# KPZ equation with a small noise, deep upper tail and limit shape

Pierre Yves Gaudreau Lamarre, Yier Lin, and Li-Cheng Tsai  Department of
Statistics, University of Chicago<EMAIL_ADDRESS>Department of
Statistics, University of Chicago<EMAIL_ADDRESS>Departments of
Mathematics, University of Utah<EMAIL_ADDRESS>
###### Abstract.

In this paper, we consider the KPZ equation under the weak noise scaling. That
is, we introduce a small parameter $\sqrt{\varepsilon}$ in front of the noise
and let $\varepsilon\to 0$. We prove that the one-point large deviation rate
function has a $\frac{3}{2}$ power law in the deep upper tail. Furthermore, by
forcing the value of the KPZ equation at a point to be very large, we prove a
limit shape of the solution of the KPZ equation as $\varepsilon\to 0$. This
confirms the physics prediction in [KK07, KK09, KMS16, MKV16, LDMRS16, HMS19].

##  1\. Introduction

The Kardar–Parisi–Zhang (KPZ) equation [KPZ86] is a non-linear stochastic PDE
which describes the random growth of an interface that has a property of
lateral growth and relaxation

$\partial_{t}h=\frac{1}{2}\partial_{xx}h+\frac{1}{2}(\partial_{x}h)^{2}+\xi.$
(1.1)

Here $\xi$ is the space-time white noise, which can be informally understood
as a Gaussian field with Dirac-delta correlation function
$\mathbb{E}[\xi(t,x)\xi(s,y)]=\delta(t-s)\delta(x-y)$. The KPZ equation has
been studied intensively over the past 35 years. We refer to [FS10, Qua11,
Cor12, QS15, CW17, CS20] for some surveys of the mathematical studies of the
KPZ equation.

Care is needed to make sense of the solution to (1.1) due to the non-linearity
and space-time white noise in the equation. One way of defining the solution
is through the _Hopf-Cole transform_. That is, we define $h:=\log Z$, where
$Z$ solves the Stochastic Heat Equation (SHE)

$\partial_{t}Z=\frac{1}{2}\partial_{xx}Z+\xi Z.$

We say that $Z$ is the _mild solution_ to the SHE if

$Z(t,x)=\int_{\mathbb{R}}p(t,x-y)Z(0,y)dy+\int_{0}^{t}\int_{\mathbb{R}}p(t-s,x-y)Z(s,y)\xi(s,y)dsdy,$
(1.2)

where $p(t,x):=\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^{2}}{2t}}$ is the heat
kernel. The solution theory of the SHE is standard; see [Qua11, Sections
2.1-2.6] for more details. Moreover, for function-valued initial data
$Z(0,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\geq 0$ that is not
identically zero, [Mue91] shows that $Z$ is always positive, i.e. almost
surely $Z(t,x)>0$ for all $t>0$ and $x\in\mathbb{R}$. This guarantees the
wellposedness of $h$. One often considered initial data is
$Z(0,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})=\delta({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$,
where $\delta({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$ is a Dirac-
delta function. We refer to this as the Dirac-delta initial data for $Z$ and
the _narrow wedge initial data_ for $h$. [MF14] shows that under the Dirac-
delta initial data, almost surely $Z$ is positive for all $t>0$ and
$x\in\mathbb{R}$. Other definitions and constructions of the solution to the
KPZ equation are given by regularity structure [Hai14], paracontrolled
distribution [GIP15] or the notion of energy solution [GJ14, GP18].

In recent years, the large deviations of the KPZ equation have received much
attention in the mathematics and physics communities. The large deviations of
the KPZ equation can be studied in two regimes: long time regime
$(t\to\infty)$ and short time regime $(t\to 0)$. For the long time regime, the
work [CG20b] rigorously proved a detailed bound for the lower tail of the KPZ
equation under the narrow wedge initial data. This bound captures a cubic to
$\frac{5}{2}$ crossover; see also the physics work [KLD18]. [CG20a] obtained
similar bounds for the KPZ equation under general initial data. Under the
narrow wedge initial data, the exact one-point lower tail large deviation rate
function was derived in the physics works [SMP17, CGK+18, KLDP18, LD20] and
was proved rigorously by [Tsa18, CC21]. [KLD19] showed that the four methods
in [SMP17, CGK+18, KLDP18, Tsa18] are closely related. For the upper tail, the
physics work [LDMS16] predicted the $\frac{3}{2}$-power law for the entire
rate function of the KPZ equation narrow wedge initial data. [DT21] gave a
rigorous proof for the upper tail Large Deviation Principle (LDP). The result
was extended to general initial data by [GL20].

For the large deviations of the KPZ equation in the short time regime, the
results are fruitful in the physics literature; see [Kra19]. In particular,
the physics literature [KK07, KK09, KMS16, MKV16] predicted that for the
narrow wedge and flat initial data, the one-point large deviation rate
function exhibits a $\frac{3}{2}$-power law in the deep upper tail, a
quadratic power law in the near-center tail and a $\frac{5}{2}$-power law in
the deep lower tail. The physics work [LDMRS16] derived the entire one-point
rate function, from which the authors are able to confirm these power laws.
Their prediction was backed by the numerical result [HLDM+18]. The one-point
large deviations were rigorously proved in [LT21]. The authors also rigorously
proved the quadratic to $\frac{5}{2}$-power law crossover in the lower tail
rate function.

Studying the KPZ equation in the short time regime is the same as studying the
KPZ equation in the _weak noise regime_. That is, we introduce a small
parameter $\sqrt{\varepsilon}$ in front of the noise,

$\partial_{t}h_{\varepsilon}=\tfrac{1}{2}\partial_{xx}h_{\varepsilon}+\tfrac{1}{2}(\partial_{x}h_{\varepsilon})^{2}+\sqrt{\varepsilon}\xi.$
(1.3)

The solution to the above equation is defined to be $h_{\varepsilon}:=\log
Z_{\varepsilon}$ where $Z_{\varepsilon}$ solves the SHE

$\displaystyle\partial_{t}Z_{\varepsilon}=\tfrac{1}{2}\partial_{xx}Z_{\varepsilon}+\sqrt{\varepsilon}\xi
Z_{\varepsilon}.$ (1.4)

Throughout the paper, we set
$Z_{\varepsilon}(0,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})=\delta({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$.
The short time regime of (1.1) is related to (1.3) through scaling, namely,
$h(\varepsilon^{2}{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\varepsilon{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})+\log\varepsilon\overset{d}{=}h_{\varepsilon}({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$.
We add $\log\varepsilon$ to guarantee that $h_{\varepsilon}$ starts from the
narrow wedge initial data.

The LDP of
$Z_{\varepsilon}=Z_{\varepsilon}({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$
under the limit $\varepsilon\to 0$ has been rigorously proven in [LT21]. The
rate function is of Freidlin-Wentzell type. In particular, by the contraction
principle, for $\lambda\geq 0$,

$\displaystyle\lim_{\varepsilon\to
0}\varepsilon\log\mathbb{P}\big{[}h_{\varepsilon}(2,0)+\log\sqrt{4\pi}\leq-\lambda\big{]}$
$\displaystyle=-\Phi(-\lambda),$ (1.5) $\displaystyle\lim_{\varepsilon\to
0}\varepsilon\log\mathbb{P}\big{[}h_{\varepsilon}(2,0)+\log\sqrt{4\pi}\geq\lambda\big{]}$
$\displaystyle=-\Phi(\lambda),$ (1.6)

where $\Phi$ is the infimum of the Freidlin-Wentzell rate function subject to
the relevant constraint. Extracting the asymptotics of $\Phi$ is non-trivial.
[LT21] proved that $\lim_{\lambda\to
0}\lambda^{-2}\Phi(\lambda)=\tfrac{1}{\sqrt{2\pi}}$ and
$\lim_{\lambda\to\infty}\lambda^{-\frac{5}{2}}\Phi(-\lambda)=\tfrac{4}{15\pi}$.

###  1.1. Main results.

The first result of the current paper concerns the deep upper tail of the rate
function $\Phi$. In other words, we look at the asymptotic of $\Phi(\lambda)$
as $\lambda\to\infty$. It has been predicted in the physics literature [KK07,
KK09, KMS16, MKV16, LDMRS16] that
$\lim_{\lambda\to\infty}\lambda^{-3/2}\Phi(\lambda)=\tfrac{4}{3}$. Our first
result gives a rigorous proof of the $\frac{3}{2}$-power law in the deep upper
tail.

######  Theorem 1.1.

We have
$\displaystyle\lim_{\lambda\to\infty}\lambda^{-3/2}\Phi(\lambda)=\tfrac{4}{3}$.

The second result of the current paper proves the limit shape of (the solution
of) the KPZ equation under the weak noise scaling and the deep upper tail
conditioning. This limit shape was predicted in the physics works [KK07, KK09,
KMS16, MKV16, HMS19].

######  Theorem 1.2.

Define
$h_{\varepsilon,\lambda}=\lambda^{-1}h_{\varepsilon}(t,\lambda^{\frac{1}{2}}x)$.
For arbitrary fixed $\delta>0$, we have

$\lim_{\lambda\to\infty}\lim_{\varepsilon\to
0}\mathbb{P}\big{[}\|h_{\varepsilon,\lambda}-\mathsf{h}_{*}\|_{L^{\infty}([\delta,2]\times[-\delta^{-1},\delta^{-1}])}<\delta\,\big{|}\,h_{\varepsilon}(2,0)+\log\sqrt{4\pi}\geq\lambda\big{]}=1.$

Here, we use
$\|{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}\|_{L^{\infty}(D)}$ to
denote the $L^{\infty}$ norm on the domain $D$. The limit shape
$\mathsf{h}_{*}$ is given by

$\mathsf{h}_{*}(t,x):=\begin{cases}-|x|+\frac{t}{2},&\text{when }|x|\leq t,\\\
-\frac{x^{2}}{2t},&\text{when }|x|\geq t.\end{cases}$ (1.7)

See Figure 1 for illustration.

Figure 1. The graph of
$\mathsf{h}_{*}(t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$, when
$t=0.5,1,1.5$.

######  Remark 1.3.

Theorem 1.1 gives the $\lambda\to\infty$ limit of $h_{\varepsilon,\lambda}$
under the upper-tail conditioning. A natural related question is to obtain the
limit under the lower-tail conditioning
$h_{\varepsilon}(2,0)+\log\sqrt{4\pi}\leq-\lambda$. The latter question has
recently been solved in [LT22]. We emphasize that the mechanisms for the large
deviations are very different in the upper- and lower-tail conditioning. In
the upper-tail conditioning, the contribution of the noise $\xi$ concentrates
around $x=0$; in the lower-tail conditioning, the contribution of the noise
spans a wide region in spacetime. The required analysis in the current paper
and in [LT22] hence differ.

######  Remark 1.4.

The limit shape $\mathsf{h}_{*}$ was predicted earlier in the physics work
[KK07, KK09, KMS16, MKV16] via the weak noise theory, and in [HMS19] by
simulations. Recently, the physics work [KLD21] solved the finite $\lambda$
limit shape of $h_{\varepsilon,\lambda}$. Using this, they confirmed the rate
function $\Phi$ discovered in [LDMRS16]. In the limit of large $\lambda$, they
discussed how the form (1.7) emerges in the exact solution.

######  Remark 1.5.

The reason that we set $h_{\varepsilon}(2,0)+\log\sqrt{4\pi}>\lambda$ in the
theorem instead of $h_{\varepsilon}(2,0)>\lambda$ is purely for the
convenience of the proof. It makes no difference since we let
$\lambda\to\infty$.

######  Remark 1.6.

Our method does not rely on exact formulas and may apply to other initial
data. In particular, our method should apply to the flat initial data
$h_{\varepsilon}(0,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})=0$. The
result in Theorem 1.1 remains the same for the flat initial data, while
Theorem 1.2 holds with a different limit shape

$\mathsf{h}^{\text{flat}}_{*}(t,x):=\begin{cases}-|x|+\frac{t}{2},&\text{when
}|x|\leq\frac{t}{2},\\\ 0,&\text{when }|x|>\frac{t}{2}.\end{cases}$

More broadly, one can consider applying our method to a function-valued,
symmetrically decreasing initial data: $h_{\text{ic}}(x)=h_{\text{ic}}(|x|)$
and $h_{\text{ic}}(|x|)$ non-increasing in $|x|$. We conjecture that the
result in Theorem 1.1 remains the same; the limit shape (in-general) needs to
be adjusted according to the initial data.

Going beyond symmetrically decreasing initial data, one may see different
behaviors of the deviations. In particular, a dynamical phase transition
triggered by a symmetry breaking has been predicted in [JKM16, SKM18] (see
also [KLD17, HMS21, KLD22]) for the Brownian initial data. For such initial
data, we do not expect our method to apply directly and new ideas are needed.

###  1.2. A review of the Freidlin-Wentzell LDP for the SHE

To motivate the proof of Theorems 1.1 and 1.2, we recall the Freidlin-Wentzell
LDP for the SHE $\\{Z_{\varepsilon}\\}_{\varepsilon>0}$. The result was
established for the SHE under function-valued initial data and the narrow
wedge initial data in [LT21, Proposition 1.7]. For our propose, we only state
the result for the narrow wedge initial data.

Let us first recall the definition of an LDP. Let $\Omega$ be a topological
space. We say that a sequence of $\Omega$-valued random variables
$\\{Y_{\varepsilon}\\}_{\varepsilon>0}$ satisfies an LDP with speed
$\varepsilon^{-1}$ and rate function $I$ if

$\displaystyle\limsup_{\varepsilon\to
0}\varepsilon\log\mathbb{P}(Y_{\varepsilon}\in F)$
$\displaystyle\leq-\inf_{x\in F}I(x)\qquad\text{ if }F\subseteq\Omega\text{ is
closed},$ $\displaystyle\liminf_{\varepsilon\to
0}\varepsilon\log\mathbb{P}(Y_{\varepsilon}\in G)$
$\displaystyle\geq-\inf_{x\in G}I(x)\qquad\text{ if }G\subseteq\Omega\text{ is
open}.$

We now state the LDP for the SHE $\\{Z_{\varepsilon}\\}_{\varepsilon>0}$. Fix
$T>0$ and $\delta\in(0,T)$. We take
$\Omega=C([\delta,T]\times[-\delta^{-1},\delta^{-1}])$ with the uniform
topology and view $Z_{\varepsilon}$ as an $\Omega$-valued random variable. The
reason that we avoid $t=0$ in our choice of $\Omega$ is because
$Z_{\varepsilon}$ starts from the Dirac-delta initial data, which is singular.
For $t>0$, the heat kernel in (1.2) smoothes out the singularity, so
$Z_{\varepsilon}$ is $\Omega$-valued for any fixed $\delta>0$; see [Qua11].

Since $Z_{\varepsilon}$ in (1.4) is driven by the space-time white noise
$\sqrt{\varepsilon}\xi$, it would be helpful to first look at the LDP of
$\\{\sqrt{\varepsilon}\xi\\}_{\varepsilon>0}$. We view $\rho\in
L^{2}([0,T]\times\mathbb{R})$ as a deviation $\sqrt{\varepsilon}\xi$. Since
$\xi$ has Dirac-delta correlation function, we have informally
$\mathbb{P}(\sqrt{\varepsilon}\xi\approx\rho)\approx\exp(-\frac{1}{2}\varepsilon^{-1}\|\rho\|^{2}_{L^{2}([0,T]\times\mathbb{R})})$
for small $\varepsilon$. We replace the noise $\sqrt{\varepsilon}\xi$ with its
deviation $\rho$ and consider the PDE

$\partial_{t}\mathsf{Z}=\frac{1}{2}\partial_{xx}\mathsf{Z}+\rho\mathsf{Z},\qquad\mathsf{Z}(0,x)=\delta(x),$
(1.8)

where $\mathsf{Z}=\mathsf{Z}(\rho;t,x)$, $t\in[0,T]$ and $x\in\mathbb{R}$.
Like the SHE, the solution to this PDE is understood in the mild form

$\mathsf{Z}(\rho;t,x)=p(t,x)+\int_{0}^{t}\int_{\mathbb{R}}p(t-s,x-y)\rho(s,y)\mathsf{Z}(\rho;s,y)dsdy.$
(1.9)

By iteration of (1.9), the solution $\mathsf{Z}(\rho)$ admits a series
expansion and we have the Feynman-Kac formula

$\mathsf{Z}(\rho;t,x)=\mathbb{E}_{0\to
x}\bigg{[}\exp\Big{(}\int_{0}^{t}\rho(s,B_{\text{b}}(s))ds\Big{)}\bigg{]}p(t,x),\qquad
t\in(0,T]\times\mathbb{R},$ (1.10)

where $B_{\text{b}}$ is a Brownian bridge such that $B_{\text{b}}(0)=0$ and
$B_{\text{b}}(t)=x$.

It is standard to see that $\mathsf{Z}(\rho)$ is a continuous function on
$(0,T]\times\mathbb{R}$, we refer to [LT21, Section 2] for more detail. Fix
$\delta>0$. We view $\mathsf{Z}:\rho\mapsto\mathsf{Z}(\rho)$ as a map from
$L^{2}([0,T]\times\mathbb{R})$ to
$C([\delta,T]\times[-\delta^{-1},\delta^{-1}])$. We now state the LDP for
$\\{Z_{\varepsilon}\\}_{\varepsilon>0}$.

######  Proposition 1.7.

Fix $T<\infty$ and $0<\delta<T$. Let $Z_{\varepsilon}$ be the solution to
(1.4) with the Dirac-delta initial data.
$\\{Z_{\varepsilon}\\}_{\varepsilon>0}$ satisfies an LDP in
$C([\delta,T]\times[-\delta^{-1},\delta^{-1}])$ with speed $\varepsilon^{-1}$
and the rate function

$I(f):=\inf\big{\\{}\tfrac{1}{2}\|\rho\|_{L^{2}([0,T]\times\mathbb{R})}^{2}\,:\,\rho\in
L^{2}([0,T]\times\mathbb{R}),\mathsf{Z}(\rho)=f\big{\\}},\quad f\in
C([\delta,T]\times[-\delta^{-1},\delta^{-1}]),$

where $\mathsf{Z}(\rho)$ is the unique solution to
$\partial_{t}\mathsf{Z}=\frac{1}{2}\partial_{xx}\mathsf{Z}+\rho\mathsf{Z}$
with the Dirac-delta initial data $\mathsf{Z}(0,x)=\delta(x)$.

###### Proof.

This is a direct consequence of [LT21, Proposition 1.7] Part (b). ∎

Since $Z_{\varepsilon}=e^{h_{\varepsilon}}$, we have
$\mathbb{P}[h_{\varepsilon}(2,0)+\log\sqrt{4\pi}\geq\lambda]=\mathbb{P}[Z_{\varepsilon}(2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}]$.
By (1.6), we have $\Phi(\lambda)=-\lim_{\varepsilon\to
0}\varepsilon\log\mathbb{P}[Z_{\varepsilon}(2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}]$.
Taking $T=2$ in Proposition 1.7 and applying contraction principle yield that
for $\lambda\geq 0$,

$\Phi(\lambda)=\inf\Big{\\{}\frac{1}{2}\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}^{2}\,:\rho\in
L^{2}([0,2]\times\mathbb{R}),\,\mathsf{Z}(\rho;2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}\Big{\\}}.$
(1.11)

###  1.3. Proof ideas

We explain the ideas for proving Theorems 1.1 and 1.2. The first step towards
proving Theorem 1.1 is to apply a scaling to the variational formula (1.11).
By the Feynman-Kac formula (1.10), we know that for arbitrary $\rho\in
L^{2}([0,2]\times\mathbb{R})$,
$\mathsf{Z}(\lambda\rho(\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda^{\frac{1}{2}}{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}});t,x)=\lambda^{\frac{1}{2}}\mathsf{Z}(\rho;\lambda
t;\lambda^{\frac{1}{2}}x).$ Using this relation and applying the scaling
$\rho\mapsto\lambda\rho(\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda^{\frac{1}{2}}{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$,
one can rewrite the rate function in (1.11) as

$\Phi(\lambda)=\lambda^{\frac{3}{2}}\inf\Big{\\{}\frac{1}{2\lambda}\|\rho\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}:\rho\in
L^{2}([0,2\lambda]\times\mathbb{R}),\,\mathsf{Z}(\rho;2\lambda,0)\geq\frac{1}{\sqrt{4\pi\lambda}}e^{\lambda}\Big{\\}}.$
(1.12)

The value of the infimum should be of constant order, this explains the
$\tfrac{3}{2}$-power in Theorem 1.1. To prove Theorem 1.1, it suffices to show
that

$\lim_{\lambda\to\infty}\inf\Big{\\{}\frac{1}{2\lambda}\|\rho\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}:\rho\in
L^{2}([0,2\lambda]\times\mathbb{R}),\,\mathsf{Z}(\rho;2\lambda,0)\geq\frac{1}{\sqrt{4\pi\lambda}}e^{\lambda}\Big{\\}}=\frac{4}{3}.$
(1.13)

We explain why the $\frac{4}{3}$ appears on the right hand side. To motivate
the discussion, let us assume that the minimizers (there might be more than
one) of (1.13) become asymptotically _time-independent_ as $\lambda\to\infty$.
Under this assumption, we only need to consider $\rho$ such that
$\rho(t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})=\varphi({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$
for $t\in[0,2\lambda]$ and some $\varphi:\mathbb{R}\to\mathbb{R}$. So (1.13)
simplifies to

$\lim_{\lambda\to\infty}\inf\Big{\\{}\|\varphi\|_{L^{2}(\mathbb{R})}^{2}:\varphi\in
L^{2}(\mathbb{R}),\,\mathsf{Z}(\varphi;2\lambda,0)\geq\frac{1}{\sqrt{4\pi\lambda}}e^{\lambda}\Big{\\}}=\frac{4}{3}.$
(1.14)

Recall that $\mathsf{Z}(\varphi)$ solves the PDE
$\partial_{t}\mathsf{Z}=\frac{1}{2}\partial_{xx}\mathsf{Z}+\varphi\mathsf{Z}$.
Since $\varphi$ does not depend on time, we treat $\mathsf{Z}(t)$ as function-
valued and view the PDE as a function-valued ODE
$\partial_{t}\mathsf{Z}(t)=A^{\varphi}\mathsf{Z}(t)$, where
$A^{\varphi}:=\frac{1}{2}\partial_{xx}+\varphi$ is a Schrödinger-type
operator. Set the largest eigenvalue of $A^{\varphi}$ to be $F(\varphi)$,
whose expression is given in (2.3). It is natural to expect that
$\mathsf{Z}(\rho;t,0)$ at large time $t$ grows as $\exp(tF(\varphi))$. In
particular, we have $\mathsf{Z}(\varphi;2\lambda,0)\sim\exp(2\lambda
F(\varphi))$ as $\lambda\to\infty$. This suggests that

$\text{LHS of
}\eqref{eq:heuristic2}=\inf\Big{\\{}\|\varphi\|^{2}_{L^{2}(\mathbb{R})}:F(\varphi)\geq\frac{1}{2}\Big{\\}}.$
(1.15)

The problem of finding the minimizers of (1.15) can be transformed into
understanding when the equality of the Gagliardo–Nirenberg-Sobolev inequality
holds, see Lemma 2.4. It turns out that the minimizers of the infimum are
given by $\operatorname{sech}^{2}(x+v)$, where $v$ is an arbitrary constant.
Since $\|\operatorname{sech}^{2}(x+v)\|_{L^{2}(\mathbb{R})}^{2}=\frac{4}{3}$
for all $v$, we know that the right hand side of (1.15) equals $\tfrac{4}{3}$.
This explains why Theorem 1.1 should hold.

We proceed to explain the idea for proving Theorem 1.2. Under the time-
independence assumption, it is natural to believe that the minimizers of the
infimum in (1.12) converges to the minimizers of the infimum in (1.15) as
$\lambda\to\infty$. However, from the previous paragraph, the minimizer of the
infimum in (1.15) is not unique and this causes a problem to our analysis. To
resolve this, we show that any minimizer of the infimum in (1.12) has to be
symmetric in the $x$ coordinate (in fact the function should reach its peak at
$0$ and decrease on both sides of $0$). The proof of this is carried out in
Section 3.1.2. Consequently, we see that the minimizer of (1.13) should
converge to $\operatorname{sech}^{2}x$ as $\lambda\to\infty$.

We denote $\rho_{*}(t,x):=\rho_{*}(x):=\operatorname{sech}^{2}x$. By the
Freidlin-Wentzell LDP stated in Proposition 1.7 and scaling, one can show that
for fixed $\lambda$, as $\varepsilon\to 0$, the space-time path
$h_{\varepsilon,\lambda}$ concentrates around
$\lambda^{-1}\log\mathsf{Z}(\rho_{*};\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$.
By the Feynman-Kac formula (1.10),

$\lambda^{-1}\log\mathsf{Z}(\rho_{*};\lambda t,\lambda
x)=\lambda^{-1}\log\mathbb{E}_{\lambda x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda
t}\rho_{*}(B_{\text{b}}(s))ds\Big{)}\bigg{]}-\frac{x^{2}}{2t}-\lambda^{-1}\log\sqrt{4\pi}.$
(1.16)

We seek to compute the $\lambda\to\infty$ limit of the first term on the right
hand side of (1.16).

Assume $x\neq 0$. The Brownian bridge starts from $\lambda x$, which is far
way from $0$. Since $\rho_{*}(x)=\operatorname{sech}^{2}(x)$ decays
exponentially as $|x|\to\infty$, the path of $B_{\text{b}}$ contributes little
to the integral until it arrives near $0$. Fix $s\in[0,t]$, the probability of
the event that the first time $B_{\text{b}}$ hits $0$ around time $\lambda s$
is approximately $\exp(-\frac{\lambda x^{2}(t-s)}{2st})$. After arriving at
$0$, we have $\mathbb{E}_{0\to 0}[\exp(\int_{\lambda s}^{\lambda
t}\rho_{*}(B_{\text{b}}(r))dr)]\approx\exp(\frac{\lambda(t-s)}{2})$ for large
$\lambda$. Optimizing over $s\in[0,t]$, we expect that

$\lim_{\lambda\to\infty}\lambda^{-1}\log\mathbb{E}_{\lambda x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda
t}\rho_{*}(B_{\text{b}}(s))ds\Big{)}\bigg{]}=\sup_{s\in[0,t]}\Big{(}-\frac{x^{2}(t-s)}{2st}+\frac{t-s}{2}\Big{)}=\mathsf{h}_{*}(t,x)+\frac{x^{2}}{2t}.$

Using this together with (1.16) shows that
$\lambda^{-1}\log\mathsf{Z}(\rho_{*};\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$
converges to $\mathsf{h}_{*}$ defined in (1.7).

###  1.4. Technical Difficulties

In this section, we emphasize some of the technical difficulties in our proof.
Denote the set of minimizers of (1.12) to be $\mathcal{K}_{\lambda}$. As
mentioned in the previous section, one important step in our proof is to show
that $\mathcal{K}_{\lambda}$ converges to $\rho_{*}$ as $\lambda\to\infty$. We
explain the proof ingredients of it in Sections 1.4.1 and 1.4.2. In Section
1.4.3, we describe a technical issue for proving the limit shape.

####  1.4.1. Properties of $\mathcal{K}_{\lambda}$

To prove that the elements of $\mathcal{K}_{\lambda}$ converge to $\rho_{*}$
as $\lambda\to\infty$, we first need to show that $\mathcal{K}_{\lambda}$ is
not empty, i.e. the minimizer of the infimum in (1.12) exists. To prove this,
we need 1). the compactness of the level sets
$\\{\rho:\|\rho\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}\leq r\\}$; 2). the
continuity of the map $\rho\mapsto\mathsf{Z}(\rho;2\lambda,0)$. Unfortunately,
the level sets $\\{\rho:\|\rho\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}\leq
r\\}$ are not compact in $L^{2}([0,2\lambda]\times\mathbb{R})$. To overcome
this, we consider an abstract Wiener space $(\mathcal{B},\mu)$ such that
$L^{2}([0,2\lambda]\times\mathbb{R})\subseteq\mathcal{B}$ is the Cameron-
Martin space. This gives us the compactness of
$\\{\rho:\|\rho\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}\leq r\\}$ in
$\mathcal{B}$ and it preserves the continuity of the map
$\rho\mapsto\mathsf{Z}(\rho;2\lambda,0)$.

We also need to show that the elements of $\mathcal{K}_{\lambda}$ are
symmetric and decreasing in space (see Section 3.1.2 for the precise
definition). We prove this by considering the symmetric and decreasing
rearrangement of $\rho$. Using the rearrangement inequalities from [BLL74,
LL01], we show that the symmetric and decreasing rearrangement of $\rho$
preserves the $L^{2}$ norm while it increases the value of
$\mathsf{Z}(\rho;2\lambda,0)$. This implies that the elements in
$\mathcal{K}_{\lambda}$ must be symmetric and decreasing in space.

####  1.4.2. $\mathcal{K}_{\lambda}$ converges to $\rho_{*}$

The key innovations for proving the convergence are 1). A near-optimal upper
bound of $\mathsf{Z}(\rho;2\lambda,0)$ in terms of an exponential integral of
the ground state $F(\rho)$, see Proposition 2.9; 2). A perturbative analysis
of $F$, see Lemma 3.9.

For 1), one can obtain an upper bound of $\mathsf{Z}(\rho;2\lambda,0)$ via
iterating (1.9) and applying the Cauchy-Schwarz inequality, see Lemma 2.3.
However, such bounds are not optimal. In Proposition 2.9, we prove a near-
optimal bound of $\mathsf{Z}(\rho;2\lambda,0)$. The proof of the proposition
relies on the time-dependent semigroup
$\partial_{t}-\frac{1}{2}\partial_{xx}-\rho$, an energy estimate when $\rho$
is smooth and compactly supported, and the method of approximation for
$\rho\in L^{2}([0,2\lambda]\times\mathbb{R})$, which is accomplished in Lemmas
2.3, 2.6 and 2.7.

We explain 2) in more detail. Using the $L^{4}$ Gagaliardo-Nirenberg-Sobolev
inequality, we obtain an optimal upper bound of $F(\varphi)$ as well as
identify its optimizers; see Lemma 2.4. The optimizer in Lemma 2.4 is only
unique up to shifting. Our perturbative analysis says that when $\varphi$ is
symmetric and decreasing and $F(\varphi)$ is near its optimal upper bound,
$\varphi$ must be close to the symmetric optimizer. The proof of this is
through a weak-convergence argument.

The proof of the convergence is carried out in Proposition 3.10. Combining
items 1) and 2), the condition $\rho\in\mathcal{K}_{\lambda}$ implies that for
most $r\in[0,2\lambda]$,
$\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{L^{2}(\mathbb{R})}$
is close to $\|\rho_{*}\|_{L^{2}(\mathbb{R})}$ and
$F(\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))$ is close to
$F(\rho_{*})$, which is the optimal upper bound. This implies the convergence
of $\mathcal{K}_{\lambda}$ to $\rho_{*}$, as $\lambda\to\infty$.

####  1.4.3. The equi-continuity

By the Freidlin-Wentzell LDP, we can show that for fixed $\lambda>0$, as
$\varepsilon\to 0$, $h_{\varepsilon,\lambda}$ concentrates around the set of
functions
$\lambda^{-1}\log\lambda^{\frac{1}{2}}\mathsf{Z}(\mathcal{K}_{\lambda};\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$.
In Section 1.4.2, we explain how to prove that $\mathcal{K}_{\lambda}$
converges to $\rho_{*}$, as $\lambda\to\infty$. To prove Theorem 1.2, we need
to show additionally that the distance between
$\lambda^{-1}\log\lambda^{\frac{1}{2}}\mathsf{Z}(\mathcal{K}_{\lambda};\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$
and
$\lambda^{-1}\log\lambda^{\frac{1}{2}}\mathsf{Z}(\rho_{*};\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$
is small as $\lambda\to\infty$. This is proved by establishing the equi-
continuity of the maps
$f_{\lambda}:\rho\mapsto\lambda^{-1}\log\lambda^{\frac{1}{2}}\mathsf{Z}(\rho;\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$,
see Proposition 4.1.

### Acknowledgments.

We thank Ivan Corwin, Alexandre Krajenbrink, Pierre Le Doussal, and Baruch
Meerson for their helpful comments on the presentation of this work. We thank
the referees for their useful comments on the manuscript, especially for
pointing out an error in Lemma 3.1 in the first version of the manuscript. The
research of LCT is partially supported by the Sloan Fellowship and the NSF
through DMS-1953407 and DMS-2153739.

### Outline of the rest of the paper

In Section 2, we prove Theorem 1.1 and confirm the $\frac{3}{2}$-power law in
the deep upper tail rate function. In Section 3, we give some detailed
characterization of the minimizers in (1.13). We also prove that the
asymptotic limit of these minimizers equals $\operatorname{sech}^{2}x$. In
Section 4, we establish some result about equi-continuity and prove the
convergence of the left hand side of (1.16) to $\mathsf{h}_{*}$, thus
completing the proof of Theorem 1.2.

##  2\. The $\frac{3}{2}$-power law

By the discussion in Section 1.3, to prove Theorem 1.1, it suffices to show
(1.13). In particular, (1.13) follows if we can show

$\displaystyle\limsup_{\lambda\to\infty}\inf\Big{\\{}\frac{1}{2\lambda}\|\rho\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}:\rho\in
L^{2}([0,2\lambda]\times\mathbb{R}),\,\mathsf{Z}(\rho;2\lambda,0)\geq\frac{1}{\sqrt{4\pi\lambda}}e^{\lambda}\Big{\\}}\leq\frac{4}{3},$
($\mathsf{LimSup}$)
$\displaystyle\liminf_{\lambda\to\infty}\inf\Big{\\{}\frac{1}{2\lambda}\|\rho\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}:\rho\in
L^{2}([0,2\lambda]\times\mathbb{R}),\,\mathsf{Z}(\rho;2\lambda,0)\geq\frac{1}{\sqrt{4\pi\lambda}}e^{\lambda}\Big{\\}}\geq\frac{4}{3}.$
($\mathsf{LimInf}$)

The rest of the section is devoted to proving ($\mathsf{LimSup}$) and
($\mathsf{LimInf}$).

For the rest of the paper, we will use $C=C(a_{1},a_{2},\dots)$ to denote a
deterministic positive finite constant. The constant may change from line to
line or even within the same line, but depends only on the designated
variables $a_{1},a_{2},\dots$. In addition, we will denote the Brownian motion
as $B$ and Brownian bridge as $B_{\text{b}}$. When we write $\mathbb{E}_{x\to
y}[f(\int_{a}^{b}B_{\text{b}}(s)ds)]$, the expectation is taken with respect
to a Brownian bridge with $B_{\text{b}}(a)=x$ and $B_{\text{b}}(b)=y$. When we
write $\mathbb{E}_{x}[f(\int_{a}^{b}B(s)ds)]$, the expectation is taken with a
Brownian motion starting from $B(a)=x$.

###  2.1. Proof of ($\mathsf{LimSup}$)

The key to the proof is to consider the exponential moment of the
$B_{\text{b}}$ subject to a time-independent potential $\varphi$. More
precisely, for $\varphi\in L^{2}(\mathbb{R})$, we consider

$\mathsf{Z}(\varphi;2\lambda,0)=\mathbb{E}_{0\to
0}\bigg{[}\exp\Big{(}\int_{0}^{2\lambda}\varphi(B_{\text{b}}(s))ds\Big{)}\bigg{]}p(2\lambda,0).$
(2.1)

To analyze (2.1), let us recall some background knowledge from [Che10, Section
4.1]. Fixing a bounded continuous function $\varphi:\mathbb{R}\to\mathbb{R}$,
we define $T^{\varphi}_{t}:L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})$ as

$(T^{\varphi}_{t}g)(x)=\mathbb{E}_{x}\bigg{[}\exp\Big{(}\int_{0}^{t}\varphi(B(s))ds\Big{)}g(B(t))\bigg{]}.$

By [Che10, page 105], $(T^{\varphi}_{t})_{t\geq 0}$ forms a strongly-
continuous, self-adjoint and non-negative semi-group on $L^{2}(\mathbb{R})$.
Let $A^{\varphi}$ be the generator of the semi-group $(T^{\varphi}_{t})_{t\geq
0}$. Denote the domain of the generator $A^{\varphi}$ by
$\mathcal{D}(A^{\varphi})$. Let $C_{c}^{\infty}(\mathbb{R})$ be the space of
compactly supported smooth function. Let $H^{1}(\mathbb{R})$ to be the Sobolev
space $\\{g\in L^{2}(\mathbb{R}):g^{\prime}\in L^{2}(\mathbb{R})\\}$. By
[Che10, Theorem 4.1.2 and Lemma 4.1.3],
$C_{c}^{\infty}(\mathbb{R})\subseteq\mathcal{D}(A^{\varphi})\subseteq
H^{1}(\mathbb{R})$. Further, for $g\in\mathcal{D}(A^{\varphi})$,

$\langle
A^{\varphi}g,g\rangle_{L^{2}(\mathbb{R})}=\int_{\mathbb{R}}\varphi(x)g(x)^{2}-\frac{1}{2}g^{\prime}(x)^{2}dx.$
(2.2)

For $\varphi\in L^{2}(\mathbb{R})$, define

$F(\varphi):=\sup\Big{\\{}\int_{\mathbb{R}}\varphi(x)g(x)^{2}-\frac{1}{2}g^{\prime}(x)^{2}dx:g\in
H^{1}(\mathbb{R}),\|g\|_{L^{2}(\mathbb{R})}=1\Big{\\}}.$ (2.3)

Assume that $\varphi$ is bounded and continuous. Since
$C_{c}^{\infty}(\mathbb{R})$ is dense in $H^{1}(\mathbb{R})$ (w.r.t. the
$H^{1}$-norm), we have

$F(\varphi)=\sup\\{\langle A^{\varphi}g,g\rangle_{L^{2}(\mathbb{R})}:g\in
C_{c}^{\infty}(\mathbb{R}),\|g\|_{L^{2}(\mathbb{R})}=1\\}.$ (2.4)

Further, we record two useful inequalities from [Che10, Eq 4.1.25 and 4.1.29].
For $g\in C_{c}^{\infty}(\mathbb{R})$,

$\exp\big{(}t\langle
A^{\varphi}g,g\rangle_{L^{2}(\mathbb{R})}\big{)}\leq\langle
T^{\varphi}_{t}g,g\rangle_{L^{2}(\mathbb{R})}\leq\exp\big{(}tF(\varphi)\big{)}.$
(2.5)

It is known that (see [Che10, Thereom 4.1.6]) for bounded continuous
$\varphi$,

$\lim_{\lambda\to\infty}{\color[rgb]{0,0,0}\frac{1}{\lambda}\log}\mathbb{E}_{0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B(s))ds\Big{)}\bigg{]}=F(\varphi).$

We prove a similar result for the Brownian bridge, allowing the starting
position to deviate from zero by an amount less than
$\mathcal{O}(\lambda^{\frac{1}{2}})$.

######  Lemma 2.1.

Fix bounded continuous $\varphi:\mathbb{R}\to\mathbb{R}$. Fix
$\alpha\in(0,1/2)$. Uniformly for $|x|\leq\lambda^{\alpha}$,

$\lim_{\lambda\to\infty}\frac{1}{\lambda}\log\mathbb{E}_{x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B_{b}(s))ds\Big{)}\bigg{]}=F(\varphi).$
(2.6)

######  Remark 2.2.

Here we only need the $x=0$ result for proving ($\mathsf{LimSup}$), and the
result for $x\neq 0$ will be needed in Section 4.

We say that $\liminf_{\lambda\to\infty}f_{\lambda}(t,x)\geq g(t,x)$ uniformly
for $(t,x)\in\mathcal{O}$, if for any $\zeta>0$, there exists $M$ such that
$f(t,x)>g(t,x)-\zeta$ for all $\lambda>M$ and $(t,x)\in\mathcal{O}.$
Similarly, we say that $\limsup_{\lambda\to\infty}f_{\lambda}(t,x)\leq g(t,x)$
uniformly for $(t,x)\in\mathcal{O}$ if for any $\zeta>0$, there exists $M$
such that $f(t,x)<g(t,x)+\zeta$ for all $\lambda>M$ and $(t,x)\in\mathcal{O}.$

###### Proof of Lemma 2.1.

The idea of the proof is from [Che10, Theroem 4.1.6]. Recall that for a
Brownian bridge $B_{\text{b}}(s)$ with $B_{\text{b}}(0)=x$ and
$B_{\text{b}}(\lambda)=0$, the random variable $B_{\text{b}}(1)$ and
$B_{\text{b}}(\lambda-1)$ has joint probability density function

$f_{B_{\text{b}}(1),B_{\text{b}}(\lambda-1)}(y,z):=\frac{p(1,x-y)p(\lambda-2,y-z)p(1,z)}{p(\lambda,x)}.$

Using the boundedness of $\varphi$ and the preceding joint density in order,
we have

$\displaystyle\mathbb{E}_{x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B_{b}(s))ds\Big{)}\bigg{]}$
$\displaystyle\geq\frac{1}{C}\mathbb{E}_{B_{\text{b}}(0)=x,B_{\text{b}}(\lambda)=0}\bigg{[}\exp\Big{(}\int_{1}^{\lambda-1}\varphi(B_{b}(s))ds\Big{)}\bigg{]}$
$\displaystyle=\frac{1}{Cp(\lambda,x)}\int_{\mathbb{R}}p(1,x-y)\mathbb{E}_{y}\bigg{[}\exp\Big{(}\int_{1}^{\lambda-1}\varphi(B(s))ds\Big{)}p(1,B(\lambda-1))\bigg{]}dy.$
(2.7)

Since $\varphi$ is fixed, we omit the dependence on $\varphi$ in the constant
$C$. For any compactly supported $g\in C_{c}^{\infty}(\mathbb{R})$, since
$|x|\leq\lambda^{\alpha}$, $\frac{1}{p(\lambda,x)}\geq C\sqrt{\lambda}$ and
$p(1,x-y)\geq\frac{1}{C(g)}|g(y)|\exp(-C\lambda^{2\alpha})$. Hence,

$\displaystyle\frac{1}{p(\lambda,x)}\int_{\mathbb{R}}p(1,x-y)\mathbb{E}_{y}\bigg{[}\exp\Big{(}\int_{1}^{\lambda-1}\varphi(B(s))ds\Big{)}p(1,B(\lambda-1))\bigg{]}dy$
$\displaystyle\geq\frac{\exp(-C\lambda^{2\alpha})}{C(g)}\int_{\mathbb{R}}g(y)\mathbb{E}_{y}\bigg{[}\exp\Big{(}\int_{1}^{\lambda-1}\varphi(B(s))ds\Big{)}g(B(\lambda-1))\bigg{]}dy$
$\displaystyle\geq\frac{\exp(-C\lambda^{2\alpha})}{C(g)}\langle
T^{\varphi}_{\lambda-2}g,g\rangle\geq\frac{\exp(-C\lambda^{2\alpha})}{C(g)}\exp\Big{(}(\lambda-2)\langle
A^{\varphi}g,g\rangle_{L^{2}(\mathbb{R})}\Big{)}.$

The last inequality follows from (2.5). Inserting the above bound into the
right hand side of (2.7) yields

$\mathbb{E}_{x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B_{b}(s))ds\Big{)}\bigg{]}\geq\frac{\exp(-C\lambda^{2\alpha})}{C(g)}\exp\Big{(}(\lambda-2)\langle
A^{\varphi}g,g\rangle_{L^{2}(\mathbb{R})}\Big{)}.$

Taking the logarithm of the both sides, dividing the result by $\lambda$ and
sending $\lambda\to\infty$, we obtain that uniformly for
$|x|\leq\lambda^{\alpha}$,

$\liminf_{\lambda\to\infty}\frac{1}{\lambda}\log\mathbb{E}_{x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B_{b}(s))ds\Big{)}\bigg{]}\geq\langle
A^{\varphi}g,g\rangle_{L^{2}(\mathbb{R})}.$

Taking the supremum of the right hand side over $g\in
C_{c}^{\infty}(\mathbb{R})$ and using (2.4), we conclude that uniformly for
$|x|\leq\lambda^{\alpha}$,

$\liminf_{\lambda\to\infty}\frac{1}{\lambda}\log\mathbb{E}_{x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B_{b}(s))ds\Big{)}\bigg{]}\geq
F(\varphi).$

To prove the reverse inequality, we write

$\displaystyle\mathbb{E}_{x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B_{b}(s))ds\Big{)}\bigg{]}$
$\displaystyle=\mathbb{E}_{x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B_{b}(s))ds\Big{)}\mathbf{1}_{\\{|B_{b}(1)|\leq\lambda^{2},|B_{b}(\lambda-1)|\leq\lambda^{2}\\}}\bigg{]}$
$\displaystyle\quad+\mathbb{E}_{x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B_{b}(s))ds\Big{)}\mathbf{1}_{\\{|B_{b}(1)|\geq\lambda^{2}\text{
or }|B_{b}(\lambda-1)|\geq\lambda^{2}\\}}\bigg{]}$
$\displaystyle\leq\mathbb{E}_{x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B_{b}(s))ds\Big{)}\mathbf{1}_{\\{|B_{b}(1)|\leq\lambda^{2},|B_{b}(\lambda-1)|\leq\lambda^{2}\\}}\bigg{]}+C\exp(-C^{-1}\lambda^{2}).$
(2.8)

Recall that $|x|\leq\lambda^{\alpha}$, $\alpha\in(0,\frac{1}{2})$. The last
inequality follows from the boundedness of $\varphi$ together with the tail
decay of $\mathbb{P}[B_{\text{b}}(1)\geq\lambda^{2}]$ and
$\mathbb{P}[B_{\text{b}}(\lambda-1)\geq\lambda^{2}]$. For the first term on
the right hand side of (2.8),

$\displaystyle\mathbb{E}_{x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B_{b}(s))ds\Big{)}\mathbf{1}_{\\{|B_{b}(1)|\leq\lambda^{2},|B_{b}(\lambda-1)|\leq\lambda^{2}\\}}\bigg{]}$
$\displaystyle\leq
C\mathbb{E}_{B_{\text{b}}(0)=x,B_{\text{b}}(\lambda)=0}\bigg{[}\exp\Big{(}\int_{1}^{\lambda-1}\varphi(B_{b}(s))ds\Big{)}\mathbf{1}_{\\{|B_{b}(1)|\leq\lambda^{2},|B_{b}(\lambda-1)|\leq\lambda^{2}\\}}\bigg{]}$
$\displaystyle=\frac{C}{p(\lambda,x)}\int_{\mathbb{R}}p(1,x-y)\mathbf{1}_{\\{|y|\leq\lambda^{2}\\}}\mathbb{E}_{y}\bigg{[}\exp\Big{(}\int_{1}^{\lambda-1}\varphi(B(s))ds\Big{)}p(1,B(\lambda-1))\mathbf{1}_{\\{|B(\lambda-1)|\leq\lambda^{2}\\}}\bigg{]}dy.$
(2.9)

The first inequality follows from the boundedness of $\varphi$ and the second
equality is due to the transition probabilities of the Brownian bridge. Let
$g_{\lambda}\in C_{c}^{\infty}(\mathbb{R})$ be such that $g_{\lambda}(y)=1$
for all $|y|\leq\lambda^{2}$, $g_{\lambda}(y)=0$ for $|y|\geq 2\lambda^{2}$
and $g_{\lambda}(y)\in[0,1]$ for all $y$. Then
$p(1,x-y)\mathbf{1}_{\\{|y|\leq\lambda^{2}\\}}\leq g_{\lambda}(y)$ and thus

$\displaystyle\int
p(1,x-y)\mathbf{1}_{\\{|y|\leq\lambda^{2}\\}}\mathbb{E}_{y}\bigg{[}\exp\Big{(}\int_{1}^{\lambda-1}\varphi(B(s))ds\Big{)}p(1,B(\lambda-1))\mathbf{1}_{\\{|B(\lambda-1)|\leq\lambda^{2}\\}}\bigg{]}dy$
$\displaystyle\leq\int
g_{\lambda}(y)\mathbb{E}_{y}\bigg{[}\exp\Big{(}\int_{1}^{\lambda-1}\varphi(B(s))ds\Big{)}g_{\lambda}(B(\lambda-1))\bigg{]}dy$
$\displaystyle=\langle
T^{\varphi}_{\lambda-2}g_{\lambda},g_{\lambda}\rangle_{L^{2}(\mathbb{R})}\leq\|g_{\lambda}\|_{L^{2}(\mathbb{R})}^{2}\exp\big{(}(\lambda-2)F(\varphi)\big{)}.$

The last equality follows from (2.5). Inserting the above bound into (2.9) and
then inserting the result to (2.8), we conclude that uniformly for
$|x|\leq\lambda^{\alpha}$,

$\limsup_{\lambda\to\infty}\frac{1}{\lambda}\log\mathbb{E}_{x\to
0}\bigg{[}\exp\Big{(}\int_{0}^{\lambda}\varphi(B_{b}(s))ds\Big{)}\bigg{]}\leq
F(\varphi).$

This concludes the lemma. ∎

We are now ready to prove ($\mathsf{LimSup}$). Recall that
$\rho_{*}(t,x)=\rho_{*}(x)=\operatorname{sech}^{2}(x)$. Fix arbitrary
$\zeta>0$ and take $x=0$ in Lemma 2.1. We have

$\displaystyle\begin{split}\lim_{\lambda\to\infty}\frac{1}{\lambda}\log\mathsf{Z}((1+\zeta)\rho_{*};2\lambda,0)&=\lim_{\lambda\to\infty}\frac{1}{\lambda}\log\mathbb{E}_{0\to
0}\bigg{[}\exp\Big{(}\int_{0}^{2\lambda}(1+\zeta)\rho_{*}(B_{b}(s))ds\Big{)}\bigg{]}{\color[rgb]{0,0,0}\cdot
p(2\lambda,0)}\\\ &=2F((1+\zeta)\rho_{*}).\end{split}$ (2.10)

Referring to (2.3) for the definition of $F$ and taking
$g=\frac{1}{\sqrt{2}}\operatorname{sech}(x)$, $\|g\|_{L^{2}(\mathbb{R})}=1$,
we have

$F((1+\zeta)\rho_{*})\geq\int_{\mathbb{R}}(1+\zeta)\rho_{*}(x)g(x)^{2}-\frac{1}{2}g^{\prime}(x)^{2}dx=\frac{1}{2}+\frac{2}{3}\zeta.$

This, together with (2.10), implies that for $\lambda$ large enough,
$\mathsf{Z}((1+\zeta)\rho_{*};2\lambda,0)\geq
e^{(1+\zeta)\lambda}>\frac{1}{\sqrt{4\pi\lambda}}e^{\lambda}$. Consequently,
for $\lambda$ large enough,

$\inf\Big{\\{}\frac{1}{2\lambda}\|\rho\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}:\mathsf{Z}(\rho;2\lambda,0)\geq\frac{1}{\sqrt{4\pi\lambda}}e^{\lambda}\Big{\\}}\leq\frac{1}{2\lambda}\|(1+\zeta)\rho_{*}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}=\frac{4}{3}(1+\zeta)^{2}.$

In the last equality, we used
$\|\operatorname{sech}({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})^{2}\|_{L^{2}(\mathbb{R})}^{2}=\frac{4}{3}$.
Taking $\zeta\to 0$ concludes ($\mathsf{LimSup}$).

###  2.2. Proof of ($\mathsf{LimInf}$)

Fix $T>0$ and $\rho\in L^{2}([0,T]\times\mathbb{R})$. Let us define the time-
dependent semigroup $P(\rho;s\to t):L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})$,
$0\leq s<t\leq T$. Given a function $f\in L^{2}(\mathbb{R})$, fix $s$ and
consider the PDE

$\partial_{r}\mathsf{Z}^{s,f}(r,x)=\frac{1}{2}\partial_{xx}\mathsf{Z}^{s,f}(r,x)+\rho(r,x)\mathsf{Z}^{s,f}(r,x),\qquad\mathsf{Z}^{s,f}(s,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})=f.$
(2.11)

We define $P(\rho;s\to
t)f:=\mathsf{Z}^{s,f}(t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$.
Note that $\mathsf{Z}^{s,f}$ is the unique solution to the integral equation

$\mathsf{Z}^{s,f}(t,x)=\int_{\mathbb{R}}p(t-s,x-y)f(y)dy+\int_{s}^{t}\int_{\mathbb{R}}p(t-r,x-y)\rho(r,y)\mathsf{Z}^{s,f}(r,y)drdy.$

Via Picard iteration, we have
$\mathsf{Z}^{s,f}(t,x)=\int_{\mathbb{\mathbb{R}}}P((s,x)\to(t,y))f(y)dy$ where
the kernel

$\displaystyle P(\rho;(s,x)\to(t,y))$
$\displaystyle:=p(t-s,x-y)+\sum_{n=1}^{\infty}\int_{s<t_{n}<\dots<t_{1}<t}\int_{\mathbb{R}^{n}}\prod_{i=1}^{n+1}p(t_{i-1}-t_{i},x_{i-1}-x_{i})\rho(t_{i},x_{i})dt_{i}dx_{i}.$
(2.12)

Here, we set $t_{0}=t$ and $x_{0}=x$, $t_{n+1}=s$ and $x_{n+1}=y$.

Next we establish four bounds for the time-dependent semigroup.

######  Lemma 2.3.

There exists a universal constant $C$ such that for all $s<t$ and $a>0$,

$\displaystyle\|P(\rho;(s,0)\to(t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))\|_{L^{2}(\mathbb{R})}\leq
C(t-s)^{-1/4}\exp(a^{2}C(t-s)+\frac{1}{a}\|\rho\|^{2}_{L^{2}([s,t]\times\mathbb{R})}),$
(2.13)
$\displaystyle\|P(\rho;(s,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\to(t,0)\|_{L^{2}(\mathbb{R})}\leq
C(t-s)^{-1/4}\exp(a^{2}C(t-s)+\frac{1}{a}\|\rho\|^{2}_{L^{2}([s,t]\times\mathbb{R})}),$
(2.14) $\displaystyle\|P(\rho;s\to t)\|_{L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R})}\leq
C\exp(a^{2}C(t-s)+\frac{1}{a}\|\rho\|^{2}_{L^{2}([s,t]\times\mathbb{R})}),$
(2.15) $\displaystyle|P(\rho;(s,y)\to(t,x))|\leq
C\exp(a^{2}C(t-s)+\frac{1}{a}\|\rho\|^{2}_{L^{2}([s,t]\times\mathbb{R})})p(t-s,x-y).$
(2.16)

###### Proof.

In the proof, we assume without loss of generality that $s=0$.

Let us first prove (2.13). View the right hand side of (2.12) as a series of
function in $x$, and bound the $L^{2}$ norm of each function. For the zeroth
function in (2.12), we have
$\|p(t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{L^{2}(\mathbb{R})}^{2}=Ct^{-\frac{1}{2}}$.
For the $n$-th functions in (2.12), applying the Cauchy-Schwarz inequality
gives

$\displaystyle\|\text{(n-th)}\|_{L^{2}(\mathbb{R})}^{2}$
$\displaystyle\leq\int_{\mathbb{R}}\int_{0<t_{n}<\dots<t_{1}<t}\int_{\mathbb{R}^{n}}\prod_{i=1}^{n+1}p(t_{i-1}-t_{i},x_{i-1}-x_{i})^{2}dt_{i}dx_{i}dx\frac{\|\rho\|^{2n}_{L^{2}([0,t]\times\mathbb{R})}}{n!}$
$\displaystyle=\frac{C^{n}}{\Gamma(n/2)}t^{\frac{n}{2}-\frac{1}{2}}\frac{\|\rho\|^{2n}_{L^{2}([0,t]\times\mathbb{R})}}{n!}.$

Hence, we have

$\displaystyle\|P(\rho;(0,0)\to(t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))\|_{L^{2}(\mathbb{R})}$
$\displaystyle\leq
Ct^{-\frac{1}{4}}\sum_{n=0}^{\infty}\frac{C^{n}t^{n/4}}{\Gamma(n/2)^{\frac{1}{2}}}\frac{\|\rho\|^{n}_{L^{2}([0,t]\times\mathbb{R})}}{(n!)^{1/2}}$
$\displaystyle\leq
Ct^{-\frac{1}{4}}\Big{(}\sum_{n=0}^{\infty}\frac{a^{n}C^{n}t^{n/2}}{\Gamma(n/2)}\Big{)}^{\frac{1}{2}}\Big{(}\sum_{n=0}^{\infty}\frac{a^{-n}\|\rho\|^{2n}_{L^{2}([0,t]\times\mathbb{R})}}{n!}\Big{)}^{\frac{1}{2}}.$

This concludes (2.13).

The left hand sides of (2.13) and (2.14) are the same upon time reversal, so
(2.14) follows.

The proof of (2.15) is similar in spirit to (2.13). Using $(P(\rho;0\to
t)f)(x)=\int_{\mathbb{R}}P((0,x)\to(t,y))f(y)dy$ and (2.12), we express
$P(\rho;0\to t)f$ as a series of functions. We bound the $L^{2}$ norm of each
function in the series. By the Cauchy-Schwarz inequality, we have

$\Big{(}\int_{\mathbb{R}}p(t,x-y)f(y)dy\Big{)}^{2}\leq\int_{\mathbb{R}}p(t,x-y)f(y)^{2}dy.$
(2.17)

Integrating both sides of (2.17) in $x$ gives that for the zero-th function,
$\|\int_{\mathbb{R}}p(t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}-y)f(y)dy\|_{L^{2}(\mathbb{R})}^{2}\leq\|f\|_{L^{2}(\mathbb{R})}^{2}$.
For the $n$-th function, applying the Cauchy-Schwarz inequality yields

$\displaystyle\|\text{(n-th)}\|_{L^{2}(\mathbb{R})}^{2}\leq\int_{\mathbb{R}}\int_{0<t_{n}<\dots<t_{1}<t}\int_{\mathbb{R}^{n}}\Big{(}\int
p(t_{n},x_{n}-y)f(y)dy\Big{)}^{2}\prod_{i=1}^{n}p(t_{i-1}-t_{i},x_{i-1}-x_{i})^{2}dt_{i}dx_{i}dx\frac{\|\rho\|_{L^{2}([0,t]\times\mathbb{R})}^{2n}}{n!}.$

Using (2.17) and $p(t,x)^{2}\leq\frac{1}{\sqrt{2\pi}t}p(t,x)$, we get

$\|\text{(n-th)}\|_{L^{2}(\mathbb{R})}^{2}\leq\frac{\|\rho\|^{2n}_{L^{2}([0,t]\times\mathbb{R})}\|f\|_{L^{2}(\mathbb{R})}}{n!}\int_{0<t_{n}<\dots<t_{1}<t}\prod_{i=1}^{n}\frac{1}{\sqrt{2\pi(t_{i-1}-t_{i})}}dt_{i}=\frac{C^{n}t^{n/2}}{\Gamma(n/2)}\frac{\|\rho\|^{2n}_{L^{2}([0,t]\times\mathbb{R})}}{n!}\|f\|^{2}_{L^{2}(\mathbb{R})}.$

Taking the square root of both sides and summing over $n$, the rest of the
proof is similar to the proof of (2.13).

To prove (2.16), we view the right hand side of (2.12) as a series of real
numbers. We bound the value of each term in the series. The zeroth term equals
$p(t-s,x-y)$. For the $n$-th term, applying the Cauchy-Schwarz inequality
yields

$\displaystyle|(\text{n-th})|^{2}\leq\int_{0<t_{n}<\dots<s_{1}<t}\int_{\mathbb{R}^{n}}\prod_{i=1}^{n+1}p(t_{i-1}-t_{i},x_{i-1}-x_{i})^{2}dt_{i}dx_{i}\frac{\|\rho\|_{L([s,t]\times\mathbb{R})}^{2n}}{n!}\leq\frac{C^{n}t^{n/2}}{\Gamma(n/2)}\frac{\|\rho\|^{2n}_{L^{2}([0,t]\times\mathbb{R})}}{n!}p(t,x-y)^{2}.$

Taking the square root of both sides and summing over $n$, the rest of the
proof is similar to the proof of (2.13).

∎

######  Lemma 2.4.

For every $\varphi\in L^{2}(\mathbb{R})$, we have

$F(\varphi)\leq\frac{1}{2}(\frac{3}{4})^{2/3}\|\varphi\|_{L^{2}(\mathbb{R})}^{\frac{4}{3}}.$
(2.18)

Moreover, the equality holds if and only if
$\varphi({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})=\alpha^{2}\operatorname{sech}^{2}(\alpha({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}-v))$
for some $\alpha\geq 0$.

###### Proof.

Referring to (2.3), we have

$\displaystyle F(\varphi)$
$\displaystyle\leq\sup\Big{\\{}\|\varphi\|_{L^{2}(\mathbb{R})}\Big{(}\int
g(x)^{4}dx\Big{)}^{\frac{1}{2}}-\frac{1}{2}\int
g^{\prime}(x)^{2}dx\Big{\\}}:g\in
H^{1}(\mathbb{R}),\|g\|_{L^{2}(\mathbb{R})}=1\Big{\\}}$ (2.19)
$\displaystyle\leq\sup\Big{\\{}3^{-\frac{1}{4}}\|\varphi\|_{L^{2}(\mathbb{R})}\|g^{\prime}\|_{L^{2}(\mathbb{R})}^{\frac{1}{2}}-\frac{1}{2}\|g^{\prime}\|_{L^{2}(\mathbb{R})}^{2}:g\in
H^{1}(\mathbb{R}),\|g\|_{L^{2}(\mathbb{R})}=1\Big{\\}}$ (2.20)
$\displaystyle\leq\frac{1}{2}(\frac{3}{4})^{2/3}\|\varphi\|_{L^{2}(\mathbb{R})}^{\frac{4}{3}}.$
(2.21)

The inequality (2.19) is due to the Cauchy-Schwarz inequality. The inequality
(2.20) follows from the one-dimensional $L^{4}$ Gagliardo-Nirenberg-Sobolev
inequality, which states that, for $g\in L^{2}(\mathbb{R})$ and $g^{\prime}\in
L^{2}(\mathbb{R})$,

$\|g\|_{L^{4}(\mathbb{R})}\leq
3^{-1/8}\|g^{\prime}\|_{L^{2}(\mathbb{R})}^{1/4}\|g\|_{L^{2}(\mathbb{R})}^{3/4}.$
(2.22)

The inequality (2.21) follows from the elementary inequality
$3^{-\frac{1}{4}}yx^{\frac{1}{2}}-\frac{1}{2}x^{2}\leq\frac{1}{2}(\frac{3}{4})^{2/3}y^{\frac{4}{3}}$.

To prove the if and only if statement, we need to investigate when the
inequalities (2.19)-(2.21) become equalities. The inequality (2.19) becomes an
equality if and only if $\varphi=\text{const}\cdot g$. By [DELL14, Proposition
3.1], the inequality (2.20), i.e. (2.22) becomes an equality if and only if
$g(x)=a\operatorname{sech}(b(x-v))$ for some $a,v$ and $b>0$; the inequality
(2.21) becomes an equality if and only if
$\|g^{\prime}\|_{L^{2}(\mathbb{R})}=(\frac{1}{2})^{2/3}3^{-1/6}\|\varphi\|_{L^{2}(\mathbb{R})}^{2/3}$.
Combining these with $\|g\|_{L^{2}(\mathbb{R})}=1$ gives that

$g({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})=a\operatorname{sech}\big{(}2a^{2}({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}-v)\big{)},\qquad\varphi({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})=4a^{4}\operatorname{sech}^{2}\big{(}2a^{2}({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}-v)\big{)}.$

Replacing $2a^{2}$ with $\alpha$ implies that
$\varphi({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})=\alpha^{2}\operatorname{sech}^{2}(\alpha({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}-v))$.
∎

######  Remark 2.5.

We only need (2.18) for this section, and the if and only if part is required
in Section 3.

The key for proving ($\mathsf{LimInf}$) is an improved upper bound (compared
with (2.15)) for $\|P(\rho;s\to t)\|_{L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})}$
stated in Proposition 2.9. To prove this upper bound, we need to first
establish the following two lemmas about continuity.

######  Lemma 2.6.

Fix $0\leq s<t$. The map
$\|P({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}};s\to
t)\|_{L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R})}:L^{2}([s,t]\times\mathbb{R})\to\mathbb{R}$ is continuous.

###### Proof.

Fix $f\in L^{2}(\mathbb{R})$. Since $\mathsf{Z}^{s,f}(t,x)=(P(\rho;s\to
t)f)(x)$ solves (2.11), by the Feynman-Kac formula,

$\mathsf{Z}^{s,f}(\rho;t,x)=\mathbb{E}_{x}\bigg{[}\exp\Big{(}\int_{s}^{t}\rho(t-r,B(r))dr\Big{)}f(B(t))\bigg{]}.$

Given $\rho,\widetilde{\rho}\in L^{2}([s,t]\times\mathbb{R})$. We write
$\rho=\rho-(1-\zeta)\widetilde{\rho}+(1-\zeta)\widetilde{\rho}$ in the
exponent above and apply the Hölder inequality to the result. We have for
$\zeta\in(0,1)$,

$\displaystyle\mathsf{Z}^{s,f}(\rho;t,x)$
$\displaystyle=\mathbb{E}_{x}\bigg{[}\exp\Big{(}\int_{s}^{t}(\rho-(1-\zeta)\widetilde{\rho})(t-r,B(r))dr\Big{)}\big{(}f(B(t))\big{)}^{\zeta}$
$\displaystyle\hskip
52.00005pt\cdot\exp\Big{(}\int_{s}^{t}(1-\zeta)\widetilde{\rho}(t-r,B(r))dr\Big{)}\big{(}f(B(t))\big{)}^{1-\zeta}\bigg{]}$
$\displaystyle\leq\mathsf{Z}^{s,f}\big{(}\zeta^{-1}(\rho-(1-\zeta)\widetilde{\rho});t,x\big{)}^{\zeta}\mathsf{Z}^{s,f}(\widetilde{\rho};t,x)^{1-\zeta}.$

Squaring both sides, integrating in $x$ and using the Hölder inequality again
gives

$\big{\|}\mathsf{Z}^{s,f}(\rho;t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\big{\|}_{L^{2}(\mathbb{R})}\leq\big{\|}\mathsf{Z}^{s,f}\big{(}\zeta^{-1}(\rho-(1-\zeta)\widetilde{\rho});t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}\big{)}\big{\|}_{L^{2}(\mathbb{R})}^{\zeta}\big{\|}\mathsf{Z}^{s,f}(\widetilde{\rho};t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\big{\|}_{L^{2}(\mathbb{R})}^{1-\zeta}.$

Taking the supremum over $\\{f:\|f\|_{L^{2}(\mathbb{R})}\leq 1\\}$ yields that

$\big{\|}P(\rho;s\to t)\big{\|}_{L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R})}\leq\big{\|}P\big{(}\zeta^{-1}(\rho-(1-\zeta)\widetilde{\rho});s\to
t\big{)}\big{\|}_{L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R})}^{\zeta}\big{\|}P(\widetilde{\rho};s\to
t)\big{\|}_{L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})}^{1-\zeta}.$ (2.23)

Set $G(\rho):=\log\|P(\rho;s\to t)\|_{L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R})}$. Our goal is to show that $G$ is continuous in $\rho$. We
do this by showing that $G$ is both upper and lower semi-continuous. Taking
the logarithm of both sides of (2.23) and applying (2.15) yields

$\displaystyle G(\rho)$ $\displaystyle\leq\zeta
G\big{(}\zeta^{-1}(\rho-(1-\zeta)\widetilde{\rho})\big{)}+(1-\zeta)G(\widetilde{\rho})$
$\displaystyle\leq
C(s,t)\zeta+C\zeta\|\zeta^{-1}(\rho-(1-\zeta)\widetilde{\rho})\|_{L^{2}([s,t]\times\mathbb{R})}^{2}+(1-\zeta)G(\widetilde{\rho}).$
(2.24)

Subtracting $(1-\zeta)G(\widetilde{\rho})+\zeta G(\rho)$ from both sides and
applying the inequality $\|f+g\|_{L^{2}}^{2}\leq
2(\|f\|^{2}_{L^{2}}+\|g\|_{L^{2}}^{2})$ yield

$\displaystyle(1-\zeta)(G(\rho)-G(\widetilde{\rho}))$ $\displaystyle\leq
C(s,t)\zeta+C\zeta^{-1}(1-\zeta)^{2}\|\rho-\widetilde{\rho}\|_{L^{2}([s,t]\times\mathbb{R})}^{2}+C\zeta\|\rho\|_{L^{2}([s,t]\times\mathbb{R})}^{2}-\zeta
G(\rho)$ $\displaystyle\leq
C\zeta^{-1}(1-\zeta)^{2}\|\rho-\widetilde{\rho}\|_{L^{2}([s,t]\times\mathbb{R})}^{2}+\zeta
C(s,t,\rho).$

Taking $\zeta=\|\rho-\widetilde{\rho}\|_{L^{2}([s,t]\times\mathbb{R})}$ and
letting $\widetilde{\rho}\to\rho$, we see that
$\limsup_{\widetilde{\rho}\to\rho}G(\widetilde{\rho})\leq G(\rho)$. Thus, $G$
is upper semi-continuous.

On the other hand, if we swap $\rho$ and $\widetilde{\rho}$ in (2.24), a
similar argument gives

$G(\widetilde{\rho})-G(\rho)\leq\zeta
C(s,t,\rho)+C\zeta^{-1}\|\widetilde{\rho}-\rho\|_{L^{2}([s,t]\times\mathbb{R})}^{2}.$

Taking $\zeta=\|\rho-\widetilde{\rho}\|_{L^{2}([s,t]\times\mathbb{R})}$ and
letting $\widetilde{\rho}\to\rho$, we see that $G$ is lower semi-continuous.
This concludes the lemma. ∎

######  Lemma 2.7.

Fix $0\leq s<t$ and $M\geq 0$. There exists a constant $C(M)>0$. For
$\varphi_{1},\varphi_{2}\in L^{2}(\mathbb{R})$ such that
$0\leq\varphi_{1},\varphi_{2}\leq M$ almost everywhere, we have

$|F(\varphi_{1})-F(\varphi_{2})|\leq
C(M)\|\varphi_{1}-\varphi_{2}\|_{L^{2}(\mathbb{R})}.$

###### Proof.

By the definition of $F(\varphi_{2})$ in (2.3), for any $\zeta>0$, there
exists $g_{\zeta}$ that $\|g_{\zeta}\|_{L^{2}(\mathbb{R})}^{2}=1$ and

$F(\varphi_{2})\leq\int_{\mathbb{R}}\varphi_{2}g_{\zeta}^{2}-\frac{1}{2}(g^{\prime}_{\zeta})^{2}+\zeta.$
(2.25)

Moreover, we have
$F(\varphi_{1})\geq\int_{\mathbb{R}}\varphi_{1}g_{\zeta}^{2}-\frac{1}{2}(g^{\prime}_{\zeta})^{2}$.
Using this and (2.25), we have

$\displaystyle
F(\varphi_{2})-F(\varphi_{1})\leq\int_{\mathbb{R}}(\varphi_{2}-\varphi_{1})g_{\zeta}^{2}+\zeta.$

Applying the Cauchy-Schwarz inequality to the right hand side above and then
applying (2.22) to the result, we have

$F(\varphi_{2})-F(\varphi_{1})\leq
3^{-\frac{1}{4}}\|\varphi_{2}-\varphi_{1}\|_{L^{2}(\mathbb{R})}\|g_{\zeta}^{\prime}\|_{L^{2}(\mathbb{R})}^{\frac{1}{2}}+\zeta.$
(2.26)

By (2.25),
$\frac{1}{2}\|g^{\prime}_{\zeta}\|_{L^{2}(\mathbb{R})}^{2}\leq\int_{\mathbb{R}}\varphi_{2}g_{\zeta}^{2}-F(\varphi_{2})+\zeta.$
Since $0\leq\varphi_{2}\leq M$ and $\|g_{\zeta}\|_{L^{2}(\mathbb{R})}=1$, we
can upper bound the first term on the right hand side by $M$ and the second
term by a universal constant $C$. Hence, we have
$\frac{1}{2}\|g^{\prime}_{\zeta}\|_{L^{2}(\mathbb{R})}^{2}\leq C(M).$
Inserting this to the right hand side of (2.26) and letting $\zeta\to 0$
yields

$F(\varphi_{2})-F(\varphi_{1})\leq
C(M)\|\varphi_{2}-\varphi_{1}\|_{L^{2}(\mathbb{R})}.$

Swapping $\varphi_{1}$ and $\varphi_{2}$ concludes the lemma. ∎

######  Corollary 2.8.

Fix $0\leq s<t$ and $M\geq 0$. There exists a constant $C(M,t,s)$. For
$\rho_{1},\rho_{2}\in L^{2}([s,t]\times\mathbb{R})$ such that
$0\leq\rho_{1},\rho_{2}\leq M$ almost everywhere, we have

$\Big{|}\int_{s}^{t}F(\rho_{1}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr-\int_{s}^{t}F(\rho_{2}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr\Big{|}\leq
C(M,t,s)\|\rho_{1}-\rho_{2}\|_{L^{2}([s,t]\times\mathbb{R})}.$

###### Proof.

It is straightforward to see that

$\displaystyle\Big{|}\int_{s}^{t}F(\rho_{1}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr-\int_{s}^{t}F(\rho_{2}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr\Big{|}\leq\int_{s}^{t}\big{|}F(\rho_{1}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))-F(\rho_{2}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))\big{|}dr.$
(2.27)

Applying Lemma 2.7 to upper bound the right hand side above and applying the
Cauchy Schwarz inequality to the result, we have

$\text{RHS of \eqref{eq:Fcont3}}\leq
C(M)\int_{s}^{t}\|\rho_{1}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})-\rho_{2}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{L^{2}(\mathbb{R})}dr\leq
C(M)(t-s)^{\frac{1}{2}}\|\rho_{1}-\rho_{2}\|_{L^{2}([s,t]\times\mathbb{R})}.$

This concludes the corollary. ∎

######  Proposition 2.9.

Fix $0\leq s<t$. For all $\rho\in L^{2}([s,t]\times\mathbb{R})$ satisfying
$\rho\geq 0$ almost everywhere, we have

$\|P(\rho;s\to t)\|_{L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R})}\leq\exp\Big{(}\int_{s}^{t}F(\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr\Big{)}.$
(2.28)

###### Proof.

Fix smooth and compactly supported $\rho$ and $f$. Since $\rho$ is smooth, the
function $\mathsf{Z}^{s,f}(t,x)=(P(s\to t)f)(x)$ solves (2.11) point-wise.
Multiply both side by $\mathsf{Z}$, integrate the result over $x\in\mathbb{R}$
and interchange the integration and differentiation. We have

$\frac{1}{2}\partial_{r}\|\mathsf{Z}^{s,f}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{L^{2}(\mathbb{R})}^{2}\leq
F(\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))\|\mathsf{Z}^{s,f}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{L^{2}(\mathbb{R})}^{2}.$

Integrating in time from $s$ to $t$ gives

$\|\mathsf{Z}^{s,f}(t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{L^{2}(\mathbb{R})}^{2}\leq\|f\|_{L^{2}(\mathbb{R})}^{2}\exp\Big{(}2\int_{s}^{t}F(\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr\Big{)}.$

For fixed smooth and compact supported $\rho$, by approximation, the above
inequality also holds for general $f\in L^{2}(\mathbb{R})$. Thus, we have
proven (2.28) for smooth and compactly supported $\rho$.

The general result follows by approximation and monotonicity. Fix arbitrary
$M\geq 0$. We first prove that (2.28) holds for all $\rho\in
L^{2}([s,t]\times\mathbb{R})$ such that almost everywhere $0\leq\rho\leq M$.
By a density argument, there exists smooth and compactly supported functions
$\\{\rho_{n}\\}_{n=1}^{\infty}$ such that $0\leq\rho_{n}\leq M$ and
$\|\rho_{n}-\rho\|_{L^{2}([s,t]\times\mathbb{R})}\to 0$ as $n\to\infty$.
(2.28) holds when $\rho$ is replaced by $\rho_{n}$. Let $n\to\infty$, by Lemma
2.6 and Corollary 2.8, (2.28) also holds for $\rho$.

Finally, we prove (2.28) for $\rho\in L^{2}([s,t]\times\mathbb{R})$ such that
almost everywhere $\rho\geq 0$. As $n\to\infty$, $\rho\mathbf{1}_{\\{\rho\leq
n\\}}\to\rho$ in $L^{2}([s,t]\times\mathbb{R})$. In the preceding paragraph,
we have proved that (2.28) holds when $\rho$ is replaced by
$\rho\mathbf{1}_{\\{\rho\leq n\\}}$, for all $n$. Using this and the
monotonicity of $F$, we have

$\|P(\rho\mathbf{1}_{\\{\rho\leq n\\}};s\to t)\|_{L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R})}\leq\exp\Big{(}\int_{s}^{t}F(\rho\mathbf{1}_{\\{\rho\leq
n\\}}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr\Big{)}\leq\exp\Big{(}\int_{s}^{t}F(\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr\Big{)}.$

Letting $n\to\infty$, by Lemma 2.6, the left hand side above converges to
$\|P(\rho;s\to t)\|_{L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})}$. Hence, we
conclude (2.28). ∎

Based on the preceding results, we now proceed to prove ($\mathsf{LimInf}$).
The proof amounts to showing that, for any $\rho\in
L^{2}([0,2\lambda]\times\mathbb{R})$ such that
$\mathsf{Z}(\rho;2\lambda,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}$, we have
$\frac{1}{2\lambda}\|\rho\|^{2}_{L^{2}([0,2\lambda]\times\mathbb{R})}\geq\frac{4}{3}-o_{\lambda}(1)$
where $\lim_{\lambda\to\infty}o_{\lambda}(1)=0$.

With loss of generality, we can assume that

$\frac{1}{2\lambda}\|\rho\|^{2}_{L^{2}([0,2\lambda])\times\mathbb{R}}\leq 2.$
(2.29)

Moreover, define $\rho_{+}:=\min(\rho,0)$. Because
$\mathsf{Z}(\rho_{+};2\lambda,0)\geq\mathsf{Z}(\rho;2\lambda,0)$ and
$\|\rho_{+}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}\leq\|\rho\|_{L^{2}([0,2\lambda]\times\mathbb{R})}$,
we can also assume that $\rho\geq 0$ almost everywhere. Note that
$\mathsf{Z}(\rho;2\lambda,0)=P(\rho;(0,0)\to(2\lambda,0))$. Using the
semigroup property, we have

$\displaystyle\mathsf{Z}(\rho;2\lambda,0)$
$\displaystyle=P(\rho;(0,0)\to(1,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))P(\rho;1\to(2\lambda-1))P(\rho;(2\lambda-1,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\to(2\lambda,0))$
$\displaystyle\leq\|P(\rho;(0,0)\to(1,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))\|_{L^{2}(\mathbb{R})}\|P(\rho;1\to(2\lambda-1))\|_{L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R})}\|P(\rho;(2\lambda-1,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\to(2\lambda,0))\|_{L^{2}(\mathbb{R})}.$

Applying (2.13)-(2.14) (take $a=\lambda^{\frac{1}{3}}$) to upper bound the
first and third term on the right hand side, using Proposition 2.9 and (2.29)
to upper bound the second term, we have

$\displaystyle\mathsf{Z}(\rho;2\lambda,0)$ $\displaystyle\leq
C\|P(\rho;1\to(2\lambda-1))\|_{L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R})}\exp\big{(}C\lambda^{\frac{2}{3}}+\lambda^{-\frac{1}{3}}\|\rho\|^{2}_{L^{2}([0,1]\times\mathbb{R})}+\lambda^{-\frac{1}{3}}\|\rho\|^{2}_{L^{2}([2\lambda-1,2\lambda]\times\mathbb{R})}\big{)}$
$\displaystyle\leq
C\exp(C\lambda^{\frac{2}{3}})\exp\Big{(}\int_{0}^{2\lambda}F(\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr\Big{)}.$
(2.30)

Here $C$ is a universal constant. For the integral above, using Lemma 2.4 to
bound $F$, together with Young’s inequality,
$\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{L^{2}(\mathbb{R})}^{\frac{4}{3}}\leq\frac{6^{1/3}}{3}(\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{L^{2}(\mathbb{R})}^{2}+\frac{2}{3})$
for every $r$, we have

$\displaystyle\mathsf{Z}(\rho;2\lambda,0)$ $\displaystyle\leq
C\exp(C\lambda^{\frac{2}{3}})\exp\Big{(}\int_{0}^{2\lambda}\frac{1}{2}\Big{(}\frac{3}{4}\Big{)}^{\frac{2}{3}}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{L^{2}(\mathbb{R})}^{\frac{4}{3}}dr\Big{)}$
$\displaystyle\leq
C\exp(C\lambda^{\frac{2}{3}})\exp\bigg{(}\int_{0}^{2\lambda}\frac{1}{4}\Big{(}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{L^{2}(\mathbb{R})}^{2}+\frac{2}{3}\Big{)}dr\bigg{)}.$

Applying
$\frac{1}{\sqrt{4\pi\lambda}}e^{\lambda}\leq\mathsf{Z}(\rho;2\lambda,0)$ to
lower bound $\mathsf{Z}(\rho;2\lambda,0)$, we have that that
$\lambda(1-o_{\lambda}(1))\leq\frac{1}{4}\|\rho\|^{2}_{L^{2}([0,2\lambda]\times\mathbb{R})}+\frac{\lambda}{3}$,
where $\lim_{\lambda\to}o_{\lambda}(1)=0$. Hence
$\frac{1}{2\lambda}\|\rho\|^{2}_{L^{2}([0,2\lambda]\times\mathbb{R})}\geq\frac{4}{3}-o_{\lambda}(1)$.
Letting $\lambda\to\infty$ concludes the desired ($\mathsf{LimInf}$).

##  3\. Minimizers of the variational formula

For a $\rho\in L^{2}([0,2\lambda]\times\mathbb{R})$, recall that
$\mathsf{Z}(\rho)=\mathsf{Z}(\rho;t,x)$ is the (unique) solution to the PDE

$\partial_{t}\mathsf{Z}=\frac{1}{2}\partial_{xx}\mathsf{Z}+\rho\mathsf{Z},\qquad(t,x)\in(0,2\lambda]\times\mathbb{R},\qquad\mathsf{Z}(\rho;0,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})=\delta_{0}({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}).$

Let $\mathcal{K}_{\lambda}$ denote the set of minimizers of

$\inf\Big{\\{}\frac{1}{2}\|\rho\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}:\rho\in
L^{2}([0,2\lambda]\times\mathbb{R}),\,\mathsf{Z}(\rho;2\lambda,0)\geq\frac{1}{\sqrt{4\pi\lambda}}e^{\lambda}\Big{\\}}.$
(3.1)

It is convenient to consider the scaled version of $\mathcal{K}_{\lambda}$:
Let $\widetilde{\mathcal{K}}_{\lambda}$ be the set of minimizers of

$\inf\Big{\\{}\frac{1}{2}\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}^{2}:\rho\in
L^{2}([0,2]\times\mathbb{R}),\,\mathsf{Z}(\rho;2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}\Big{\\}}.$
(3.2)

Recall the scaling property
$\mathsf{Z}(\lambda\rho(\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda^{\frac{1}{2}}{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}});2,0)=\lambda^{\frac{1}{2}}\mathsf{Z}(\rho;2\lambda,0)$.
We have

$\widetilde{\mathcal{K}}_{\lambda}=\\{\lambda\rho(\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda^{\frac{1}{2}}{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}):\rho\in\mathcal{K}_{\lambda}\\}.$
(3.3)

To prepare for the proof of Theorem 1.2, in this section, we develop a few
properties of $\mathcal{K}_{\lambda}$ for fixed $\lambda$. We also study the
asymptotic behavior of $\mathcal{K}_{\lambda}$ as $\lambda\to\infty$.

###  3.1. Properties of $\mathcal{K}_{\lambda}$

####  3.1.1. $\mathcal{K}_{\lambda}$ is not empty.

We first show that $\mathcal{K}_{\lambda}$ is not empty. By the scaling (3.3),
it suffices to show that $\widetilde{\mathcal{K}}_{\lambda}$ is not empty,
i.e. a minimizer of (3.2) exists.

Let us first introduce a few facts that would be useful throughout Section 3.
We take a Banach space $\mathcal{B}$ with Gaussian measure $\mu$ such that
$L^{2}([0,2]\times\mathbb{R})$ is the Cameron-Martin space of
$(\mathcal{B},\mu)$. We further assume the embedding
$L^{2}([0,2]\times\mathbb{R})\subseteq\mathcal{B}$ is dense. For a concrete
choice of $(\mathcal{B},\mu)$, see [LT21, Section 2.1.1]. The advantage of
introducing $\mathcal{B}$ is that it brings us _compactness_. Since
$L^{2}([0,2]\times\mathbb{R})$ is the Cameron-Martin space of $\mathcal{B}$,
for arbitrary $r\geq 0$, the set
$\\{\rho:\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq r\\}$ is compact in
$\mathcal{B}$. The compactness would be crucial for us to prove the non-
emptiness of $\mathcal{K}_{\lambda}$.

We are going to view $\mathsf{Z}$ as a map from $\rho\in
L^{2}([0,2]\times\mathbb{R})$ to the space-time function
$\mathsf{Z}(\rho;{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$.
If we restrict the time and space coordinates in
$(t,x)\in[\delta,2]\times[-\delta^{-1},\delta^{-1}]$, it follows from [LT21,
Section 2.1] that $\mathsf{Z}$ maps $L^{2}([0,2]\times\mathbb{R})$ to
$C([\delta,2]\times[-\delta^{-1},\delta^{-1}])$.

We use $L^{{\color[rgb]{0,0,0}2}}_{\mathcal{B}}([0,2]\times\mathbb{R})$ to
denote the same space $L^{2}([0,2]\times\mathbb{R})$ with the topology induced
by the topology of the Banach space $\mathcal{B}$. Even though
$\mathsf{Z}:L^{2}_{\mathcal{B}}([0,2]\times\mathbb{R})\to
C([\delta,2]\times[-\delta^{-1},\delta^{-1}])$ is not continuous in general,
its restriction onto $\\{\rho:\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq
r\\}$ is continuous for any $r<\infty$, as shown in the following lemma.

######  Lemma 3.1.

Fix arbitrary $\delta>0$ and $r<\infty$. The map
$\mathsf{Z}:\rho\mapsto\mathsf{Z}(\rho;{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$
is continuous from
$L_{\mathcal{B}}^{2}([0,2]\times\mathbb{R}){\color[rgb]{0,0,0}\cap\\{\rho:\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq
r\\}}$ to $C([\delta,2]\times[-\delta^{-1},\delta^{-1}])$.

###### Proof.

Define the extension of $\mathsf{Z}$

$\mathsf{Z}^{\prime}:\mathcal{B}\to
C([\delta,2]\times[\delta^{-1},\delta^{-1}]),\qquad\mathsf{Z}^{\prime}(\rho):=\begin{cases}\mathsf{Z}(\rho),&\text{
when }\rho\in L^{2}([0,2]\times\mathbb{R}),\\\ 0,&\text{
otherwise.}\end{cases}$ (3.4)

By the proof of [LT21, Lemma 3.7] (see the paragraph above Eq (3.18’)
therein), there exists a sequence of continuous functions
$\varphi_{N}:\mathcal{B}\to C([\delta,2]\times[-\delta^{-1},\delta^{-1}])$
such that for all $r<\infty$,

$\lim_{N\to\infty}\sup_{\frac{1}{2}\|\rho\|^{2}_{L^{2}([0,2]\times\mathbb{R})}\leq
r}\|\mathsf{Z}^{\prime}(\rho)-\varphi_{N}(\rho)\|_{L^{\infty}([\delta,2]\times[-\delta^{-1},\delta^{-1}])}=0.$

Using this, we conclude that the map
$\mathsf{Z}^{\prime}:\mathcal{B}{\color[rgb]{0,0,0}\cap\\{\rho:\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq
r\\}}\to C([\delta,2]\times[-\delta^{-1},\delta^{-1}])$ is continuous. By
(3.4), we have the desired continuity of
$\mathsf{Z}:L_{\mathcal{B}}^{2}([0,2]\times\mathbb{R}){\color[rgb]{0,0,0}\cap\\{\rho:\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq
r\\}}\to C([\delta,2]\times[-\delta^{-1},\delta^{-1}])$. ∎

Denote the infimum in (3.2) by $q_{\lambda}$.

######  Proposition 3.2.

Fix $\lambda>0$, the set $\mathcal{K}_{\lambda}$ is not empty.

###### Proof.

By the scaling (3.3), it suffices to prove that
$\widetilde{\mathcal{K}}_{\lambda}$ is not empty, i.e. the minimizer of (3.2)
exists. Let $\\{\rho_{n}\\}_{n\in\mathbb{Z}_{\geq 1}}\subseteq
L^{2}([0,2]\times\mathbb{R})$ be such that
$\mathsf{Z}(\rho_{n};2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}$ and
$\frac{1}{2}\|\rho_{n}\|_{L^{2}([0,2]\times\mathbb{R})}^{2}\downarrow
q_{\lambda}$. Since for arbitrary fixed $r{\color[rgb]{0,0,0}<\infty}$, the
set $\\{\rho:\frac{1}{2}\|\rho\|^{2}_{L^{2}([0,2]\times\mathbb{R})}\leq r\\}$
is compact in $\mathcal{B}$, after passing to a subsequence, $\rho_{n}$
converges to some $\rho\in L^{2}([0,2]\times\mathbb{R})$ in the topology of
$\mathcal{B}$. By Lemma 3.1, $\rho\mapsto\mathsf{Z}(\rho;2,0)$ is a continuous
map from
$L_{\mathcal{B}}([0,2]\times\mathbb{R}){\color[rgb]{0,0,0}\cap\\{\rho:\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq
r\\}}$ to $\mathbb{R}$. This implies that
$\mathsf{Z}(\rho;2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}$. By the compactness
of $\\{\rho:\frac{1}{2}\|\rho\|^{2}_{L^{2}([0,2]\times\mathbb{R})}\leq r\\}$,
we have
$\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq\liminf_{n\to\infty}\|\rho_{n}\|_{L^{2}([0,2]\times\mathbb{R})}$.
Hence, $\rho$ is a minimizer of (3.2). ∎

####  3.1.2. Symmetric decreasing function in space

In this subsection, we show that every $\rho\in\mathcal{K}_{\lambda}$ is
symmetric decreasing in space.

We say that $f$ is _symmetric decreasing_ if there exists a non-increasing
function $g:[0,\infty)\to[0,\infty)$ such that $f(x)=g(|x|)$ holds for almost
every $x\in\mathbb{R}$. We say that $f$ is _strictly symmetric decreasing_ if
$g$ is strictly decreasing. We say that a measurable function
$\rho:[0,T]\times\mathbb{R}\to[0,\infty)$ is _symmetric decreasing in space_
if for almost every $0\leq s\leq T$,
$\rho(s,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$ is symmetric
decreasing.

######  Proposition 3.3.

Fix $\lambda>0$, every element $\rho\in\mathcal{K}_{\lambda}$ is symmetric
decreasing in space.

Let $A\subseteq\mathbb{R}$ be a measurable set with finite Lebesgue measure
$|A|$. We define the _symmetric rearrangement_ of $A$, denoted $A^{*}$, as the
interval $[-|A|/2,|A|/2]$. Given a measurable function
$f:\mathbb{R}\to[0,\infty)$, we can express $f$ by the layer-cake
representation
$f(x)=\int_{0}^{\infty}\mathbf{1}_{\\{y\in\mathbb{R}:f(y)>\ell\\}}(x)d\ell$.
We define its _symmetric decreasing rearrangement_ as

$f^{*}(x):=\int_{0}^{\infty}\mathbf{1}_{\\{y\in\mathbb{R}:f(y)>\ell\\}^{*}}(x)d\ell.$

It is straightforward to show that for arbitrary fixed $\ell\geq 0$, the sets
$\\{x:f(x)\geq\ell\\}$ and $\\{x:f^{*}(x)\geq\ell\\}$ have the same Lebesgue
measure. As a consequence,
$\|f\|_{L^{p}(\mathbb{R})}=\|f^{*}\|_{L^{p}(\mathbb{R})}$ for every $p\geq 1$.

Let $\rho:[0,T]\times\mathbb{R}\to[0,\infty)$. We define the _Steiner
symmetrization_ of $\rho$ along the time-axis, denoted $\rho^{\mathsf{s}}$, as
follows: For every fixed $0\leq s\leq T$, we define the function
$\rho^{\mathsf{s}}(s,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}):=\rho^{*}(s,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$.

To prove Proposition 3.3, we rely on the following two rearrangement
inequalities known respectively as the Hardy-Littlewood inequality and the
Brascamp-Lieb-Luttinger inequality.

######  Lemma 3.4 ([LL01, Theorem 3.4]).

For any non-negative $f,g\in L^{2}(\mathbb{R})$, one has

$\int_{\mathbb{R}}f(x)g(x)dx\leq\int_{\mathbb{R}}f^{*}(x)g^{*}(x)dx.$

Moreover, if $f=f^{*}$ is strictly symmetric decreasing, then the inequality
above becomes an equality if and only if $g=g^{*}$ almost everywhere.

######  Lemma 3.5 ([BLL74, Theorem 1.2]).

Let $f_{j},1\leq j\leq k$, be non-negative measurable functions on
$\mathbb{R}$, and let $a_{jm}$, $1\leq j\leq k$, $1\leq m\leq n$ be real
numbers, then

$\int_{\mathbb{R}^{n}}\prod_{i=1}^{k}f_{j}(\sum_{m=1}^{n}a_{jm}x_{m})dx_{1}\dots
dx_{n}\leq\int_{\mathbb{R}^{n}}\prod_{i=1}^{k}f_{j}^{*}(\sum_{m=1}^{n}a_{jm}x_{m})dx_{1}\dots
dx_{n}.$

The following result contains the major step for proving Proposition 3.3.

######  Lemma 3.6.

If $\rho\in L^{2}([0,2]\times\mathbb{R})$ is non-negative, then
$\mathsf{Z}(\rho;2,0)\leq\mathsf{Z}(\rho^{\mathsf{s}};2,0)$. In addition, the
equality holds if and only if $\rho$ is symmetric decreasing in space.

###### Proof.

Note that

$\mathsf{Z}(\rho;2,0)=p(2,0)+\sum_{n=1}^{\infty}\int_{0<t_{n}<\dots<t_{1}<t_{0}}\int_{\mathbb{R}^{n}}\prod_{i=1}^{n+1}p(t_{i-1}-t_{i},x_{i-1}-x_{i})\prod_{i=1}^{n}\rho(t_{i},x_{i})dt_{i}dx_{i},$
(3.5)

where $t_{0}=2$ and $x_{0}=t_{n+1}=x_{n+1}=0$. For the $n=1$ term in the sum
above, by Lemma 3.4,

$\int_{0}^{2}\int_{\mathbb{R}}p(2-s,x)p(s,x)\rho(s,x)dsdx\leq\int_{0}^{2}\int_{\mathbb{R}}p(2-s,x)p(s,x)\rho^{\mathsf{s}}(s,x)dsdx.$
(3.6)

Since
$p(2-s,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})p(s,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$
is strictly symmetric decreasing for every fixed $s\in(0,2)$, the above
becomes an equality if and only if $\rho$ is symmetric decreasing in space.

For the $n\geq 2$ term. View the integrand as product of functions in space
for fixed $t_{n}<t_{n-1}<\dots<t_{1}$ and note that
$p(t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})=p^{\mathsf{s}}(t,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$.
By Lemma 3.5,

$\displaystyle\int_{0<t_{n}<\dots<t_{1}<t_{0}}\int_{\mathbb{R}^{n}}\prod_{i=1}^{n+1}p(t_{i-1}-t_{i},x_{i-1}-x_{i})\prod_{i=1}^{n}\rho(t_{i},x_{i})dt_{i}dx_{i}$
$\displaystyle\leq\int_{0<t_{n}<\dots<t_{1}<t_{0}}\int_{\mathbb{R}^{n}}\prod_{i=1}^{n+1}p(t_{i-1}-t_{i},x_{i-1}-x_{i})\prod_{i=1}^{n}\rho^{\mathsf{s}}(t_{i},x_{i})dt_{i}dx_{i}.$
(3.7)

Combining (3.5)-(3.7), we conclude that
$\mathsf{Z}(\rho;2,0)\leq\mathsf{Z}(\rho^{\mathsf{s}};2,0)$. Furthermore, the
equality holds if and only if $\rho$ is symmetric decreasing in space. ∎

###### Proof of Proposition 3.3.

By the scaling (3.3), it suffices to prove Proposition 3.3 for
$\widetilde{\mathcal{K}}_{\lambda}$. First, we claim that if
$\rho\in\widetilde{\mathcal{K}}_{\lambda}$, then $\rho\geq 0$ almost
everywhere. Define $\rho_{+}=\max(\rho,0)$. The claim follows from
$\mathsf{Z}(\rho;2\lambda,0)\leq\mathsf{Z}(\rho_{+};2\lambda,0)$ and
$\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}>\|\rho_{+}\|_{L^{2}([0,2]\times\mathbb{R})}$,
if $\rho$ is not almost everywhere non-negative. Next, suppose that there
exists $\rho\in\widetilde{\mathcal{K}}_{\lambda}$ that is not symmetric
decreasing in space. By Lemma 3.6, we have
$\mathsf{Z}(\rho^{\mathsf{s}};2,0)>\mathsf{Z}(\rho;2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}$.
By the continuity of $\rho\mapsto\mathsf{Z}(\rho;2,0)$ in
$L^{2}([0,2]\times\mathbb{R})$, there exists some $0<\theta<1$ such that
$\mathsf{Z}(\theta\rho^{\mathsf{s}};2,0)=\frac{1}{\sqrt{4\pi}}e^{\lambda}$.
However,

$\|\theta\rho^{\mathsf{s}}\|_{L^{2}([0,2]\times\mathbb{R})}<\|\rho^{\mathsf{s}}\|_{L^{2}([0,2]\times\mathbb{R})}=\|\rho\|_{L^{2}([0,2]\times\mathbb{R})},$

which contradicts with $\rho\in\widetilde{\mathcal{K}}_{\lambda}$. ∎

###  3.2. The $\varepsilon\to 0$ limit of $h_{\varepsilon,\lambda}$ for fixed
$\lambda$

For $f\in C([\delta,2]\times[-\delta^{-1},\delta^{-1}])$ and
$\mathcal{A}\subseteq C([\delta,2]\times[-\delta^{-1},\delta^{-1}])$, define

$\mathrm{dist}_{\delta}(f,\mathcal{A}):=\inf\\{\|f-g\|_{L^{\infty}([\delta,2]\times[-\delta^{-1},\delta^{-1}])}:g\in\mathcal{A}\\}.$

Recall that we denote the infimum in (3.2) by $q_{\lambda}$.

######  Lemma 3.7.

Consider $\\{\rho_{n}\\}_{n\in\mathbb{Z}_{\geq 1}}\subseteq
L^{2}([0,2]\times\mathbb{R})$ such that
$\mathsf{Z}(\rho_{n};2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}$. If
$\lim_{n\to\infty}\frac{1}{2}\|\rho_{n}\|_{L([0,2]\times\mathbb{R})}^{2}=q_{\lambda}$,
then we have

$\lim_{n\to\infty}\mathrm{dist}_{\delta}(\mathsf{Z}(\rho_{n}),\mathsf{Z}(\widetilde{\mathcal{K}}_{\lambda}))=0.$
(3.8)

###### Proof.

The proof is similar to the proof of Proposition 3.2. Fix arbitrary $r\geq 0$.
The set $\\{\rho:\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq r\\}$ is compact
in $\mathcal{B}$. Using this, after passing to a subsequence, $\rho_{n}$
converges to some $\rho\in L^{2}([0,2]\times\mathbb{R})$ in the topology
$\mathcal{B}$. In addition,
$\mathsf{Z}(\rho;2,0)=\lim_{n\to\infty}\mathsf{Z}(\rho_{n};2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}$
and
$\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq\liminf_{n\to\infty}\|\rho_{n}\|_{L^{2}([0,2]\times\mathbb{R})}$.
Hence $\rho\in\widetilde{\mathcal{K}}_{\lambda}$. This implies that

$\lim_{n\to\infty}\inf\\{\|\rho_{n}-\rho\|_{\mathcal{B}}:\rho\in\widetilde{\mathcal{K}}_{\lambda}\\}=0.$
(3.9)

Further, by Lemma 3.1, the map
$\mathsf{Z}:L_{\mathcal{B}}^{2}([0,2]\times\mathbb{R}){\color[rgb]{0,0,0}\cap\\{\rho:\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq
r\\}}\to C([\delta,2]\times[-\delta^{-1},\delta^{-1}])$ is continuous. Since
$\\{\rho:\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq r\\}$ is compact in
$\mathcal{B}$, the map

$\mathsf{Z}:\\{\rho:\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq r\\}\to
C([\delta,2]\times[-\delta^{-1},\delta^{-1}])$

is uniformly continuous, where we endow
$\\{\rho:\|\rho\|_{L^{2}([0,2]\times\mathbb{R})}\leq r\\}$ with the topology
induced by $\mathcal{B}$. Using the uniform continuity of $\mathsf{Z}$ and the
convergence in (3.9), we conclude (3.8). ∎

Recall that $h_{\varepsilon,\lambda}=\lambda^{-1}\log
Z_{\varepsilon}(t,\lambda^{\frac{1}{2}}x)$. Define
$\mathsf{h}_{\lambda}(\rho;t,x):=\lambda^{-1}\log\big{(}\lambda^{\frac{1}{2}}\mathsf{Z}(\rho;\lambda
t,\lambda x)\big{)}$.

######  Proposition 3.8.

Fix $\lambda,\delta>0$. We have

$\lim_{\varepsilon\to
0}\mathbb{P}\big{[}\mathrm{dist}_{\delta}(h_{\varepsilon,\lambda},\mathsf{h}_{\lambda}(\mathcal{K}_{\lambda}))<\delta\,\big{|}\,Z_{\varepsilon}(2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}\big{]}=1.$

###### Proof.

Set
$\widetilde{\mathsf{h}}_{\lambda}(\rho;t,x):=\lambda^{-1}\log\mathsf{Z}(\rho;t,\lambda^{\frac{1}{2}}x)$.
By the scaling (3.3) and
$\mathsf{Z}(\lambda\rho(\lambda{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda^{\frac{1}{2}}{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}});t,x)=\lambda^{\frac{1}{2}}\mathsf{Z}(\rho;\lambda
t;\lambda^{\frac{1}{2}}x)$, we have
$\mathsf{h}_{\lambda}(\mathcal{K}_{\lambda})=\widetilde{\mathsf{h}}_{\lambda}(\widetilde{\mathcal{K}}_{\lambda})$.
Hence, we need to prove

$\lim_{\varepsilon\to
0}\mathbb{P}\big{[}\mathrm{dist}_{\delta}(\lambda^{-1}\log
Z_{\varepsilon}({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda^{\frac{1}{2}}{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}),\lambda^{-1}\log\mathsf{Z}(\widetilde{\mathcal{K}}_{\lambda};{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},\lambda^{\frac{1}{2}}{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))\leq\delta\,\big{|}\,Z_{\varepsilon}(2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}\big{]}=1.$

Since $\lambda$ is fixed, we can drop the $\lambda^{-1}$ in front of the
$\log$ and $\lambda^{\frac{1}{2}}$ in front of the $x$-variable. It suffices
to prove

$\lim_{\varepsilon\to 0}\mathbb{P}\big{[}\mathrm{dist}_{\delta}(\log
Z_{\varepsilon},\log\mathsf{Z}(\widetilde{\mathcal{K}}_{\lambda}))<\delta\,\big{|}\,Z_{\varepsilon}(2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}\big{]}=1.$
(3.10)

Let us first show that for fixed $\lambda,\delta>0$,

$\lim_{\varepsilon\to
0}\mathbb{P}\big{[}\mathrm{dist}_{\delta}(Z_{\varepsilon},\mathsf{Z}(\widetilde{\mathcal{K}}_{\lambda}))<\delta\,\big{|}\,Z_{\varepsilon}(2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}\big{]}=1,$
(3.11)

and then explain how (3.11) implies (3.10).

To prove (3.11), consider the set $\mathsf{G}_{\lambda,\delta}:=\\{f\in
C([\delta,2]\times[-\delta^{-1},\delta^{-1}]):f(2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}\text{
and
}\mathrm{dist}_{\delta}(f,\mathsf{Z}(\widetilde{\mathcal{K}}_{\lambda}))\geq\delta\\}$.
Since $\mathsf{G}_{\lambda,\delta}$ is a closed set in
$C([\delta,2]\times[-\delta^{-1},\delta^{-1}])$, by Proposition 1.7, we have

$\limsup_{\varepsilon\to
0}\varepsilon\log\mathbb{P}\big{[}Z_{\varepsilon}\in\mathsf{G}_{\lambda,\delta}\big{]}\leq-\inf_{f\in\mathsf{G}_{\lambda,\delta}}I(f).$
(3.12)

By (1.11), we also have

$\lim_{\varepsilon\to
0}\varepsilon\log\mathbb{P}\big{[}Z_{\varepsilon}\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}\big{]}=-q_{\lambda},$
(3.13)

recall that $q_{\lambda}$ is equal to (3.2). Using

$\mathbb{P}\big{[}\mathrm{dist}_{\delta}(Z_{\varepsilon},\mathsf{Z}(\widetilde{\mathcal{K}}_{\lambda}))\geq\delta\,\big{|}\,Z_{\varepsilon}(2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}\big{]}=\frac{\mathbb{P}[Z_{\varepsilon}\in\mathsf{G}_{\lambda,\delta}]}{\mathbb{P}[Z_{\varepsilon}(2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}]}$

and applying (3.12)-(3.13), we have

$\limsup_{\varepsilon\to
0}\varepsilon\log\mathbb{P}\big{[}\mathrm{dist}_{\delta}(Z_{\varepsilon},\mathsf{Z}(\widetilde{\mathcal{K}}_{\lambda}))\geq\delta\,\big{|}\,Z_{\varepsilon}(2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}\big{]}\leq
q_{\lambda}-\inf_{f\in\mathsf{G}_{\lambda,\delta}}I(f).$

By Lemma 3.7, there exists $\zeta>0$ such that
$\inf_{f\in\mathsf{G}_{\lambda,\delta}}I(f)\geq q_{\lambda}+\zeta$. The above
inequality implies

$\lim_{\varepsilon\to
0}\mathbb{P}\big{[}\mathrm{dist}_{\delta}(Z_{\varepsilon},\mathsf{Z}(\widetilde{\mathcal{K}}_{\lambda}))\geq\delta\,\big{|}\,Z_{\varepsilon}(2,0)\geq\frac{1}{\sqrt{4\pi}}e^{\lambda}\big{]}=0.$

Therefore, We conclude (3.11).

Finally, we show how (3.11) implies (3.10). Recall that
$\mathsf{Z}(\rho;t,x)=P(\rho;(0,0)\to(t,x))$. Referring to (2.12), due to the
non-negativity of $\rho$, we have the lower bound $\mathsf{Z}(\rho;t,x)\geq
p(t,x)$. The upper bound of $\mathsf{Z}(\rho;t,x)$ is given by (2.16). This
implies the existence of a constant $M$ such that for all
$\rho\in\widetilde{\mathcal{K}}_{\lambda}$,

$M^{-1}\leq\|\mathsf{Z}(\rho)\|_{L^{\infty}([\delta,2]\times[-\delta^{-1},\delta^{-1}])}\leq
M.$ (3.14)

(3.10) follows from (3.11), (3.14) and the uniform continuity of the $\log$
function on the interval $[(2M)^{-1},2M]$. ∎

###  3.3. The $\lambda\to\infty$ asymptotic of $\mathcal{K}_{\lambda}$

For simplicity, we adopt the shorthand notation
$\|{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}\|_{2}:=\|{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}\|_{L^{2}(\mathbb{R})}$.
The functional $F$ defined in (2.3) enjoys a scaling property: Letting
$\varphi_{\alpha}(x)=\alpha^{2}\varphi(\alpha x)$, we have
$F(\varphi_{\alpha})=\alpha^{2}F(\varphi)$. Choosing
$\alpha=\|\varphi\|_{2}^{-2/3}$ to normalize the $L^{2}$ norm of $\varphi$, we
have

$F(\varphi)=\|\varphi\|_{2}^{\frac{4}{3}}F(\varphi_{\|\varphi\|_{2}^{-2/3}}),\qquad\|\varphi_{\|\varphi\|_{2}^{-2/3}}\|_{2}=1.$
(3.15)

We set

$\mathsf{r}_{\star}(x):=(\frac{3}{4})^{2/3}\operatorname{sech}^{2}((\frac{3}{4})^{1/3}x).$
(3.16)

Also, let $\mathsf{SD}$ denote the space of symmetric decreasing functions
(see Section 3.1.2 for the definition).

######  Lemma 3.9.

For any $\zeta>0$, there exists $\delta>0$ such that for all
$\varphi\in\mathsf{SD}$,

$F(\varphi)\geq\frac{1}{2}(\frac{3}{4})^{\frac{2}{3}}\|\varphi\|_{2}^{\frac{4}{3}}(1-\delta)\
\text{ implies }\
\|\varphi_{\|\varphi\|_{2}^{-2/3}}-\mathsf{r}_{\star}\|_{2}<\zeta.$

###### Proof.

By the scaling (3.15) we assume without loss of generality that
$\|\varphi\|_{2}=1$. The desired statement is equivalent to the following: Any
sequence $\\{\varphi_{n}\\}\subseteq\mathsf{SD}$ with $\|\varphi_{n}\|_{2}=1$
such that $F(\varphi_{n})\to\frac{1}{2}(\frac{3}{4})^{\frac{2}{3}}$ has a
subsequence that converges in $L^{2}$ to $\mathsf{r}_{\star}$. Fix any such
$\varphi_{n}$. Let $f_{n}\in H^{1}(\mathbb{R})$ be such that $\|f_{n}\|_{2}=1$
and $\int\varphi_{n}f_{n}^{\prime 2}-\frac{1}{2}f_{n}^{\prime
2}dx\to\frac{1}{2}(\frac{3}{4})^{2/3}$. Since
$\\{(\varphi_{n},f_{n},f_{n}^{\prime})\\}$ is bounded in
$(L^{2}(\mathbb{R}))^{3}$, by the Banach-Alaoglu theorem, after passing to a
subsequence, $(\varphi_{n},f_{n},f_{n}^{\prime})$ converges weakly to
$(\mathsf{r}_{*},f_{*},g_{*})$ with $\|r_{*}\|_{2}\leq 1$, $\|f_{*}\|_{2}\leq
1$ and $\|g_{*}\|\leq\liminf_{n\to\infty}\|f_{n}^{\prime}\|_{2}$. Since
$(f_{n},f_{n}^{\prime})$ converges weakly to $(f_{*},g_{*})$, we must have
$f^{\prime}_{*}=g_{*}$. Moreover, since $f_{n}$ and $f_{n}^{\prime}$ are
bounded in $L^{2}(\mathbb{R})$, $f_{n}$ is locally uniformly bounded and equi-
continuous. By the Arzela-Ascoli theorem, we know that $f_{n}\to f_{*}$ on
compact sets. Next, because $\|\varphi_{n}\|_{2}=1$ and because
$\varphi_{n}\in\mathsf{SD}$, $\varphi_{n}(\pm M)\leq M^{-\frac{1}{2}}$.
Therefore, $\int_{|x|>M}\varphi_{n}f_{n}^{2}dx\leq
M^{-\frac{1}{2}}\|f_{n}\|_{2}^{2}=M^{-\frac{1}{2}}$. This gives

$\limsup_{n\to\infty}\int_{\mathbb{R}}\varphi_{n}f_{n}^{2}\mathbf{1}_{\\{|x|\leq
M\\}}-\frac{1}{2}f_{n}^{\prime
2}dx\geq\frac{1}{2}\Big{(}\frac{3}{4}\Big{)}^{\frac{2}{3}}-M^{-\frac{1}{2}}.$

The left hand side is bounded above by
$\int_{\mathbb{R}}\mathsf{r}_{*}f_{*}^{2}-\frac{1}{2}{f_{*}^{\prime}}^{2}dx$.
Sending $M\to\infty$ gives
$\int_{\mathbb{R}}\mathsf{r}_{*}f_{*}^{2}-\frac{1}{2}{f^{\prime}_{*}}^{2}dx\geq\frac{1}{2}(\frac{3}{4})^{2/3}$.
This together with $\|\mathsf{r}_{*}\|_{2}\leq 1$ and $\|f_{*}\|_{2}\leq 1$
implies that $F(\mathsf{r}_{*})\geq\frac{1}{2}(\frac{3}{4})^{2/3}$. By the if
and only if part in Lemma 2.4 and the fact that
$\mathsf{r}_{*}\in\mathsf{SD}$, we know that
$\mathsf{r}_{*}=\mathsf{r}_{\star}$. Since $\varphi_{n}$ weakly converges to
$\mathsf{r}_{\star}$ and
$\lim_{n\to\infty}\|\varphi_{n}\|_{2}=1=\|\mathsf{r}_{\star}\|_{2}$, we
conclude that $\varphi_{n}$ converges to $\mathsf{r}_{\star}$ in $L^{2}$. ∎

Recall that $\rho_{*}(t,x)=\rho_{*}(x)=(\operatorname{sech}x)^{2}$. The main
result of this section says that the elements of $\mathcal{K}_{\lambda}$ tend
to $\rho_{*}$ as $\lambda\to\infty$.

######  Proposition 3.10.

We have
$\lim_{\lambda\to\infty}\sup\\{\frac{1}{2\lambda}\|\rho-\rho_{*}\|^{2}_{L^{2}([0,2\lambda]\times\mathbb{R})}:\rho\in\mathcal{K}_{\lambda}\\}=0$.

###### Proof.

Set $\zeta>0$. According to (1.13), which is proved in Section 2, for large
enough $\lambda$ and all $\rho\in\mathcal{K}_{\lambda}$,

$\frac{1}{2\lambda}\|\rho\|^{2}_{L^{2}([0,2\lambda]\times\mathbb{R})}\leq\frac{4}{3}+\zeta.$
(3.17)

By $\mathsf{Z}(\rho;2\lambda,0)\geq\frac{1}{\sqrt{4\pi\lambda}}e^{\lambda}$
and (2.30),

$\displaystyle\frac{1}{\sqrt{4\pi\lambda}}e^{\lambda}\leq
C\exp\Big{(}C\lambda^{2/3}+\int_{0}^{2\lambda}F(\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr\Big{)}.$

Taking the logarithm of both sides and then applying Lemma 2.4 to the right
hand side yields that for all $\lambda$ large enough,

$\lambda(1-\zeta)\leq\int_{0}^{2\lambda}F(\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr\leq\int_{0}^{2\lambda}\frac{1}{2}\Big{(}\frac{3}{4}\Big{)}^{\frac{2}{3}}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{\frac{4}{3}}dr.$

Young’s inequality gives
($(\frac{3}{4}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2})^{2/3}\leq\frac{1}{3}+\frac{2}{3}(\frac{3}{4}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2})$).
Furthermore, there exists a strictly increasing function
$\psi:\mathbb{R}_{\geq 0}\to\mathbb{R}$ such that $\psi(0)=0$,
$(\frac{3}{4}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2})^{2/3}\leq\frac{1}{3}+\frac{2}{3}(\frac{3}{4}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2})-\psi(|1-\frac{3}{4}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2}|)$.
Applying this to the right hand side above yields

$\displaystyle\lambda(1-\zeta)\leq\int_{0}^{2\lambda}F(\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr\leq\int_{0}^{2\lambda}\frac{1}{2}\Big{(}\frac{3}{4}\Big{)}^{\frac{2}{3}}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{\frac{4}{3}}dr$
$\displaystyle\leq\int_{0}^{2\lambda}\frac{1}{6}+\frac{1}{4}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2}-\frac{1}{2}\psi(1-\frac{3}{4}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2})dr$
$\displaystyle\leq\lambda(1+\zeta)-\int_{0}^{2\lambda}\frac{1}{2}\psi(1-\frac{3}{4}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2})dr.$

The last inequality is due to (3.17). Using the inequalities above, we have
for all $\lambda$ large enough,

$\displaystyle\int_{0}^{2\lambda}\psi(|1-\frac{3}{4}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2}|)dr\leq
4\lambda\zeta,$ (3.18)
$\displaystyle\int_{0}^{2\lambda}\frac{1}{2}\Big{(}\frac{3}{4}\Big{)}^{\frac{2}{3}}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{\frac{4}{3}}-F(\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}))dr\leq
4\lambda\zeta.$ (3.19)

Since $\psi$ is strictly increasing with $\psi(0)=0$. For any
$\zeta^{\prime}>0$, we have
$\psi(z)\geq\psi(\zeta^{\prime})=:C(\zeta^{\prime})$ when
$z\geq\zeta^{\prime}$. Applying this to (3.18) gives that for all $\lambda$
large enough,

$\text{Leb}\bigg{[}\Big{\\{}r\in[0,2\lambda]:\Big{|}1-\frac{3}{4}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2}\Big{|}>\zeta^{\prime}\Big{\\}}\bigg{]}\leq\frac{4\lambda\zeta}{C(\zeta^{\prime})}.$
(3.20)

Here, Leb denotes the Lebesgue measure. Since $\rho\in\mathcal{K}_{\lambda}$,
by Proposition 3.3,
$\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\in\mathsf{SD}$ for
almost all $r\in[0,2\lambda]$. Using (3.19) and Lemma 3.9 we conclude that for
any $\zeta^{\prime}>0$, there exists $C(\zeta^{\prime})>0$ such that

$\text{Leb}\bigg{[}\Big{\\{}r\in[0,2\lambda]:\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})_{\|\rho(r,{\raisebox{-1.50694pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{-2/3}}-r_{\star}\|_{2}>\zeta^{\prime}\Big{\\}}\bigg{]}\leq\frac{\lambda\zeta}{C(\zeta^{\prime})},$
(3.21)

where we denote
$\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})_{\alpha}=\alpha^{2}\rho(r,\alpha{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$.
We have

$\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})_{\|\rho(r,{\raisebox{-1.50694pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{-2/3}}-r_{\star}\|_{2}=\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{-1}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})-(r_{\star})_{\|\rho(r,{\raisebox{-1.50694pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2/3}}\|_{2}.$

Applying this to (3.21) yields

$\text{Leb}\bigg{[}\Big{\\{}r\in[0,2\lambda]:\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})-(r_{\star})_{\|\rho(r,{\raisebox{-1.50694pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2/3}}\|_{2}>\zeta^{\prime}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}\Big{\\}}\bigg{]}\leq\frac{\lambda\zeta}{C(\zeta^{\prime})}.$
(3.22)

Taking $\zeta\leq\min(C(\zeta^{\prime}),1)\zeta^{\prime}$, the right hand side
of (3.22) is upper bounded by $\lambda\zeta^{\prime}$. By (3.20), we know that
for most $r\in[0,2\lambda]$,
$\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}$ is around
$\sqrt{\frac{4}{3}}$. Applying this and
$(r_{\star})_{(\sqrt{\frac{4}{3}})^{2/3}}=\rho_{*}$ to (3.22) gives that for
any $\zeta^{\prime}>0$,

$\text{Leb}\bigg{[}\Big{\\{}r\in[0,2\lambda]:\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})-\rho_{*}\|_{2}>\zeta^{\prime}\Big{\\}}\bigg{]}\leq
C\lambda\zeta^{\prime}$ (3.23)

holds for all $\lambda$ large enough. The inequality (3.23) together with the
fact $\|\rho_{*}\|_{2}^{2}=\frac{4}{3}$ implies

$\frac{1}{2\lambda}\int_{0}^{2\lambda}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2}\mathbf{1}_{\\{\|\rho(r,{\raisebox{-1.50694pt}{\scalebox{1.8}{$\cdot$}}})-\rho_{*}\|_{2}\leq\zeta^{\prime}\\}}dr\geq\frac{4}{3}-C\zeta^{\prime}.$

This, together with (3.17) (we have taken $\zeta\leq\zeta^{\prime}$) imply
that

$\frac{1}{2\lambda}\int_{0}^{2\lambda}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2}\mathbf{1}_{\\{\|\rho(r,{\raisebox{-1.50694pt}{\scalebox{1.8}{$\cdot$}}})-\rho_{*}\|_{2}>\zeta^{\prime}\\}}dr\leq
C\zeta^{\prime}.$ (3.24)

In addition, by
$\|\rho_{*}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2}=\frac{4}{3}$
and (3.23),

$\frac{1}{2\lambda}\int_{0}^{2\lambda}\|\rho_{*}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|_{2}^{2}\mathbf{1}_{\\{\|\rho(r,{\raisebox{-1.50694pt}{\scalebox{1.8}{$\cdot$}}})-\rho_{*}\|_{2}>\zeta^{\prime}\\}}dr\leq
C\zeta^{\prime}.$ (3.25)

Combining (3.24) and (3.25), we conclude that

$\displaystyle\frac{1}{2\lambda}\|\rho-\rho_{*}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}$
$\displaystyle=\frac{1}{2\lambda}\int_{0}^{2\lambda}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})-\rho_{*}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|^{2}_{2}\mathbf{1}_{\\{\|\rho(r,{\raisebox{-1.50694pt}{\scalebox{1.8}{$\cdot$}}})-\rho_{*}\|_{2}>\zeta^{\prime}\\}}dr$
$\displaystyle\quad+\frac{1}{2\lambda}\int_{0}^{2\lambda}\|\rho(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})-\rho_{*}(r,{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})\|^{2}_{2}\mathbf{1}_{\\{\|\rho(r,{\raisebox{-1.50694pt}{\scalebox{1.8}{$\cdot$}}})-\rho_{*}\|_{2}\leq\zeta^{\prime}\\}}dr\leq
C\zeta^{\prime}.$

Letting $\lambda\to\infty$ and then $\zeta^{\prime}\to 0$ concludes the
proposition. ∎

##  4\. The Limit shape

The goal of the section is to prove Theorem 1.2.

###  4.1. Equi-continuity of $\mathsf{h}_{\lambda}$

Recall from Section 3.2 that
$\mathsf{h}_{\lambda}(\rho;t,x)=\lambda^{-1}\log\big{(}\lambda^{\frac{1}{2}}\mathsf{Z}(\rho;\lambda
t,\lambda x)\big{)}$. Let $L_{\geq 0}^{2}([0,2\lambda]\times\mathbb{R})$
denote the set of non-negative functions which belong to
$L^{2}([0,2\lambda]\times\mathbb{R})$. Consider the normalized norm
$\lambda^{-\frac{1}{2}}\|{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}$.
The following proposition settles the equi-continuity of
$\mathsf{h}_{\lambda}({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},t,x):L_{\geq
0}^{2}([0,2\lambda]\times\mathbb{R})\to\mathbb{R}$ with respect to this norm.

######  Proposition 4.1.

There exists a constant $C$ such that for all $\lambda>0,\rho_{1},\rho_{2}\in
L_{\geq 0}^{2}([0,2\lambda]\times\mathbb{R})$ and
$(t,x)\in(0,2]\times\mathbb{R}$, if
$\lambda^{-\frac{1}{2}}\|\rho_{1}-\rho_{2}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}<1$,

$|\mathsf{h}_{\lambda}(\rho_{1};t,x)-\mathsf{h}_{\lambda}(\rho_{2};t,x)|\leq
C\lambda^{-\frac{1}{2}}\|\rho_{1}-\rho_{2}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}(1+\lambda^{-1}\|\rho_{1}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}+\lambda^{-1}\|\rho_{2}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}).$

###### Proof.

Set $\widetilde{\mathsf{Z}}(\rho;t,x)=\mathsf{Z}(\rho;t,x)/p(t,x)$. Since

$\displaystyle\lambda^{-1}\log\widetilde{\mathsf{Z}}(\rho_{1};\lambda
t,\lambda x)-\lambda^{-1}\log\widetilde{\mathsf{Z}}(\rho_{2};\lambda t,\lambda
x)$ $\displaystyle=\lambda^{-1}\log\mathsf{Z}(\rho_{1};\lambda t,\lambda
x)-\lambda^{-1}\log\mathsf{Z}(\rho_{2};\lambda t,\lambda x)$
$\displaystyle=\mathsf{h}_{\lambda}(\rho_{1};\lambda t,\lambda
x)-\mathsf{h}_{\lambda}(\rho_{2};\lambda t,\lambda x),$

it suffices to prove that for $\lambda>0$ and $(t,x)\in(0,2]\times\mathbb{R}$,

$\displaystyle\lambda^{-1}\big{|}\log\widetilde{\mathsf{Z}}(\rho_{1};\lambda
t,\lambda x)-\log\widetilde{\mathsf{Z}}(\rho_{2};\lambda t,\lambda x)\big{|}$
$\displaystyle\leq
C\lambda^{-\frac{1}{2}}\|\rho_{1}-\rho_{2}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}(1+\lambda^{-1}\|\rho_{1}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}+\lambda^{-1}\|\rho_{2}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}^{2}).$
(4.1)

Note that $\mathsf{Z}(\rho;t,x)=P(\rho;(0,0)\to(t,x))$. We apply (2.16) with
$a=1$ to obtain

$\widetilde{\mathsf{Z}}(\rho;\lambda t,\lambda x)\leq C\exp(C\lambda
t+\|\rho\|^{2}_{L^{2}([0,\lambda t]\times\mathbb{R})}).$ (4.2)

The constant $C$ here is universal. Further, by the Feynman-Kac formula,

$\widetilde{\mathsf{Z}}(\rho;\lambda t,\lambda x)=\mathbb{E}_{0\to\lambda
x}\Big{[}\exp\Big{(}\int_{0}^{\lambda
t}\rho(s,B_{\text{b}}(s))\Big{)}\Big{]}.$

Let us use (4.2) and the Feynman-Kac formula to prove (4.1). Fix
$\zeta\in(0,1)$. Similar to the proof of Lemma 2.6, by the Feynman-Kac formula
and the Hölder inequality,

$\displaystyle\widetilde{\mathsf{Z}}(\rho_{1};\lambda t,\lambda x)$
$\displaystyle\leq\widetilde{\mathsf{Z}}(\zeta^{-1}(\rho_{1}-(1-\zeta)\rho_{2});\lambda
t,\lambda x)^{\zeta}\widetilde{\mathsf{Z}}(\rho_{2};\lambda t,\lambda
x)^{1-\zeta}.$

Taking the logarithm of both sides yields

$\displaystyle\log\widetilde{\mathsf{Z}}(\rho_{1};\lambda t,\lambda x)$
$\displaystyle\leq\zeta\log\widetilde{\mathsf{Z}}(\zeta^{-1}(\rho_{1}-(1-\zeta)\rho_{2});\lambda
t,\lambda x)+(1-\zeta)\log\widetilde{\mathsf{Z}}(\rho_{2};\lambda t,\lambda
x)$
$\displaystyle\leq\zeta\log\widetilde{\mathsf{Z}}(\zeta^{-1}(\rho_{1}-(1-\zeta)\rho_{2});\lambda
t,\lambda x)+\log\widetilde{\mathsf{Z}}(\rho_{2};\lambda t,\lambda x).$ (4.3)

The last inequality is due to $\widetilde{\mathsf{Z}}(\rho_{2};t,x)\geq 1$,
since $\rho_{2}$ is non-negative.

Subtracting $\log\widetilde{\mathsf{Z}}(\rho_{2};\lambda t,\lambda x)$ from
both sides of (4.3) and applying (4.2) to the resulting right hand side, we
get

$\displaystyle\log\widetilde{\mathsf{Z}}(\rho_{1};\lambda t,\lambda
x)-\log\widetilde{\mathsf{Z}}(\rho_{2};\lambda t,\lambda x)$
$\displaystyle\leq\zeta\log\widetilde{\mathsf{Z}}(\zeta^{-1}(\rho_{1}-(1-\zeta)\rho_{2});\lambda
t,\lambda x)$ $\displaystyle\leq C\zeta\lambda
t+C\zeta^{-1}\|\rho_{1}-(1-\zeta)\rho_{2}\|^{2}_{L^{2}([0,\lambda
t]\times\mathbb{R})}.$

Applying $t\leq 2$ and the inequality $\|f+g\|^{2}_{L^{2}}\leq
2(\|f\|_{L^{2}}^{2}+\|g\|_{L^{2}}^{2})$ to the right hand side leads to

$\displaystyle\log\widetilde{\mathsf{Z}}(\rho_{1};\lambda t,\lambda
x)-\log\widetilde{\mathsf{Z}}(\rho_{2};\lambda t,\lambda x)\leq
C\zeta\lambda+C\zeta^{-1}\|\rho_{1}-\rho_{2}\|^{2}_{L^{2}([0,\lambda
t]\times\mathbb{R})}+C\zeta\|\rho_{2}\|^{2}_{L^{2}([0,\lambda
t]\times\mathbb{R})}.$ (4.4)

The constant $C$ is universal and does not depend on $\lambda,t,x$. Swapping
$\rho_{1},\rho_{2}$ in (4.4) gives that

$\big{|}\log\widetilde{\mathsf{Z}}(\rho_{1};\lambda t,\lambda
x)-\log\widetilde{\mathsf{Z}}(\rho_{2};\lambda t,\lambda x)\big{|}\leq
C\Big{(}\zeta^{-1}\|\rho_{1}-\rho_{2}\|^{2}_{L^{2}([0,\lambda
t]\times\mathbb{R})}+\zeta(\lambda+\|\rho_{1}\|^{2}_{L^{2}([0,\lambda
t]\times\mathbb{R})}+\|\rho_{2}\|^{2}_{L^{2}([0,\lambda
t]\times\mathbb{R})})\Big{)}.$

Dividing both sides by $\lambda$ and using $t\leq 2$, we have

$\displaystyle\lambda^{-1}\big{|}\log\widetilde{\mathsf{Z}}(\rho_{1};\lambda
t,\lambda x)-\log\widetilde{\mathsf{Z}}(\rho_{2};\lambda t,\lambda x)\big{|}$
$\displaystyle\leq
C\Big{(}\zeta^{-1}\lambda^{-1}\|\rho_{1}-\rho_{2}\|^{2}_{L^{2}([0,2\lambda]\times\mathbb{R})}+\zeta\lambda^{-1}(\lambda+\|\rho_{1}\|^{2}_{L^{2}([0,2\lambda]\times\mathbb{R})}+\|\rho_{2}\|^{2}_{L^{2}([0,2\lambda]\times\mathbb{R})})\Big{)}.$
(4.5)

Taking
$\zeta=\lambda^{-\frac{1}{2}}\|\rho_{1}-\rho_{2}\|_{L^{2}([0,2\lambda]\times\mathbb{R})}\in(0,1)$,
we conclude (4.1). ∎

###  4.2. Proof of Theorem 1.2

We begin with a reduction. Recall that Proposition 3.8 states, for arbitrary
fixed $\lambda,\delta>0$,

$\lim_{\varepsilon\to
0}\mathbb{P}\big{[}\mathrm{dist}_{\delta}(h_{\varepsilon,\lambda},\mathsf{h}_{\lambda}(\mathcal{K}_{\lambda}))<\delta\,\big{|}\,h_{\varepsilon}(2,0)+\log\sqrt{4\pi}\geq\lambda\big{]}=1.$

Combining Propositions 3.10 and 4.1 gives
$\lim_{\lambda\to\infty}\mathrm{dist}_{\delta}(\mathsf{h}_{\lambda}(\mathcal{K}_{\lambda}),\mathsf{h}_{\lambda}(\rho_{*}))=0$.
Given these results, it suffices to prove

$\lim_{\lambda\to\infty}\mathrm{dist}_{\delta}(\mathsf{h}_{\lambda}(\rho_{*}),\mathsf{h}_{*})=0,\qquad\text{for
any }\delta>0.$ (4.6)

The proof of (4.6) starts with the Feynman-Kac formula. Set $\varphi=\rho_{*}$
in (2.1) and take logarithm on both sides to get

$\displaystyle\mathsf{h}_{\lambda}(\rho_{*};t,x)$
$\displaystyle=\lambda^{-1}\log\mathbb{E}_{\lambda x\to
0}\Big{[}\exp\Big{(}\int_{0}^{\lambda
t}\rho_{*}(B_{\text{b}}(s))ds\Big{)}\Big{]}-\frac{x^{2}}{2t}-\lambda^{-1}\log\sqrt{4\pi}.$
(4.7)

For $U_{i}=U_{i}(t,x,\lambda)>0$, we write $U_{1}\sim U_{2}$ if
$\lim_{\lambda\to\infty}\lambda^{-1}\log(U_{1}/U_{2})=0$ uniformly over
$(t,x)\in[\delta,2]\times[-\delta^{-1},\delta^{-1}]$. Our goal is to estimate
the expectation in (4.7). Fix a mesoscopic scale $\lambda^{a}$. Any
$a\in(0,\frac{1}{2})$ will do, and we fix $a=\frac{1}{4}$ for the sake of
concreteness. The Brownian bridge in (4.7) starts at $B_{\text{b}}(0)=\lambda
x$ and returns to $B_{\text{b}}(\lambda t)=0$. Consider the first time the
bridge enters the region $[-\lambda^{\frac{1}{4}},\lambda^{\frac{1}{4}}]$,
namely $\tau:=\inf\\{s\geq 0:|B_{\text{b}}(s)|\leq\lambda^{\frac{1}{4}}\\}$.
Decompose the integral in (4.7) into $\int_{0}^{\tau}+\int_{\tau}^{\lambda
t}$. In the first integral, we have $|B_{\text{b}}(s)|>\lambda^{\frac{1}{4}}$,
which gives $|\rho_{*}(B_{\text{b}}(s))|\leq\exp(-\lambda^{1/4}/C)$. This
shows that the contribution of $\rho_{*}(B_{\text{b}}(s))$ within
$s\in[0,\tau]$ is negligible, so

$\displaystyle\mathbb{E}_{\lambda x\to 0}\Big{[}\exp\Big{(}\int_{0}^{\lambda
t}\rho_{*}(B_{\text{b}}(s))\mathrm{d}s\Big{)}\Big{]}\sim\mathbb{E}_{\lambda
x\to 0}\Big{[}\exp\Big{(}\int_{\tau}^{\lambda
t}\rho_{*}(B_{\text{b}}(s))\mathrm{d}s\Big{)}\Big{]}.$ (4.8)

To proceed, condition on $\tau$ in (4.8), namely $\mathbb{E}_{\lambda x\to
0}[{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}]=\mathbb{E}_{\lambda x\to
0}[\mathbb{E}[{\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}}|\tau]]$, and
recognize the inner expectation as $\mathbb{E}[\exp(\int_{\tau}^{\lambda
t}\rho_{*}(B_{\text{b}}(s))\mathrm{d}s)|\tau]=U(\tau,t)$, where
$U(s^{\prime},t):=\mathbb{E}_{\lambda^{1/4}\to 0}[\exp(\int_{0}^{\lambda
t-s^{\prime}}\rho_{*}(B_{\text{b}}(s))\mathrm{d}s)]$. Fix any
$\zeta\in(0,\delta)$. By Lemma 2.1 and 2.4,
$U(s^{\prime},t)\sim\frac{1}{2}\exp(\lambda t-s^{\prime})$, uniformly over
$t\in[\delta,2]$ and $s^{\prime}\in[0,\lambda t-\lambda\zeta]$. This
approximation extends to $s^{\prime}\in[0,\lambda t]$. To see why, note that
when $s^{\prime}\in(\lambda t-\lambda\zeta,\lambda t]$, we have
$|\int_{\tau}^{\lambda
t}\rho_{*}(B_{\text{b}}(s))\mathrm{d}s|\leq\lambda\zeta$, and note that
$\zeta$ can be taken to be arbitrarily small. Therefore,

$\displaystyle\mathbb{E}_{\lambda x\to 0}\Big{[}\exp\Big{(}\int_{0}^{\lambda
t}\rho_{*}(B_{\text{b}}(s))\mathrm{d}s\Big{)}\Big{]}\sim e^{\frac{\lambda
t}{2}}\mathbb{E}_{\lambda x\to 0}[e^{-\frac{1}{2}\tau}].$ (4.9)

Next we bound the right side of (4.9). By symmetry it suffices to consider
$x\geq 0$, which we assume hereafter. Let $\mathsf{T}(s,u)$ be the first
hitting time of $0$ of the Brownian bridge that starts from $u$ at time $0$
and returns to $0$ at time $s$. For fixed $u>0$, $\mathsf{T}(s,u)$ is
stochastically increasing in $s$. This fact can be proven by expressing the
bridge as a drifted Brownian motion via Doob’s h transform. Recall that $\tau$
is the first time $B_{\text{b}}$ enters $[-\lambda^{1/4},\lambda^{1/4}]$.
After the first entrance, consider the excess amount of time it takes for
$B_{\text{b}}$ to hit $0$, namely $\sigma:=\inf\\{s\geq
0:B_{\text{b}}(s+\tau)=0\\}$. Indeed, $\tau+\sigma=\mathsf{T}(\lambda
t,\lambda x)$, so $\mathbb{E}_{\lambda x\to
0}[e^{-\frac{1}{2}\tau}]=\mathbb{E}[e^{-\frac{1}{2}\mathsf{T}(\lambda
t,\lambda x)}e^{\frac{1}{2}\sigma}]$. Conditioned on $\tau$, $\sigma$ is equal
in law to $\mathsf{T}(\lambda t-\tau,\lambda^{1/4})$, which is stochastically
bounded above by $\mathsf{T}(\lambda t,\lambda^{1/4})$. Using Hölder’s
inequality and the stochastic bound, we have

$\displaystyle\mathbb{E}[e^{-\frac{1}{2}\mathsf{T}(\lambda t,\lambda
x)}]\leq\mathbb{E}_{\lambda x\to
0}[e^{-\frac{1}{2}\tau}]\leq\big{(}\mathbb{E}[e^{-\frac{n+1}{2n}\mathsf{T}(\lambda
t,\lambda
x)}]\big{)}^{\frac{n}{n+1}}\big{(}\mathbb{E}[e^{\frac{n+1}{2}\mathsf{T}(\lambda
t,\lambda^{1/4})}]\big{)}^{\frac{1}{n+1}}.$ (4.10)

Given (4.10), we seek to estimate $\mathbb{E}[\exp(-\beta\mathsf{T}(\lambda
t,\lambda x))]$. The first step is to derive the probability density function
of $\mathsf{T}(\lambda t,\lambda x)$. Express the Brownian bridge
$B_{\text{b}}$ by a Brownian motion as $B_{\text{b}}(s)=(1-\frac{s}{\lambda
t})(\lambda x-\sqrt{\lambda t}B(\frac{s}{\lambda t-s}))$; relate
$\mathsf{T}(\lambda t,\lambda x)$ to a hitting time of $B$; use the known
density function of the hitting time of $B$. The result reads $(\text{density
function of }\mathsf{T}(\lambda t,\lambda
x))(s)=(\sqrt{\lambda^{3}tx^{2}}/\sqrt{2\pi s^{3}(\lambda
t-s)})\exp(-\tfrac{\lambda t-s}{2\lambda ts}(\lambda x)^{2}).$ Use this
density function to express $\mathbb{E}[\exp(-\beta\mathsf{T}(\lambda
t,\lambda x))]$ as an integral, and perform a change of variables
$s\mapsto\lambda ts$. We have

$\displaystyle\mathbb{E}[e^{-\beta\mathsf{T}(\lambda t,\lambda
x)}]=\int_{0}^{1}\frac{\sqrt{\lambda x^{2}}}{\sqrt{2\pi
ts^{3}(1-s)}}\exp\big{(}\lambda\,V_{\beta}(s,t,x)\big{)}\mathrm{d}s,$ (4.11)

where $V_{\beta}(s,t,x):=-\beta ts-\frac{1-s}{2st}x^{2}$. This integral can be
analyzed by Laplace’s method. Differentiating $V$ in $s$, one finds that
$V({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}},x)$ attains its unique
maximum at $s=\min\\{\frac{x}{\sqrt{2\beta}\,t},1\\}$, and
$\partial^{2}_{s}V_{\beta}=\frac{x^{2}}{ts^{3}}$. Using these properties in
(4.11), it is not hard to show that $\mathbb{E}[\exp(-\beta\mathsf{T}(\lambda
t,\lambda x))]\sim\exp(-\lambda
tV_{\beta}(\min\\{\frac{x}{\sqrt{2\beta}\,t},1\\},t,x))$, for fixed $\beta>0$
and uniformly over $[t,x]\in[\delta,2]\times[0,\delta^{-1}]$. Insert this
estimate into (4.10); take
$\lambda^{-1}\log({\raisebox{-2.15277pt}{\scalebox{1.8}{$\cdot$}}})$ of the
result; send $\lambda\to\infty$ first and $n\to\infty$ later. We obtain

$\displaystyle\lim_{\lambda\to\infty}\lambda^{-1}\log\mathbb{E}_{\lambda x\to
0}[e^{-\frac{1}{2}\tau}]=V_{1/2}(\min\\{\tfrac{x}{t},1\\},t,x)=\left\\{\begin{array}[]{l@{,}l}\frac{x^{2}}{2t}-x&\text{
when }x\in[0,t],\\\ -\frac{t}{2}&\text{ when }x>t,\end{array}\right.$

uniformly over $[t,x]\in[\delta,2]\times[0,\delta^{-1}]$. Inserting this into
(4.9) and then inserting the result into (4.7) completes the proof of (4.6)
and hence the proof of Theorem 1.2.

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  *[KPZ]: Kardar–Parisi–Zhang
  *[SHE]: Stochastic Heat Equation
  *[LDP]: Large Deviation Principle