text:
\begin{document}

\title{Analysis of the gradient of the solution to a stochastic
	heat equation via fractional Brownian motion
	\thanks{Research supported in part by NSF grants DMS-1006903
	and DMS-1307470 (D.K. and P.M.), EPSRC grant EP/J017418/1 (M.F.),
	the Simons Foundation (Simons Visiting Professorship, D.K.),
	Mathematisches Forschungsinstitut Oberwolfach (D.K.),
	and TU-Dresden (D.K.).}
}

\author{Mohammud Foondun\\Loughborough University
	\and Davar Khoshnevisan\\University of Utah
	\and Pejman Mahboubi\\University of Utah
}

\date{June 11, 2014}
\maketitle

\begin{abstract}
	Consider the stochastic partial differential equation
	$\partial_t u = Lu+\sigma(u)\xi$, where
	$\xi$ denotes space-time white noise and $L:=-(-\Delta)^{\alpha/2}$
	denotes the fractional Laplace operator of index $\alpha/2\in(\nicefrac12\,,1]$.
	We study the detailed behavior of the approximate spatial gradient
	$u_t(x)-u_t(x-\varepsilon)$ at fixed times $t>0$, as $\varepsilon\downarrow0$.
	We discuss a few applications of this work to the study of the sample functions
	of the solution to the KPZ equation as well.\\

	\noindent{\it Keywords:} The stochastic heat equation.\\

	\noindent{\it \noindent AMS 2010 subject classification:}
	Primary. 60H15, 60G17; Secondary. 60H10, 47B80.
\end{abstract}

\section{Introduction and main results}
We consider the stochastic partial differential equation
\begin{align}\label{SHE}
	\frac{\partial}{\partial t} u_t(x) =-(-\Delta)^{\alpha/2}
	u_t(x) + \sigma(u_t(x)) \,\xi,
	\tag{SHE}
\end{align}
where $\alpha\in(1\,,2]$ is a fixed ``spatial scaling''
parameter, $-(-\Delta)^{\alpha/2}$ denotes
the fractional Laplacian to the power $\alpha/2$, and
$\xi$ denotes space-time white noise.
Throughout, we assume that the initial function $u_0$
is non random and bounded.
We assume also that $\sigma$ is Lipschitz continuous; that is,
$\lip_\sigma<\infty$, where
\begin{equation}
	\lip_\sigma:=\sup_{-\infty<x<y<\infty}\left|
	\frac{\sigma(y)-\sigma(x)}{y-x}\right|.
\end{equation}
It is well known that, under the present conditions,  (SHE) has
a continuous solution that is uniquely defined by the moment
condition
\begin{equation}\label{eq:moments}
	\sup_{x\in\R}\sup_{t\in[0,T]}\E\left(\left| u_t(x)\right|^k\right)<\infty
	\qquad\text{for all $T>0$},
\end{equation}
for one, hence all, $k\in[2\,,\infty)$; see  \cite{DebbiDozzi}.

The main objective of this paper is to study the spatial gradient of the
random function $x\mapsto u_t(x)$ where the time parameter $t>0$
is held fixed. Our results will have the \emph{a priori} [unsurprising]
consequence that
$\partial u_t(x)/\partial x$ does not exist. Therefore, instead of
studying the gradient itself we consider an approximate un-normalized
gradient $u_t(x)-u_t(x-\varepsilon)$, where $\varepsilon\approx 0$.

Our main result is the following:
\begin{theorem}\label{th:Differentiate}
	For every fixed $t>0$,
	there exists an {\rm fBm} $F:=\{F(x)\}_{x\in\R}$ with Hurst index
	$H:=\frac12(\alpha-1)\in(0\,,\nicefrac12]$ such that
	\begin{equation}\label{eq:RSDE}
		\lim_{\varepsilon\downarrow0}\sup_{x\in\R}\P\left\{
		\left| \frac{u_t(x) - u_t(x-\varepsilon)}{F(x)-F(x-\varepsilon)} -
		\mathfrak{A}_\alpha\sigma(u_t(x))
		\right| > \lambda\right\} =0,
	\end{equation}
	for all $\lambda>0$, where $\mathfrak{A}_\alpha$ is the
	following numerical constant:
	\begin{equation}\label{A(alpha)}
		\mathfrak{A}_\alpha := \left\{ 2\Gamma(\alpha)\left\vert
		\cos\left(\alpha\pi/2\right)\right\vert\right\}^{-1/2}.
	\end{equation}
\end{theorem}

Theorem \ref{th:Differentiate} basically asserts that $x\mapsto u_t(x)$
solves the following rough stochastic differential equation:
$$
	\frac{\d u_t(x) }{\d x}
	= \mathfrak{A}_\alpha\sigma(u_t(x))\,
	\frac{\d F(x)}{\d x}\qquad(x\in\R),
	\eqno{\textnormal{(R-SDE)}}
$$
where $F$ denotes an fBm  with Hurst index $H:=\frac12(\alpha-1)\in(0\,,\nicefrac12]$.
It has recently been shown in \cite{KSXZ}, using other methods,
that the \emph{temporal} approximate gradient of
(SHE) solves a rough differential equation that
is driven by a fractional Brownian motion
with Hurst index $H/\alpha\in(0\,,\nicefrac14]$.

When $H\in(\nicefrac12\,,1]$, the
random function $F$ is smoother than a H\"older-continuous
function of index $>\nicefrac12$, and hence (R-SDE) can be solved
by using classical theory of Young integrals. Therefore, Theorem
\ref{th:Differentiate} shows the existence of solutions to all
remaining possible rough differential equations that are driven by
fractional Brownian motion. However, we caution that (R-SDE)
is an anticipative stochastic differential equation, even in the case
that $\alpha=2$ where $F$ simplifies to a standard Brownian motion.

Stochastic differential equations
such as (R-SDE) have been the subject of intense recent activity
\cite{AlosLeonNualart,AlosMazetNualart,CoutinFritzVictoir,
ErramiRusso,FritzVictoir2010a,GradinaruRussoVallois,Gubinelli,KSXZ,
LyonsQian,Lyons:95,Lyons:94,NualartTindel,RussoVallois,Unterberger}.
As such, Theorem \ref{th:Differentiate} can be viewed
as a non-trivial contribution to the existence theory of very rough
[anticipative] stochastic
differential equations. More significantly, the proof of Theorem \ref{th:Differentiate} also
teaches us about the local structure of
the stochastic heat equation (SHE). For instance, one can conclude fairly easily
from the method we employed in the proof of Theorem \ref{th:Differentiate} a local law of the iterated logarithm, as the following suggests.

\begin{corollary}\label{cor:LIL}
	Choose and fix $t>0$ and $x\in\R$. Then with probability one,
	\[
		\limsup_{\varepsilon\downarrow 0}\frac{u_t(x)-u_t(x-\varepsilon)}{
		\sqrt{2\varepsilon^{\alpha-1}\log\log(1/\varepsilon)}}
		=-\liminf_{\varepsilon\downarrow 0}\frac{u_t(x)-u_t(x-\varepsilon)}{
		\sqrt{2\varepsilon^{\alpha-1}\log\log(1/\varepsilon)}}
		= \mathfrak{A}_\alpha |\sigma(u_t(x))|.
	\]
\end{corollary}

One cannot improve the convergence-in-probability assertion in
\eqref{eq:RSDE}
to a statement with almost-sure convergence. This is because
there are infinitely-many small random values of $\varepsilon>0$ such that
the denominator $F(x)-F(x-\varepsilon)$ of the difference quotient in
\eqref{eq:RSDE} is zero. However, the following almost-sure-in-density
improvement does hold:
\begin{corollary}\label{a.s.}
	Fix $t>0$ and $x\in\R$. With probability one,
	\begin{equation}\label{eq:a.s}
		\lim_{s\to0}\frac{1}{s}\int_0^s\left|\frac{u_t(x)-u_t(x-\varepsilon)}
		{F(x)-F(x-\varepsilon)}-\mathfrak{A}_\alpha\sigma(u_t(x))\right|\d\varepsilon
		=0.
	\end{equation}
\end{corollary}

We can also deduce an interesting ``central limit theorem,'' whose limit
law is a mixture of mean-zero normal distributions.

\begin{corollary}\label{cor:CLT}
	Choose and fix $t>0$ and $x\in\R$. Then, for all $a\in\R$,
	\begin{equation}
		\lim_{\varepsilon\downarrow 0}\P\left\{
		\frac{u_t(x)-u_t(x-\varepsilon)}{\varepsilon^{(\alpha-1)/2}} \le a\right\}
		= \P\left\{ |\sigma(u_t(x))| \times
		\mathcal{N}\le \frac{a}{\mathfrak{A}_\alpha}\right\},
	\end{equation}
	where $\mathcal{N}$ denotes a standard Gaussian random variable, independent
	of $u_t(x)$.
\end{corollary}

Let us also mention a corollary about the variations of the solution to (SHE).
This result extends the recent work of Posp\`\i{}\v{s}il and Tribe \cite{PospisilTribe}.

\begin{corollary}\label{cor:VAR}
	If $\phi:\R\to\R$ is Lipschitz continuous, then
	for all non random reals $b>a$ and $t>0$,
	\begin{equation}\begin{split}
		&\lim_{n\to\infty} \sum_{a2^n\le j \le b2^n} \phi( u_t
			(j 2^{-n})) \left| u_t( (j+1)2^{-n})- u_t\left( j2^{-n}\right)
			\right|^{2/(\alpha-1)}\\
		&\hskip1.7in
			=\mathfrak{B}_\alpha \int_a^b \phi(u_t(x))
			\left[ \sigma(u_t(x))\right]^{2/(\alpha-1)}\d x,
	\end{split}\end{equation}
	almost surely and in $L^2(\P)$, where $\mathfrak{B}_\alpha$ is
	the following numerical constant:
	\begin{equation}\label{B(alpha)}
		\mathfrak{B}_\alpha
		:= \frac{1}{\sqrt\pi}\left| \frac{1}{\Gamma(\alpha)\cos(\alpha\pi/2)}
		\right|^{1/(\alpha-1)} \Gamma\left( \frac{1}{2}+\frac{1}{\alpha-1}\right).
	\end{equation}
\end{corollary}

In Section \ref{sec:VS} we describe further consequences of these
Corollaries to the analysis of the KPZ equation of statistical mechanics. See,
in particular, Corollary \ref{cor:KPZ}. The latter corollary is deeply connected to,
and complements,
the recent works \cite{Hairer,QuastelRemenik} on the quadratic variation, in
the space variable, of the Hopf--Cole solution to the KPZ equation.

Throughout, we write
\begin{equation}\label{eq:nabla}
	(\nabla_\varepsilon f)(x) := f(x)-f(x-\varepsilon),
\end{equation}
as substitute for the approximate spatial gradient of any real
function $f$. We also adopt the following notation
\begin{equation}
	\|X\|_k:=\left\{ \E(|X|^k) \right\}^{1/k},
\end{equation}
 for every $k\in [1,\,\infty)$ and $X\in L^k(\Omega)$.
\section{Preliminaries}
Throughout this paper, we make use of the Walsh
theory of SPDEs \cite{Walsh} to interpret  \eqref{SHE} as the following integral equation,
\begin{equation}\label{mild}
	u_t(x) = (p_t*u_0)(x) + \int_{(0,t)\times\R} p_{t-s}(y-x)\sigma(u_s(y))\, \xi(\d s\,\d y),
\end{equation}
where $\{p_t(x)\}_{t>0,x\in\R}$ denotes the heat kernel of the fractional Laplacian
and the stochastic integral is a Walsh--It\^o integral.
The function $(t\,,x)\mapsto p_t(x)$
also describes the transition density functions
of an isotropic $\alpha$-stable L\'evy
process $\{X_t\}_{t\ge 0}$,
and are determined via their Fourier transforms that are normalized as follows:
\begin{equation}\label{pihat}
	\hat p_t(\chi)=\e^{-t|\chi|^\alpha}\qquad(t> 0,\, \chi\in\R).
\end{equation}

The theory of Dalang \cite{Dalang} ensures the existence of a
solution $u$ to the SPDE (SHE), which is unique among all
solutions that satisfy  \eqref{eq:moments}
for all $T>0$ and $k\in[2\,,\infty)$.

The proof of the main result of this paper will rely on some perturbation
arguments which will require the study of the following
linear stochastic heat equation.  The main results of this section
will be devoted to the study of this linear equation.  First, let us
fix some notation.

Let us consider the following linearization of (SHE):
\begin{align}\label{SHE-Linear}
	\frac{\partial}{\partial t} Z_t(x) =-(-\Delta)^{\alpha/2}
	Z_t(x) + \xi,
	\tag{L-SHE}
\end{align}
subject to $Z_0(x):=0$ for all $x\in\R$.
In keeping with \eqref{mild},
the solution to (L-SHE) can be written as the Wiener-integral process,
\begin{equation}\label{Z}
	Z_t(x) := \int_{(0,t)\times\R} p_{t-s}(y-x)\, \xi(\d s\,\d y)
	\qquad(t>0,\, x\in\R).
\end{equation}

It is very well known that $x\mapsto Z_t(x)$ has a version that is
H\"older continuous of any index $<(\alpha-1)/2$. This
fact relies on another well-known bound of the form
$\E(|Z_t(x)-Z_t(x-\varepsilon)|^2)=O(\varepsilon^{\alpha-1})$,
as $\varepsilon\downarrow 0$.
Our first result is basically an improvement of such an estimate.
It will be of central importance to our later needs.

\begin{lemma}\label{(0,t)timesR}
	Choose and fix a $T>0$. Then,
	\begin{equation}
		\E\left( \left| Z_t(x) - Z_t(x-\varepsilon)\right|^2\right)
		=\mathfrak{A}_\alpha^2\varepsilon^{\alpha-1}
		+ O(\varepsilon^2)
		\qquad(\varepsilon\downarrow 0),
	\end{equation}
	uniformly for all $t\in[0\,,T]$ and $x\in\R$, where $\mathfrak{A}_\alpha$ is defined in \eqref{A(alpha)}.
\end{lemma}

\begin{proof}
	To aid the presentation of the proof, we define
	\begin{equation}
		Q := \E\left(\left| Z_t(x) - Z_t(x-\varepsilon) \right|^2\right)
		\qquad(t,\varepsilon>0,\,x\in\R).
	\end{equation}
	The mild formulation of the solution together with Wiener's isometry and Plancherel's formula yield
	\begin{equation}\label{D2}\begin{split}
		Q &=\int_0^t\d s\int_{-\infty}^\infty\d y\
			\left[ p_s(y)-p_s(y-\varepsilon)\right]^2\\
		&= \frac{1}{2\pi}\int_0^t\d s\int_{-\infty}^\infty\d\chi\
			\e^{-2s|\chi|^\alpha}\left| 1- \e^{-i\chi\varepsilon}\right|^2\\
		&=\frac{1}{\pi}\int_0^t\d s\int_{-\infty}^\infty\d\chi\
			\e^{-2s|\chi|^\alpha}[1-\cos(\chi\varepsilon)],
	\end{split}\end{equation}
	where we have also used the simple fact that $|1-\exp(i\theta)|^2=2(1-\cos\theta)$
	for every $\theta\in\R$. We now use Fubini's theorem in order
	to interchange the order of integration, and then use symmetry to deduce
	the following:
	\begin{equation}\label{D3}\begin{split}
		Q & = \frac1\pi
			\int_0^\infty\frac{1-\e^{-2t|\chi|^\alpha}}{\chi^\alpha}
			[1-\cos(\chi\varepsilon)]\,\d\chi\\
		&=\frac1\pi\int_0^\infty\frac{1-\cos(\chi\varepsilon)}{\chi^\alpha}\,\d\chi
			-\frac1\pi\int_0^\infty\frac{\e^{-2t|\chi|^\alpha}}{\chi^\alpha}
			[1-\cos(\chi\varepsilon)]\,\d\chi.
	\end{split}\end{equation}
	Next, we separately compute the preceding integrals.
	In order to calculate the first, we use scaling and then
	evaluate the resulting integral using \cite[\S4, p.\ 13]{K-SN}:
	\begin{equation}\label{eq:1-cos}
		\int_0^\infty\frac{1-\cos(\chi\varepsilon)}{\chi^\alpha}\,\d\chi
		=\varepsilon^{\alpha-1} \cdot\int_0^\infty\frac{1-\cos z}{z^\alpha}\,\d z
		=\pi \mathfrak{A}_\alpha^2\varepsilon^{\alpha-1}.
	\end{equation}
	Since $1-\cos \theta \le \theta^2$ for all $\theta\in\R$,
	the second integral on the last line of \eqref{D3} is bounded from above by
	\begin{equation}\label{eq:e(1-cos)}
		\int_0^\infty\frac{\e^{-2t|\chi|^\alpha}}{\chi^\alpha}
		[1-\cos(\chi\varepsilon)]\,\d\chi \le \varepsilon^2\cdot
		\int_0^\infty \e^{-2t\chi^\alpha}\chi^{2-\alpha}\,\d\chi=O(\varepsilon^2),
	\end{equation}
	uniformly for all $t\in[0\,,T]$. The lemma follows from combining the
	preceding two displays with \eqref{D3}.
\end{proof}

Lemma \ref{(0,t)timesR} has a number of immediate, though still useful,
consequences. We list two of those that we shall need. The
first computes the correct power that ``essentially linearizes''
the spatial increments of the process $Z$.

\begin{corollary}\label{cor:Mean:Linear}
	Choose and fix a $T>0$. Then,
	\begin{equation}
		\E\left( \left| Z_t(x) - Z_t(x-\varepsilon)\right|^{2/(\alpha-1)}\right)
		=\mathfrak{B}_\alpha\varepsilon + O\left(
		\varepsilon^{2/(\alpha-1)} \right) \qquad(\varepsilon\downarrow 0),
	\end{equation}
	uniformly for all $t\in[0\,,T]$ and $x\in\R$.
\end{corollary}

\begin{proof}
	We appeal to the fact that every positive moment of a
	centered Gaussian random variable $X$ is determined by the variance
	of $X$.  In particular,
		\begin{equation}\label{theabove}
			\E\left(|X|^{2/(\alpha-1)}\right) =
			c \left[\E\left(X^2\right)\right]^{1/(\alpha-1)},
		\end{equation}
		where
		\begin{equation}
			c  := \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty|z|^{2/(\alpha-1)}
			\e^{-z^2/2}\,\d z = \frac{2^{1/(\alpha-1)}}{\sqrt\pi}\Gamma\left(
			\frac12+\frac{1}{\alpha-1}\right).
		\end{equation}
	For every fixed $t\in [0\,, T]$ and $x\in \R$,
	the increment $Z_t(x) - Z_t(x-\varepsilon)$ is a Gaussian random variable with
	mean zero.
	Therefore we may apply \eqref{theabove}
	[with $X:=Z_t(x) - Z_t(x-\varepsilon)$],
	together with Lemma \ref{(0,t)timesR} and \eqref{B(alpha)},
	in order to conclude the proof of the corollary.
\end{proof}

Next we mention the following consequence of Lemma
\ref{(0,t)timesR}; it states the increments of the solution $Z$ behaves much like those of the solution $u$ to the non-linear (SHE).

\begin{corollary}\label{cor:Modulus:x}
	Let $u_t(x)$ denote the solution to \eqref{SHE}. Then, for all real numbers
	$T>0$ and $k\in[2\,,\infty)$
	there exists a finite constant $C:=C_{k,T}$, such that
	\begin{equation}
		\E\left( \left| u_t(x)-u_t(y)
		\right|^k\right)\le C|x-y|^{(\alpha-1)k/2},
	\end{equation}
	simultaneously for every $x,y\in\R$ and $t\in[0,T]$.
\end{corollary}

\begin{proof}
	Thanks to \eqref{eq:moments}, it suffices to consider the case that
	$|x-y|\le 1$, which we do from here on.

	Since $u_0$ is bounded, $(p_t*u_0)(x)$ is smooth for every fixed $t>0$.
	Indeed, \eqref{pihat} ensures that
	$\hat p_t(\chi)$---whence also $p_t$---are in the Schwartz space
	of rapidly decreasing test functions.  Because $u_0$ is a tempered distribution
	a standard fact  \cite[Theorem 3.13]{S-W} implies that
	$(p_t*u_0)(x)$ is smooth.
	An application of
	a sharp form of the Burkholder--Davis--Gundy inequality---as in \cite{FK}---
	ensures that, uniformly for all
	$x,y\in\R$ and $t\in[0\,,T]$,
	\begin{align}
		&\|u_t(x)-u_t(y)\|^2_k\\\notag
		&\le c_2\left[|x-y|^2
			+\int_0^t\d s\int_{-\infty}^\infty\d z\
			|p_{t-s}(z-x)-p_{t-s}(z-y)|^2\|\sigma(u_s(z))\|_k^2\right],
	\end{align}
	where $c_2$ is positive and finite, and depends only on $k$ and $T$.
	[See \cite{FK} for the details of this sort
	of argument.]

	Because $\sigma$ has at-most-linear growth, \eqref{eq:moments}
	ensures that $\|\sigma(u_s(y))\|_k^2$ is bounded uniformly in
	$s\in[0\,,T]$ and $y\in\R$, and therefore we can find
	$c_3:=c_{k,T}\in(0\,,\infty)$ such that, uniformly for $x,y\in\R$ and
	$t\in[0\,,T]$,
	\begin{align}\notag
		\|u_t(x)-u_t(y)\|^2_k
			&\le c_3\left[|x-y|^2
			+\int_0^t\d r\int_{-\infty}^\infty\d z\
			|p_r(z-x)-p_r(z-y)|^2\right]\\
		&=c_3\left[|y-x|^2+ \E\left( \left| Z_t(x)-Z_t(y) \right|^2\right)\right].
	\end{align}
	Because we consider only the case that $|x-y|\le1$,
	Lemma \ref{(0,t)timesR} shows us that
	$\|u_t(x)-u_t(y)\|^2_k\le\text{const}\cdot |x-y|^{\alpha-1}$, as desired.
\end{proof}

Let us conclude this section with another continuity estimate.
This bound follows essentially from the Appendix of \cite{FK};
see also \cite{DebbiDozzi}. Therefore, we will not describe a proof.
\begin{proposition}\label{pro:Modulus:t}
	Suppose that $u_t(x)$ be as in the previous Corollary, then for all $T>0$
	and $k\in[2\,,\infty)$, there exists a finite constant $C:=C_{k, T}$ such that
	\begin{equation}
		\|u_s(x)-u_t(x)\|_k\leq C|t-s|^{(\alpha-1)/2\alpha},
	\end{equation}
	simultaneously for every $s, t\in [0, T]$ and $x\in \R$.
\end{proposition}

\section{Fractional Brownian motion}
Recall that a random field
$\Phi:=\{\Phi(x)\}_{x\in\R}$ is called a \emph{fractional
Brownian motion with Hurst index $H\in(0\,,1)$} if $\Phi$ is a mean-zero
Gaussian process with
\begin{equation}
	\E\left( \left| \Phi(x) - \Phi(y) \right|^2\right)
	= |x-y|^{2H}\qquad\text{for all $x,y\in\R$}.
\end{equation}
We abbreviate this by saying that $\Phi$ is a $\textnormal{fBm}(H)$.
In particular, fBm($\nicefrac12$) is easily seen to be ordinary Brownian motion.

\begin{proposition}\label{pr:fBm}
	Fix some $t>0$. Then, the solution $Z_t$ to the linear stochastic  heat equation
	\textnormal{(L-SHE)}, at time $t$, can be decomposed as
	\begin{equation}\label{eq:ZFS}
		Z_t(x) = \mathfrak{A}_\alpha
		F(x) + S(x)\qquad\text{for all $x\in\R$},
	\end{equation}
	where $F$ is $\textnormal{fBm}((\alpha-1)/2)$ and $S$ is a centered
	Gaussian process with $C^\infty$ sample functions.
\end{proposition}

\begin{proof}
	The random field $S$ is defined explicitly as the following
	Wiener integral process:
	\begin{equation}
		S(x) :=
		\int_{(t,\infty)\times\R} \left[ p_s(y)-p_s(y-x)\right] \, \xi(\d s\,\d y)
		\qquad(x\in\R).
	\end{equation}
	In order to verify that $S$ is a well-defined Gaussian random field
	we proceed by applying the Plancherel theorem, such as we did when
	we developed \eqref{D2}, and find that
	\begin{equation}\begin{split}
		\E\left( \left| S(x) \right|^2\right) &= \int_t^\infty \d s\int_{-\infty}^\infty
			\d y\ \left| p_s(y) - p_s(y-x)\right|^2\\
		&=\frac{2}{\pi}\int_t^\infty\d s\int_0^\infty\d\chi\
			\e^{-2s\chi^\alpha}\left[ 1-\cos(x\chi)\right]\\
		&=\frac{1}{\pi}\int_0^\infty \e^{-2t\chi^\alpha}
			\left(\frac{1-\cos(x\chi)}{\chi^\alpha}\right)\d\chi<\infty.
	\end{split}\end{equation}
	This proves that $S$ is a mean-0 Gaussian random field. Similarly,
	one shows that its $n$th generalized
	derivative $\d^n S/\d x^n$ is the mean-0
	Gaussian random field
	\begin{equation}
		\frac{\d ^n S}{\d x^n}(x) = (-1)^{n+1}\int_{(t,\infty)\times\R}
		\frac{\d ^n p_s}{\d x^n} (y-x)\, \xi(\d s\,\d y).
	\end{equation}
	And we verify that the preceding is a well-defined bona fide stochastic
	process by checking that
	\begin{equation}\begin{split}
		\E\left( \left| \frac{\d^n S}{\d x^n}(x)\right|^2\right)
			&= 	\int_t^\infty\d s\int_{-\infty}^\infty\d y\
			\left|\frac{\d ^n p_s}{\d x^n} (y)\right|^2\\
		&=\frac{1}{2\pi}\int_t^\infty\d s\int_{-\infty}^\infty\d\chi\
			|\chi|^{2n}\e^{-2s|\chi|^\alpha},
	\end{split}\end{equation}
	thanks to Plancherel's theorem. Therefore, for every $x\in\R$
	and $n\ge 1$,
	\begin{equation}\label{eq:E(dS2)}
		\E\left( \left| \frac{\d^n S}{\d x^n}(x)\right|^2\right)
		= \frac{1}{2\pi}\int_0^\infty
		\e^{-2t\chi^\alpha}\chi^{2n-\alpha}\,\d\chi<\infty.
	\end{equation}
	The remaining details involve making a few routine
	computations that are similar to some of the calculations that were
	made earlier in this section. We only discuss the case when $n=1$.
	Since, by the Wiener isometry, for any $x_2>x_1$,
	\begin{align}
		\E\left(\left|\frac{\d S}{\d x}(x_2)-\frac{\d S}{\d x}(x_1)\right|^2\right)
		&=\int_t^\infty\d s\int_\R\d y\,|p'_s(y-x_2)-p'_s(y-x_1)|^2\notag\\

		&=\frac{1}{\pi}\int_0^\infty\e^{-2t|\chi|^\alpha}
		\frac{1-\cos\chi(x_2-x_1)}{\chi^{\alpha-1}}\d\chi\\
		&=O(|x_2-x_1|^2),\notag
	\end{align}
	 by~\eqref{eq:e(1-cos)}. Therefore, by the Kolmogorov's continuity theorem,
	 the distributional derivative $(\d S/\d x)$ of
	 $S$ is a continuous  function [up to a modification, which we may
	 choose to adopt without further mention], whence
	$S\in C^1(\R)$ a.s. And a very similar estimate shows that
	if $S\in C^k(\R)$ for some integer $k\ge 1$, then
	$S\in C^{k+1}(\R)$ a.s.\ [up to a modification, once again]. This
	proves that $S$ is a.s.\ $C^\infty$.

	Now that we have defined $S$, let us observe that $Z_t$ and $S$ are
	totally independent from one another [this has to do only with the independence
	properties of white noise]. Therefore, it follows
	that
	\begin{equation}\label{eq:B=Z-S}
		B(x) := Z_t(x)-S(x) \qquad(x\in\R)
	\end{equation}
	defines a mean-zero Gaussian random field
	whose distribution is computed as follows: For every $x,x'\in\R$,
	\begin{equation}\begin{split}
		\E\left( \left| B(x)-B(x')\right|^2\right) &=
			\E\left( \left| Z_t(x)-Z_t(x') \right|^2\right) +
			\E\left( \left| S(x)-S(x') \right|^2\right)\\
		&=\int_0^\infty\d s\int_{-\infty}^\infty\d y\
			\left[ p_s(y-x) - p_s(y-x')\right]^2\\
		&= \frac{1}{\pi}\int_0^\infty\d s\int_{-\infty}^\infty\d\chi\
			\e^{-2s|\chi|^\alpha}\left[ 1-\cos((x-x')\chi)\right],
	\end{split}\end{equation}
	thanks to Plancherel's theorem. This and Fubini's theorem together yield
	\begin{equation}\begin{split}
		\E\left( \left| B(x)-B(x')\right|^2\right)
			&= \frac1{\pi}\int_0^\infty
			\left(\frac{ 1-\cos(|x-x'|\chi)}{\chi^\alpha}\right)\d\chi\\
		&=\pi^{-1}|x-x'|^{\alpha-1}\int_0^\infty
			\left(\frac{ 1-\cos z}{z^\alpha}\right)\d z\\
		&= \frac{|x-x'|^{\alpha-1}}{2\Gamma(\alpha)\vert\cos(\alpha\pi/2)\vert}
			=[\mathfrak{A}_\alpha]^2|x-x'|^{\alpha-1};
	\end{split}\end{equation}
	see \eqref{A(alpha)} and \eqref{eq:1-cos}. In particular,
	$F(x) := B(x)/\mathfrak{A}_\alpha$ $(x\in\R)$
	defines a fractional Brownian motion with Hurst index
	$(\alpha-1)/2$; and property \eqref{eq:ZFS} follows from the
	construction in \eqref{eq:B=Z-S}.
\end{proof}

\section{Localization}
Throughout, let us choose and fix a parameter
\begin{equation}\label{eq:beta:gamma}
	\beta>1,\text{ and set }\gamma := 1 + \beta^{3/2}.
\end{equation}
Then, we define a family of  space-time boxes as follows: For every $x\in\R$
and $\varepsilon>0$,
\begin{equation}\label{B}
	\mathbf{B}_\beta(x\,,t\,;\varepsilon) := \left[ t-\beta\varepsilon^\alpha,t \right]\times
	[x-\varepsilon\gamma\,,x+\varepsilon\gamma].
\end{equation}
When $\varepsilon\approx0$, the preceding describes a very small box
in space-time; see the figure below.

\begin{figure}[h!]
 \begin{tikzpicture}[line/.style={>=latex}]
\fill [white] (14.5,-.26) coordinate (0,1) rectangle (21,1.4);
\fill [orange!30] (17.57,1) coordinate (0,1) rectangle(19.47,1.37);
   \draw [->] (17,-1.5) -- (17,2) node [right] {};
      \draw [white, thick] (17,-1.5) -- (17,-.5) node [right] {};
            \draw [white, thick] (17,-1.5) -- (17,-.5) node [right] {};
      \draw  [->](13,-.27) -- (21,-.27) node [above right] {};
            \draw  [->, white, very thick](13,-.27) -- (14.5,-.27) node [above right] {};
              \draw  [->, white, thick](13,-.27) -- (14.5,-.27) node [above right] {};
            \draw  [dashed](17,1.4) -- (19.5,1.4) node [right] {};
\draw  [dashed](17,1) -- (19.5,1) node [right] {};
             \draw [dashed] (17.55,-.2) -- (17.55,1.3) node [right] {};
               \draw [dashed] (18.55,-.2) -- (18.55,1) node [right] {};
        \draw [dashed] (19.5,-.2) -- (19.5,1.3) node [right] {};
\draw(16.6,1.4) node {$t$};
\draw(16.4,1) node {$t-\beta\varepsilon^\alpha$};
\draw(17.6,-.6) node {$x-\varepsilon\gamma$};
\draw(19.6,-.6) node {$x+\varepsilon\gamma$};
\draw(18.55,1.4) node {$\bullet$};
\draw[](18.55,1.8) node {$(t\,,x)$};
\draw[](18.55,1.19) node {\textcolor{darkgray}{${\mathbf B}_\beta(x\,,t\,;\varepsilon)$}};
\end{tikzpicture}
\caption{A depiction of the localization region}
\end{figure}

Let us recall the solution $Z$ to the linear equation (L-SHE), and
the [approximate] gradient operator $\nabla_\varepsilon$ from \eqref{eq:nabla}.
We may observe that
\begin{equation}\label{eq:Z:SI}
	(\nabla_\varepsilon Z_t)(x) = Z_t(x) - Z_t(x-\varepsilon)=
	\int_{(0,t)\times\R}
	(\nabla_\varepsilon p_{t-s})(y-x)\, \xi(\d s\,\d y).
\end{equation}
The following is the main technical computation of this section.

\begin{proposition}[Localization of the gradient]\label{pr:localize:Z}
	There exists a finite and positive constant $A$---depending only on the
	value of $\alpha$---such that
	\begin{equation}
		\E\left( \left| (\nabla_\varepsilon Z_t)(x) - \int_{\mathbf{B}_\beta(x,t;\varepsilon) }
		(\nabla_\varepsilon p_{t-s})(y-x)\, \xi(\d s\,\d y)\right|^2\right) \le
		A\varepsilon^{\alpha-1}\beta^{-1/2},
	\end{equation}
	simultaneously for all $x\in\R$, $t>0$,
	$\varepsilon\in(0\,,1)$ and $\beta>1$.
\end{proposition}

In the case that $\alpha=2$,
the preceding is similar to Lemma 3.6 of Hairer, Maas, and Weber \cite{HMW},
which has been used as a core of the solution theory of rough Burgers-like equations
\cite{H,Hairer,HM}. The novelty here is
the sharp description of the localization estimates in terms of the auxilliary parameter $\beta$.

\begin{proof}
	The Wiener isometry allows us to write
	\begin{equation}\begin{split}
		Q:=&\ \E\left(\left| (\nabla_\varepsilon Z_t)(x) - \int_{\mathbf{B}_\beta(x,t;\varepsilon) }
			(\nabla_\varepsilon p_{t-s})(y-x)\, \xi(\d s\,\d y)\right|^2\right)\\
		=&\ \int_{[(0,t)\times\R]\setminus \mathbf{B}_\beta(0,t;\varepsilon)}
			\left| (\nabla_\varepsilon p_{t-s})(y)
			\right|^2\, \d s\,\d y.
	\end{split}\end{equation}
	Let us  decompose $Q$ as
	\begin{equation}
		Q = Q_1 + Q_2,
	\end{equation}
	where
	\begin{equation}
		Q_1 := \int_0^{t-\beta\varepsilon^\alpha}\d s\int_{-\infty}^\infty
		\d y \left| \left(\nabla_\varepsilon p_{t-s}\right)(y)\right|^2,
	\end{equation}
	and
	\begin{equation}
		Q_2 := \int_{t-\beta\varepsilon^\alpha}^t\d s\int_{|y|>\varepsilon\gamma}
		\d y \left| \left(\nabla_\varepsilon p_{t-s}\right)(y)\right|^2.
	\end{equation}
	We estimate $Q_1$ and $Q_2$ in this order, since it is easier to bound $Q_1$.

	Because  $({\nabla_\varepsilon p_s})^{\widehat{}}\, (\chi) = \e^{-s|\chi|^\alpha}
	(1-\e^{-i\chi\varepsilon}),$ Plancherel's theorem implies that
	\begin{align}\notag
		Q_1 &= \frac{1}{\pi}\int_{\beta\varepsilon^\alpha}^t\d s
			\int_{-\infty}^\infty\d\chi\ \e^{-2s|\chi|^\alpha}
			\left[ 1-\cos(\chi\varepsilon)\right]
			\le\frac{\varepsilon^2}{\pi}\int_{\beta\varepsilon^\alpha}^\infty\d s
			\int_{-\infty}^\infty\d\chi\ \e^{-2s|\chi|^\alpha}\chi^2\\
		&=\frac{\varepsilon^2}{2\pi}
			\int_{-\infty}^\infty\e^{-2\beta\varepsilon^\alpha|\chi|^\alpha}
			|\chi|^{2-\alpha}\,\d\chi
			= \frac{\varepsilon^{\alpha-1}(2\beta)^{-(3-\alpha)/\alpha}}{\pi\alpha}
			\,\Gamma\left(\frac{3-\alpha}{\alpha}\right).
	\end{align}
	This is the desired bound for $Q_1$. The estimation of  $Q_2$ requires
	a little more effort.

	First, we write
	\begin{align}
		Q_2 &=\int_0^{\beta\varepsilon^\alpha} \d s\int_{|y|>\varepsilon\gamma}
			\d y \left| (\nabla_\varepsilon p_s)(y)\right|^2\\\notag
		&\le 2\int_0^{\beta\varepsilon^\alpha} \d s\int_{|y|>\varepsilon\gamma}
			\d y \left| p_s(y)\right|^2 +
			2\int_0^{\beta\varepsilon^\alpha} \d s\int_{|y|>\varepsilon\gamma}
			\d y \left| p_s(y-\varepsilon)\right|^2,
	\end{align}

	Next, let us recall that the inversion formula and symmetry together
	imply that
	$p_s(z) := \pi^{-1}\int_0^\infty\cos(z\chi)\exp\{-s\chi^\alpha\}\,\d\chi$,
	whence
	\begin{equation}\label{eq:p}
		\sup_{z\in\R}p_s(z) = p_s(0)
		=\frac{\Gamma(1/\alpha)}{\alpha}\, s^{-1/\alpha}.
	\end{equation}
	In particular, if $0\le \eta\le1$ then
	\begin{equation}
		\int_{|y|>\varepsilon\gamma} |p_s(y-\eta\varepsilon)|^2\,\d y
		\le\frac{\Gamma(1/\alpha)}{\alpha}\, s^{-1/\alpha}
		\cdot\P\{|X_s|>\varepsilon(\gamma-\eta)\}.
	\end{equation}
	Since $\gamma=1+\beta^{3/2}$ [see \eqref{eq:beta:gamma}],
	we apply the triangle inequality to see that
	\begin{equation}
		\sup_{0\le\eta\le1}
		\int_{|y|>\varepsilon\gamma} |p_s(y-\eta\varepsilon)|^2\,\d y
		\le\frac{\Gamma(1/\alpha)}{\alpha}\, s^{-1/\alpha}
		\cdot\P\{|X_s|>\varepsilon\beta^{3/2}\}.
	\end{equation}
	By scaling, $\P\{|X_s|>\varepsilon\beta^{3/2}\}=
	\P\{|X_1|>\varepsilon\beta^{3/2}s^{-1/\alpha}\}$;
	and a well-known bound
	 on the tail of stable distributions shows that
	 $\P\{|X_1|>\lambda\}\le\text{const}\cdot\lambda^{-\alpha}$
	 for all $\lambda>0$, and hence
	 \begin{equation}
	 	\sup_{0\le\eta\le1}
		\int_{|y|>\varepsilon\gamma} |p_s(y-\eta\varepsilon)|^2\,\d y
		\le\text{const}\cdot s^{(\alpha-1)/\alpha} \varepsilon^{-\alpha}\beta^{-3\alpha/2}.
	 \end{equation}
	 We integrate this quantity from $s=0$ to $s=\beta\varepsilon^\alpha$
	 in order to see that
	 \begin{equation}
	 	Q_2 \le\text{const}\cdot\varepsilon^{\alpha-1}\beta^{2-(1/\alpha)-(3\alpha/2)}.
	 \end{equation}

	 Finally, we can combine our bounds for $Q_1$ and $Q_2$
	 in order to deduce the inequality,
	 \begin{equation}
	 	Q \le \text{const}\cdot\varepsilon^{\alpha-1}\left[ \beta^{-(3-\alpha)/\alpha}
		+\beta^{2-(1/\alpha)-(3\alpha/2)}\right].
	 \end{equation}
	 Elementary analysis of the exponent of $\beta$ shows that
	 \begin{equation}
	 	\min\left( \frac{3-\alpha}{\alpha} \,, -2+\frac{1}{\alpha}+\frac{3\alpha}{2}\right)
		\ge\frac12,
	 \end{equation}
	 because $1<\alpha\le 2$. Since $\beta>1$, it follows that
	 $Q\le \text{const}\cdot \varepsilon^{\alpha-1}\beta^{-1/2}$,
	 as desired.
\end{proof}
Since Lemma \ref{(0,t)timesR} says that
$\E(|(\nabla_\varepsilon Z_t)(x)|^2)\approx \text{const}\cdot\varepsilon^{\alpha-1}$
when $\varepsilon\ll1$,
Proposition \ref{pr:localize:Z} shows that
\begin{equation}
	(\nabla_\varepsilon Z_t)(x) \approx \int_{\mathbf{B}_\beta(x\,,t\,;\varepsilon) }
	(\nabla_\varepsilon p_{t-s})(y-x)\, \xi(\d s\,\d y)
	\ \text{when $\beta\gg1$ and $\varepsilon\ll1$},
\end{equation}
to first order, where the approximation holds in $L^2(\P)$ and
has good uniformity in $(t\,,x\,,\varepsilon\,,\beta)$.
A quick glance at \eqref{eq:Z:SI} then reveals the meaning of
Proposition \ref{pr:localize:Z}: If $\beta\gg1$ and $\varepsilon\ll1$,
then most of the contribution to
the stochastic integral in \eqref{eq:Z:SI} comes from the region
$\mathbf{B}_\beta(x\,,t\,;\varepsilon) $. Because
$\mathbf{B}_\beta(x\,,t\,;\varepsilon) $ is a very small subset of $(0\,,t)\times\R$,
Proposition \ref{pr:localize:Z} is describing a \emph{strong
localization} property of the spatial gradient of $u$.
We will prove that the subsequent results of this paper are
consequences of this localization property.
Let us illustrate the potential usefulness of Proposition \ref{pr:localize:Z}
by showing how it immediately implies a strong localization result for the
spatial gradient of the solution
$u$ to the fully nonlinear heat equation (SHE).

\begin{corollary}\label{cor:localize:u}
	Choose and fix real numbers $T>0$ and $k\in[2\,,\infty)$. Then,
	there exists a finite constant $A$---depending only on $(\alpha\,,\beta\,,T,k)$---such
	that
	\begin{align}\notag
		&\E\left( \left| (\nabla_\varepsilon u_t)(x)
		 	- \int_{\mathbf{B}_\beta(x,t;\varepsilon) }(\nabla_\varepsilon p_{t-s})(y-x)
			\sigma(u_s(y))\, \xi(\d s\,\d y)\right|^k\right)\\
		&\hskip3in\le A\varepsilon^{(\alpha-1)k/2}\beta^{-k/4},
	\end{align}
	simultaneously for all $x\in\R$, $\varepsilon\in(0\,,1)$, $\beta>1$,
	and $t\in(0\,,T]$.
\end{corollary}

\begin{proof}

We begin by noting that
	\begin{equation}\begin{split}
		&(\nabla_\varepsilon u_t)(x)\\
		&=[\nabla_\varepsilon (p_{t-s}*u_0 )](x)+
			\int_{(0,t)\times\R}(\nabla_\varepsilon p_{t-s})(y-x)
			\sigma(u_s(y))\, \xi(\d s\,\d y).
	\end{split}\end{equation}

	Next we observe that  $| ( \nabla_\varepsilon p_{t-s}*u_0 )(x) |
		\le \text{const}\cdot\varepsilon$; see the discussion at the
		beginning of the proof of Corollary~\ref{cor:localize:u}.
	Therefore, because $\beta>1$, it suffices to prove that
	\begin{align}\notag
		&\E\left( \left| (\nabla_\varepsilon u_t)(x) -\left(\nabla_\varepsilon p_t*u_0
			\right)(x) - \int_{\mathbf{B}_\beta(x,t;\varepsilon) }(\nabla_\varepsilon p_{t-s})(y-x)
			\sigma(u_s(y))\, \xi(\d s\,\d y)\right|^k\right)\\
		&\hskip3in\le A\varepsilon^{(\alpha-1)k/2}\beta^{-k/2}.
	\end{align}
	Define
	\begin{equation}\begin{split}
		I_1 &:= \int_{(0,t)\times\R}(\nabla_\varepsilon p_{t-s})(y-x)
			\sigma(u_s(y))\, \xi(\d s\,\d y),\\
		I_2 &:= \int_{\mathbf{B}_\beta(x,t;\varepsilon) }(\nabla_\varepsilon p_{t-s})(y-x)
			\sigma(u_s(y))\, \xi(\d s\,\d y).
	\end{split}\end{equation}
	It remains to prove that
	\begin{equation}\label{goal:localize:u}
		\| I_1-I_2\|_k \le \text{const}\cdot \varepsilon^{(\alpha-1)/2}\beta^{-1/2}.
	\end{equation}

	An application of the Burkholder--Davis--Gundy inequality shows that
	\begin{equation}
		\|I_1-I_2\|_k^2 \le c\int_{[(0,t)\times\R]\setminus\mathbf{B}_\beta(x,t;\varepsilon) }
		\left| (\nabla_\varepsilon p_{t-s})(y-x)\right|^2
		\left\| \sigma(u_s(y))\right\|_k^2\, \d s\,\d y,
	\end{equation}
	uniformly for all $x\in\R$, $\beta>1$, $\varepsilon\in(0\,,1)$,
	and $t\in[0\,,T]$, where $c$ depends only on $k$ and $T$.
	[See Foondun and Khoshnevisan \cite{FK}
	for the details of this sort of argument.]
	Thanks to \eqref{eq:moments} and the Lipschitz continuity
	of the function $\sigma$,  $C_k:=\sup_{s\in[0,T]}\sup_{y\in\R}
	\E(|\sigma(u_s(y))|^k)$ is finite. Consequently,
	\begin{equation}
		\E\left(\left| I_1-I_2\right|^k\right)
		\le c^{k/2}C_k\mathcal{T}^{k/2},
	\end{equation}
	where
	\begin{equation}\begin{split}
		\mathcal{T} &:=
			\int_{[(0,t)\times\R]\setminus\mathbf{B}_\beta(x,t;\varepsilon) }
			\left| (\nabla_\varepsilon p_{t-s})(y-x)\right|^2\,\d s\,\d y\\
		&=\E\left(\left|(\nabla_\varepsilon Z_t)(x) - \int_{\mathbf{B}_\beta(x,t;\varepsilon) }
			(\nabla_\varepsilon p_{t-s})(y-x)\, \xi(\d s\,\d y)\right|^2\right).
	\end{split}\end{equation}
	Therefore, Proposition \ref{pr:localize:Z} implies \eqref{goal:localize:u},
	whence the corollary.
\end{proof}

\section{Proof of Theorem \ref{th:Differentiate}}
Our proof of Theorem \ref{th:Differentiate} requires only one
more technical result.

\begin{lemma}\label{lem:localize:u:1}
	Choose and fix $T>0$ and $k\in[2\,,\infty)$. Then there exists a finite
	constant $A$ such that
	\begin{equation}\begin{split}\label{not-predictable}
		&\E\left(\left| \int_{\mathbf{B}_\beta(x,t;\varepsilon) }
			(\nabla_\varepsilon p_{t-s})(y-x)\left[ \sigma(u_s(y))-\sigma(u_t(\tilde x))\right]
			\xi(\d s\,\d y)\right|^k\right) \\
		&\hskip2.7in\le A\varepsilon^{(\alpha-1)k}\beta^{3(\alpha-1)k/4},
	\end{split}\end{equation}
	simultaneously for all $x\in\R$,
	$\tilde x\in[x-\gamma\varepsilon\,, x+\gamma\varepsilon],$
	$t\in[0\,,T]$, $\varepsilon\in(0\,,1)$,
	and $\beta>1$.
\end{lemma}

\begin{proof}
	This lemma is based on an application of the Burkholder--Davis--Gundy
	inequality. However, a small technical problem crops up when we try to
	apply that inequality. Namely, that the quantity $\sigma(u_t(\tilde{x}))$
	that appears in \eqref{not-predictable} truly
	depends on the white noise by time $t$. At the same time,
	the stochastic integral of interest is computed over the set
	$\mathbf{B}_\beta(x,t;\varepsilon)\subset[0\,,t]\times\R$,
	and therefore also depends on the white noise by time $t$. In other words,
	\begin{equation}
		\int_{\mathbf{B}_\beta(x,t;\varepsilon)}\sigma(u_t(\tilde{x}))\,\xi(\d s\,\d y)
	\end{equation}
	is \emph{not} a Walsh integral of a predictable process [viewed as a function of
	$t$]; rather it is merely equal to
	\begin{equation}
		\sigma(u_t(\tilde{x}))\cdot
		\int_{\mathbf{B}_\beta(x,t;\varepsilon)}\xi(\d s\,\d y);
	\end{equation}
	that is, a product of two correlated quantities.
	Thus, we consider first a related quantity $Q_1^{k/2}$, where
	\begin{equation}
		Q_1 :=\left\| \int_{\mathbf{B}_\beta(x,t;\varepsilon) }
			(\nabla_\varepsilon p_{t-s})(y-x)\left[ \sigma(u_s(y))-
			\sigma(u_{t-\beta\varepsilon^\alpha}(\tilde x))\right]
			\xi(\d s\,\d y)\right\|_k^2.
	\end{equation}
	Since $u_{t-\beta\varepsilon^\alpha}(\tilde{x})$ is measurable with
	respect to the white noise of $[0\,,t-\beta\varepsilon^\alpha]\times\R$,
	it is independent of the white noise of $\mathbf{B}_\beta(x,t;\varepsilon)$.
	Therefore, we may apply the Burkholder--Davis--Gundy inequality---
	see Foondun and Khoshnevisan \cite{FK} for the details of this application---in
	order to see that
	\begin{equation}\begin{split}
		Q_1 &\le c\int_{\mathbf{B}_\beta(x,t;\varepsilon) }
			\left|(\nabla_\varepsilon p_{t-s})(y-x)\right|^2
			\left\|\sigma(u_s(y))-\sigma(u_{t-\beta\varepsilon^\alpha}(\tilde x))\right\|_k^2
			\d s\,\d y\\
		&\le c\lip^2_\sigma\int_{\mathbf{B}_\beta(x,t;\varepsilon) }
			\left|(\nabla_\varepsilon p_{t-s})(y-x)\right|^2
			\left\|u_s(y)-u_{t-\beta\varepsilon^\alpha}(\tilde x)\right\|_k^2
			\d s\,\d y,
	\end{split}\end{equation}
	where $c$ is some constant depending on $k$ and $T$.
	Let $\tilde x\in[x-\gamma\varepsilon\,,x+\gamma\varepsilon]$.
	Corollary \ref{cor:Modulus:x} and Proposition
	\ref{pro:Modulus:t} together imply the bound
	\begin{align}\notag
		Q_1&\le K\int_{\mathbf{B}_\beta(x,t;\varepsilon) }
			\left|(\nabla_\varepsilon p_{t-s})(y-x)\right|^2
			\left[|\tilde x-y|^{\alpha-1}+|t-\beta\varepsilon^\alpha-s|^{(\alpha-1)/\alpha}
			\right] \d s\,\d y\\
		&\le K\varepsilon^{\alpha-1}[2\gamma+\beta^{1/\alpha}]^{\alpha-1}
			\int_{\mathbf{B}_\beta(x,t;\varepsilon)}
			\left|(\nabla_\varepsilon p_{t-s})(z)\right|^2
			\d s\,\d z,
	\end{align}
	where $K$ depends only on $\alpha\,,k$ and $T$. In the preceding, the $\d z$-integral
	ranges over $z\in[x-\varepsilon\gamma\,,x+\varepsilon\gamma]$.
	If we replace $\mathbf{B}_\beta(x\,,t\,;\varepsilon) $ by $(0\,,t)\times\R$,
	then we obtain the bound
	\begin{equation}
		Q_1 \le K\varepsilon^{\alpha-1}[2\gamma+\beta^{1/\alpha}]^{\alpha-1}\E\left(
		\left| (\nabla_\varepsilon Z_t)(0)\right|^2\right)
		\le K'\varepsilon^{(\alpha-1)}[2\gamma+\beta^{1/\alpha}]^{\alpha-1},
	\end{equation}
	thanks to Lemma~\ref{(0,t)timesR}.

	Next we estimate the cost of estimating $u_t(\tilde{x})$
	by $u_{t-\beta\varepsilon^\alpha}(\tilde{x})$. Indeed,
	by the H\"older inequality,
	\begin{align}\notag
		Q_2&:=\left\|\left[\sigma(u_t(\tilde x))-
			\sigma(u_{t-\beta\varepsilon^\alpha}(\tilde x))\right]
			\int_{\mathbf{B}_\beta(x,t;\varepsilon) }(\nabla_\varepsilon p_{t-s})(y-x)
			\xi(\d s\,\d y)\right\|_k^2\\\notag
		&\leq\lip_\sigma^2\left\|u_t(\tilde x) -
			u_{t-\beta\varepsilon^\alpha}(\tilde x)\right\|_{2k}^2
			\left\|\int_{\mathbf{B}_\beta(x,t;\varepsilon) }(\nabla_\varepsilon p_{t-s})(y-x)
			\xi(\d s\,\d y)\right\|_{2k}^2\\\notag
		&\leq K\beta^{(\alpha-1)/\alpha}\varepsilon^{(\alpha-1)}
			\int_{\mathbf{B}_\beta(x,t;\varepsilon)}
			\left|(\nabla_\varepsilon p_{t-s})(z)\right|^2
			\d s\,\d z\\
		&\leq K\beta^{(\alpha-1)/\alpha}\varepsilon^{2(\alpha-1)}.
	\end{align}
	Since $\gamma = 1+\beta^{3/2}\le2\beta^{3/2}$
	and $\beta^{1/\alpha}\leq \beta^{3/2}$, we can conclude that
	\begin{align}
		\sqrt{Q_1}+\sqrt{Q_2}\leq c\,\varepsilon^{(\alpha-1)}\beta^{3(\alpha-1)/4}.
	\end{align}
	This and Minkowski's inequality together imply the lemma.
\end{proof}

Now we conclude our first main effort.

\begin{proof}[Proof of Theorem \ref{th:Differentiate}]
	We begin the proof by looking at the following quantity:
	\begin{equation}\label{eq:E(error^k)}
			\sup_{x\in\R}\sup_{t\in[0,T]}
			\| (\nabla_\varepsilon u_t)(x)
			-\mathfrak{A}_\alpha\sigma(u_t(\tilde x))(\nabla_\varepsilon F)(x)
			\|_k;
		\end{equation}
	where $\tilde x\in[x-\gamma\varepsilon\,,x+\gamma\varepsilon]$.
	Let us first split the preceding expectation in two parts as follows:
	\begin{eqnarray*}
		\| (\nabla_\varepsilon u_t)(x)
		-\mathfrak{A}_\alpha\sigma(u_t(x))(\nabla_\varepsilon F)(x)\|_k\le I_1+I_2;
	\end{eqnarray*}
	where
	\begin{equation}\label{eq:error}
		I_1 := \sup_{x\in\R}\sup_{t\in[0,T]}\left\| (\nabla_\varepsilon u_t)(x)
		-\sigma(u_t(\tilde x))(\nabla_\varepsilon Z_t)(x) \right\|_k;
	\end{equation}
	and
	\begin{equation}\label{I_2}
		I_2 :=\sup_{x\in\R}\sup_{t\in [0, T]}\|\sigma(u_t(\tilde x)) (\nabla_\varepsilon Z_t)(x)
		-\mathfrak{A}_\alpha\sigma(u_t(\tilde x))(\nabla_\varepsilon F)(x)\|_k.
	\end{equation}

	Consider first the quantity $I_1$. The cost of replacing
	$\sigma(u_t(\tilde x))$ by $\sigma(u_t(x))$ is controlled
	by the triangle inequality, which yields the bound
	\begin{equation}\label{holder}\begin{split}
		\|(\nabla_\varepsilon Z_t)(x)[\sigma(u_t(\tilde x))&-\sigma(u_t(x))]\|_k\\
		&\leq\|(\nabla_\varepsilon Z_t)(x)\|_{2k}\lip_\sigma\|u_t(\tilde x)-u_t(x)\|_{2k}\\
		&\leq A\varepsilon^{\alpha-1}.
	\end{split}\end{equation}
	[The last inequality follows from Lemma \ref{(0,t)timesR} and Corollary
	\ref{cor:Modulus:x}.]
	Lemma \ref{lem:localize:u:1} and Corollary \ref{cor:localize:u} together
	imply that
	\begin{equation}\label{b4opt}\begin{split}
		&\left\| (\nabla_\varepsilon u_t)(x)
		-\sigma(u_t(x))(\nabla_\varepsilon Z_t)(x) \right\|_k \\
		&\hskip1.7in\le c'\,\varepsilon^{\alpha-1}\beta^{3(\alpha-1)/4} +
		c'\,\varepsilon^{(\alpha-1)/2}\beta^{-1/4},
	\end{split}\end{equation}
	where $c'$ denotes a finite constant that does not depend on
	the values of $x\in\R$,
	$t\in[0\,,T]$, and $\varepsilon\in(0\,,1)$.
	Define $\beta := \varepsilon^{-2b}>1$. Since the
	left-hand side of \eqref{b4opt} does not depend on $\beta$,
	we can optimize the right-hand side over $b>0$,
	to find that the best bound in~\eqref{b4opt} is attained when
	\begin{equation}\label{b}
		b :=\frac{\alpha-1}{3\alpha-2}.
	\end{equation}
	This particular choice yields
	\begin{equation}\label{eq:error1}
		\left\| (\nabla_\varepsilon u_t)(x)
		-\sigma(u_t(x))(\nabla_\varepsilon Z_t)(x) \right\|_k
		\le c\,\varepsilon^{(\alpha-1+b)/2}.
	\end{equation}
	Because $(\alpha-1+b)/2<\alpha-1$, we deduce from
	\eqref{holder} and \eqref{eq:error1} the following bound for $I_1$:
	\begin{equation}\label{eq:error}
		I_1\le c\,\varepsilon^{(\alpha-1+b)/2}.
	\end{equation}

	In order to bound $I_2$, we appeal to Proposition \ref{pr:fBm} and write
	\begin{equation}
		Z_t(x) = \mathfrak{A}_\alpha
		F(x) + S(x)\qquad(x\in\R),
	\end{equation}
	where $F$ is fBm($(\alpha-1)/2$) and $S$ is a mean-zero Gaussian process
	with $C^\infty$ trajectories. In accord with \eqref{eq:E(dS2)}, there
	exists a finite constant $A$ such that $\|(\nabla_\varepsilon S)(x)\|_2\le A\varepsilon$
	simultaneously for all $x\in\R$ and $\varepsilon>0$. Because the variance of a mean-zero
	Gaussian random variable determines all of its moments, we can find for all $k\in[2\,,\infty)$
	a finite constant $A_k$ such that
	\begin{equation}
		\sup_{x\in\R}\E \left(  | (\nabla_\varepsilon S)(x)  |^k \right) \le A_k
		\varepsilon^k,
	\end{equation}
	for all $\varepsilon>0$.
	Equivalently,
	\begin{equation}
		\sup_{x\in\R}\E\left( \left| (\nabla_\varepsilon Z_t)(x)
		-\mathfrak{A}_\alpha(\nabla_\varepsilon F)(x)
		\right|^k\right) \le A_k \varepsilon^k,
	\end{equation}
	uniformly for all $\varepsilon>0$. Since $\sup_{t\in[0,T]}\sup_{x\in\R}
	\E(|\sigma(u_t(x))|^2 )<\infty$ for all $T>0$, the Cauchy--Schwarz inequality
	yields the following bound: For all $k\in[2\,,\infty)$ and $T>0$, there exists
	a finite constant $A'$ such that
	$I_2 \le A' \varepsilon$,
	uniformly for all $\varepsilon>0$.
	Because $(\alpha-1+b)/2<1$, we then deduce from \eqref{eq:error} that
	for all $\varepsilon\in(0\,,1)$,
	\begin{equation}
		I_1+I_2\leq  A'\,\varepsilon^{(\alpha-1+b)/2}\qquad
		(0<\varepsilon<1).
	\end{equation}
	Equivalently,
	\begin{equation}\label{eq:E(error^k)}
		\sup_{x\in\R}\sup_{t\in[0,T]}
		\E\left( \left| (\nabla_\varepsilon u_t)(x)
		-\mathfrak{A}_\alpha\sigma(u_t(\tilde x))(\nabla_\varepsilon F)(x)
		\right|^k\right) \le A'\,\varepsilon^{k(\alpha-1+b)/2},
	\end{equation}
	where $\tilde x\in[x-\gamma\varepsilon\,,x+\gamma\varepsilon]$,
	and the finite constant $A'$ does not depend on $\varepsilon$.

	We are now ready to complete the proof of Theorem \ref{th:Differentiate}.
	Set $\tilde x:=x$. Thanks to \eqref{eq:E(error^k)} and
	Chebyshev's inequality, for all $\zeta\in(0\,, b)$ and $\varepsilon\in(0\,,1)$,
	\begin{equation}
		\sup_{x\in\R}
		\P\left\{\left| \frac{(\nabla_\varepsilon u_t)(x)}{(\nabla_\varepsilon F)(x)}
		- \mathfrak{A}_\alpha\sigma(u_t(x))
		\right|>\frac{\varepsilon^{(\alpha-1+\zeta)/2}}{|(\nabla_\varepsilon F)(x)|}\right\}
		\le A'\varepsilon^{k(b-\zeta)/2}.
	\end{equation}
	The preceding makes sense because $(\nabla_\varepsilon F)(x)$ is a.s.\  non zero,
	as it is a centered Gaussian random variable. This concludes the proof because
	it is easy to see that $(\nabla_\varepsilon F)(x)$ has the same distribution as
	$F(\varepsilon)$, regardless of the value of $x\in\R$,
	and the latter random variable has the same
	law as $\varepsilon^{(\alpha-1)/2}$ times a  standard Gaussian random variable $\mathcal N$, whence
	\begin{equation}
		\lim_{\varepsilon\downarrow 0}
		\P\left\{\frac{\varepsilon^{(\alpha-1+\zeta)/2}}{|(\nabla_\varepsilon F)(x)|}
		>\lambda\right\} = \lim_{\varepsilon\downarrow 0}
		\P\left(|\mathcal N|<\frac{\varepsilon^{\zeta/2}}{\lambda}\right)
		= 0,
	\end{equation}
	uniformly in $x\in\R$ and for all $\lambda>0$.
\end{proof}

\section{Proof of Corollary \ref{cor:LIL}}
Fix $x\in \R$ and $t>0$. We shall prove the following
``strong approximation'' result:
Choose and fix $q\in(0\,,b)$, where $b$ is defined by~\eqref{b}. Then, with probability one,
\begin{equation}\label{eq:SA}
	\left| (\nabla_\varepsilon u_t)(x) - \mathfrak{A}_\alpha \sigma(u_t(x))(
	\nabla_\varepsilon F)(x)\right|= o\left( \varepsilon^{(\alpha-1+q)/2}\right),
\end{equation}
as $\varepsilon\downarrow 0$.

Corollary \eqref{cor:LIL}
is then a ready consequence of \eqref{eq:SA} and the law of the iterated logarithm for
fractional Brownian motion, which itself follows fairly readily
from Theorem 1.1 of \cite{QuallsWatanabe}.

Let us define a stochastic process $G$ via
\begin{align}
	G(\varepsilon)=u_t(x-\varepsilon)-\mathfrak{A}_\alpha\sigma(u_t(x))F(x-\varepsilon)
	\qquad(0\le \varepsilon\le x).
\end{align}
For all $x\ge \nu\ge \varepsilon\geq0$, we let $r:=\nu-\varepsilon$ and see that
\begin{align}
	|G(\varepsilon)-G(\nu)|&=
	\big|\nabla_{r}u_t(x-\varepsilon)-
	\mathfrak{A}_\alpha\sigma(u_t(x))\nabla_{r}F(x-\varepsilon)\big|.
\end{align}
Therefore, if we let $z:=x-\varepsilon$, then
\begin{equation}\begin{split}
	\|G(\varepsilon)-G(\nu)\|_k &=
		\big\|\nabla_{r}u_t(z)-
		\mathfrak{A}_\alpha\sigma(u_t(z+\varepsilon))\nabla_{r}F(z)\big\|_k\\
	&\leq A'r^{(\alpha-1+b)/2},
\end{split}\end{equation}
thanks to \eqref{eq:E(error^k)}. By the
Kolmogorov's continuity theorem, $G$ is locally a H\"older-continuous process
with any index $a\in(0\,,\frac12(\alpha-1)+b-k^{-1})$.
Recall from~\eqref{b} that $b>0$ for all $\alpha>1$, and that $k$ is arbitrary.
Therefore, by choosing $k$ sufficiently large, we find $q:= b-k^{-1}>0$.
Next we combine the law of the iterated logarithm for the
fBm $F$ with~\eqref{eq:SA} to finish the proof.\qed

\section{Proof of Corollary \ref{a.s.}}
Define, for all $s>0$,
\begin{equation}\label{X_s}
	X_s:=\int_0^s\left| \frac{(\nabla_\varepsilon u_t)(x)}{(\nabla_\varepsilon F)(x)}
	 - \mathfrak{A}_\alpha \sigma(u_t(x))\right|\d\varepsilon.
\end{equation}
Choose and fix $q\in(0\,,b)$, where $b$ is defined by~\eqref{b}.
Then, Minkowski's inequality and \eqref{eq:SA} together imply that,
with probability one,
\begin{equation}
	X_s =o\left(\int_0^s\frac{\varepsilon^{(\alpha-1+q)/2}}{
	|(\nabla_\varepsilon F)(x)|}
	\,\d\varepsilon\right)\qquad(s\downarrow0).
\end{equation}
Since $\varepsilon^{(\alpha-1)/2}\nabla_\varepsilon F$ has a
standard normal distribution, Minkowski's inequality
guarantees that for every $r\in(0\,,1)$ and $s>0$,
\begin{equation}
	\E(|X_s|^r)
	\leq \E(\mathcal{N}^{-r})\left|\int_0^s\varepsilon^{q/2}
	\,\d\varepsilon\right|^r
	= c s^{r(2+q)/2},
\end{equation}
where $c\in(0\,,\infty)$ depends only on $r$, and
$\mathcal{N}$ is a random variable with the standard normal distribution.
Therefore, by the Chebyshev inequality,
\begin{equation}
	\P\left\{|X_{2^{-n}}|>\lambda 2^{-n\tau}\right\}
	\leq c2^{-nr(1+(q/2)-\tau)}\lambda^{-r},
\end{equation}
for all $\lambda,\tau>0$ and integers $n\ge 0$.
If we let $\tau<1+(q/2)$, then we can see from the Borel-Cantelli lemma that
\begin{equation}\label{dyadicLim}
	 \lim_{n\to\infty}2^{n\tau}X_{2^{-n}}=0
	 \qquad\text{a.s.}
\end{equation}
If $s$ is a real number in $[2^{-n-1},\,2^{-n}]$
for some integer $n\ge 0$, then $s^{-\tau}X_s \le 2^{\tau(n+1)}X_{2^{-n}}$.
Therefore,
\eqref{dyadicLim} implies that $X_s=o(s^\tau)$ a.s.,
as was claimed.
\qed

\section{Proof of Corollary \ref{cor:CLT}}
Throughout this proof we write
\begin{equation}
	\beta := \beta(\varepsilon) := \log(1/\varepsilon),
\end{equation}
and
\begin{equation}
	\mathcal{I}_\varepsilon(x\,,t) := \int_{\mathbf{B}_\beta(x,t;\varepsilon)}
	(\nabla_\varepsilon p_{t-s})(y-x)\,\xi(\d s\,\d y),
\end{equation}
for the sake of notational simplicity.

The proof of Corollary \ref{cor:CLT} hinges on
the work that has been developed so far, together with a rather crude estimate
on the temporal modulus of continuity of $u$, namely~\ref{pro:Modulus:t}.
See also \cite[Theorem 2]{DebbiDozzi}.
We  need to know only that the following holds for every $k\in[2\,,\infty)$ and $T>0$ fixed:
Uniformly for all $\varepsilon\in(0\,,1)$,
\begin{equation}
	\left\| \left\{\sigma(u_t(x))-\sigma(u_{t- \varepsilon^\alpha}(x))\right\}\cdot
	\mathcal{I}_\varepsilon(x\,,t)\right\|_k
	\le \text{const}\cdot \varepsilon^{(\alpha-1)/2}
	\left\| \mathcal{I}_\varepsilon(x\,,t)\right\|_{2k},\label{that}
\end{equation}
thanks to the Cauchy--Schwarz inequality
and the already-mentioned fact that $\sup_{x\in\R}\sup_{t\in[0,T]}
\E(|\sigma(u_t(x))|^k)<\infty$. We apply Proposition \ref{pr:localize:Z} next in order
to see that
\begin{equation}\begin{split}
	\left\| \mathcal{I}_\varepsilon(x\,,t)\right\|_{2k}
		&\le\textnormal{const}\cdot\sqrt{\frac{\varepsilon^{\alpha-1}}{\log(1/\varepsilon)}}
		+ \left\|(\nabla_\varepsilon Z_t)(x)\right\|_k\\
	&\le \text{const}\cdot\varepsilon^{(\alpha-1)/2}.
	\label{that2}
\end{split}\end{equation}
In the last two inequalities, we have used Lemma \ref{(0,t)timesR} and
also the fact that the $L^2$-norm of a Gaussian random variable
determines all of its $L^k$-norms. This can be combined with \eqref{that},
Corollary \ref{cor:localize:u},
and Lemma \ref{lem:localize:u:1} in order to show that
\begin{equation}\label{eq:localize:u:2}
	\left\| (\nabla_\varepsilon u_t)(x) - \sigma(u_{t-\beta\varepsilon^\alpha}(x))
	\mathcal{I}_\varepsilon(x\,,t)\right\|_k
	\le \text{const}\cdot \frac{\varepsilon^{(\alpha-1)/2}}{\left|\log(1/\varepsilon)\right|^{1/4}},
\end{equation}
uniformly for $x\in\R$, $t\in[0\,,T]$, and $\varepsilon\in(0\,,1)$.
Consequently, we see that $\varepsilon^{-(\alpha-1)/2}(\nabla_\varepsilon u_t)(x)$
and $\varepsilon^{-(\alpha-1)/2}\sigma(u_{t-\beta\varepsilon^\alpha}(x))
\mathcal{I}_\varepsilon(x\,,t)$ have the same asymptotic
behavior.

Next let us observe that, because of the independence properties of space-time
white noise, $\sigma(u_{t-\beta\varepsilon^\alpha}(x))$ is independent of
$\varepsilon^{-(\alpha-1)/2}\mathcal{I}_\varepsilon(x\,,t)$, and converges to
$\sigma(u_t(x))$ a.s.\ as $\varepsilon\downarrow 0$, by continuity. [The requisite
continuity is assured by Proposition~\ref{pro:Modulus:t} and the Kolmogorov continuity theorem.]
Therefore, the asymptotic behavior of $\varepsilon^{-(\alpha-1)/2}
(\nabla_\varepsilon p_{t-s})(y-x)$ is the same as the asymptotic behavior of
$\sigma(\tilde{u}_t(x))\times \varepsilon^{-(\alpha-1)/2}\mathcal{I}_\varepsilon
(x\,,t)$, where $\tilde{u}$ is independent
of $\mathcal{I}_\varepsilon(x\,,t)$ and has the same law as $u$. Proposition
\ref{pr:localize:Z} assures that the asymptotic behavior of
$\varepsilon^{-(\alpha-1)/2}\mathcal{I}_\varepsilon(x\,,t)$ is
the same as that of $\varepsilon^{-(\alpha-1)/2}(\nabla_\varepsilon Z_t)(x)$.
According to Proposition \ref{pr:fBm},
\begin{equation}
	\varepsilon^{-(\alpha-1)/2}(\nabla_\varepsilon Z_t)(x)=\mathfrak{A}_\alpha
	\varepsilon^{-(\alpha-1)/2}(\nabla_\varepsilon F)(x) + O\left( \varepsilon^{(3-\alpha)/2}\right)
	\qquad\text{a.s.},
\end{equation}
and Corollary \ref{cor:CLT} follows from the defining property of fBm.
In this case, that property implies that $\varepsilon^{-(\alpha-1)/2}
(\nabla_\varepsilon F)(x)$ has a standard normal law.\qed

\section{Proof of Corollary \ref{cor:VAR}}
Throughout this proof, we use the short-hand notation:
\begin{equation}\label{epsbeta}
	\varepsilon := \varepsilon(n) := 2^{-n}
	\quad\text{and}\quad
	\beta :=\beta(n) := n^{16/(\alpha-1)}.
\end{equation}

We wish to prove that
\begin{equation}\begin{split}
	&\lim_{n\to\infty} \sum_{a\le j\varepsilon \le b}
		\phi( u_t (j \varepsilon)) \left| (\nabla_\varepsilon u_t)((j+1)\varepsilon)
		\right|^{2/(\alpha-1)}\\
	&\hskip1.7in
		=\mathfrak{B}_\alpha \int_a^b \phi(u_t(x))
		\left[ \sigma(u_t(x))\right]^{2/(\alpha-1)}\d x,
\end{split}\end{equation}
a.s.\ and in $L^2(\P)$. This is done via a series of 1-step reductions.
First, let us apply the Cauchy-Schwarz inequality to see that
\begin{equation}\begin{split}
	&\left\| \left| \phi( u_t (j \varepsilon)) - \phi(
		u_{t-\beta\varepsilon^\alpha}(j\varepsilon))\right|
		\times \left| (\nabla_\varepsilon u_t)((j+1)\varepsilon)
		\right|^{2/(\alpha-1)}\right\|_k\\
	&\le \sup_{x\in\R} \left\| \phi( u_t (x)) - \phi(
		u_{t-\beta\varepsilon^\alpha}(x)\right\|_{2k}\times
		\sup_{y\in\R}\left\| (\nabla_\varepsilon u_t)(y)
		\right\|_{4k/(\alpha-1)}^{2/(\alpha-1)}\\
	&\le\text{const}\cdot n^{16\delta/(\alpha-1)}2^{-n(1+\delta)},
\end{split}\end{equation}
thanks to \eqref{pro:Modulus:t} and Corollary \ref{cor:Modulus:x}.
Therefore, the Borel--Cantelli reduces our problem to verifying that,
almost surely and in $L^2(\P)$,
\begin{equation}\begin{split}
	&\lim_{n\to\infty} \sum_{a\le j\varepsilon \le b}
		\phi( u_{t-\beta\varepsilon^\alpha} (j \varepsilon)) \left| (\nabla_\varepsilon u_t)((j+1)\varepsilon)
		\right|^{2/(\alpha-1)}\\
	&\hskip1.7in
		=\mathfrak{B}_\alpha \int_a^b \phi(u_t(x))
		\left[ \sigma(u_t(x))\right]^{2/(\alpha-1)}\d x.
\end{split}\end{equation}
Similarly, we apply \eqref{eq:localize:u:2} in order to reduce the problem further
to one about showing that, almost surely and in $L^2(\P)$,
\begin{equation}\begin{split}
	&\lim_{n\to\infty} \sum_{a\le j\varepsilon \le b}
		\phi( u_{t-\beta\varepsilon^\alpha} (j \varepsilon))
		\left|\sigma(u_{t-\beta\varepsilon^\alpha} (j \varepsilon)) \right|^{2/(\alpha-1)}
		\left| \mathcal{I}_\varepsilon(j\varepsilon\,;t)\right|^{2/(\alpha-1)}\\
	&\hskip1.7in
		=\mathfrak{B}_\alpha \int_a^b \phi(u_t(x))
		\left[ \sigma(u_t(x))\right]^{2/(\alpha-1)}\d x.
\end{split}\end{equation}

Let us define the function
\begin{equation}
	\mathcal{H}(x) := \phi(x)\left|\sigma(x) \right|^{2/(\alpha-1)}
	\qquad(x\in\R),
\end{equation}
in order to simplify some of the typography. We emphasize once
again that:
\textbf{(i)} $\sup_{x\in\R}\sup_{t\in[0,T]}\E(|\mathcal{H}(u_t(x))|^k)<\infty$
for all $T>0$; and \textbf{(ii)} Our goal is to prove that,
a.s.\ and in $L^k(\P)$,
\begin{equation}\label{H:0}
	\lim_{n\to\infty} \sum_{a\le j\varepsilon \le b}
	\mathcal{H}( u_{t-\beta\varepsilon^\alpha} (j \varepsilon))
	\left| \mathcal{I}_\varepsilon(j\varepsilon\,;t)\right|^{2/(\alpha-1)}
	=\mathfrak{B}_\alpha \int_a^b \mathcal{H}(u_t(x))\,\d x.
\end{equation}
Because $\mathcal{I}_\varepsilon(j\varepsilon\,;t)$ is independent of
$\{u_{t-\beta\varepsilon^\alpha}(i\varepsilon)\}_{i\le j}$,
\textbf{(i)} implies that
\begin{align}
	&\text{Var}\sum_{a\le j\varepsilon \le b}
		\mathcal{H}( u_{t-\beta\varepsilon^\alpha} (j \varepsilon))
		\left| \mathcal{I}_\varepsilon(j\varepsilon\,;t)\right|^{2/(\alpha-1)}\\\notag
	&\le\text{const}\cdot\sum_{a\le j\varepsilon\le b}
		\text{Var}\left(\left| \mathcal{I}_\varepsilon(j\varepsilon\,;t)
		\right|^{2/(\alpha-1)}\right) \le\text{const}\cdot\sum_{a\le j\varepsilon\le b}
		\E\left(\left| \mathcal{I}_\varepsilon(j\varepsilon\,;t)
		\right|^{4/(\alpha-1)}\right),
\end{align}
uniformly for all $t\in[0\,,T]$ and $n\ge 1$ [whence also $\varepsilon\in(0\,,1)$
and $\beta>1$, as defined in \eqref{epsbeta}]. The
right-most quantity is $O(n^{-2})$; this is shown very much as \eqref{that2}
was, but the parameter $\beta$ has to now be adjusted.
In this way we reduce our problem \eqref{H:0}---
thanks to the Borel--Cantelli lemma---to one
about proving  that, a.s.\ and in $L^2(\P)$,
\begin{equation}\label{H:1}
	\lim_{n\to\infty} \sum_{a\le j\varepsilon \le b}
	\mathcal{H}( u_{t-\beta\varepsilon^\alpha} (j \varepsilon))
	\E\left(\left| \mathcal{I}_\varepsilon(j\varepsilon\,;t)\right|^{2/(\alpha-1)}\right)
	=\mathfrak{B}_\alpha \int_a^b \mathcal{H}(u_t(x))\,\d x.
\end{equation}
In accord with Proposition \ref{pr:localize:Z},
\begin{equation}
	\E\left( \left| (\nabla_\varepsilon Z_t)(x) - \mathcal{I}_\varepsilon(x\,;t)\right|^{2/
	(\alpha-1)}\right) \le\text{const}\cdot  n^{-2}2^{-n},
\end{equation}
uniformly for all $t\in[0\,,T]$ and $n\ge 1$.
Therefore, we may appeal to Corollary \ref{cor:Mean:Linear} in order to see
that
\begin{equation}
	\E\left(\left| \mathcal{I}_\varepsilon(j\varepsilon\,;t)\right|^{2/(\alpha-1)}\right)
	= \mathfrak{B}_\alpha 2^{-n}(1+o(1))
	\qquad(n\to\infty).
\end{equation}
This and the continuity of the random function $(t\,,x)\mapsto u_t(x)$ together
show that \eqref{H:1} holds, thanks to a Riemann--sum approximation.
This completes our proof of the corollary.

\section{Comments on variance stabilization}\label{sec:VS}

Corollary \ref{cor:CLT} implies that if $\varepsilon$ is small, then
$(\nabla_\varepsilon u_t)(x)$ behaves as $\varepsilon^{(\alpha-1)/2}$
times $\sigma(u_t(x))Z$ where $Z$ is an independent standard normal
random variable.
The aim of this section is to show how one
can apply a conditional form of what statisticians call a
``variance stabilizing transformation''  in order to simplify the preceding
into a bona fide central limit theorem with a Gaussian limit.

Let us first recall a classical definition that has also been recently adapted
to the potential theory of stochastic PDEs
\cite{DalangNualart}.

\begin{definition}
	A Borel set $E\subset\R$ is \emph{polar} for $u$ if
	\begin{equation}
		\P\left\{ u_t(x) \in E \text{ for some $t\ge 0$ and $x\in\R$}\right\}=0.
	\end{equation}
\end{definition}
Our definition of polar sets is slightly different from its progenitor in
probabilistic potential theory: Note that $t=0$ {\it is} included.
\begin{lemma}
	If $\sigma^{-1}\{0\}:=\{z\in\R:\ \sigma(z)=0\}$
	is polar for $u$, then the following is a continuous
	random field indexed by $\R_+\times\R$:
	\begin{equation}\label{X}
		X_t(x) := \int_{u_t(0)}^{u_t(x)}\frac{\d y}{\sigma(y)}
		\qquad(t\ge 0,\,x\in\R).
	\end{equation}
\end{lemma}

\begin{proof}
	Recall that $u$ is a continuous random function of $(t\,,x)\in\R_+\times\R$.
	Define $J(t\,,x)$ as the closed random subinterval of $\R$ whose endpoints are
	$u_t(0)$ and $u_t(x)$.
	Since $\sigma^{-1}\{0\}$ is polar for $u$, there is a $\P$-null set off which
	$J(t\,,x)\cap\sigma^{-1}\{0\}=\varnothing$ simultaneously for all $x\in\R$
	and $t\ge 0$; this holds because of the mean value theorem.  Therefore,
	the continuity of $\sigma$ ensures that, in addition, $\inf_{y\in J(t\,,x)}|\sigma(y)|>0$
	a.s. The rest of the proof is easy.
\end{proof}

Let $X$ denote the random field that was defined in \eqref{X}. If
$\sigma^{-1}\{0\}$ were polar for $u$, then $X$ would be well defined and
\begin{equation}
	X_t(x) - X_t(x-\varepsilon) = \frac{u_t(x)-u_t(x-\varepsilon)}{\sigma(u_t(x))}+
	O\left(\left| u_t(x)-u_t(x-\varepsilon)\right|^2\right),
\end{equation}
uniformly for every $x\in[a\,,b]$ and $t\in[0\,,T]$. This observation
and Theorem \ref{th:Differentiate} together yield the following.

\begin{theorem}\label{th:X:Differentiate}
	If $\sigma^{-1}\{0\}$ is polar for $u$,
	then for all $t>0$ and $x\in\R$: (i)
	\begin{equation}
		\lim_{\varepsilon\downarrow0}\sup_{y\in\R}\P\left\{\left|
		\frac{X_t(y)-X_t(y-\varepsilon)}{F(y)-F(y-\varepsilon)}-\mathfrak{A}_\alpha
		\right|>\lambda\right\}=0\qquad\text{for all $\lambda>0$};
	\end{equation}
	(ii)	With probability one,
	\begin{equation}
		\limsup_{\varepsilon\downarrow0}\frac{X_t(x)-X_t(x-\varepsilon)}{\sqrt{2
		\varepsilon^{\alpha-1}\log\log(1/\varepsilon)}}= - \liminf_{\varepsilon\downarrow0}
		\frac{X_t(x)-X_t(x-\varepsilon)}{\sqrt{2
		\varepsilon^{\alpha-1}\log\log(1/\varepsilon)}}=\mathfrak{A}_\alpha;
	\end{equation}
	(iii) For all $a\in\R$,
	\begin{equation}
		\lim_{\varepsilon\downarrow0}\P\left\{
		\frac{X_t(x)-X_t(x-\varepsilon)}{\varepsilon^{(\alpha-1)/2}}\le a\right\}
		=\P\left\{ \mathcal{N} \le a/\mathfrak{A}_\alpha\right\},
	\end{equation}
	where $\mathcal{N}$ denotes a standard Gaussian random variables; and
	(iv) For all Lipschitz-continuous functions $\varphi:\R\to\R$ and $b>a$ and $t>0$,
	all non random, we have
	\begin{equation}\begin{split}
		&\sum_{a2^n\le j\le 2^nb}\varphi(X_t(j2^{-n}))\left[
			X_t((j+1)2^{-n}) - X_t(j2^{-n})\right]^{2/(\alpha-1)}\\
		&\hskip2.5in=\mathfrak{B}_\alpha\int_a^b\varphi(X_t(x))\,\d x,
	\end{split}\end{equation}
	almost surely and in $L^2(\P)$.
\end{theorem}

Part  (iv) of the preceding requires a real-variable argument that we leave
to the interested reader. The rest are immediate corollaries of Theorem \ref{th:Differentiate}.

For an interesting example, let us consider the ``parabolic Anderson model
for the Laplacian, driven by space-time white noise,''
$$
	\frac{\partial}{\partial t} u_t(x) = \frac{\partial^2}{\partial x^2}u_t(x) + u_t(x)\xi.
	\eqno{(\text{PAM})}
$$
 That is, (PAM) is the specialization of
the stochastic heat equation (SHE) to the case that
$\alpha=2$ and $\sigma(x):= x$.
In this case, Mueller (see \cite{Mueller} for a related result,
and \cite[Theorem 5.1, p.\ 130]{Minicourse} for the one that is applicable here)
has proved that, if in addition $u_0(x)>0$ for all $x\in\R$, then
$\sigma^{-1}\{0\}=\{0\}$ is polar for $u$. The process $X$ is thus well defined,
and is given by
\begin{equation}
	X_t(x) = h_t(x) - h_t(0)\qquad(t\ge 0,\,x\in\R),
\end{equation}
where
\begin{equation}
	h_t(x) := \log u_t(x)\qquad(t\ge 0,\,x\in\R)
\end{equation}
is the so-called Hopf--Cole solution to the KPZ equation \cite{KPZ},
$$
	\frac{\partial}{\partial t} h_t(x) = \frac{\partial^2}{\partial x^2} h_t(x)
	+\left( \frac{\partial}{\partial x}h_t(x)\right)^2 + \xi.
	\eqno{(\text{KPZ})}
$$
[This is an ill-posed stochastic PDE.]
Since fBm($\nicefrac12$) is standard Brownian motion,
$\mathfrak{A}_2=1/\sqrt 2$, and $\mathfrak{B}_2=1/2$
[see \eqref{A(alpha)} and \eqref{B(alpha)}], we arrive at the following
ready consequence of Theorem \ref{th:X:Differentiate}.

\begin{corollary}\label{cor:KPZ}
	Consider the Hopf--Cole solution $h$ to \textnormal{(KPZ)},
	subject to an initial profile $h_0:\R\to\R$ that is non random,
	uniformly bounded from above,
	and Lipschitz continuous. Then, $h$
	satisfies the following for all $t>0$ and $x\in\R$:
	(i) There exists a Brownian motion $\{B(y)\}_{y\in\R}$ such that
		\begin{equation}
			\lim_{\varepsilon\downarrow0}\sup_{y\in\R}\P\left\{\left|
			\frac{h_t(y)-h_t(y-\varepsilon)}{B(y)-B(y-\varepsilon)}- \frac{1}{\sqrt 2}
			\right|>\lambda\right\}=0\qquad\text{for all $\lambda>0$};
		\end{equation}
	\begin{enumerate}
		\item[(ii)] With probability one,
		\begin{equation}
			\limsup_{\varepsilon\downarrow0}\frac{h_t(x)-h_t(x-\varepsilon)}{
			\sqrt{\varepsilon\log\log(1/\varepsilon)}}= - \liminf_{\varepsilon\downarrow0}
			\frac{h_t(x)-h_t(x-\varepsilon)}{\sqrt{
			\varepsilon\log\log(1/\varepsilon)}}=1;
		\end{equation}
		\item[(iii)] For all $a\in\R$,
		\begin{equation}
			\lim_{\varepsilon\downarrow0}\P\left\{
			\frac{h_t(x)-h_t(x-\varepsilon)}{\sqrt{\varepsilon/2}}\le a\right\}
			=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^a\exp(-x^2/2)\,\d x;
		\end{equation}
		\item[(iv)] For all Lipschitz-continuous functions $\varphi:\R\to\R$ and $b>a$ and $t>0$,
		all non random, the following holds almost surely and in $L^2(\P)$:
		\begin{align}
			&\sum_{a2^n\le j\le 2^nb}\varphi(h_t(j2^{-n}))\left[
				h_t((j+1)2^{-n}) - h_t(j2^{-n})\right]^2\\\notag
			&\hskip3in= \frac12\int_a^b\varphi(h_t(x))\,\d x.
		\end{align}
	\end{enumerate}
\end{corollary}
\begin{small}


text:
\section{Basic definitions and preliminaries}
\subsection{Chain recurrence classes and filtrating neighbourhoods }

The following notions and theorems are due to Conley \cite{Co} and they can be found in several other references (for example \cite{AN}).
\begin{itemize}
  \item We say that pair of sequences $\set{x_i}_{0\leq i\leq k}$ and  $\set{t_i}_{0\leq i\leq k-1}$, $k\geq 1$,  is an \emph{ $\varepsilon$-pseudo orbit from $x_0$ to $x_k$} for a flow $\phi$,
  if for every $0\leq i \leq k-1$ one has
  $$ t_i\geq 1 \mbox{ and }d(x_{i+1},\phi^{t_i}(x_i))<\varepsilon.$$

  \item A compact invariant set $\Lambda$ is called \emph{chain transitive} if    for any $\varepsilon > 0$, and for any $x, y \in\Lambda$
there is an $\varepsilon$-pseudo orbit from $x$ to $y$.
  \item We say that $x, y \in M$ are chain related if,  for every $\varepsilon>0$, there are $\varepsilon$-pseudo orbits form $x$ to $y$ and from $y$ to $x$. This is an equivalence relation.
\item  We say that $x\in M$ is \emph{chain recurrent} if  for every $\varepsilon>0$, there is a non trivial $\varepsilon$-pseudo orbit from $x$ to $x$. We call the set of chain recurrent points, the\emph{ Chain recurrent set}
and we note it  $\cR(f)$. The equivalent classes of this equivalence relation are called \emph{ chain recurrence classes}.
\end{itemize}

\begin{defi}
\begin{itemize} \item An \emph{attracting region} (also called \emph{trapping region} by some authors) is a compact set $U$ so that $\phi^t(U)$ is contained in the interior
of $U$ for every $t>0$. The maximal invariant set in an attracting region is called an \emph{attracting  set}.  A repelling region is an attracting region for $-X$, and the maximal invariant set is called a repeller.

\item A \emph{filtrating region} is the intersection of an attracting region with a repelling region.

\item Let $C$ be a chain recurrence class of $M$ for the flow $\phi$.
A \emph{filtrating neighbourhood } of $C$ is a (compact) neighbourhood which is a filtrating region.

\end{itemize}
\end{defi}

The following is a corollary of the fundamental theorem of dynamical systems  \cite{Co}.
\begin{coro}\cite{Co}
Let $X$  be a $C^1$-vector field on a compact manifold $M$. Every chain class $C$  of $X$ admits a  basis of filtrating neighbourhoods, that is, every neighbourhood
of $C$ contains a filtrating neighbourhood of $C$.
\end{coro}

\begin{defi}
Let $C$ be a chain recurrence class of $M$ for the vector field $X$.
We say that $C$  is \emph{robustly chain transitive} if there exist a filtrating neighbourhood $U$ of $C$, and $C^1$ neighbourhood of $X$ called $\mathcal{U}$ such that for every $Y\in\mathcal{U}$, the maximal invariant set for $Y$ ($C_Y$) in $U$ is a unique  chain class.
\end{defi}

\begin{defi}
Let $C$ be a robustly chain transitive class of $M$ for the vector field $X$.
We say that $C$  is robustly transitive if there is a $C^1$ neighbourhood of $X$ called $\mathcal{U}$ such that for every $Y\in\mathcal{U}$, there is an orbit for $Y$ which is dense in $C_Y$.
\end{defi}

\subsection{Linear cocycles}
Let  $\phi=\{\phi^t\}_{t\in\RR}$ be a topological flow on a compact metric space  $K$.

Consider as well:
\begin{itemize}
\item A $d$ dimensional  linear bundle $E$ over $K$ with $\pi:E\to K$
\item A continuous map $A_t:(x,t)\in K\times \RR \mapsto GL(E_x,E_{\phi^t(x)})$  that satisfies the following \emph{cocycle relation }:
for any $x\in K$  and $t,s\in\RR$ one has:  $$A_{t+s}(x)=A_t(\phi^s(x))A_s(x)$$

\end{itemize}
We define a \emph{linear cocycle over   $(K,\phi)$}   as  the associated morphism   $A^t\colon K\times \RR\to K$
defined by $$A^t(x, v) = (\phi^t(x), A_t(x)v)\,.$$

Note that $\cA=\{A^t\}_{t\in\RR}$ is a flow on the space $E$ which projects on $\phi^t$.
$$\begin{array}[c]{ccc}
E &\stackrel{A^t}{\longrightarrow}&E\\
\downarrow&&\downarrow\\
K&\stackrel{\phi^t}{\longrightarrow}&K
\end{array}$$

If $\Lambda\subset K$ is a $\phi$-invariant subset,  then $\pi^{-1}(\Lambda)\subset E$ is $\cA$-invariant, and  we  call
\emph{the restriction of $\cA$ to $\Lambda$}  the restriction of $\{A^t\}$ to $\pi^{-1}(\Lambda)$.

 \subsection{Hyperbolicity and dominated splitting on linear cocycles}

 \begin{defi}
 Let $\phi$ be a topological flow on a compact metric space $M$ and a $\phi$-invariant connected compact subset $\La$. We consider a vector bundle  $\pi\colon E\to \La$ and
   a linear cocycle $\cA$ over $(\La,X)$.

 We say that $\cA$ admits a \emph{dominated splitting over $\Lambda$} if
 \begin{itemize}\item There exists a splitting $E=E^1\oplus\dots\oplus E^k$ over $\Lambda$ into $k$ sub-bundles
 \item  The dimension of the sub-bundles is constant, i.e. $dim(E^i_x)=dim(E^i_y)$ for all $x,y\in\Lambda$ and $i\in \set{1\dots k}$,
     \item The splitting is invariant, i.e. $A^t(x)(E^i_x)=E^i_{\phi^t(x)}$ for all $i\in \{1\dots k\}$,

\item There exists a $t>0$ such that for every $x\in \Lambda$ and any pair of non vanishing vectors $u\in E^i_x$ and $v\in E^j_x$, $i<j$ one has
 \begin{equation}\label{e.dom}
 \frac{\norm{A^t(u)}}{\norm u}\leq \frac{1}{2}\frac{\norm{ A^t(v)}}{\norm v}
 \end{equation}

 We denote $E^1\oplus_{_\prec}\dots\oplus_{_\prec} E^k$, or $E^1\oplus_{{_\prec}_t}\dots\oplus_{{_\prec}_t} E^k$ if one wants to emphasis the role of $t$: in that case one says that
 the splitting is \emph{$t$-dominated}.
\end{itemize}
 \end{defi}

 A classical result (see for instance \cite[Appendix B]{BDV})  asserts that the bundles of a dominated splitting are always continuous. A given cocycle may admit several dominated
  splittings.  However, the dominated splitting is unique if one prescribes the dimensions $dim(E^i)$.

 One says that one of the bundle $E^i$ is \emph{(uniformly) contracting} (resp. \emph{expanding}) if there is $t>0$ so that for every
 $x\in\La$ and every non vanishing vector   $u\in E^i_x$ one has $\frac{\norm{A^t(u)}}{\norm u}<\frac 12$
 (resp. $\frac{\norm{A^t(u)}}{\norm u}<\frac 12$). In both cases one says that $E^i$ is \emph{hyperbolic}.

 Notice that if $E^j$ is contracting (resp. expanding) then the same holds for any $E^i$, $i<j$  (reps. $j<i$).

\begin{defi}
 We say that the linear cocycle  $\cA$ is \emph{hyperbolic over $\Lambda$} if
  there is a dominated splitting $E=E^s\oplus E^u$ over $\Lambda$ into $2$  hyperbolic sub-bundles so that $E^s$ is uniformly contracting and $E^u$ is
  uniformly expanding.

  One says that $E^s$ is the \emph{stable bundle}, and $E^u$ is the \emph{unstable bundle}.
 \end{defi}

 The existence of a dominated splitting or of a hyperbolic splitting is a robust property

 \subsection{Linear Poincar\'e flow}

 Let $X$ be a $C^1$ vector field on a compact manifold $M$.  We denote by $\phi^t$ the flow of $X$.

 \begin{defi} The \emph{normal bundle} of $X$ is the vector bundle $N_X $ over $M\setminus Sing(X)$ defined as follows: the fiber $N_X(x)$ of $x\in M\setminus Sing(X)$ is
 the quotient space of $T_xM$ by the  line $\RR.X(x)$.
 \end{defi}
 Note that, if $M$ is endowed with a Riemannian metric, then $N_X(x)$ is canonically identified with the orthogonal space of $X(x)$:
 $$N_X=\{(x,v)\in TM, v\perp X(x)\} $$

Consider $x\in M\setminus Sing(M)$ and $t\in \RR$.  Thus $D\phi^t(x):T_xM\to T_{\phi^t(x)}M$ is a linear automorphism mapping $X(x)$ onto $X(\phi^t(x))$. Therefore
$D\phi^t(x)$ passes to the quotient as an linear map $\psi^t(x)\colon N_X(x)\to N_X(\phi^t(x))$:

$$\begin{array}[c]{ccc}
T_xM&\stackrel{D\phi^t}{\longrightarrow}&T_{\phi^t(x)}M\\
\downarrow&&\downarrow\\
N_X(x)&\stackrel{\psi^t}{\longrightarrow}&N_X(\phi^t(x))
\end{array}$$
where the vertical arrows are the canonical projections of the tangent space to the normal space. We call this map as the \emph{linear Poincar\'e flow}.

\begin{defi}
 We say that a vector field $X$ is \emph{hyperbolic over $\Lambda$} if
  there is a dominated splitting $TM=E^s\oplus\RR .X(x)\oplus E^u$ over $\Lambda$ into $2$  hyperbolic sub-bundles so that $E^s$ is uniformly contracting and $E^u$ is
  uniformly expanding.

  One says that $E^s$ is the \emph{stable bundle}, and $E^u$ is the \emph{unstable bundle}.
 \end{defi}

 Note that if $X$ is non singular the linear Poincar\'e flow is a linear cocycle.

 Notice that the notion of dominated splitting for non-singular flows is sometimes better expressed in terms of the linear Poincar\'e flow: for instance,
 the linear Poincar\'e flow of a robustly transitive non singular vector field
 always admits a dominated splitting, when the flow by itself may not admit any dominated splitting.
 An example of a diffeomorphism with a robustly transitive set having dominated splitting into two bundles, that none of them is contracting or expanding is exhibited in \cite{BV}. The suspension of this diffeomorphism would not have a dominated splitting of the tangent space.

 \subsection{Extended linear Poincar\'e flow}

 We are dealing with singular flows and the linear Poincar\'e flow is not defined on the singularities of the vector field $X$.
 However we can extend the linear Poincar\'e flow  as defined in \cite{GLW}.

 This flow will be a linear co-cycle define on some linear bundle over $M$ (even over the singularities of $X$), that we define now.

 \begin{defi}Let $M$ be a  manifold of dimension $d$.
 \begin{itemize}
 \item We call \emph{the projective tangent bundle of $M$}, and denote by $\Pi_\PP\colon \PP M\to M$, the fiber bundle whose fiber $\PP_x$ is
 the projective space of the tangent space $T_xM$: in other word, a point $L_x\in \PP_x$ is a $1$-dimensional vector subspace of $T_xM$. There is a natural projection  $\pi:TM\to \PP M$ than takes a $1$-subspace of $TM$ to the corresponding element of $\PP M$

  \item We call \emph{normal bundle of $\PP M $} and we denote by $\Pi_\cN\colon \cN M\to \PP M$, the $d-1$-dimensional vector bundle over $\PP M$ whose fiber $\cN_{L}$ over
  $L\in \PP_xM$
  is the quotient space $T_x M/L$.

  If we endow $M$ with riemannian metric, then $\cN_L$  is identified with the orthogonal hyperplane of $L$ in $T_xM$.
 \end{itemize}
 \end{defi}

 Let $X$ be a $C^r$ vector field on a compact manifold  $M$, and $\phi^t$ its flow. The natural actions of the derivative of $\phi^t$ on $\PP M$ and $\cN M$ define
 $C^{r-1}$ flows on these manifolds.  More precisely, for any $t\in \RR$,

 \begin{itemize}
  \item We denote by $\phi_{\PP}^t\colon\PP M\to \PP M$ the flow defined by $$\phi_{\PP}^t(L_x)= D\phi^t(L_x)\in \PP_{\phi^t(x)}.$$

  \item We denote by $\psi_{\cN}^t\colon\cN \to \cN $ the $C^{r-1}$ flow  whose restriction to a fiber $\cN_L$, $L\in \PP_x$,
  is the linear map onto
  $\cN_{\phi^t_{\PP}(L)}$ defined as follows: $D\phi^t(x)$ is a linear map from $T_xM$ to $T_{\phi^t(x)}M$, which maps the line $\pi^{-1}(L)\subset T_xM$
  onto the line $\pi^{-1}(\phi^t_{\PP}(L))$.  Therefore it pass to the quotient in the announced linear map.
  $$\begin{array}[c]{ccc}
T_xM &\stackrel{D\phi^t}{\longrightarrow}&T_{\phi^t(x)}M\\
\downarrow&&\downarrow\\
\cN_L&\stackrel{\psi^t_{\cN}}{\longrightarrow}&\cN_{\phi^t_{\PP}(L)}
\end{array}$$

 \end{itemize}

 The one-parameter family   $\psi^t_\cN$ defines a flow on $\cN $,  which is a linear co-cycle over $\phi^t_\PP$.
 We call  $\psi^t_\cN$ the \emph{extended linear Poncar\'e flow}.
 We can summarize  by the following diagrams:

$$\begin{array}[c]{ccc}
\cN &\stackrel{\psi^t_{\cN}}{\longrightarrow}&\cN \\
\downarrow&&\downarrow\\
\PP M &\stackrel{\phi_{\PP}^t}{\longrightarrow}&\PP M\\
\downarrow&&\downarrow\\
M&\stackrel{\phi^t}{\longrightarrow}&M
\end{array}$$

\begin{rema} The extended linear Poincar\'e flow is really an extension of the linear Poincar\'e flow defined in the previous section; more precisely:

 Let $S_X\colon M\setminus Sing(X)\to \PP M$ be the section of the projective bundle defined as $S_X(x)$ is the line $\langle X(x)\rangle\in \PP_x$ generated by $X(x)$.
 Then $N_X(x)= \cN_{S_X(x)}$ and the linear automorphisms $\psi^t\colon N_X(x)\to N_X(\phi^t(x))$ and $\psi^t_\cN\colon \cN_{S_X(x)}\to \cN_{S_X(\phi^t(x))}$ are the same.
\end{rema}

\subsection{Strong stable, strong unstable and center spaces associated to a hyperbolic singularity.}

Let $X$ be a vector field and $\sigma\in Sing (X)$ be a hyperbolic singular point of $X$.
Let
 $\lambda^s_k\dots\lambda^s_2<\lambda^s_1<0<\lambda^u_1<\lambda^u_2\dots \lambda^u_l$ be the Lyapunov exponents of $\phi_t$ at $\sigma$ and let
 $E^ s_k\oplus_{_<}\cdots E^s_2\oplus_{_<}E^s_1\oplus_{_<}E^u_1\oplus_{_<}E^u_2\oplus_{_<}\cdots \oplus_{_<}E^s_l$ be the corresponding (finest) dominated
 splitting over $\sigma$.

 A subspace $F$ of $T_\sigma M$ is called a \emph{center subspace} if it is of one of the possible form below:
 \begin{itemize}
  \item Either $F=E^ s_i\oplus_{_<}\cdots E^s_2\oplus_{_<}E^s_1$
  \item Or $F=E^u_1\oplus_{_<}E^u_2\oplus_{_<}\cdots \oplus_{_<}E^u_j$
  \item Or else $F=E^ s_i\oplus_{_<}\oplus_{_<}E^s_1\oplus_{_<}E^u_1\oplus_{_<}\cdots \oplus_{_<}E^s_j$
 \end{itemize}

 A subspace  of $T_\sigma M$ is called a \emph{strong stable space}, and we denote it  $E^{ss}_{i}(\sigma)$,  is there in $i\in\{1,\dots, k\}$ such that:
 $$E^{ss}_{i}(\sigma)=E^ s_k\oplus_{_<}\cdots E^s_{j+1}\oplus_{_<}E^s_i$$

 A classical result from hyperbolic dynamics asserts that for any $i$ there is a unique infectively immersed manifold $W^{ss}_i(\sigma)$, called a
 \emph{strong stable manifold}
 tangent at $E^{ss}_i(\sigma)$ and invariant by the flow of $X$.

 We define analogously the \emph{strong unstable spaces} $E^{uu}_j(\sigma)$ and the \emph{strong unstable manifolds} $W^{uu}_j(\sigma)$ for $j=1,\dots ,l$.

We can also define the strong stable and unstable manifolds in an analogue way, for regular points $x$ in an invariant set $\La$.

config: arxiv


TITLE: How to calculate the sum $\sum_{i=1}^{\infty} \frac{6}{n(n+3)}$
QUESTION [0 upvotes]: This is a telescoping series whose sum is
$$\sum_{i=1}^{\infty} \frac{6}{n(n+3)} = \frac{11}{3}$$
I calculated it as $$\left(\frac{2}{1} - \frac{2}{1+3}\right) - \left(\frac{2}{\infty} - \frac{2}{\infty+3}\right) = 2 - \frac{2}{4} = \frac{3}{2}$$
However, the answer was
$$2\left(1+\frac{1}{2}+ \frac{1}{3}\right)= \frac{11}{3}$$
How was this calculated? It seems like they just added a few numbers out of the whole series.

REPLY [2 votes]: Generalizing Marty Cohen's answer, and assuming $0\le a<b$,
if
$$s(a,b)=\sum_{n=1}^{\infty} \dfrac1{(n+a)(n+b)},
$$
consider the partial sum
$$s(a, b, m)
=\sum_{n=1}^{m} \dfrac1{(n+a)(n+b)}, \text{where $m>b$}
$$
Then
$$s(a,b)
=\lim_{m \to \infty} s(a, b, m)
=\dfrac1{b-a} \sum_{n=a+1}^{b} \dfrac1{n}
$$
Proof.
If $m > b$ then
$$\begin{array}\\
s(a, b, m)
&=\sum_{n=1}^{m} \dfrac1{(n+a)(n+b)}\\
&=\sum_{n=1}^{m} \dfrac1{b-a}\left(\dfrac1{n+a}-\dfrac1{n+b}\right)\\
&=\dfrac1{b-a}\left(\sum_{n=1}^{m} \dfrac1{n+a}-\sum_{n=1}^{m} \dfrac1{n+b}\right)\\
&=\dfrac1{b-a}\left(\sum_{n=a+1}^{a+m} \dfrac1{n}-\sum_{n=b+1}^{b+m} \dfrac1{n}\right)\\
&=\dfrac1{b-a}\left(\sum_{n=a+1}^{b} \dfrac1{n}+\sum_{n=b+1}^{a+m} \dfrac1{n}-\sum_{n=b+1}^{a+m} \dfrac1{n}-\sum_{n=a+m+1}^{b+m} \dfrac1{n}\right)\\
&=\dfrac1{b-a}\left(\sum_{n=a+1}^{b} \dfrac1{n}-\sum_{n=a+m+1}^{b+m} \dfrac1{n}\right)\\
d(a, b, m)
&=s(a,b,m)-\dfrac1{b-a}\sum_{n=a+1}^{b} \dfrac1{n}\\
&=-\dfrac1{b-a}\sum_{n=a+m+1}^{b+m} \dfrac1{n}\\
|d(a,b, m)|
&=\dfrac1{b-a}\sum_{n=a+m+1}^{b+m} \dfrac1{n}\\
&\le\dfrac1{b-a}\sum_{n=a+m+1}^{b+m} \dfrac1{a+m+1}\\
&=\dfrac1{b-a} \cdot\dfrac{b-a}{a+m+1}\\
&=\dfrac1{a+m+1}\\
&{\to 0} \ \ \text{for $m\to +\infty$}\\
\end{array}
$$
Examples:
$$\color{red}{s(a,b)
=\sum_{n=1}^{\infty} \dfrac1{(n+a)(n+b)}
=\dfrac1{b-a} \sum_{n=a+1}^{b} \dfrac1{n}
}\\
s(0,3)=\sum_{n=1}^{\infty} \dfrac1{n(n+3)}=\dfrac13 \sum_{n=1}^{3} \dfrac1{n}=\frac{11}{18}\\
s(1,3)=\sum_{n=1}^{\infty} \dfrac1{(n+1)(n+3)}=\dfrac12 \sum_{n=2}^{3} \dfrac1{n}=\frac{5}{12}\\
$$


TITLE: Using the answer from one quadratic to solve another quadratic
QUESTION [0 upvotes]: a)Factorise the expression $2x^2-3x-35=(2x+7)(x-5)$
b)Use your answer to part a to solve the  equation $2(2x-1)^2-3(2x-1)-35$
How would I use my answer from a to solve b? I don't see  how it helps me solve it.

REPLY [1 votes]: The second expression is similar to first, just x has been replaced by 2x-1, so in your initial answer{ (2x+7)(x-5) },  replace x with 2x-1 and you should get the answer.


TITLE: If we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period/frequency
QUESTION [1 upvotes]: An Introduction to Information Theory: Symbols, Signals and Noise, by John R. Pierce, says the following:

With the very surprising property of linearity in mind, let us return to the transmission of signals over electrical circuits. We have noted that the output signal corresponding to most input signals has a different shape or variation with time from the input signal. Figures II-1 and II-2 illustrate this. However, it can be shown mathematically (but not here) that, if we use a sinusoidal signal, such as that of Figure II-4, as an input signal to a linear transmission path, we always get out a sine wave of the same period, or frequency. The amplitude of the output sine wave may be less than that of the input sine wave; we call this attenuation of the sinusoidal signal. The output sine wave, may rise to a peak later than the input sine wave; we call this phase shift, or delay of the sinusoidal signal.

I'm trying to find the aforementioned proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency.
During my research, the closest thing to this that I have come across is slide 30 of this presentation:

I would greatly appreciate it if people could please take the time to either prove this or redirect me somewhere that has the proof.

REPLY [2 votes]: $y=G(x)$ is translation invariant, $G\Bigl(T_sx\Bigr)(t)=(T_sy)(t)=y(t+s)$ Together with the linearity this has the consequence that also differential operators are preserved,
$$\dot y(t)=\lim_{s\to 0}\frac{(T_sy)(t)-y(t)}s=\lim_{s\to 0}\frac{G\bigl(T_sx\bigr)(t)-G\bigl(x\bigr)(t)}{s}=G\left(\lim_{s\to 0}\frac{T_sx-x}s\right)(t)=G\bigl(\dot x\bigr)(t).$$
Now you can also apply this to the oscillator equation, $G(\ddot x+\omega^2x)=\ddot y+ω^2y$ and if $x$ is sinusoid with frequency $ω$, then so is $y$.
With $$G(\cos(ω\,\cdot\,))(t)=a\cos(ωt)+b\sin(ωt)$$ you also get the shifted
$$
G(\sin(ω\,\cdot\,))=G(\cos(ω\,\cdot\,-\tfrac\pi2))(t)=a\sin(ωt)-b\cos(ωt)
$$
so that indeed there are only two free parameters per frequency. To get the attenuation and phase, you only need to compute the polar coordinates $(A,\varphi)$ of the point $(a,-b)$.


TITLE: Why does the polarization type ($s$ or $p$) of electromagnetic waves effect the reflectivity and reflectivity of the light differently
QUESTION [1 upvotes]: So the Fresnel equations can describe the reflectivity and transmission of $s$ and $p$ polarized light and show that there is a different relationship between angle of incidence and the power reflection coefficient for each type of polarization (shown in the the graph).
What i am wondering is why that is, Why is it that parallel polarization and perpendicular polarization react differently when reflecting off a medium?

REPLY [0 votes]: Well, I encourage you to work through the derivation of the Fresnel Equations. What you’ll find is that the difference between $s$ and $p$ polarizations comes from the boundary conditions. Due to Gauss’ Law, the parallel component of the electric field is continuous across the boundary, while the normal component is discontinuous. The $p$ polarization has a component of field normal to the interface, while $s$ does not.


TITLE: Is $x = -x$ a contradiction?
QUESTION [4 upvotes]: So I was doing this problem. And I got to what I thought was a weird result. can anyone explain to me why this makes sense?
$$\begin{align}
5e^x &= 5e^{-x} \\
e^x &= e^{-x} \\
\ln (e^x) &= \ln (e^{-x}) \\
x &= -x
\end{align}$$
This last step is what I'm wondering about. Although this is algebraically correct isn't it impossible for a number to be equal to its opposite? Don't both sides have to be equal? Note that I am not asking for help on how to solve it from here...

REPLY [5 votes]: But why so difficult: $5e^x=5e^{-x} \Leftrightarrow e^x=e^{-x} \Leftrightarrow e^{2x}=1 \Leftrightarrow 2x=0 \Leftrightarrow x=0$


TITLE: Question about closure.
QUESTION [2 upvotes]: This seems like a simple question but for some reason I cannot wrap my head around it. Let $V$ be a fixed subset of $\textbf{R}^n$ and let $A$ be a subset of $V$.
I am trying to show that $V\cap \overline{V\setminus A}=V\cap\overline{\mathbf{R}^n\setminus A}$. It is clear that $V\cap \overline{V\setminus A}\subseteq V\cap\overline{\mathbf{R}^n\setminus A}$, but I am having trouble with the other direction. Specifically, if $x\in V\cap\overline{\mathbf{R}^n\setminus A}$, then $x$ is a limit point of $\mathbf{R}^n\setminus A$. How can I show that it is a limit point of $V\setminus A$?

REPLY [1 votes]: The statement is not true. Take $A=V$ for example. Or take $V=[0,2]$ and $A=[1,2]$.


TITLE: Concrete example that projective representation of symmetry group occurs in a quantum system except the case of spin half integer?
QUESTION [7 upvotes]: I for the first time learnt there exist projective representation of symmetry group in Weinberg's QFT vol1. In Weinberg's textbook, the definition of symmetry of a system is that a symmetry is a bijection of Hilbert space such that for any normalized states $|\Psi\rangle$, $|\Phi\rangle$ and the mapped states $|\Psi'\rangle$, $|\Phi'\rangle$
$$|\langle \Psi|\Phi\rangle| = |\langle \Psi'|\Phi'\rangle|.$$
With this definition, we see symmetry group of the system admits a projective representation:
$$U(T_2)U(T_1)=e^{i\phi(T_2,T_1)}U(T_2T_1).$$
For example, spin-$1/2$ is a projective representation of $SO(3)$ group. I want to consider whether there are other symmetry except internal symmetry which can occur projective representation in a system, but I can't construct one.
In quantum mechanics, the definition of a symmetry $S$ is a operator which commutes with Hamitonian $H$, i.e
$$[S,H]=0.$$
Then we see in the eigenenergy $E_i$ subspace,
$$H\psi^i_\mu=E_i \psi^i_\mu$$
with $\mu=1,2,\cdots,f_i$. That is the eigenspace of energy $E_i$ with dimension $f_i$.
$$HS\psi^i_\mu=SH\psi^i_\mu=E_i S\psi^i_\mu.$$
So $S\psi^i_\mu$ still have the same energy $E_i$ and it must can be expanded by $\psi^i_\nu$ with $\nu=1,2,\cdots,f_i$. Define the expansion coefficients as $D^i_{\nu\mu}(S)$,
$$S\psi^i_\mu= \sum_\nu D^i_{\nu\mu}(S)\psi^i_\nu$$
For any two symmetries $R,S$ with $[R,H]=0=[S,H]$, then $RS\equiv Q$ as a whole must be a symmetry because $[RS,H]=[R,H]S+R[S,H]=0$.
On the one hand,
$$RS\psi^i_\nu=R\sum_\mu D^i_{\mu\nu}(S)\psi^i_\mu =\sum_{\mu\gamma}D^i_{\mu\nu}(S) D^i_{\gamma \mu}(R)\psi^i_\gamma= \sum_{\mu\gamma} D^i_{\gamma \mu}(R)D^i_{\mu\nu}(S)\psi^i_\gamma$$
On the other hand,
$$RS\psi^i_\nu=Q \psi^i_\nu=\sum_\gamma D^{i}_{\gamma\nu}(Q)\psi^i_\gamma==\sum_\gamma D^{i}_{\gamma\nu}(RS)\psi^i_\gamma. $$
So $$D^i(RS)=D^i(R)D^i(S).$$
That is we can only get the representation of symmetry group other than the projective representation.
My question:

Does it mean only representation of symmetry group can occur in quantum system? According to above derivation, if it's true, what's the meaning to discuss the projective representation?
If the answer of 1st question is No. What's the loophole in my arguement. And except the case of spin half integer under rotation (or Lorentz group, Poincare group), give me a concrete example  projective representation of symmetry group can occur?

REPLY [4 votes]: The confusion here arises from two different notions of "symmetry":

Symmetry in the sense of Wigner's theorem: This is Weinberg's notion of symmetry - a ray transformation that leaves all inner products invariant. Wigner's theorem says that such ray transformations have (anti-)unitary representants determined up to a phase, and when dealing with a whole group of such transformation, this ambiguity in phase is essentially where the notion of projective representations arises from. For a comprehensive review, see also my Q&A on projective representations, central extensions and universal covers.
Dynamical symmetry: Physically, the first notion of "symmetry" is too weak: If the (anti-)unitary operators corresponding to the ray transformations do not commute with the Hamiltonian, then applying the symmetry before time evolution yields different results from applying  it afterwards. In particular, the expectation values of the "symmetry" operators are not conserved in time, something we want from symmetries. So a dynamical symmetry - what one usually simply calls a symmetry of the system - is an operator (or a set of operators) that commutes with the Hamiltonian.

Now, if you start with the notion of "a symmetry group" as a group of operators on a Hilbert space that commute with the Hamiltonian, then you are correct that there are no projective representations - a group of linear operators is clearly linearly represented on the space the operators are acting on.
The projective representations arise if we know that abstractly, there should be a representation of an abstract symmetry group $G$ (like the rotation group, or the Lorentz group, or any other symmetry group the classical system we are quantizing has) on the space of states. Then we know that:

To each $g\in G$ there is a ray transformation $T(g)$ that is a symmetry in the sense of Wigner.
For each $T(g)$, the (anti-)unitary representatives commute with the Hamiltonian. If this holds for one representative, it holds for all since phases commute with everything. A projective representation of $G$ is a consistent choice of these representatives.


TITLE: $\lim_{n\to \infty}\frac{(3^4)^{n+1}((n+1)!)^4}{(4(n+1))!}\cdot\frac{(4n)!}{(3^4)^n(n!)^4}$
QUESTION [1 upvotes]: EDIT: This really is just a question about factorials, see @Angelo's answer
If it helps, I was originally trying to prove the convergence of $$\sum_{n=1}^{\infty}\frac{(3^4)^n(n!)^4}{(4n)!}$$ by using the ratio test...
I have to find:$$\lim_{n\to \infty}\frac{(3^4)^{n+1}((n+1)!)^4}{(4(n+1))!}\cdot\frac{(4n)!}{(3^4)^n(n!)^4}$$I'm pretty sure the issue comes down to how I simplified it:
$$\lim_{n\to \infty}\frac{(3^4)((n+1)!)^4}{(4(n+1))!}\cdot\frac{(4n)!}{(n!)^4}$$
$$\lim_{n\to \infty}\frac{(3^4)((n+1)n!)^4}{(4n+4)!}\cdot\frac{(4n)!}{(n!)^4}$$
$$\lim_{n\to \infty}\frac{(3^4)(n+1)^4}{(4n+4)(4n)!}\cdot(4n)!$$
$$\lim_{n\to \infty}\frac{3^4}{4}\cdot(n+1)^3$$=$\infty$, so the series diverges.
But according to my homework, that's incorrect.
Any help is appreciated

REPLY [3 votes]: Actually
$$\lim_{n\to \infty}\frac{(3^4)\big((n+1)n!\big)^4}{(4n+4)!}\cdot\frac{(4n)!}{(n!)^4}=$$
$$=\lim_{n\to \infty}\frac{(3^4)(n+1)^4}{(4n+4)(4n+3)(4n+2)(4n+1)(4n)!}\cdot(4n)!=$$
$$=\lim_{n\to \infty}\frac{3^4(n+1)^4}{4^4(n+1)\left(n+\frac34\right)\left(n+\frac12\right)\left(n+\frac14\right)}=\left(\frac34\right)^4<1\;.$$
So the series converges.

REPLY [1 votes]: You made a mistake in your calculation. It should be:$$(4n+4)!=(4n+4)(4n+3)(4n+2)(4n+1)(4n!)$$


TITLE: How is it possible a quadratic Hermitian Hamiltonian $H = b^\dagger b^\dagger + b b$, with $b$ boson, cannot be diagonalized?
QUESTION [5 upvotes]: How can a Hermitian Hamiltonian $H = b^\dagger b^\dagger + b b$, with $b$ boson, cannot be diagonlized?
Given a Hamiltonian
$$\hat H = b^\dagger b^\dagger + b b \tag{1}$$
with $b, b^\dagger$ boson operators and $[b, b^\dagger] = 1$, this Hamiltonian is obviously Hermitian ${\hat H}^\dagger = {\hat H}$.
Let's compute the eigenvalue of this Hamiltonian, we need to do Bogoliubov transformation,
$$b = u a + v a^\dagger\tag{2}$$
$$b^\dagger = u^* a^\dagger + v^* a$$
The commutation relation requires:
$$[b,b^\dagger] = (|u|^2 - |v|^2) [a, a^\dagger]$$
that is
$$|u|^2 - |v|^2 = 1 \tag{3}$$
Take (2) into (1):
$$H = (u^* a^\dagger + v^* a)(u^* a^\dagger + v^* a）+ (u a + v a^\dagger)(u a + v a^\dagger)$$
$$H = ((u^*)^2 + v^2) a^\dagger a^\dagger + ((u)^2 + (v^*)^2) a a + (v^* u^* + v u) (a a^\dagger + a^\dagger a)$$
The diagonalizing of $H$ requires
$$(u^*)^2 + v^2 = 0 \tag{4}$$
$(3)\implies$
$$u = \cosh r e^{i \theta_1}$$
$$v = \sinh r e^{i \theta_2} \tag{5}$$
Take $(5)$ into $(4)$, we get
$$\cosh^2 r e^{-2 i \theta_1} + \sinh^2 r e^{ 2 i \theta_2 } = 0\tag{6}$$
i.e.
$$\cosh^2 r \cos 2\theta_1 + \sinh^2 r \cos 2\theta_2 = 0$$
$$\cosh^2 r \sin 2\theta_1 - \sinh^2 r \sin 2\theta_2 = 0$$
We can prove there is no real solutions of above equations.
How is it possible a Hermitian Hamiltonian (especially a quadratic Hamiltonian) cannot be diagonalized?
My questions:

Is there any loophole in above argument? I can't believe it is possible. If $H$ can be diagonalized, how to get the eigenvalue and eigenvector of this specific Hamiltonian?
If it's really true that $H$ cannot be diagonalized, please give me the reason why it contradicts the common sense like Hermitian matrix/operator can be diagonalized with real eigenvalue, quadratic Hamiltonian can be exactly solvable, etc.

REPLY [10 votes]: Essentially,  you are trying to prove that your initial Hamiltonian can be rewritten as
$$H = r a^\dagger a + sI\tag{1}$$
where $r$ and $s$ are reals and $a^\dagger$ and $a$ satisfy bosonic ccrs and there is a vacuum vector for $a$.
This is a much stronger condition than diagonalizability!
However, it is possible to prove that the requirement above cannot be satisfied.  In fact, it immediately arises from (1)  that $H$  is either bounded below or above depending on the sign of $r$, because
$$\langle \psi| H \psi \rangle = r \langle a \psi| a \psi \rangle + s \langle \psi |\psi \rangle = r ||a\psi||^2 + s||\psi||^2$$
Instead
(with some technical  hypothesis on the domain and some other technicality as  Stone-von Neumann's theorem) one can prove that
 $b^\dagger b^\dagger + bb$ is not bounded below nor above. Therefore operators $a$ and $a^\dagger$ cannot exist.
To prove that the initial Hamiltonian cannot be bounded, observe that defining $X = b + b^\dagger$ and $P = i(b-b^\dagger)$, these operators satisfy standard ccr up to a real factor and $H$ is proportional to $P^2-X^2$ which is unbouded below and above (here technicalities should be used). In fact,  this is the Hamiltonian of a repulsive oscillator so it is unbouded below, but a unitary transformation which therefore preserves the spectrum swaps $X$ and $P$ changing the sign of $H$, hence $H$ is also unbounded above.
Finally, all that does not mean that $H$ is not "diagonalizable". $H$ is at least Hermitian, presumably selfadjoint on some domain (it depends on details) thus it admits a PVM over the reals. However,  the spectrum is continuos as it happens for the repulsive oscillator, instead of discrete.


TITLE: Gaussian integration by parts, Wick's / Isserlis' theorem?
QUESTION [3 upvotes]: Wikipedia's entry on Isserlis' theorem has the following identity:
$$ \mathbb{E}\Big[ X_1 f(X_1, ..., X_n) \Big] = \sum_{i=1}^n \mathbb{E}[X_1 X_i] \mathbb{E}\Big[ \frac{\partial}{\partial X_i} f(X_1, ..., X_n) \Big],$$
where $X_1, ..., X_n $ is a zero-mean multivariate Gaussian random vector. My questions:

What is this theorem called? Is this some incarnation of Wick's theorem? (I do not understand the notation on this page at all).
What are the conditions required  on $f$ for this identity to hold?

REPLY [5 votes]: Happily, I found the answer to 2. myself by proving the result.
Proof:
First, we prove the statement when $X_1, ..., X_n$ are independent with standard deviations $1$. Stein's lemma says
$$ \mathbb{E}[X_1 g( X_1)] =  \mathbb{E}[g'( X_1)],$$
for all $g$ such that $\mathbb{E}| X_1 g( X_1)| < \infty $ and $\mathbb{E}| g'( X_1)| < \infty $. Now for all $i$
\begin{align}
\mathbb{E}[X_i f(X_1, ..., X_n)]  &= \mathbb{E} \Big[ \mathbb{E}\big[ X_i f(X_1, ..., X_n) \mid X_2, ...X_n\big] \Big]  \\
& = \mathbb{E} \Big[ (\partial/\partial X_i) f(X_1, ..., X_n)  \Big],
\end{align}
or in vector notation,
$$ \text{Cov}[\mathbf{X}, f(\mathbf{X}) ] = \mathbb{E}[\nabla f(\mathbf{X}) ].$$
Now apply an affine transform to $\mathbf{X}$, $\mathbf{Z} = \Sigma^{(1/2)}\mathbf{X} + \boldsymbol{\mu}$. Also "absorb" the affine transform into $f$, $f(\mathbf{X})=h(\Sigma^{(1/2)} \mathbf{X} + \boldsymbol{\mu})$ for some $h$. We have
\begin{align}
\text{Cov} [\mathbf{Z}, h(\mathbf{Z}) ] &= \text{Cov} [\Sigma^{(1/2)}\mathbf{X}+ \boldsymbol{\mu}, f(\mathbf{X})]\\
&=\text{Cov} [\Sigma^{(1/2)}\mathbf{X}, f(\mathbf{X})] \\
&=\Sigma^{(1/2)}\text{Cov} [\mathbf{X}, f(\mathbf{X})] \\
&=\Sigma^{(1/2)}\mathbb{E}[\nabla f(\mathbf{X})] \\
&=\Sigma\,\mathbb{E}[\nabla h(\mathbf{Z})].
\end{align}
In particular, we may extract the first entry of the vector, yelding
$$\text{Cov} [Z_1 h(\mathbf{Z})] = \sum_{i=1}^n \mathbb{E}[Z_i Z_1] \mathbb{E} [ (\partial/\partial Z_i) h(\mathbf{Z})].$$
Answer to 2. In particular, in applying Stein's lemma we used that $h$ and its first partial derivatives have bounded absolute first moment.