k}^{(0)}-\frac{1}{2}(x_{n})_{+}^{2}\right\|_{L^{\infty}(\mathscr{C}_{1})}\leq\varepsilon_{\circ}.$
(13.29)

Suppose also that

$\partial\big{\\{}u(2^{-j}\,\cdot\,,4^{-j}\,\cdot\,)=0\big{\\}}\cap\big{(}B_{N_{\circ}}\times\big{(}-N_{\circ},{\textstyle-\frac{1}{N_{\circ}}}\big{)}\big{)}\subset\big{\\{}|x_{n}|\leq\varepsilon_{\circ}\big{\\}}.$
(13.30)

Then, at the next scale $j+1$, there exists a new Ansatz
$\mathscr{P}_{k}^{(j+1)}\in\mathcal{N}_{\delta_{k}}$ such that

$\bigg{(}\int_{\Omega^{\beta}_{N_{\circ}}\cap\mathscr{C}_{1}}\left(u-\mathscr{P}_{k}^{(j+1)}\right)^{2}\big{(}2^{-(j+1)}x,4^{-(j+1)}t\big{)}\,dx\,dt\bigg{)}^{1/2}\leq\varepsilon_{\circ}\big{(}2^{-(j+1)}\big{)}^{k+\alpha}.$
(13.31)

###### Proof.

Let $\theta_{\circ}\in(0,1)$ be chosen so that
$\Omega^{\beta}_{\theta_{\circ}R}$ is mapped to $\Omega^{\beta}_{R}$ under the
parabolic doubling of the scaling $(x,t)\mapsto(2x,4t)$. More precisly,

$(x,t)\in\Omega^{\beta}_{\theta_{\circ}R}\ \Leftrightarrow\
|x|\leq\theta_{\circ}R(-t)^{\frac{1}{2+\beta}}\ \Leftrightarrow\
|2x|\leq\theta_{\circ}2^{\frac{\beta}{2+\beta}}R(-4t)^{\frac{1}{2+\beta}}\
\Leftrightarrow\ (2x,4t)\in\Omega^{\beta}_{R}$

leads to

$\theta_{\circ}:=2^{-\frac{\beta}{2+\beta}}.$

Given $M_{\circ}$ large, we define $N_{\circ}=N_{\circ}(\beta,M_{\circ})$ as

$N_{\circ}:=M_{\circ}\theta_{\circ}^{-M_{\circ}}\gg M_{\circ}.$

Defining $N_{\circ}$ in this way we guarantee that

$B_{M_{\circ}}\times(-2,-1)\subset\Omega^{\beta}_{M_{\circ}}=\big{\\{}(2^{-M_{\circ}}x,4^{-M_{\circ}}t)\
:\ (x,t)\in\Omega^{\beta}_{N_{\circ}}\big{\\}}.$ (13.32)

We divide the proof in four steps.

$\bullet$ Step 1. We will first show that assumptions (13.28) and (13.29)
imply

$\mathscr{P}_{k}^{(m)}\in\mathcal{N}_{\delta_{k}},\qquad\bigg{\|}\,\mathscr{P}_{k}^{(m)}-\frac{1}{2}(x_{n})_{+}^{2}\,\bigg{\|}_{L^{\infty}(\mathscr{C}_{1})}\leq\delta\quad\mbox{
for all }0\leq m\leq j,$ (13.33)

where $\delta\leq\delta_{k}$ can be made arbitrarily small (by decreasing
$\varepsilon_{\circ}$).

Recall that, thanks to Lemma 13.9, the Ansatz
$\mathscr{P}^{(m)}_{k}=\mathscr{P}^{(m)}_{k}\big{[}(Q_{\ell}^{(m)})_{2\leq\ell\leq
k-1},\sigma^{(m)},\tau^{(m)},\boldsymbol{R}^{(m)}\big{]}$ is determined by its
coefficients of degree $\leq k$ (since these determine the spatial and time
translations, the rotations, and the $Q_{\ell}$’s).

Set
$K_{\circ}:=\mathscr{C}_{1/2}\cap\\{x_{n}>\frac{1}{10},\,t\leq-\frac{1}{10}\\}$
and note that
$K_{\circ}\subset\subset\\{x_{n}>0\\}\cap\Omega^{\beta}_{M_{\circ}}$. Then,