text:
\begin{document}
\title{Gaussian Limit for Determinantal Random Point Fields
\footnote{ AMS 2000 subject classification: 60G55 (primary), 60F05 (secondary);
keywords and phrases: determinantal random point fields, Central Limit Theorem,
self-similar random processes}}
\author{Alexander Soshnikov\\
University of
California, Davis
\\Department of Mathematics\\Davis, CA  95616, USA\\
soshniko@math.ucdavis.edu}
\date{}
\maketitle
\begin{abstract}
We prove that, under fairly general conditions, a properly
rescaled determinantal random point field converges to
a generalized Gaussian random process.
\end{abstract}

\section{Introduction and Formulation of Results}

Let $E$ be a locally compact Hausdorff space satisfying the
second axiom of countability, $B-\sigma$-algebra of Borel
subsets and $\mu \ \ \ $a $\sigma$-finite measure on $(E,B)$, such
that $\mu (K)<\infty$ for any compact $K\subset E$.  We
denote by $X$ the space of locally finite configurations
of particles in $E$:  $X=\{\xi =(x_i)^\infty_{i=-\infty}:
x_i\in E\ \forall i$, and for any compact $K\subset E$
$\#_K(\xi ):=\#(x_i:x_i\in K)<+\infty\}$.  A $\sigma$-algebra
$\f$ of measurable subsets of $X$ is generated by the
cylinder sets $C^B_n=\{\xi\in X:\#_B(\xi )=n\}$, where $B$
is a Borel set with a compact closure and $n\in\bbZ^1_+=
\{0,1,2,\dots\}$.  Let $P$ be a probability measure on
$(X,\f )$.  A triple $(X,\f ,P)$ is called a random point
field (process) (see [DVJ], [Le1-Le3]).  In this paper we
will be interested in a special class of random point fields
called determinantal random point fields.  It should be
noted that most, if not all the important examples
of determinantal point fields arise when $E=\coprod^k_{i=1}
E_i \ $ (here we use the notation $ \coprod $  for the disjoint union),
$\ E_i\cong\bbR^d$ or $\bbZ^d$ and $\mu$ is either the
Lebegue or the counting measure.  We will, however, develop our
results in the general setting ( our arguments
will not require significant changes).

Let $dx_i,\ i=1,\dots ,n$ be disjoint infinitesimally small subsets
around the $x_i$'s.  Suppose that a probability to find a particle in
each $dx_i$ (with no restrictions outside of $\coprod^n_{i=1}
dx_i$) is proportional to $\prod^n_{i=1}\mu (dx_i)$, i.e.
\begin{equation}
P(\#(dx_i)=1,\ i=1,\dots ,n)=\rho_n(x_1,\dots ,x_n)\mu (dx_1)\dots
\mu (dx_n)
\end{equation}
The function $\rho_n(x_1,\dots ,x_n)$ is then called the $n$-point
correlation function.  The equivalent definition is given by the
equalities
$$E\prod^m_{i=1}\frac{(\#_{B_i})!}{(\#_{B_i}-n_i)!}=\int_{B^{n_1}_1
\times\dots\times B^{n_m}_m}
\rho_n(x_1,\dots ,x_n) d\mu (x_1)\dots
d\mu (x_n)$$
where $B_1,\dots ,B_m$ are disjoint Borel sets with
compact closures, $m\geq 1,\ n_i\geq 1,\ i=1,\dots ,m, n_1+\dots
+n_m=n$.

A random point field is called determinantal if
\begin{equation}
\rho_n(x_1,\dots ,x_n)=\det\bigl (K(x_i,x_j)\bigr )_{1\leq
i,j\leq n},
\end{equation}
where $K(x,y)$ is a kernel of an integral operator $K: L^2(E,d\mu )
\rightarrow L^2(E,d\mu )$ and $K(x,y)$ satisfies some natural
regularity conditions discussed below. Such a kernel
$K(x,y)$ is  called a correlation kernel.

It follows from 2) and the non-negativity of the $n$-point
correlation functions that $K$ must have non-negative minors, and
in particular if $K$ is Hermitian it must be a non-negative operator.
In this paper we shall always restrict ourselves to the Hermitian
case.

Determinantal (also known as fermion) random point fields were introduced
by Macchi in the early seventies (see [Ma1], [Ma2], [DVJ]).  A recent
survey of the subject with applications to random matrix theory,
statistical mechanics, quantum mechanics, probability theory and
representation theory is given in [So1].  Diaconis and Evans in [DE1]
introduced a generalization of determinantial random point processes,
called immanantal point processes.

Let $K$  be a Hermitian,
locally trace class, integral operator on $L^2(E,d\mu )$.  Suppose
that we can choose a kernel $K(x,y)$ in such a way that for any
Borel set $B$ with compact closure
\begin{equation}
\tr (K\x_B)=\int_BK(x,x)dx,
\end{equation}
where $\x_B$ denotes the multiplication operator by the indicator of
$B\ (\equiv$projector on the subspace of the functions supported in
$B$).

Since it is always true that
\begin{gather*}
\tr (K\x_{B_1}\dots K\x_{B_n})=
\\ \int_{B_1\times\dots\times B_n}
K(x_1,x_2)K(x_2,x_3)\dots K(x_n,x_1)
d\mu (x_1)\dots d\mu (x_n)
\tag{3'}
\end{gather*}
for $n>1$ and Borel sets $B_1,\dots ,B_n$ with compact closure, (3)
implies that (3$'$) holds for all $n$.

(3) can always be achieved for $E=\bbR^d$ (see e.g. [So1], Lemmas
1,2).  From now on we will assume that both (2) and (3) are satisfied.

The main goal of our paper is to study the behavior of linear
statistics
$$S_f(\xi )=\sum_i f(x_i),\ \xi =(x_i),$$
for sufficiently ``nice" test functions in a scaling limit.
The moments of $S_f$ can be calculated from (2).  For instance,
\begin{equation}
ES_f=\int f(x) K(x,x) d\mu (x),
\end{equation}
\begin{equation}
{\rm Var}\ S_f=\int f^2(x)K(x,x)d\mu (x)-\int f(x)f(y)\vert K(x,y)
\vert^2d\mu (x)d\mu (y).
\end{equation}
Taking $E=\bbR^1$ and $K(x,y)=\tfrac{\sin\pi (x-y)}{\pi (x-y)}$, a
so-called sine kernel, we obtain a random point field well known
in the theory of random matrices.  It can be viewed as a limit
$n\rightarrow\infty$ of the distribution of the appropriately scaled
eigenvalues of $n\times n$ random Hermitian matrices with Gaussian
entries (see e.g. [D], chapter 5).  It was proven by Spohn in [Sp]
(see also [So2]), that if $K$ is the sine kernel and a test
function $f$ is sufficiently smooth and fast decaying at infinity,
then $\sum^\infty_{i=-\infty}f(\tfrac{x_i}{L})
-L\int^\infty_{-\infty}f(x)dx$ converges in distribution
to the normal law $N(0,\int^\infty_{-\infty}\vert\hat f
(k)\vert^2\cdot\vert k\vert dk)$, where $\hat f$ is the
Fourier transform of $f$, $\hat f(k)=\int^\infty_{-\infty}
f(x)e^{-2\pi ikx}dx$.  In other words we can say that the
random signed measure
$$\sum^\infty_{i=-\infty}\delta (x-\tfrac{x_i}{L})-Ldx$$
converges as $L\rightarrow\infty$ to the generalized
self-similar Gaussian random process with the spectral
density $\vert k\vert$ (see e.g., [Dob1], [Dob2] \S 3,
[S], and, for the introduction to the theory of generalized
random processes, [GV]).  The fact that the variance of the
linear statistics $\sum^\infty_{i=-\infty}f(\tfrac{x_i}
{L})$ does not grow to infinity for Schwartz functions
is the manifestation of the strong repulsiveness of the
distribution of the eigenvalues of random matrices.  Similar
results for other ensembles of random matrices have been
obtained in [DS], [Jo1], [Jo2], [B], [BW], [So2], [W],[DE2].  The
kernels appearing in these ensembles are, in some respect,
very much like the sine kernel.  In particular, the variance
of the number of particles in an interval grows as a
logarithm of the mathematical expectation of the number of
particles.  The following result was established by Costin
and Lebowitz for the sine kernel ([CL]):  let $f$ be an
indicator of an interval, $f=\x_I,\ I=(a,b)$, then
$$E\sum^\infty_{i=-\infty}f(x_i/L)=E(\#(x_i: aL<x_i
<bL))=L(b-a),$$
$${\rm Var}\ (\sum^\infty_{i=-\infty}f(
x_i/L))=\frac{1}{\pi^2}\log L+O(1),$$
and
$$\frac{\#(x_i:
aL<x_i<bL)-L(b-a)}{\sqrt{\frac{1}{\pi^2}\log L}}$$
converges
in distribution as $L\rightarrow\infty$ to the normal law
$N(0,1)$.  The proof of the Costin-Lebowitz theorem holds, quite
remarkably, for arbitrary determinantal random point fields
with Hermitian kernel.
\medskip

\noindent{\bf Theorem} ([So3]) {\it Let $(X,\f ,P_L), \ L\geq 0,$
be a family of
determinantal random point fields with  Hermitian locally trace class kernels
$K_L$ and $\{I_L\}_{L\geq 0}$ be a family of Borel subsets
of $E$ with compact closure.  Then if ${\rm Var}_L\ (\#(x_i:
x_i\in I_L))\underset{L\rightarrow\infty}{\rightarrow}
\infty$, the normalized random variable $\tfrac{\#(x_i:
x_i\in I_L)-E_L\#_{I_L}}{\sqrt{{\rm Var}_L\ \#_{I_L}}}$
converges in distribution to $N(0,1)$.}
\medskip

Here and below we denote by $E_L, \ \ {\rm Var}_L$  the mathematical
expectation and the variance with respect to $P_L$.
One can also establish a similar result for the step
functions (finite linear combinations of indicators).
\medskip

\noindent{\bf Theorem} {\it Let $(X,\f ,P_L)$ be a family of
determinantal random point fields with  Hermitian locally trace class kernels
$K_L$ and $\{I^{(1)}_L,\dots ,I^{(k)}_L\}_{L\geq 0}$ be a
family of Borel subsets of $E$, disjoint for any fixed $L$,
with compact closure.  Then if for some $\alpha_1,\dots ,
\alpha_k\in\bbR^1$, the variance of the linear statistics
$\sum^\infty_{i=-\infty}f_L(x_i)$ with $f_L(x)=\sum^k_{j=1}
\alpha_j\cdot\x_{I^{(j)}_L}(x)$,  grows to infinity in such a way
that ${\rm Var}_L (\#_{I^{(j)}_L})=O({\rm Var}_L\ (\sum^\infty_{i=
-\infty}f_L(x_i)))$ for any $1\leq j\leq k$, the Central
Limit Theorem holds:
$$\frac{\sum^k_{j=1}\alpha_j^{(L)}\cdot\#_{I^{(j)}_L}-E_L\left (
\sum^k_{j=1}\alpha_j\cdot\#_{I^{(j)}_L}\right )}{\sqrt
{{\rm Var}_L\ (\sum^k_{j=1}\alpha_j\cdot\#_{I^{(j)}_L})}}
\overset{w}{\longrightarrow}N(0,1).$$}
\medskip

\noindent{\bf Remark 1} We use standard notations $f=O(g)$
and $f=o(g)$ when $\tfrac{f}{g}$ stays bounded or $\tfrac
{f}{g}\rightarrow 0$.
\medskip

\noindent{\bf Remark 2}  The last theorem has been
explicitly stated in [So3] only in the special case of
the Airy and Bessel kernels and the kernels arising in the
classical compact groups (see Theorems 1, 2, 4, 6), however
the key Lemmas 7 and 8 proven there allow rather
straightforward generalization to the case of an arbitrary
Hermitian kernel.  A result close to our Theorem 6 from [So3] was
also established by K. Wieand ([W]).

We recall that a Hermitian kernel $K(x,y)$ defines a
determinantal random point field if and only if the
integral operator $K$ is non-negative and bounded from
above by the identity,
\begin{equation}
0\leq K\leq\id
\end{equation}
([So1], Theorem 3).  For the translation-invariant
kernels $K(x-y)$ and $ E= \bbR^d$  or $\bbZ^d $
this is equivalent to $0\leq \hat K(t)\leq
1$, where
\begin{equation}
K(x)=\int e^{2\pi i(x\cdot t)}\hat K(t)dt
\end{equation}
The sine kernel $K(x-y)=\tfrac{\sin\pi (x-y)}{\pi (x-y)}$
corresponds to $\hat K(t)=\x_{\left
[-\tfrac{1}{2},\tfrac{1}{2}\right ]}
(t)$, the indicator of $[-\tfrac{1}{2},\tfrac{1}{2}]$.
It might be worth noting and  actually is not very difficult
to see, that the logarithmic rate of the growth of Var $(\#(x_i:\vert
x_i\vert\leq L))$ for the sine kernel is the slowest among
all translation-invariant kernels corresponding to projectors,
$\hat K = \x_B $, for which $inf(B)$ and $sup(B)$ are the density points
of $B$.
For the generic translation-invariant kernel
$K(x-y)$ ( $\hat K$ is not an indicator) Var $(\#(
x_i:\vert x_i\vert\leq L))$ is proportional to Vol $(x_i:\vert x_i
\vert\leq L)\sim E(\#(x_i:\vert x_i\vert\leq L))$ ([So1], section 3).

In our main result we prove CLT for the linear statistics
when the variance grows faster than some arbitrary
small, but fixed, power of the mathematical expectation.

\medskip

\noindent{\bf Theorem 1} {\it Let $(X, \f, P_L), \ L\geq 0 $ be a family of
determinantal random point fields with Hermitian correlation kernels $K_L$.
Suppose that
$f_L, \ L\geq 0$ are
bounded measurable  functions with precompact support
(i.e.  $supp(f_L) $ has a compact
closure for any $L \geq 0$), such that

\begin{equation}
{\rm Var}_L\ S_{f_L}\rightarrow\infty\text{ as }L\rightarrow
\infty
\end{equation}
and
\begin{equation}
sup |f_L(x)| =o ({\rm Var}_L)^{\epsilon} , \ \
E_LS_{\vert f\vert_L}=O\bigl (({\rm Var}_L\ S_{f_L})^\delta
\bigr ),
\end{equation}
for any $\epsilon >0$  and some  $\delta >0$.  Then the normalized linear
statistics
$\tfrac{S_{f_L}-E_LS_{f_L}}{\sqrt{{\rm Var}_L\ S_{f_L}}}$
converges in distribution to the standard normal law
$N(0,1)$.}
\medskip

As a very important special case of  Theorem 1 one can consider
$f_L(x):=f(T_Lx)$, where
$\{T_L\},\ L\in\bbR^1_+$, is a one-parameter family
of measurable transformations $T_L:E\rightarrow E$ such
that $T^{-1}_LD$ has compact closure for any compact $D$.
If for a sufficiently rich class of test
functions $f$ (e.g., continuous functions with compact support)
(8),(9) are satisfied, and
the rate of the growth of ${\rm Var}_L  \ (S_{f_L})$ is the same,
$${\rm Var}_L\ (S_{f_L})=B(f)\cdot V_L\cdot (1+o(1)),$$
where $B(f)$ is some functional on a space of test
functions, Theorem 1 implies that the random signed measure
$$V^{-\frac{1}{2}}_L\left (\sum^\infty_{i=-\infty}\delta
(x-T_Lx_i)-T_L (K_L(x,x)d\mu (x))\right )$$
converges as $L\rightarrow\infty$ to the generalized
Gaussian process with the correlation functional $B(f,f)=
B(f)$ ( we denote by
$T_L (K_L(x,x)d\mu (x)) $ the image of the measure
$K_L(x,x)d\mu (x) $ under $T_L$).

Let us consider a Euclidean one-particle space
$E=\bbR^d \ $, a one-parameter family of dilations
$$ T_L :\bbR^d \rightarrow \bbR^d, \ \ T_L x= x/L, $$
and a  correlation kernel
\begin{equation}
K_L(x,y)=A_L(x-y)+R_L(x,y),
\end{equation}

where
\begin{equation}
\vert R_L(x,y)\vert\leq Q(x_{abs}+y_{abs}),
\end{equation}
$x_{abs}=(\vert x_1\vert,\dots ,\vert x_d\vert ),\ Q\in
L^2(\bbR^d_+)\cap L^\infty (\bbR^d_+)$
It follows from (6),
(10) and (11) that $0\leq A_L\leq\id$, which implies
$0\leq\hat A_L(k)\leq 1$, $0\leq\int_{\bbR^d}\hat A_L(k)-(\hat
A_L(k))^2dk=A_L(0)- \int_{\bbR^d} |A_L(x)|^2 dx =:\sigma_L^2$, and  $\sigma_L =0$ if and only if $\hat A_L$
is an indicator.
\medskip

\noindent{\bf Theorem 2} {\it Let the kernel $K_L$ satisfy
(10), (11) and there exist constants $const, \ \sigma >0$ and
$\kappa_L \to \infty$ as $L \to \infty$ such that
$$ \sigma_L \to \sigma \ \ {\rm as} \  L \to \infty ,$$
$$ |A_L(0)| <const, $$
and
$$ \int_{|x|>L/\kappa_L} |A_L(x)|^2 dx \to 0.$$
Then for any real-valued
function $f\in L^1(\bbR^d)\cap L^2(\bbR^d)$ the
normalized linear statistics
$$\frac{1}{L^{\frac{d}{2}}\sigma}\left (\sum^\infty_{
i=-\infty}f\left (\frac{x_i}{L}\right )-A_L(0)\cdot L
\int_{\bbR^d}f(x)dx\right )$$
converges in distribution to the Gaussian random variable
$N\left (0,\int_{\bbR^d}\left (f(x)\right )^2dx\right )$.}
\medskip

\noindent{\bf Remark 3} Theorem 2 says that under the stated
conditions the random
signed measure
$$\frac{1}{L^{\frac{d}{2}}\sigma}\left (\sum^\infty_{i=-
\infty}\delta\left (x-\frac{x_i}{L}\right )-A_L(0)\cdot Ldx
\right )$$
converges to the white noise  as $L\rightarrow
\infty$ (for the definition of the white noise see e.g.
[H]).  Similar results hold in the discrete case.

Let us now restrict our attention to the
translation-invariant kernels $K(x,y)=A(x-y)$.  We will
use the notation
$$m(\lambda ):=\int\hat A(k)-\hat A(k)\hat A(k-\lambda )
dk.$$
Observe that $\sigma^2=m(0)$ and
\begin{equation}
{\rm Var}\ \left (\sum^\infty_{i=-\infty}f(x_i)\right )
=\int\vert\hat f(\lambda )\vert^2m(\lambda )d\lambda .
\end{equation}
In particular
\begin{equation}
{\rm Var}\ \left (\#_{[-L,L]^d}\right )={\rm Vol}\ \left
([-L,L]^d\right )\cdot\left (m(0)+o(1)\right ).
\end{equation}
It follows from (12) that the rate of the growth of the
variance of $S_{f_L}$ depends on the asymptotics of $m(
\lambda )$ near the origin.  In the next theorem we consider
the degenerate case $\sigma^2=0$ in one dimension.
\medskip

\noindent{\bf Theorem 3} {\it Let $K(x,y)=A(x-y)$ be a
translation-invariant kernel in $\bbR^1$ and $m(\lambda )
=\vert\lambda\vert^\alpha\varphi (\lambda )$, where $\varphi
(\lambda )$ is a slowly varying function at the origin and
$0<\alpha <1$.  Then for any Schwartz function $f$ :
$E S_{f_L}=LA(0)\int f(x)dx$, Var $S_{f_L}=L^{1-\alpha}
\varphi (L^{-1})\int\vert\hat f(k)\vert^2\vert k
\vert^\alpha dk (1+o(1))$, and
$$\frac{S_{f_L}-ES_{f_L}}
{(L^{1-\alpha}\varphi (L^{-1}))^{\tfrac{1}{2}}}$$
converges
in distribution to $N(0,\int\vert\hat f(k)\vert^2\vert
k\vert^\alpha dk$.}
\medskip

\noindent{\bf Remark 5}  We recall that $\varphi (\lambda) \geq 0$
is slowly varying at the origin if $\lim_{\lambda\rightarrow
0}\tfrac{\varphi (a\lambda )}{\varphi (\lambda )}=1$ for
any $a\neq 0$ (see [Se]).
\medskip

\noindent{\bf Remark 6}  The result of Theorem 3 can be
interpreted as the convergence in distribution of the
random signed measure
$$\biggl (L^{1-\alpha }\varphi (L^{-1})\biggr )
^{-\tfrac{1}{2}}\left (
\sum\delta \left (x-\frac{x_i}{L}\right )-A(0)Ldx\right )$$
to the self-similar (also called automodel in the Russian literature)
generalized Gaussian random
process with the spectral density $\vert k\vert^\alpha, \ 0<\alpha <1.$
Self-similarity  means
that the distribution of the process is invariant under the
action of the renorm-group $\xi (x)\rightarrow\xi (ax)a^\gamma$,
$\gamma =\tfrac{1+\alpha}{2}$ . The self-similar generalized Gaussian random
process corresponding to $\alpha =0$ is exactly the white noise (see Remark 3
above). It was proven by Dobrushin that the only self-similar random processes
in $\bbR^1$ are the ones with the spectral density
$\vert k\vert^\alpha, \ 0\leq\alpha \leq 1 $.
A self-similar generalized random process with the spectral density $|k| \ $
appeared in the Spohn's results ([Sp]) discussed above after the formulas
$ (4), \ (5)$ (see also [Jo1], [B], [So2]).
For additional information
on self-similar random processes we refer the reader to [Dob1], [Dob2],[S].
\medskip

\noindent{\bf Example}  Let $\hat A$ be the indicator of
$\sqcup_{n\geq 1}[n,n+n^{-\beta}],\ \beta >1$.  Then
$m(\lambda )=\const\cdot\vert\lambda\vert^{1-\tfrac{1}
{\beta}}(1+o(1))$. On the other hand if the length
$l_n$  of the $n-$th interval $ [n,n+l_n]$ decays
sufficiently fast, say $ 0 \leq l_{n+1} \leq  l_n^{1+\epsilon}, \ \
\epsilon >0 \ $ than $m(\lambda)$  is not regularly varying
at the origin.

Finally we consider the case when $\hat A$ is the indicator
of a union of $1\leq\ell <\infty$ disjoint intervals.
It is straightforward to see that then $m(\lambda )=\ell
\vert\lambda\vert$ near the origin.
\medskip

\noindent{\bf Theorem 4} {\it Let $\hat A$ be the indicator
of $I$, $I=\sqcup^\ell_{i=1}[a_i,b_i],\ a_1<b_1<a_2<b_2<
\dots <a_\ell <b_\ell$. Then for any Schwartz function $f \ \ $
$\sum^\infty_{i=-\infty}f\left (\tfrac{x_i}{L}\right )
-A(0)\cdot L\int^\infty_{-\infty}f(x)dx$ converges in
distribution to $N(0,\ell\cdot\int^\infty_{-\infty}\vert
\hat f(k)\vert^2\vert k\vert dk)$.}
\medskip

The proofs of Theorems 1--3 will be given in the next three
sections.  The proof of Theorem 4 is the same, modulo
trivial alterations, as the one given for the sine kernel
in [So2].

The author would like to thank P.Deift and Ya. Sinai for useful discussions,
T. Shirai
for making available the preprint [ST] prior to its publication and the referee
for useful comments.

\section{Proof of Theorem 1}

We are going to prove Theorem 1 by the method of moments.
Let us denote by $C_n(S_f)$ the $n^{th}$ cumulant of $S_f$.
We remind the reader that for a random variable $\eta $ with all finite moments
, the cumulants $ C_n (\eta), \ n=1,2,\ldots$ are defined through the Taylor
coefficients of the logarithm of the characteristic function :
$$ \log E( \exp(it\eta)) = \sum_{n=1}^{\infty} C_n(\eta) (it)^n/n! .$$
We show that the $n^{th}$ cumulant of the normalized linear
statistics $\tfrac{S_{f_L}-ES_{f_L}}{({\rm Var}\ S_{f_L}
)^{\tfrac{1}{2}}}$ converges to zero as $L\rightarrow
\infty$ for sufficiently large $n$ ($n>\max (2\delta ,2))$.
The Lemma 3 from the Appendix then asserts that all
cumulants of $\tfrac{S_{f_L}-ES_{f_L}}{({\rm Var}\ S_{f_L}
)^{\tfrac{1}{2}}}$ converge to the cumulants of the standard
normal distribution, which implies the weak convergence.

We recall the lemma established in [So2] (see formula
(2.7)).
\medskip

\noindent{\bf Lemma 1}
\begin{equation}
\begin{split}
&C_n(S_f)=\sum^n_{m=1}\sum_{\overset{(n_1,\dots ,n_m):
n_1+\dots +n_m=n.}{n_1\geq 1,\ i=1,\dots ,m}}\frac
{(-1)^{m-1}}{m}\frac{n!}{n_1!\dots n_m!}
\int f^{n_1}(x_1)\\
&K(x_1,x_2)f^{n_2}(x_2)K(x_2,x_3)\dots
f^{n_m}(x_m)K(x_m,x_1)d\mu (x_1)\dots d\mu (x_m)
\end{split}
\end{equation}
\medskip

Using Lemma 1 we will be able to estimate the cumulants
of $S_{f_L}$.  We claim the following result to be true.
\medskip

\noindent{\bf Lemma 2} {\it Under the assumptions of
Theorem 1
\begin{equation}
C_n(S_{f_L})=O(({\rm Var}_LS_{ f_L})^{\delta +\epsilon}),\ n\geq 1,
\end{equation}
 where $ \epsilon$ is arbitrarily small.}
\medskip

\noindent{\bf Proof of Lemma 2} It follows from (14) that
$C_n(S_{f_L})$ is a linear combination of
\begin{equation*}
\begin{split}
&\int f^{n_1}_L(x_1)K_L(x_1,x_2)f^{n_2}_L(x_2)K_L(x_2,x_3)\dots
f^{n_m}_L(x_m)K_L(x_m,x_1)\\
&d\mu (x_1)\dots d\mu (x_m)=\tr (f^{n_1}_LK_Lf^{n_2}_LK_L\dots
f^{n_m}_LK_L),
\end{split}
\end{equation*}

where $n_i\geq 1,\ i=1,\dots ,m,\ m\geq 1.$

We claim that each term is $O(({\rm Var}_LS_{ f_L})^{\delta +\epsilon}).$
Indeed, if $m=1$, then \\
$\vert\tr f^n_LK_L\vert=\vert\int
f^n_L(x)K_L(x,x)d\mu (x)\vert \leq \Vert f_L\Vert^{n-1}_\infty
\int\vert f_L(x)\vert K_L(x,x)d\mu (x)=\Vert f_L
\Vert^{n-1}_\infty ES_{\vert f\vert_L} =
O(( {\rm Var}_LS_{ f_L})^{\delta +\epsilon}).$

If $m>1$,
represent $\tr (f^{n_1}_LK_Lf^{n_2}_L
K_L\dots f^{n_m}_LK_L)$ as a linear combination of $\tr
(f^{n_1}_{\pm ,L}K_Lf^{n_2}_{\pm ,L}K_L
\dots f^{n_m}_{\pm ,L}
K_L)$, where we use the notations $f_+=\max (f,0)$, $f_-=\max (-f
,0)$. Let us fix the
choice of $\pm$ in each of the factors.  Using the cyclicity
of the trace and the inequality $\vert\tr (AB)\vert\leq (
\tr (AA^*))^{\tfrac{1}{2}}(\tr (BB^*))^{\tfrac{1}{2}}$
for the Hilbert-Schmidt operators ([RS], section VI.6),
we obtain
\begin{equation}
\begin{split}
&\vert\tr (f^{n_1}_{\pm ,L}K_Lf^{n_2}_{\pm ,L}K_L\dots
f^{n_m}_{\pm ,L}K_L)\vert =
\vert\tr (f^{\frac{n_1}{2}}_{\pm ,L}K_Lf^{n_2}_{\pm ,L}K_L
\dots f^{n_m}_{\pm ,L}K_Lf^{\frac{n_1}{2}}_{\pm ,L})\vert
\leq\\
&[\tr ((f^{\frac{n_1}{2}}_{\pm ,L}K_Lf^{\frac{n_2}{2}
}_{\pm ,L})(f^{\frac{n_1}{2}}_{\pm ,L}K_Lf^{\frac{n_2}{2}
}_{\pm ,L})^*)]^{\frac{1}{2}}[\tr ((f^{\frac{n_2}{2}
}_{\pm ,L}K_Lf^{n_3}_{\pm ,L}K_L\dots f^{n_m}_{\pm ,L}K_L
f^{\frac{n_1}{2}}_{\pm ,L})\\
&(f^{\frac{n_2}{2}}_{\pm ,L}K_Lf^{n_3}_{\pm ,L}K_L\dots
f^{n_m}_{\pm ,L}K_Lf^{\frac{n_1}{2}}_{\pm ,L})^*)]^{\frac
{1}{2}}.
\end{split}
\end{equation}
The first factor at the r.h.s. of (16) is equal (again
by the cyclicity of the trace) to $[\tr (f^{n_1}_{\pm ,L}
K_Lf^{n_2}_{\pm ,L}K_L)]^{\tfrac{1}{2}}$ (in particular we
note that $\tr (g_1Kg_2K)\geq 0$ for non-negative $g_1,\
g_2$).

Since
\begin{equation*}
\begin{split}
&\tr ((f^{n_1}_{\pm ,L}+f^{n_2}_{\pm ,L})^2K)-\tr
((f^{n_1}_{\pm ,L}+f^{n_2}_{\pm ,L})K(f^{n_1}_{\pm ,L}
+f^{n_2}_{\pm ,L})K)=\\
&{\rm Var}\ (S_{f^{n_1}_{\pm ,L}+f^{n_2}_{\pm ,L}})\geq 0,
\end{split}
\end{equation*}
we have
\begin{equation}
\begin{split}
&0\leq\tr (f^{n_1}_{\pm ,L}K_Lf^{n_2}_{\pm ,L}K_L)
\leq\frac{1}{2}\biggl (\tr ((f^{n_1}_{\pm ,L}+f^{n_2}_{\pm ,L})^2K_L)-
\tr (f^{n_1}_{\pm ,L}K_Lf^{n_1}_{\pm ,L}K_L)-\\
&\tr (f^{n_2}_{\pm ,L}K_Lf^{n_2}_{\pm ,L}K_L)\biggr )\leq\frac{1}{2}
\tr ((f^{n_1}_{\pm ,L}+f^{n_2}_{\pm ,L})^2K_L))=O(\tr (\vert f_L
\vert K_L)) o(({\rm Var}_L S_{f_L})^{\epsilon})\\
&=O(({\rm Var}_L S_{f_L})^{\delta +\epsilon}).
\end{split}
\end{equation}

As for the second term in (16), one can rewrite $\tr\biggl ((
f^{\tfrac{n_2}{2}}_{\pm ,L}K_Lf^{n_3}_{\pm ,L}K_L\dots
\break f^{n_m}_{\pm ,L}K_Lf^{\tfrac{n_1}{2}}_{\pm ,L})
(f^{\tfrac{n_2}{2}}_{\pm ,L}K_Lf^{n_3}_{\pm ,L}K_L\dots
f^{n_m}_{\pm ,L}K_Lf^{\tfrac{n_1}{2}}_{\pm ,L})^*\biggr )$ as
\begin{equation}
\tr\biggl (f^{\frac{n_2}{2}}_{\pm ,L}K_Lf^{n_3}_{\pm ,L}K_L\dots
f^{n_m}_{\pm ,L}K_Lf^{n_1}_{\pm ,L}K_Lf^{n_m}_{\pm ,L}\dots
K_Lf^{n_3}_{\pm ,L}K_Lf^{\frac{n_2}{2}}_{\pm ,L}\biggr )=\tr (CDD^*),
\end{equation}
where $C=f^{\tfrac{n_3}{2}}_{\pm}K_Lf^{n_2}_{\pm ,L}K_L
f^{\tfrac{n_3}{2}}_{\pm ,L}$, $D=f^{\tfrac{n_3}{2}}_{\pm ,L}
K_Lf^{n_4}_{\pm ,L}K_L\dots f^{n_m}_{\pm ,L}K_L f^{\tfrac{n_1}{2}}_{\pm ,L}$.
  Note that $C\geq 0$ and
$\tr (C)=\tr (f^{n_3}_{\pm ,L}K_Lf^{n_2}_{\pm ,L}K)=O
( ({\rm Var}_L S_{f_L})^{\delta +\epsilon})$ by  arguments similar to
(17).  Using $\vert\tr (CDD^*)\vert\leq\tr (C)\cdot\Vert
DD^*\Vert=\tr (C)\cdot\Vert D\Vert^2$ ([RS], Section
VI.6) and $\Vert D\Vert\leq\Vert K\Vert^m\cdot\Vert f_L
\Vert^\aleph_\infty$, where $\aleph =(\sum^m_{i=1}n_1)-n_2$,
we conclude that (18) is $O( ({\rm Var}_L S_{f_L})^{\delta +\epsilon})$
Together with (16) and (17) this concludes the proof of
the lemma.\hfill$\dashv$

Let us now apply Lemma 2 to estimate the cumulants of the
normalized linear statistics.  We have
$$C_1\left (\frac{S_{f_L}-ES_{f_L}}{\sqrt{\var_L S_{f_L}}}
\right )=0,$$
$$C_2\left (\frac{S_{f_L}-ES_{f_L}}{\sqrt{\var_L\ S_{f_L}}}
\right )=1,$$
and, for $n>2$,
\begin{equation}
C_n\left (\frac{S_{f_L}-ES_{f_L}}{\sqrt{\var_L\ S_{f_L}}}
\right )=\frac{C_n(S_{f_L})}{(\var_L\ S_{f_L})^{\frac{n}{2}}}
=O\left (\frac{E\left (S_{\vert f\vert_L}\right )}
{\left (\var_L\ S_{f_L}\right )^{\frac{n}{2}}}\right )
\end{equation}
It follows from the Lemma 2 and (19)  that $C_n\left (
\tfrac{S_{f_L}-ES_{f_L}}{\sqrt{\var\ S_{f_L}}}\right )$
goes to zero if $n>2\delta$.

Lemma 3 from the Appendix then implies that
all cumulants of the normalized linear statistics converge
to the cumulants of the standard normal random variable,
and weak convergence of the distributions follows.

Theorem 1 is proven.\hfill$\Box$

\section{Proof of Theorem 2}

Let $(E,d\mu )$ be $(\bbR^d,dx)$ and $T_Lx=\tfrac{x}{L}$.
Consider a real-valued function $f\in L^1(\bbR^d)\cap L^2
(\bbR^d)$.  The mathematical expectations of $S_{f_L}$
is equal to
\begin{equation*}
\begin{split}
&ES_{f_L}=\int_{\bbR^d}f(x /L )K_L(x,x)dx=
\int_{\bbR^d}f(x/L )A_L(0)dx\\
&+\int_{\bbR^d}f(x/L )R_L(x,x)dx=A_L(0)\cdot
L^d\int f(x)dx+\int f(x/L )R_L(x,x)dx.
\end{split}
\end{equation*}
By (11) the absolute value of the second integral is bounded
by the sum of the integrals
\begin{equation*}
\begin{split}
&\int_{\bbR^d_+}\left \vert f(\pm x_1/L,\dots
,\pm x_d/L )\right \vert Q\left (2x_1,\dots ,
2x_d\right )dx\\
&\leq\left (\int_{\bbR^d_+}f^2(\pm x_1/L,
\dots ,\pm x_d/L)dx\right )^{\frac{1}{2}}
\left (\int_{\bbR^d_+}Q^2(2x)dx\right )^{\frac{1}{2}}
=O\left (L^{\frac{d}{2}}\right )
\end{split}
\end{equation*}
Therefore,
\begin{equation}
ES_{f_L}=A_L(0)\cdot L^d\cdot\int_{\bbR^d}f(x)dx+O
(L^{\frac{d}{2}}).
\end{equation}
The variance of $S_{f_L}$ is given by
\begin{equation}
\begin{split}
&\var\ S_{f_L}=\int f^2(x/L)K_L(x,x)dx
-\int f (x/L )f(y/L )
\vert K_L(x,y)\vert^2dxdy=\\
&A_L(0)L^d\int f^2(x)dx-\int f(x/L )
f(y/L )\vert A_L(x-y)\vert^2dxdy+ r(L),
\end{split}
\end{equation}
where
\begin{equation*}
\begin{split}
&r(L)=\int f^2(x/L)R_L(x,x)dx-
2\int f(x/L)f(y/L )
A_L(x-y)R_L(y,x)dxdy-\\
&\int f(x/L )f(y/L )
\vert R_L(x,y)\vert^2dxdy=r_1(L)+r_2(L)+r_3(L).
\end{split}
\end{equation*}
It follows from the assumptions of the theorem that the second term at the
r.h.s. of (21) is equal to
\begin{equation*}
\begin{split}
&L^d\int\vert\hat f(k)\vert^2\widehat{\vert A_L\vert^2}
(k/L )dk=L^d\widehat{\vert A_L\vert^2}
(0)\int\vert\hat f(k)\vert^2dk\cdot (1+o(1))=\\
&L^d\int\vert A_L(x)\vert^2dx\int f^2(x)dx(1+o(1)).
\end{split}
\end{equation*}
Indeed,
\begin{equation*}
\begin{split}
&\vert\int\vert\hat f(k)\vert^2 ( \widehat{\vert A_L\vert^2}
(k/L ) -\widehat{\vert A_L\vert^2}(0))  dk \vert \leq \\
& \vert \int_{|k|> \kappa_L} \vert +\vert \int_{|k|\leq \kappa_L} \vert.
\end{split}
\end{equation*}
Since
\begin{equation*}
\begin{split}
&\vert \widehat{\vert A_L\vert^2}(t)\vert = \vert \int \hat{A_L}(k)
\hat{A_L}(k-t)dk \vert \\
& \leq \int \hat{A_L}(k) dk =A_L(0) \leq const
\end{split}
\end{equation*}
we note that the first integral is bounded from above by
$$ const \int_{|k|>(\kappa_L)^{1/2}} |\hat{f}(k)|^2 dk \ \ \to 0 \ \ {\rm as}
\ \ L \to \infty .$$
To deal with the second integral we estimate from above
\begin{equation*}
\begin{split}
&\vert \widehat{\vert A_L\vert^2}
(k/L ) -\widehat{\vert A_L\vert^2}(0)\vert  \leq \\
& \vert \int |A_L|^2(t) (\exp(2\pi i t k/L) -1) dt \vert \\
& \vert \int_{|t|\geq L/\kappa_L}  + \int_{ |t| < L/\kappa_L} \vert \leq \\
& \int_{|t|\geq L/\kappa_L}  |A_L|^2(t) dt + O(1/\sqrt(\kappa_L))= \\
& o(1) + O(1/\sqrt(\kappa_L))=o(1).
\end{split}
\end{equation*}
Therefore
\begin{equation}
\begin{split}
&\var\ S_{f_L}=\left (A_L(0)-\int\vert A_L(x)\vert^2dx\right )
L^d\int f^2(x)dx+o(L^d)+r(L)=\\
&\sigma^2L^d\int f^2(x)dx+o(L^d)+r(L).
\end{split}
\end{equation}
We claim that
\begin{equation}
r(L)=o(L^d).
\end{equation}
Consider first $r_1(L)$.  By (11) it is bounded by the
integrals
$$\int_{\bbR^d_+}f^2 (\pm x_1/L,\dots
,\pm x_d/L )Q(2x)dx$$
All of these integrals are estimated in the same way.
For example,
\begin{equation*}
\begin{split}
&\int_{\bbR^d_+}f^2 (x/L )Q(2x)dx=L^d
\int f^2(x)Q(2Lx)dx=\\
&L^d\int f^2(x)Q(2Lx)\x_{\{Q(2Lx)>\frac{1}{\sqrt{L}}\}}
dx+L^d\int f^2(x)Q(2Lx)\x_{\{Q(2Lx)\leq\frac{1}{\sqrt{L}}
\}}\\
&\leq L^d\Vert Q\Vert_\infty\int f^2(x)\x_{\{Q(2Lx)>\frac
{1}{\sqrt{L}}\}}dx+L^{d-\frac{1}{2}}\int f^2 (x)dx=o(L^d),
\end{split}
\end{equation*}
since
$$\ell (x:Q(2Lx)>\frac{1}{\sqrt{L}})\underset
{L\rightarrow\infty}{\longrightarrow}0.$$
To estimate $r_3(L)$ we need to estimate the integrals
of the form
\begin{equation}
\int_{\bbR^d_+}\vert f(x/L )
\vert \ \vert f(y/L )
\vert  Q^2(x+y)dxdy=L^d\int_{\bbR^d_+}g (
z/L)Q^2(z)dz,
\end{equation}
where $g(z)=\int\vert f(x)\vert\ \vert f(z-x)\vert
\x_{\bbR^d_+}(x)\x_{\bbR^d_+}(z-x)dx$.

Since $g(z)$ is bounded, continuous, and zero at the origin,
we have
$$(24)=L^dg(0)\int Q^2(z)dz (1+o(1))=o(L^d).$$
Finally,
\begin{equation*}
\begin{split}
&\vert r_2(L)\vert=\vert \ \int f (x/L
)f (y/L)A_L(x-y)R(y,x)dxdy
\vert\\
&\leq\left [\int \ \vert f(x/L )
\vert \ \vert f(y/L )
\vert\ \vert A_L(x-y)\vert^2dxdy\right ]^{\frac{1}{2}}\\
&\left [\int \ \vert f(x/L )
\vert\ \vert f(y/L )\vert\
\vert R_L(y,x)\vert^2dxdy\right ]^{\frac{1}{2}}=O\left (
L^{\frac{d}{2}}\right )\ o\left (L^{\frac{d}{2}}\right )
=o\left (L^d\right ).
\end{split}
\end{equation*}
Combining the above estimates, we prove (23), which implies
\begin{equation}
\var\ S_{f_L}=\sigma^2L^d\int f^2(x)dx(1+o(1)).
\end{equation}
If $f$ is bounded, the Central Limit Theorem then follows
from Theorem 1 (compactness of the support of $f$ is not
needed since $f\in L^1(\bbR^d)\cap L^\infty (\bbR^d)$
guarantees that all moments of $S_{f_L}$ are finite).  The
proof in the case of the unbounded $f$ follows by a rather
standard approximation argument.  We choose $N>0$ to be
sufficiently large and consider a truncated function
$$\tilde f(x)=\begin{cases}f(x),&\text{if }\vert f(x)
\vert\leq N\\N,&\text{if }f(x)>N\\-N&\text{if }
f(x)<-N.\end{cases}$$
Observe that
$$E\left (\frac{S_{f_L}-ES_{f_L}}{\sigma L^{\frac{d}{2}}
}-\frac{S_{\tilde f_L}-ES_{\tilde f_L}}{\sigma L^{\frac
{d}{2}}}\right )^2=\frac{\var\ S_{(f-\tilde f)_L}}
{\sigma^2L^d}=\frac{\int_{\vert x\vert\geq N}f^2(x)dx}
{\sigma^2}+o(1)$$
can be made arbitrarily small by choosing $N$ and $L$
sufficiently large.

Since
$$\frac{S_{\tilde f_L}-ES_{\tilde f_L}}{\sigma L^{\frac{d}
{2}}}\overset{w}{\underset{L\rightarrow\infty}
{\longrightarrow}}N\left (0,\int_{\vert x\vert\leq N}f^2(x)dx
\right )$$
and
$$\lim_{N\rightarrow\infty}\int_{\vert x\vert\leq N}f^2
(x)dx=\int f^2(x)dx,$$
the  result follows.

Theorem 2 is proven.\hfill$\Box$

\section{Proof of Theorem 3}
We now turn to the proof of Theorem 3.
It is enough to establish that
\begin{equation}
\var\ S_{f_L}=L^{1-\alpha}\varphi (L^{-1})\int\vert\hat f
(k)\vert^2\ \vert k\vert^\alpha dk(1+o(1)).
\end{equation}
The result then will follow from Theorem 1.  We have (see (12))
\begin{equation}
\begin{split}
&\var\ S_{f_L}=\int\vert\hat f (L\lambda )\vert^2L^2m
(\lambda )d\lambda =L\int\vert\hat f(k)\vert^2m(kL^{-1})
dk=\\
&L\int\vert\hat f(k)\vert^2\ \vert k\vert^\alpha L^{-\alpha}
\varphi (kL^{-1})dk=L^{1-\alpha}\varphi (L^{-1})\\
&\int\vert
\hat f(k)\vert^2\ \vert k\vert^\alpha\frac{\varphi
(kL^{-1})}{\varphi (L^{-1})}dk
\end{split}
\end{equation}
It was proven by Karamata ([K1], [K2]) that any slowly
varying function at the origin can be represented in some
interval $(0,b]$ as
\begin{equation}
\varphi (x)=\exp\left \{\eta (x)+\int^{x^{-1}}_{b^{-1}}\frac
{\epsilon (t)}{t}dt\right \},
\end{equation}
where $\eta$ is a bounded measurable function on $(0,b]$,
such that $\eta (x)\rightarrow c$ as $x\rightarrow 0$
($\vert c\vert <\infty$), and $\epsilon (x)$ is a continuous
function on $(0,b]$ such that $\epsilon (x)\rightarrow 0$
as $x\rightarrow 0$.  (for a modern day reference we refer the reader to [Se],
Theorem 1.2; of course a similar representation
holds for $\varphi$ also on some interval $[b', 0)$ of the
negative semi-axis).  In particular
\begin{equation}
\frac{\varphi\left (\frac{k}{L}\right )}
{\varphi\left (\frac{1}{L}\right )}
\underset{L\rightarrow\infty}{\longrightarrow}1

\end{equation}
uniformly in $k$ on compact subsets of $\bbR^1\setminus\{0\}$,
and the following estimates hold uniformly in $k$ for
sufficiently large $L$
\begin{equation}
\const_1k^{-n}\leq\varphi (kL^{-1})/\varphi (L^{-1})\leq
\const_2k^n,\text{ for }1\leq k\leq L,
\end{equation}
\begin{equation}
\const_3k^{-\frac{1}{2}}\leq\varphi (kL^{-1})/\varphi
(L^{-1})\leq\const_4k^{\frac{1}{2}}\text{ for }0<k\leq 1,
\end{equation}
where $\const_i,\ i=1,\dots 4,\ n>0$ are some constants.

The estimates (28)-(31) imply
$$\int^L_{-L}\vert\hat f(k)\vert^2\ \vert k\vert^2\frac
{\varphi (kL^{-1})}{\varphi (L^{-1})}dk
\underset{L\rightarrow\infty}{\longrightarrow}
\int^\infty_{-\infty}\vert\hat f(k)\vert^2\ \vert k
\vert^\alpha dk.$$
From the other side, the integral over $\vert k\vert\geq
L$ is o(1) since $f$ is a Schwartz function and $m$ is
bounded.

Theorem 3 is proven.\hfill$\Box$

\noindent{\bf Remark 7}
We learned very recently that similar results to our Theorem 2
have been independently
obtained (in the discrete case) by Tomoyuki Shirai and Yoichiro
Takahashi in the
preprint [ST].

\section*{Appendix}

For the convenience of the reader we give here the proof
of a rather standard fact.
\medskip

\noindent{\bf Lemma 3} {\it Let $\{\eta_L\}$ be a family
of random variables such that $c_1(\eta_L)=0,\ c_2
(\eta_L)=1$ and $c_n(\eta_L)$ converges to zero as $L
\rightarrow\infty$ for all $n\geq N$, where $N<\infty$.
Then $\lim_{L\rightarrow\infty}c_n(\eta_L)=0$ for all
$n>2$ and $\eta_L$ converges in distribution to $N(0,1)$.}
\medskip

\noindent{\bf Proof}  Denote $d_L=\max (\vert c_j(\eta_L)
\vert^{\tfrac{1}{j}},\ 1\leq j\leq N-1)$.  It is clear that
$d_L\geq 1$.  Consider the random variable
$$\tilde\eta_L=\eta_L/d_L.$$
Since $c_n(\tilde\eta_L)=c_n(\eta_L)/d^n_L$ we have
$\vert c_n(\tilde\eta_L)\vert\leq 1$ for all $n$ and
$c_n(\tilde\eta_L)\rightarrow 0$ for $n\geq N$.
Consider $(N-1)$-dimensional vector $(c_1(\tilde\eta_L),
\dots ,c_{N-1}(\tilde\eta_L))$.  Let $(c_1,c_2,\dots ,
c_{N-1})$ be a limit point.  The Marcinkiewicz theorem (see e.g. [L])
states that if all but a finite number of cumulants of a random variable
are non-zero then  the random variable must either have a Gaussian distribution
or be a constant. In both cases
we have $c_j=0$ for $j>2$.  Therefore $d_L=(c_2(\eta_L
))^{\tfrac{1}{2}}=1$ for sufficiently large $L$ and
$c_n(\eta_L)\underset{L\rightarrow\infty}{\longrightarrow}
0$ for $n>2$.  Convergence of the cumulants of $\eta_L$
to the cumulants of $N(0,1)$ is equivalent to the
convergence of the moments which in turn implies convergence
in distribution.

\def\am{{\it Ann. of Math.} }
\def\ap{{\it Ann. Probab.} }
\def\temf{{\it Teor. Mat. Fiz.} }
\def\jmp{{\it J. Math. Phys.} }
\def\cmp{{\it Commun. Math. Phys.} }
\def\jsp{{\it J. Stat. Phys.} }
\def\prl{{\it Phys. Rev. Lett.} }
\def\arma{{\it Arch. Rational Mech. Anal.} }


TITLE: Exact sequence of sheaves and associated sequence of graded modules
QUESTION [2 upvotes]: Let $(X,\mathcal{O}_X)$ with $X=\mathbb{P}^n$ and consider a exact sequence of sheaves of $\mathcal{O}_X$-modules
$$0 \to \mathcal{F} \to \mathcal{H} \to \mathcal{G} \to 0 $$
Suppose that we apply the functor $\Gamma^*(\mathcal{F})=\bigoplus_{m \in \mathbb{Z}}\Gamma(X,\mathcal{F}(n))$ to the sequence (the functor in general is only left exact) and it remains exact and splits.
What can we say about the original sequence?  Does the sequence split?

REPLY [3 votes]: I will give an outline of the proof of this fact if all the sheaves involved are quasi-coherent. I claim that the original exact sequence does split.
First recall the Proj construction. Let $A=\oplus_{n\geq 0}A_n$ be a graded commutative ring, then $\operatorname{Proj} A$ is as a set the set of homogeneous prime ideas of $A$ not containing the trivial ideal $A_+=\oplus_{n>0}A_n$. For any homogeneous $f\in A$, $\operatorname{Proj}A$ has distinguished affine open $\operatorname{Spec}(A_f)_0$ (where $A_f$ is the localization by $f$ and $(A_f)_0$ means take the degree zero part of that). For a graded $A$ module $M$, we associate a quasicoherent sheaf $\tilde{M}$ by having its value on $(A_f)_0$ be $(M_f)_0$.
First I claim for for any quasi-coherent sheaf $\mathscr{M}$ on $\mathbb{P}^n$, $\widetilde{\Gamma^*(\mathscr{M})}$ is canonically isomorphic to $\mathscr{M}$. To show this you must use that the graded ring defining $\mathbb{P}^n$ is finitely generated and generated in degree $1$.
Let $A$ still be a graded ring as above. Then if $N$ and $M$ are graded $A$ modules, and $N\to M$ a morphism, it induces a map $\tilde{N}\to \tilde{M}$. One shows that if $N\to M$ is injective (resp surjective, bijective) in degrees $n\gg 0$, then $\tilde{N}\to \tilde{M}$ is an injection (resp surjection, isomorphism).
Together the last two paragraphs are enough to imply your desired statements, as the isomorphisms $\widetilde{\Gamma^*(\mathscr{M})}\cong \mathscr{M}$ get you a splitting map $\mathcal{G}\to \mathcal{H}$ from that on the level of $\Gamma^*$. Then by the last paragraph, this identifies $\mathcal{G}$ with $\mathcal{H}\oplus \mathcal{F}$ as it does on the level of $\Gamma^*$.


TITLE: Solution to the Maxwell's equation
QUESTION [0 upvotes]: Question goes as

Show that the solutions to the Maxwell's equations
  $$ \nabla \times \vec H = \frac 1 c \frac{\partial \vec E}{\partial t}+\frac {4\pi} c \vec J, \hspace{ 2 mm} \nabla \times \vec E = -\frac 1 c \frac{\partial \vec H}{\partial t}, \hspace{ 2 mm} \nabla \cdot \vec H = 0, \hspace{ 2 mm} \nabla \cdot \vec E = 4 \pi \rho $$  are given by
  $$ \vec E = -\nabla \phi - \frac 1 c \frac{\partial \vec A}{\partial t}, \hspace{2mm} \vec H = \nabla \times  \vec A,$$
  where $ \phi$  and  $\vec A$ are scalar potential and vector potential respectively which satisfy the equations
  $$ 1) \nabla \cdot \vec A + \frac 1c \frac{\partial \phi}{\partial t} = 0 $$
  $$2) \nabla^2 \phi - \frac 1 {c^2} \frac{\partial ^2 \phi}{\partial t^2} = -4\pi \rho $$
  $$3) \nabla^2 \vec A - \frac{1}{c^2}\frac{\partial ^2 \vec A}{\partial t^2}= -\frac {4\pi} c \vec J$$

Hint requested thank you!!
Also what kind of regions is described by
$$ \nabla \cdot \vec E = 4 \pi \rho?$$
It is uniformly charged region?

REPLY [1 votes]: Unless I'm missing something, all you have to do is substitute the solution you're given into Maxwell's equations and confirm the solution satisfies the equations.
For example you're given the solution $\vec H = \nabla \times  \vec A$. Substitute this into $\nabla \cdot \vec H = 0$ and you get:
$$ \nabla \cdot (\nabla \times  \vec A) = 0 $$
and this is true because div of curl is always zero.
Re comment: we make a point of not doing homework for you because your professor wouldn't be amused. You need to start working away at the equations and post here if you get stuck, giving the details of what's blocking you.
I showed the solution you're given satisfies $\nabla \cdot \vec H = 0$ because that's pretty trivial. I suggest you next tackle $\nabla \cdot \vec E = 4 \pi \rho$. Note that you're given:
$$ \nabla^2 \phi - \frac 1 {c^2} \frac{\partial ^2 \phi}{\partial t^2} = -4\pi \rho  $$
so you just have to show that:
$$ -\nabla \cdot \vec E = \nabla^2 \phi - \frac 1 {c^2} \frac{\partial ^2 \phi}{\partial t^2} $$
You can do this using the information you're given in the question. A clue is that div of grad is del squared.


TITLE: 2 x 2 Phase Portrait for 2 x 2 Linear System with Real Coefficients. Boyce, p395, Figure 7.5.4a
QUESTION [0 upvotes]: The given general solution for some linear system is
  $
x= c_{2} \mathbf{ x^{(1)} }(t) + c_{2} \mathbf{ x^{(2)} }(t) =
c_{1}\
\left(\begin{array}{l}
1\\
\sqrt{2}
\end{array}\right)\
e^{-t}+
c_{2}
\left(\begin{array}{l}
-\sqrt{2}\\
1
\end{array}\right)\
e^{-4t} \qquad (21)
$.
Graphs of the solution (21) for several values of $c_{1}$ and $c_{2}$ are shown in Figure 7. $5.4a$.
  The solution $x^{(1)}(t)$ approaches the origin along the line $x_{2}=\sqrt{2}x_{1}$, and the solution $x^{(2)}(t)$ approaches the origin along the line $x_{1}=-\sqrt{2}x_{2}$....

(1) Please explain this last sentence? I know $\color{brown}{ x_2 = \sqrt{2}x_{1} \iff }
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
= x_1 \begin{bmatrix}
1 \\
\sqrt{2}
\end{bmatrix}$
and
$ \color{green} { x_{1}=-\sqrt{2}x_{2} } \iff
 \begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
= x_2 \begin{bmatrix}
-\sqrt{2} \\
1
\end{bmatrix}$

As $ t\rightarrow\infty$, the solution $x^{(2)}(t)$ is negligible compared to $x^{(1)}(t)$ .

(2) Why 'negligible'? I know when $k > 0$, $\lim_{t \to \infty} e^{-kt} = 0 $

Thus, unless $c_{1}=0$, the solution (21) approaches the origin tangent to the line $x_{2}=\sqrt{2}x_{1}$.

(3) Please explain this last sentence? I know if $c_1 =0 $ then any solution must be
$ c_{2} \mathbf{ x^{(2)} }(t)$

... If the eigenvalues were positive rather than negative, then the trajectories would be similar but traversed in the outward direction.

(4)  Please explain this last sentence?

REPLY [0 votes]: The first equation tells us (implicitly) $$\mathbf{x}^{(1)}(t)= \begin{pmatrix} 1 \\ \sqrt{2}\end{pmatrix}e^{-t}, \quad \mathbf{x}^{(2)}(t)= \begin{pmatrix} -\sqrt{2}\\ 1\end{pmatrix}e^{-4t}.$$

This means that, starting with $t=0$ and using $e^0 = 1$, we have initial values
$$\mathbf{x}^{(1)}(0)= \begin{pmatrix} 1 \\ \sqrt{2}\end{pmatrix}, \quad \mathbf{x}^{(2)}(0)= \begin{pmatrix} -\sqrt{2}\\ 1\end{pmatrix}.$$
As $t$ increases from $0$ towards $\infty$, the factors $e^{-k}$ and $e^{-4k}$ decrease from $1$ to $0$, i.e. the solutions $\mathbf{x}^{(1)}(t)$ and $\mathbf{x}^{(2)}(t)$ run along the red/green line. In view of this, the black arrows are misleadingly positioned.
This argument of Boyce is a bit handwavy here, and my explanation is going to be similarly handwavy - I hope that this makes it more understandable. The idea is that $e^{-4t}$ converges faster to $0$ than $e^{-t}$ does. I would not call it negligible, but the two solutions converge to zero at different rates. One way to see this is to write $c_1 \mathbf{x}^{(1)}(t)+ c_2 \mathbf{x}^{(2)}(t)$ as
$$e^{-t}\left(c_1\begin{pmatrix} 1 \\ \sqrt{2}\end{pmatrix} + c_2\begin{pmatrix} -\sqrt{2}\\ 1\end{pmatrix}e^{-3t}\right) .$$
As $t$ increases, the second summand in the brackets goes to $0$, while the first one remains.
Because of this difference in convergence speed, a linear combination of the two solutions gets very close to $\mathbf{x}^{(1)}(t)$, as the part with $\mathbf{x}^{(2)}(t)$ 'converges away'. You can see this behavior in the blue arrows, which are the general solutions of the system. All of them approach the origin along the direction of $\pm \begin{pmatrix} 1 \\ \sqrt{2}\end{pmatrix}$ eventually - the part of $\mathbf{x}^{(2)}(t)$ dimishes away. I am aware that this explanation is similarly handwavy as the original one.
This is lacking context. How are the solutions related to eigenvalues (of which matrix)?


TITLE: Gerbes for a cyclic group. (or maybe G_m too)
QUESTION [8 upvotes]: Let μn be the group scheme of n-th roots of unity. If X is a scheme and L is a line bundle on X, then I can construct a μn-gerbe Y over X by letting the S-points of Y be a S-point of X, a line bundle M on S and an isomorphism between the n-th tensor power of M and the pullback of L to S. Can anyone provide examples not of this form?
Commentary: It looks like I'm taking the image of an element of H1(X,Gm) in H2(X,μn) under the long exact sequence associated to 0-->μn-->Gm-->Gm-->0. Thus examples of gerbes for the multiplicative group Gm will likely be relevant, so people providing such examples will also be appreciated.

REPLY [8 votes]: A bit of a response to your "Commentary":
As you point out, the failure of your construction to hit all $\mu_n$-gerbes is governed by the exact sequence
$H^1(X, \mathbb{G}_m) \to H^2(X, \mu_n) \to H^2(X,\mathbb{G}_m)[n] \to 0$ so answering your question is related to producing torsion $\mathbb{G}_m$-torsors.
The question of doing so has been studied as part of the theory of Brauer groups:
Let $Br(X)$ ("Brauer group") denote the group of Azumaya algebras, which are a generalization of the central simple algebras over a field (that is the classical Brauer group).
Let $Br'(X)$ ("Cohomological Brauer group") denote the torsion part of $H^2(X, \mathbb{G}_m)$.
In the case of $X = Spec k$, $k$ a field, the equivalence of these two groups is classical and they can be computed in various cases of number-theoretic interest (e.g., number fields/local fields/finite fields).  In this case, $H^1(X, \mathbb{G}_m)=0$ by Hilbert's Theorem 90, and yet there are plenty of examples where $Br(X)$ is very much non-trivial.  Grothendieck studied the relation between $Br(X)$ and $Br'(X)$ in general in Dix Exposes.  The upshot is that there is an injective map $Br(X) \to Br'(X)$ and it is an isomorphism in reasonable cases (e.g., I think $X$ quasi-projective over a field).  (See Dix Exp, or Ch. IV of Milne's "Etale Cohomology".)

I won't say more about the general picture, but I'll work out in detail the simple case of $X = Spec \mathbb{R}$.  In this case, $Br(Spec \mathbb{R}) = \mathbb{Z}/2\mathbb{Z}$ generated by the class of the usual quaternions, viewed as a central simple algebra over $\mathbb{R}$.  We can give a geometric description of the resulting $\mathbb{G}_m$-torsor:
Start with the smooth plane conic $C = Proj \mathbb{R}[x,y,z]/(x^2+y^2+z^2)$.  It's a smooth genus $0$ curve, but has no $\mathbb{R}$-points and so is not isomorphic to $\mathbb{P}^1_{\mathbb{R}}$. However, after base-change to $\mathbb{C}$ it attains a point and so becomes isomorphic to $\mathbb{P}^1_{\mathbb{C}}$; such a Galois-twisted form of $\mathbb{P}^n_{\mathbb{C}}$ is known as a Brauer-Severi variety and the elements of the Brauer-group (of a field) can also be thought of as corresponding to them (the group structure is then a bit strange).  Since $Aut(\mathbb{P}^n) = PGL_{n+1}$, these correspond to $PGL_{n+1}$-torsors and the relation to $\mathbb{G}_m$-torsors is via the exact sequence for $PGL_{n+1}=GL_{n+1}/\mathbb{G}_m$.  So, a $T$-point of the corresponding $\mathbb{G}_m$-torsor for a $\mathbb{R}$-scheme $T$ consists of the following data:
It is a rank $2$ vector bundle $V$ over $T$, together with an isomorphism of $T$-schemes $C_T \simeq \mathbb{P}(V)$ where $C_T = C \times_{Spec \mathbb{R}} T$ is the pullback of our genus $0$ curve to $T$, and $\mathbb{P}(V)$ is the associated projective space (here $\mathbb{P}^1$) bundle of our vector bundle $V$.
Why is this a $\mathbb{G}_m$-gerbe? Well, $\mathbb{P}(V) \simeq \mathbb{P}(V')$ iff $V$ and $V'$ differ by tensoring by a line bundle.  The gerbe is non-trivial since it has no $\mathbb{R}$-points, since $C$ itself is not isomorphic to projective space.  It has $\mathbb{C}$-points because the base-change is isomorphic to projective space.


text:
\begin{document}

\subjclass[2010]{Primary 35R11. Secondary 45K05, 45M05, 35C15, 26A33, 35B40, 33C60}

\keywords{non-local diffusion, Harnack inequality, Riemann-Liouville derivative, fractional Laplacian, fundamental solution, $H$ functions, asymptotics}

\begin{abstract}
We settle the open question concerning the Harnack inequality for globally positive solutions to non-local in time diffusion equations by constructing a counter-example for dimensions \(d\ge\beta\), where \(\beta\in(0,2]\) is the order of the equation with respect to the spatial variable. The equation can be non-local both in time and in space but for the counter-example it is important that the equation has a fractional time derivative. In this case, the fundamental solution is singular at the origin for all times \(t>0\) in dimensions \(d\ge\beta\). This underlines the markedly different behavior of time-fractional diffusion compared to the purely space-fractional case, where a local Harnack inequality is known.

The key observation is that the memory strongly affects the estimates. In particular, if the initial data $u_0 \in L^q_{loc}$ for $q$ larger
than the critical value $\tfrac d\beta$ of the elliptic operator \((-\Delta)^{\beta/2}\), a non-local version of
the Harnack inequality is still valid as we show.

We also observe the {\em critical dimension phenomenon} already known from other contexts: the diffusion behavior is substantially different in higher dimensions than $d=1$ provided \(\beta>1\), since we prove that the local Harnack inequality holds  if \(d<\beta\).
\end{abstract}

\maketitle

\section{Introduction}

We construct a counter-example for the local Harnack inequality for globally positive solutions to the time-fractional heat equation
\begin{equation}\label{equation}
\partial_t^\alpha(u-u_0) +(- \Delta)^{\beta/2} u = 0, \quad 0 < \alpha <1,\quad 0<\beta\le 2,
\end{equation}
in $(0,T) \times \Rn$ for $d \ge \beta$. We also show that the result is
still valid in one dimension provided \(\beta>1\), at least for global solutions. Instead of the local formulation, a non-local version of
the Harnack inequality  is true also in higher dimensions. Indeed, the initial data plays a role, as
the result remains true if the superharmonic function induced by the nonnegative initial
data satisfies the Harnack principle. Thus, the memory has a substantial effect on
the properties of the solutions, a phenomenon not seen in the case of the
space-fractional diffusion equation
\begin{equation}\label{space_fractional}
\partial_t u + (-\Delta)^{\beta/2} u =0.
\end{equation}

In particular, if $u_0 \in L^q_{loc}$ for  $q > \tfrac d\beta$, we prove a version of the
classical local Harnack principle with the constant depending on $u_0$. We also observe the {\em critical dimension
phenomenon}, which is not seen in the case of  equation~\eqref{space_fractional}, but is
known for time-fractional diffusion from~\cite{KempSiljVergZach16, KempSiljZach16}. These
results underline the fact that equation~\eqref{equation} has a substantially different
behavior from that of~\eqref{space_fractional}. The phenomena we see, as well as the
methods used, are often completely different from the ones for
equation~\eqref{space_fractional}.

In the purely space-fractional case the Harnack inequality is known and was first proven by heat kernel estimates by Bass and Levin in~\cite{BassLevi02} as well as Chen and Kumagai in~\cite{ChenKuma03}. Very recently a new proof for viscosity solutions was given by Chang-Lara and D\'avila in~\cite{ChLaDavi16}. In the variational formulation the question has been studied for kernels vanishing outside the local diagonal in~\cite{BarlBassChenKass09} and for viscosity solutions when the Harnack constant is allowed to depend on the non-local tail in~\cite{KimLee13}. For elliptic theory we refer to the work of Caffarelli and Silvestre in~\cite{CaffSilv07, CaffSilv09} and Kassmann in~\cite{Kass11} and the references given therein. See also~\cite{BonfVazq13, CaffSilv12}.

The study of equation~\eqref{equation} is firmly rooted in applications. The
time-fractional diffusion equation is closely related to a so-called continuous time random
walk (CTRW) model for particle diffusion and it has become one of the standard physics
approaches to model anomalous diffusion
processes~\cite{HenrLangWear06,MeerSiko10,MetzKlaf00,MeerBensScheBeck02}. For a
detailed derivation of this equation from physics principles and for further applications
of such models we refer to the expository review article of Metzler and Klafter
in~\cite{MetzKlaf00}. For the connection of~\eqref{equation} to the related
stochastistic process we refer to monograph~\cite{MeerSiko10} and the extensive bibliography
therein. Moreover, equations of the form (\ref{equation}) and nonlinear variants of them
appear also in modelling
of dynamic processes in materials with {\em memory}. Examples are given by the theory of heat
conduction with memory, see~\cite{Prus13} and the references therein, and the
diffusion of fluids in porous media with memory~\cite{Capu99, Jaku02}.

We point out that the non-local in time term in (\ref{equation}), with $\partial_t^\alpha$ being the Riemann-Liouville
fractional derivation operator, coincides (for sufficiently smooth $u$) with the Caputo fractional derivative of $u$,
see (\ref{caputo}) below. The formulation with the Riemann-Liouville fractional
derivative has the
advantage that a priori less regularity is required on $u$ to define the nonlocal operator. In particular, our formulation is exactly the one which naturally arises from physics applications, see for instance~\cite[equation (40)]{MetzKlaf00}.

\section{Preliminaries and main results} \label{section:preliminaries}

\subsection{Notations and definitions}

Let us first fix some notations. The Riemann-Liouville fractional integral of order \(\alpha \ge 0\) is defined for $\alpha = 0$ as \(J^0:=I\), where \(I\) denotes the identity operator, and for $\alpha >0$ as

\[
J^\alpha
f(t)=\frac{1}{\Gamma(\alpha)}\int_0^t(t-\tau)^{\alpha-1}f(\tau)d\tau = (g_\alpha * f)(t),
\]

where
\[
g_\alpha(t)=\frac{t^{\alpha-1}}{\Gamma(\alpha)}
\]
is the Riemann-Liouville kernel and $*$ denotes the convolution in time on the positive half-line. We denote the
convolution in space by $\star$.

The Riemann-Liouville fractional derivative of order $0<\alpha  < 1$ is
defined by

\[
\partial_t^{\alpha}f(t)=\frac{d}{dt} J^{1-\alpha}f(t)=\frac{d}{dt}\frac{1}{\Gamma(1-\alpha)}
\int_0^ { t } (t-\tau)^ { -\alpha } f(\tau)d\tau.
\]

Observe that for sufficiently smooth $f$ and $\alpha\in (0,1)$
\begin{equation} \label{caputo}
\partial_t^{\alpha}(f-f(0))(t)=(J^{1-\alpha} f')(t)=:{}^cD_t^\alpha f(t),
\end{equation}
the so-called Caputo fractional derivative of $f$.
In case $\alpha=1$, we have the standard time derivative.

Let
\[
\widehat{u}(\xi)=\mathcal{F}(u)(\xi)=(2\pi)^{-d/2}\int_{\Rn}\mathrm{e}^{
-ix\cdot\xi}f(x)dx
\]
and
\[
\mathcal{F}^{-1}(u)(\xi):=\mathcal{F}(u)(-\xi)
\]
denote the Fourier and inverse Fourier transform of $u$, respectively.

\subsection{The fundamental solution}
We recall that under suitable conditions on the initial data the classical solution of the Cauchy problem
\begin{equation}\label{cauchy_problem}
\left\{\begin{array}{r@{\,\,=\,\,}l}
\partial_t^\alpha(u-u_0) +(-\Delta)^{\beta/2} u & 0 \\
u(0,x) & u_0(x)
\end{array}\right.
\end{equation}
is given by the representation formula
\begin{equation}\label{solution_repr}
u(t,x) = \int_{\Rn} Z(t, x-y ) u_0 (y) \, dy

\end{equation}
where $Z$ is the fundamental solution of the problem. We call the function \(u\) defined
by~\eqref{solution_repr} the {\em mild solution} of the Cauchy
problem~\eqref{cauchy_problem} whenever the integral in~\eqref{solution_repr} is well
defined. We remark that under appropriate regularity conditions on the data, existence and
uniqueness of strong $L^p$-solutions of the Cauchy problem~\eqref{cauchy_problem} follows
from the results in~\cite{Zach05}, which are formulated in the framework
of abstract parabolic Volterra
equations, see also the monograph~\cite{Prus13}.

The fundamental solution \(Z\) can be expressed in terms of Fox $H$-functions which are quite a general class of special functions. Without going to excessive details concerning the theory of Fox
$H$-functions, for which  we refer the reader to~\cite{KilbSaig04}, we just give a very brief review, see also~\cite{KempSiljZach16}.

To simplify the notation we introduce
\[
(a_i,\alpha_i)_{1,p}:=((a_1,\alpha_1),(a_{2},\alpha_{2}),\dots,(a_p,
\alpha_p))
\]
for the set of parameters appearing in the definition of Fox $H$-functions, which are defined via Mellin-Barnes type
integrals as

\[
H^{mn}_{pq} (z):= H^{mn}_{pq}\big[ z \big| \begin{smallmatrix}
                           (a_i,\alpha_i)_{1,p}\\
(b_j,\beta_j)_{1,q}
                                           \end{smallmatrix}
\big] =\frac{1}{2\pi i}\int_{\mathcal{L}}\mathcal{H}^{mn}_{pq}(s)z^{-s}
ds,
\]

where

\[
\mathcal{H}^{mn}_{pq}(s)=\frac{\prod_{j=1}^m \Gamma(b_j+\beta_j s)\prod_{i=1}^n
\Gamma(1-a_i-\alpha_i s)}{\prod_{i=n+1}^p\Gamma(a_i+\alpha_i
s)\prod_{j=m+1}^q\Gamma(1-b_j-\beta_j s)}
\]

is the Mellin transform of \(H_{pq}^{mn}\) and
\(\mathcal{L}\) is an infinite contour in the complex plane which separates
the poles

\[
b_{jl}=\frac{-b_j-l}{\beta_j}\quad (j=1,\dots,m;\, l=0,1,2,\dots)
\]

of the Gamma function \(\Gamma(b_j+\beta_j s)\) to the left of \(\mathcal{L}\)
and the poles

\[
a_{ik}=\frac{1-a_i+k}{\alpha_i} \quad (i=1,\dots,n;\, k=0,1,2,\dots)
\]

of $\Gamma(1-a_i-\alpha_i s)$ to the right of \(\mathcal{L}\).

The fundamental solution satisfying the Cauchy problem~\eqref{cauchy_problem} with the
initial datum \(u_0\) given by the Dirac delta distribution, \(u_0(x)=\delta(x)\), can be
represented in terms of a Fox $H$-function as
\begin{equation}\label{fund_sol_repr}
Z(t,x)=\pi^{-d/2}|x|^{-d}H_{23}^{21}\bigl[ 2^{-\beta}
t^{-\alpha}|x|^{\beta} \big| \begin{smallmatrix} (1,1),
&(1, \alpha)\\
(\frac{d}{2},\frac{\beta}{2}),& (1,1 ), &(1, \frac{\beta}{2})
\end{smallmatrix} \big].
\end{equation}

Our analysis requires the sharp asymptotics of the fundamental solution, which is collected into the following lemma. For the proof we refer to~\cite[Lemma 3.3]{KempSiljZach16}. See also~\cite[equation (3.7)]{EideKoch04}
and~\cite[Theorem 1.10]{KilbSaig04}.

\begin{lemma}\label{fund_sol_asympt}

Let $\alpha\in(0,1]$, $\beta\in(0,2]$ and $d \in \mathbb{Z}_+$. Denote \(R:=2^{-\beta}|x|^\beta
t^{-\alpha}\). The function $Z$ has the following asymptotic behavior:
\begin{itemize}

\item[(i)] If \(R\leq 1\), then
\[
Z(t,x)\sim \begin{cases}
             t^{-\alpha d/\beta}, &\textrm{if $\alpha=1$, or $\beta > d$ and
$0<\alpha<1$},\\
t^{-\alpha}(|\log(|x|^\beta t^{-\alpha})|+1), &\textrm{if $\beta=d$ and
$0<\alpha <1$},\\
t^{-\alpha}|x|^{-d+\beta} &\textrm{if $0<\beta < d$ and $0<\alpha<1$}.

            \end{cases}
\]

\item[(ii)] If \(R\ge 1\), then
\[
Z(t,x)\sim
\begin{cases}
t^{\alpha}|x|^{-d-\beta}, &\textrm{if \(\beta<2\)},\\
t^{-\frac{\alpha d}2}R^{\frac{d(\alpha-1)}{2(2-\alpha)}}e^{-\sigma
R^{\frac1{2-\alpha}}}, &\textrm{if \(\beta=2\)},
\end{cases}
\]
where \(\sigma=(2-\alpha)\left(\alpha^\alpha/4\right)^{\frac{1}{2-\alpha}}\).
\end{itemize}

\end{lemma}
In Lemma~\ref{fund_sol_asympt} and in the sequel we use the notation
\[
a\sim b\quad\textrm{if there exists a constant $c>0$ such that}\quad \frac{1}{c} a\le b\le c a.
\]
Lemma~\ref{fund_sol_asympt} guarantees that we have not only upper bounds but also lower bounds for the fundamental solution \(Z\). The upper and the lower bounds are the same up to some multiplicative constants. In our analysis we are not interested in the exact constants. Therefore we give our estimates in a form \(a\le C b\), or even more briefly in a form \(a\lesssim b\).

Although the Fox $H$-functions are rather complicated objects, the Fourier transform with
respect to the spatial variable of the fundamental solution \(Z(t,\cdot)\) takes a much simpler form. In fact,
\begin{equation}\label{fund_fourier_trans}
\widehat{Z}(t,\xi)=(2\pi)^{-d/2} E_{\alpha}(-\abs{\xi}^\beta t^{\alpha}),
\end{equation}
where $E_{\alpha}$ is the Mittag-Leffler function, which is defined by
\[
E_{\alpha}(z)=\sum_{k=0}^\infty \frac{z^k}{\Gamma(1+\alpha k)},\quad z\in \mathbb{C}.
\]
It is known that for \(0<\alpha<1\) the Mittag-Leffler function behaves as
\begin{equation}\label{ML2_estimate}
E_{\alpha}(-x) \sim \frac{1}{1+x}, \quad x\in\mathbb{R}_+.
\end{equation}
For details
we refer to~\cite[Chapter 18]{Erdelyi53}.

\subsection{Main results}

The question about a full, local Harnack inequality
\begin{equation}\label{local_harnack}
\sup_{Q^-} u \le C \inf_{Q^+} u,
\end{equation}
for nonnegative solutions of equations of type~\eqref{equation} and for some time-lagged properly scaled cylinders $Q^-$ and $Q^+$ has been a somewhat
longstanding open question. The
constant $C>0$ in the classical Harnack estimate~\eqref{local_harnack} should neither depend on the
initial datum $u_0$ and the solution $u$ nor on the scaling. It turns out that for dimensions \(d\ge \beta\) this is not the case
for the solution of~\eqref{equation}. Our first result is the construction of a
counter-example for such a Harnack principle. We have the following theorem.

\begin{theorem}\label{counter_example}
Let $\alpha\in(0,1)$, $\beta\in(0,2]$ and $d \ge \beta$. There exists a sequence $(u_n)_{n=1}^\infty$ of smooth, positive
solutions to equation~\eqref{equation} in $(0,\infty)\times \Rn $ (with $u_n|_{t=0}$ instead of $u_0$) such that the ratio
\begin{equation*}
\frac{u_n(t_1, 0)}{u_n(t_2, x_0)} \to \infty \quad \text{as } n \to \infty,
\end{equation*}
for each pair of time levels $t_1, t_2 >0$ and for each $x_0\neq 0$. This
contradicts the local Harnack inequality~\eqref{local_harnack}.

\end{theorem}

The proof of Theorem~\ref{counter_example} is based on the detailed asymptotics of the fundamental solution \(Z\). We also provide another counter-example, which uses the Fourier transform.

\begin{theorem}\label{counter_example_boundedness}
Let $\alpha\in(0,1)$ and $\beta\in(0,2]$. There exists a sequence \((u_n)_{n=1}^\infty\) of smooth, positive solutions to equation~~\eqref{equation} in
$(0,\infty)\times \Rn $ (with $u_n|_{t=0}$ instead of $u_0$)
such that for all \(t_1,t_2>0\) there holds
\[
\frac{u_n(t_1,0)}{\|u_n(t_2,\cdot)\|_{L^p(\Rn)}}\to\infty\quad\text{as } n\to\infty,
\]
if
\begin{itemize}
\item[(i)] \(d>\beta\) and \(1\le p<\frac{d}{\beta}\), or
\item[(ii)] \(d=\beta\) and \(p=1\).
\end{itemize}
\end{theorem}

In the classical local case the proof of the Harnack inequality consists of the following steps:
\begin{itemize}
\item[(a)] boundedness of subsolutions ('\(L^\infty-L^p\)-- estimate'),
\item[(b)] the weak Harnack inequality for supersolutions ('\(L^p-L^{-\infty}\)-- estimate').
\end{itemize}

In~\cite{Zach12} the weak Harnack inequality is proved in the case \(\beta=2\) for all spatial dimensions \(d\)
and all positive $p<\frac{2+d\alpha}{2+d\alpha-2\alpha}$, with the upper bound for $p$ being optimal.
In~\cite{JiaPengYang17} the weak Harnack inequality is proved in the case \(\frac{d}{2}<\beta<2\) for the spatial dimensions \(d=2,3\). Theorem~\ref{counter_example_boundedness} reveals that the boundedness estimate not only of subsolutions but also
of solutions fails in the cases
\begin{itemize}
\item[(1)] \(1<\beta\le 2\) and \(d\ge 2\),
\item[(2)] \(0<\beta\le 1\) and \(d\ge 1\).
\end{itemize}
The boundedness of solutions is an open problem in the case \(1<\beta\le 2\) and \(d=1\). Hence for subdiffusion there are cases in which the weak Harnack inequality is true but the boundedness of solutions fails. Apart from the cases \(\beta=2\) and \(d\ge 1\) or \(\frac{d}{2}<\beta<2\) and \(d=2,3\) even the weak Harnack inequality is open.

Concerning the Harnack inequality for non-local diffusion equations, there are some positive results. In~\cite{ChLaDavi16} the Harnack inequality is proved for a generalization of~\eqref{space_fractional} in the case \(1\le\beta <2\). For the non-local in time equations, the only positive result up to now is the Harnack inequality for the Riemann-Liouville fractional derivation operator proved in~\cite{Zach11}, which can be regarded as the case '\(d=0\)'. The case \(0<\alpha<1\) in dimensions \(d\ge 1\) has been an open problem until now. We show in Theorem~\ref{harnack_local_in_time} that the Harnack inequality remains to hold for~\eqref{equation} for \(0<\alpha<1\) in the one-dimensional case provided \(\beta>1\). Having positive results, one could think that the Harnack inequality could hold also for the whole range of parameters \(\alpha\) and \(\beta\), but this is not the case as Theorem~\ref{counter_example} shows in dimensions \(d\ge\beta\).

The local Harnack inequality being settled for dimensions $d \ge \beta$, it is a natural
question, what is a proper replacement of this result. Unlike for most non-local
equations, the non-local memory term in the time-fractional diffusion equation has a
substantial effect on the behavior of solutions. Indeed, large enough historical events
affect the solution indefinitely for all positive times. From the construction of
Theorem~\ref{counter_example} it is evident that if the initial data is pathological
enough, then the solution will never satisfy a local Harnack inequality. If the initial
data, however, is good enough, then a Harnack principle still holds.

Moreover, the theory is also very different depending on the dimension. In one dimension, we are still able to prove the classical Harnack inequality for global solutions, but in higher dimensions the counter-example prevails. This corresponds to the {\em critical dimension phenomenon} observed in the case of the decay estimates for the equation~\eqref{equation} in~\cite{KempSiljVergZach16}. See also~\cite{KempSiljZach16}.

For dimensions $d \ge\beta$ the equation~\eqref{equation} preserves the
properties of the initial data in the sense that if the superharmonic function induced by
the nonnegative initial data satisfies a Harnack principle, also the solution of the
Cauchy problem
will have a similar behavior. For making this rigorous, we consider the
potential
\[
u\mapsto G\star u
\]
with the kernel
\begin{align}\label{riesz_kernel}
G(t,x) =
\frac{1}{c(d,\beta)}\begin{cases}
|x|^{\beta-d}, \quad &\text{for } d > \beta \\
\log|x|, \quad &\text{for } d =\beta,
\end{cases}
\end{align}
where \(c(d,\beta)\) is a normalization constant. Note that for \(\beta=2\) the function \(G\) is nothing but the Newtonian kernel and the corresponding potential is the Newtonian potential. Now $G
\star u_0$ induces a superharmonic function in $\Rn$, that is, setting $v=G\star u_0$ for a nonnegative and sufficiently
smooth initial data $u_0$ we obtain that $(-\Delta)^{\beta/2}v\ge 0$ in $\Rn$.

We have the following theorem. Here by $\chi_A(x)$ we denote the characteristic function of the set $A\subset \Rn$.

\begin{theorem}\label{main_theorem}  Let $\alpha\in(0,1)$, $\beta \in(0,2]$ and $r >0$.
Suppose $u$ is a mild solution of the Cauchy problem~\eqref{cauchy_problem} in
$[0,T]\times \Rn$ with the nonnegative and sufficiently smooth initial data $u_0\not\equiv0$.  Suppose further that $t_1$ and $t_2$ satisfy condition
\begin{equation}\label{time_cond}
(2r)^{\beta/\alpha} \le  t_1 < t_2 \le t_1+(2r)^{\beta/\alpha} \le T.
\end{equation}
Then
there exists a constant $C = C(d, \alpha,\beta ,t_2/t_1) >0$ such that for all
$x_1, x_2 \in B_{r}(0)$ there holds
\begin{equation*}
u(t_1,x_1) \le C \left[1+\frac{[G \star (u_0\chi_{B_{r_1}(x_1)})](x_1)}{[G \star
(u_0\chi_{B_{r_2}(x_2)})](x_2)}\right] u(t_2,x_2), \quad \text{if }  d>\beta,
\end{equation*}
and, if \(d=\beta\),
\begin{equation*}
u(t_1,x_1)\le C\left[1+\frac{\big[(c+\alpha \log t_1+\beta c(d,\beta)G) \star
\big(\chi_{B_{r_1}(x_1)}u_0\big)\big](x_1)}{\big[(c+\alpha\log t_2+\beta c(d,\beta)G) \star
\big(\chi_{B_{r_2}(x_2)}u_0\big)\big](x_2)}
\right] u(t_2, x_2),
\end{equation*}

where \(c=1+\beta\log 2\), \(c(d,\beta)\) is the constant in~\eqref{riesz_kernel} and $r_i =2 t_i^{\alpha/\beta}, i=1,2$. Moreover, the constant \(C\) blows up, as \(t_2\to t_1\), when \(\beta=2\).
\end{theorem}

\begin{remark}\label{time_lag_remark}
Observe that by elliptic regularity theory, if $u_0 \in L^q_{loc}(\Rn)$ with $q >
\tfrac d\beta$, then $G \star (u_0\chi_{B_{r_1}(x_1)})$ is locally bounded and continuous. In this case the convolution factor above is always bounded
as long as $B_{r_1}(x_1) \subset B_{r_2}(x_2)$ which is true for all $(t_1, x_1), (t_2,
x_2)$ satisfying
\begin{equation*}
t_2^{\alpha/\beta} \ge t_1^{\alpha/\beta} + \frac{|x_1-x_2|}{2}
\end{equation*}
and~\eqref{time_cond}. Therefore, the Harnack inequality is always satisfied for such an initial data after a proper time-lag corresponding to the scaling of the equation.
\end{remark}

We would also like to point out that the proof of the above Theorem is rather sharp consisting almost exclusively of identities rather than estimates. Indeed, the crucial estimates rely on the asymptotic behavior of the fundamental solution and these estimates provide sharp behavior not only as upper bounds, but also as lower bounds.

In the above formulation the effect of the initial data is readily observable. The
superharmonic function induced by the initial data fully determines the Harnack constant.
Moreover, it is easy to see that if this superharmonic function satisfies a proper Harnack
inequality, then this is true also for the solution of the evolution equation. In
particular, if one is willing to make some more additional assumptions on the initial
data, we are able to obtain a more classical formulation of the Harnack inequality in the
form of the following corollary.

\begin{corollary}\label{main_corollary}
Let the assumptions of Theorem~\ref{main_theorem} hold. Suppose also that the initial data satisfies the Harnack inequality
\begin{equation}\label{initial_harnack}
\sup_{B_{r_1}(x_1)}u_0 \le H_0 \inf_{B_{r_2}(x_2 )} u_0,
\end{equation}
for $x_1, x_2 \in B_r(0)$ and for some constant $H_0$. Then there exists a constant $C =
C(d, \alpha,\beta ,H_0, t_2/t_1) >0$ such that there holds
\[
u(t_1,x_1) \le C u(t_2,x_2).
\]
\end{corollary}

In the case \(\beta>d\) the situation is simpler as the counter-example we build in Theorem~\ref{counter_example} is not valid anymore. In this case the Harnack inequality~\eqref{standard_Harnack} takes the standard form.
We have the following theorem. We formulate the result in a form, which covers also the space-fractional case~\eqref{space_fractional}, since $\alpha=1$ is included in Lemma~\ref{fund_sol_asympt}.

\begin{theorem}\label{harnack_local_in_time}
Let $\alpha\in(0,1],\beta\in(0,2),r>0$ and \(x_1,x_2\in B_r(0)\). Assume that either
\begin{itemize}
\item[(a)] $\alpha\neq 1$ and $1=d<\beta$, or
\item[(b)] $\alpha=1$, $\beta\neq 2$ and $d\in\mathbb{Z}_+$.
\end{itemize}
Suppose $u$ is a mild solution of the Cauchy
problem~\eqref{cauchy_problem} in $[0,T]\times \Rn$.
Then for all \(t_1,t_2\) satisfying

\[
(2r)^{\beta/\alpha} \le t_1 < t_2  \le t_1+(2r)^{\beta/\alpha} \le T.
\]

there exists a constant $C=C(t_2/t_1,\alpha,\beta)>0$ such that
\begin{equation}\label{standard_Harnack}
u(t_1,x_1)\le C u(t_2,x_2),
\end{equation}

Moreover, the constant blows up, as \(t_2\to t_1\), when \(\beta=2\).
\end{theorem}

\begin{remark}

\begin{itemize}

\item[(i)] The proofs of Theorems~\ref{main_theorem} and~\ref{harnack_local_in_time} reveal that the time-lag is needed only for handling the exponential term, which is missing in the case \(\beta\neq 2\). In particular, the time lag is not needed in the case \(\alpha=1\) and \(\beta\neq 2\), which was observed also in~\cite{BonfSireVazq17}. Heuristically this means that the diffusion is so fast that  a lot of heat is diffused far away from the source at all times \(t>0\). In the probabilistic framework this is a consequence of a fat tail at infinity of the probability distribution \(Z(t,\cdot)\).

\item[(ii)] In the fully non-local case \(\alpha\neq 1\) and \(\beta\neq 2\) the diffusion is at the same time slow and fast. It is slow, since the fundamental solution has singularity at \(x=0\) for all times \(t>0\). On the other hand, it is fast since the fundamental solution \(Z(t,\cdot)\) has a fat tail at infinity.

\end{itemize}
\end{remark}

\section{The counter-examples}

We begin with the proof of Theorem~\ref{counter_example}. The proof relies on the
properties of the fundamental solution. Indeed, the fundamental solution is not a smooth
function and is merely superparabolic instead of being a proper solution of the problem.
This resembles the elliptic Newtonian potential and the case of some nonlinear parabolic equations, such as the porous medium
equation (and the corresponding Barenblatt solution), which have been used to model
similar phenomena as equation~\eqref{equation}. Equation~\eqref{equation} can be viewed as interpolation between the parabolic and elliptic cases.

We need the sharp behavior
of
the fundamental solution given in Lemma~\ref{fund_sol_asympt}. It seems that in the literature only upper bounds (see, e.g.~\cite[Proposition 1]{EideKoch04} and~\cite{KimLim15}) are available
although the sharp asymptotics of the Fox $H$-functions is known.
For the convenience of the reader we provide here an easy argument that the fundamental solution is
actually strictly positive on the whole space \(\mathbb{R}^d\) for all times \(t>0\). It
is well-known that the fundamental solution is a probability density and hence
nonnegative everywhere for all \(t>0\) but it does not exclude the possibility of the fundamental solution having zeros at some points.

We first prove the following auxiliary result. The proof is
based on the Mellin transform of the Fox $H$-function appearing in the representation of
the fundamental solution and the calculus of residues. For details we refer
to~\cite{KempSiljZach16,KilbSaig04}.

\begin{lemma}\label{lemma_Z_decreasing}
Let $\alpha\in(0,1]$ and $\beta\in(0,2]$. The fundamental solution \(Z(t,\cdot)\) given by~\eqref{fund_sol_repr} is radially decreasing.
\end{lemma}

\begin{proof}
We proceed similarly as in the proof of~\cite[Lemma 4.7]{KempSiljZach16}.

The fundamental solution given by~\eqref{fund_sol_repr} is clearly radial. With a slight
abuse of notation we denote \(Z(t,x)=Z(t,r)\), where \(r=|x|\). Using the differentiation
rule~\cite[Lemma 2.14]{KempSiljZach16} of the Fox $H$-functions we have
\begin{equation}\label{Fox_derivative_r}
\frac{\partial}{\partial r} Z(t, r)= \pi^{-d/2}r^{-d-1}[\beta
H_{34}^{22}(2^{-\beta}t^{-\alpha}r^\beta)-dH_{23}^{21}(2^{-\beta}t^{-\alpha}r^\beta)],
\end{equation}
where we have noted that \(Z\) is a radial function and denoted \(r=|x|\), and we have omitted the parameters appearing in the H-functions.

We may combine the linear combination
\begin{equation}\label{Fox_linear_combination}
\beta H^{22}_{34}(z)-d H^{21}_{23}(z)
\end{equation}
as follows. Let \(\mathcal{H}^{21}_{23}\) and \(\mathcal{H}^{22}_{34}\) be the Mellin
transforms of the functions \(H^{21}_{23}\) and \(H^{22}_{34}\) appearing
in~\eqref{Fox_derivative_r}. Using the definition of the Fox
$H$-function~\cite[Formulae (2.10) and (2.11)]{KempSiljZach16} we have
\[
\beta \mathcal{H}_{34}^{22}(s)-d\mathcal{H}_{23}^{21}(s)=\beta \frac{\Gamma(\frac{d}{2}+\frac{\beta}{2}s)\Gamma(1+s)\Gamma(1-s)}{\Gamma(1+\alpha
s)\Gamma(-\frac{\beta}{2}s)}
-d\frac{\Gamma(\frac{d}{2}+\frac{\beta}{2}s)\Gamma(1+s)\Gamma(-s)}{\Gamma(1+\alpha
s)\Gamma(-\frac{\beta}{2}s)}.
\]
Using the property \(\Gamma(z+1)=z\Gamma(z)\) of the Gamma function we may proceed as
\[
\begin{split}
\beta \mathcal{H}_{34}^{22}(s)-d\mathcal{H}_{23}^{21}(s)&=-2\l(\frac{d}{2}+\frac{\beta}{2}s\r)\frac{\Gamma(\frac{d}{2}+\frac{\beta}{2}
s)\Gamma(1+s)\Gamma(-s)}{\Gamma(1+\alpha s)\Gamma(-\frac{\beta}{2}s)} \\
&=-2\frac{\Gamma(\frac{d+2}{2}+\frac{\beta}{2}s)\Gamma(1+s)\Gamma(-s)}{\Gamma(1+\alpha
s)\Gamma(-\frac{\beta}{2}s)} \\
&=-2\mathcal{H}_{23}^{21}\big[ s \big| \begin{smallmatrix}(1,1),
&(1, \alpha)\\
(\frac{d+2}{2},\frac{\beta}{2}),& (1,1 ), &(1, \frac{\beta}{2})&  \end{smallmatrix} \big],
\end{split}
\]
which implies
\[
\frac{\partial}{\partial r} Z(t, r)= -2\pi^{-d/2}r^{-d-1}H_{23}^{21}\big[
2^{-\beta}
t^{-\alpha}r^{\beta} \big| \begin{smallmatrix}(1,1),
&(1, \alpha)\\
(\frac{d+2}{2},\frac{\beta}{2}),& (1,1 ), &(1, \frac{\beta}{2})&  \end{smallmatrix} \big].
\]
which is nothing but
\[
-2\pi r Z_{d+2}(t,r),
\]
where the subscript \(d+2\) means that \(Z_{d+2}\) is the fundamental solution
\(Z(t,\cdot)\) in dimensions \(d+2\ge 3\). Since \(Z(t,\cdot)\) is a probability density,
the result follows.
\end{proof}

\begin{corollary}\label{corol_no_pos_zero}
The fundamental solution \(Z(t,r)\) has no positive zeros \(r\in \R_+\).
\end{corollary}

\begin{proof}
Since \(r\mapsto Z(t,r)\) is decreasing and nonnegative in \(\R_+\), it cannot have
zeros on the positive real axis. Otherwise \(Z(t,\cdot)\) would have compact support,
which would imply that the analytic function \(H^{21}_{23}(z)\), \(z\neq 0\), would have
zeros in a continuum, which is impossible.
\end{proof}

Having Lemma~\ref{lemma_Z_decreasing} and Corollary~\ref{corol_no_pos_zero}
in our hands, we can use the sharp behavior of the fundamental solution given by the
asymptotics formulated in terms of the similarity variable
\begin{equation*}
R= \frac{|x|^\beta}{2^\beta t^\alpha}.
\end{equation*}
Note that also the classical heat kernel contains a function depending on \(R\) with \(\alpha=1\) and \(\beta=2\).

\begin{proof}[Proof of Theorem~\ref{counter_example}]
First of all, we note that by scaling it is enough to consider \(x_0\in\partial B(0,1)\). Let $\psi(x)$ be the standard mollifier
\begin{equation*}
\psi(x) :=
\begin{cases}
c_d e^{-\frac{1}{1-|x|^2}}, \quad \text{for } |x| <1 \\
0, \quad \text{for } |x| \ge1,
\end{cases}
\end{equation*}
where $c_d$ is a dimension-dependent normalization constant to guarantee the integral of
$\psi$ to equal one. We choose $u_0^\eps(x) = \eps^{-d}\psi(\tfrac x\eps)$ as the initial
data for our problem. We will find a lower bound for the left hand side
of~\eqref{local_harnack}, which will tend to infinity as \(\eps\to 0\), and an upper
bound of the right hand side of~\eqref{local_harnack}, which stays bounded uniformly in
\(\eps\). We start with the lower bound, where we consider the cases \(d> \beta\) and
\(d=\beta\) separately, since the asymptotics of the fundamental solution is different in these cases.

\noindent {\bf The lower bound. Case: $d> \beta$.} Let \(\psi\) and \(u_0^\epsilon\) be as above and let \(t>0\) be fixed. Then
\[
u^\eps(t,0)=\int_{\mathbb{R}^d} Z(t,y)u_0^\eps(y) dy=\int_{\mathbb{R}^d}Z(t,\eps
z)\psi(z) dz,
\]
where we made the change of variables \(y\leftrightarrow \eps z\). Take now
\[
R_0=\min\{t^{\alpha/\beta},\frac12\},
\]
whence for \(|z|\le R_0\) there holds
\[
R:=\frac{|z|^\beta}{2^\beta t^\alpha}\le \frac{1}{2^\beta}\quad\text{and}\quad \psi(z)\ge\mathrm{e}^{-\frac43}c_d.
\]
Note that \(R\le \frac{1}{2^\beta}\) implies \(R_\eps:=\frac{|\eps z|^\beta}{2^\beta t^\alpha}\le \frac{1}{2^\beta}\)
for all
\(\eps\le 1\). Hence the estimate
\[
Z(t,x)\sim t^{-\alpha}|x|^{\beta-d},\quad R\le\frac{1}{2^\beta}
\]
given by Lemma~\ref{fund_sol_asympt} allows us to estimate \(u^\epsilon\) from below as
\[
u^\eps(t,0)\gtrsim\int_{|z|\le R_0} Z(t,\eps z) dz\gtrsim \eps^{\beta-d}t^{-\alpha}\int_{|z|\le R_0} |z|^{\beta-d} dz.
\]
Since \(z\mapsto |z|^{\beta-d}\) is locally integrable,
\[
u^\eps(t,0)\gtrsim \eps^{\beta-d}\to\infty,\quad\eps\to 0.
\]

\noindent {\bf The lower bound. Case: \(d=\beta\).} We proceed as above. Again, Lemma~\ref{fund_sol_asympt} gives the estimate

\[
Z(t,z)\sim t^{-\alpha}(1+|\log(R)|)\sim -t^{-\alpha}\log(R),\quad R\le \frac{1}{2^\beta},
\]
which implies
\[
u^\eps (t,0)\gtrsim\int_{|z|\le R_0} Z(t,\eps z)d z\gtrsim t^{-\alpha}\int_{|z|\le
R_0}\left|\log\left(\frac{|\eps z|^\beta}{2^\beta t^\alpha}\right)\right|d z.
\]
Substituting \(y=\frac{\eps z}{2t^{\alpha/\beta}}\) in the last integral gives the lower bound
\[
u^\eps(t,0)\gtrsim \eps^{-d}\int_{|y|\le\frac{\eps
R_0}{2t^{\alpha/\beta}}}|\log(|y|)|d y.
\]
But the integral on the right hand side behaves like
\[
\eps^d|\log\eps|\quad\textrm{as $\eps\to 0$},
\]
whence
\[
u^\eps(t,0)\to\infty\quad\textrm{as $\eps\to 0$}.
\]

\noindent {\bf The upper bound.} We consider a point $x_0 \in \partial
B_1(0)$. We have
\begin{align*}
u^\eps(t,x_0) &= \int_{\Rn} Z(t, x_0 -y) u_0^\eps(y) \, dy \\
& = \eps^{-d}\int_{B_\eps(0)} Z(t, x_0 -y)\psi\big(\tfrac{y}{\eps}\big) \, dy \\
& =\int_{B_1(0)} Z(t, x_0 -\eps z)\psi(z) \, dz.
\end{align*}

For $0< \eps \le \tfrac12$, we have that $x_0 -\eps z \in B_{1/2}^c(0)$ for $z \in
\partial B_1(0)$. Since $Z(t,\cdot)$ is continuous and bounded in
$B_{1/2}^c(0)$ uniformly with respect to \(\eps\) and \(\psi\) is integrable, we have by the Lebesgue dominated convergence
theorem that
\begin{equation*}
\lim_{\eps \to 0} u^\eps(t,x_0) =\int_{B_1(0)} \lim_{\eps \to 0} Z(t, x_0 -\eps z)\psi(z)
\, dz = Z(t, x_0) < \infty,
\end{equation*}
for all $t>0$. Since for all $t>0$ we have that $u^\eps(t,0) \to \infty$ as $\eps \to 0$, the Harnack inequality cannot hold even after an arbitrarily long time lag. This finishes the proof.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{counter_example_boundedness}]
Here we do not need the exact form of the fundamental solution in terms of a quite complicated $H$-functions. Instead, working in the frequency domain we can use the simpler Mittag-Leffler function \(E_\alpha\).

Let \(u_0(x)=\mathrm{e}^{-|x|^2}\) and \(u_0^n(x)=n^{d/p}u_0(nx)\). Then by~\cite[Theorem 2.12]{KempSiljZach16} the function
\[
u_n(t,x)=\left(Z(t,\cdot)\star u_0^n\right)(x)=\int_{\Rn}Z(t,x-y)u_0^n(y)\,dy
\]
is a classical solution of~\eqref{equation}. By using the Fourier transform and~\eqref{fund_fourier_trans} we may represent \(u_n(t,0)\) in the form
\[
\begin{split}
u_n(t,0)\,=& (2\pi)^{-d/2}\mathcal{F}^{-1}\left(E_\alpha(-|\xi|^\beta t^\alpha)\widehat{u_0^n}(\xi)\right)(0)\\
=&(2\pi)^{-d} n^{\frac{d}{p}-d}\int_{\Rn} E_\alpha(-|\xi|^\beta t^\alpha)\widehat{u_0}(\xi/n)\,d\xi\\
=&(2\pi)^{-d} n^{\frac{d}{p}}\int_{\Rn} E_\alpha(-|n\xi|^\beta t^\alpha)\widehat{u_0}(\xi)\,d\xi.
\end{split}
\]
By~\eqref{ML2_estimate} we have a lower bound
\[
E_\alpha(-\rho)\ge \frac{c_1}{1+\rho},\quad \rho\ge 0,
\]
for some positive constant \(c_1\). Moreover, since the Fourier transform of the Gaussian is also Gaussian, we have \(\widehat{u_0}>0\) in \(B(0,2\delta)\) with \(\delta=t^{-\alpha/\beta}\) and fixed \(t>0\). Hence for some positive constant \(c=c(t,\alpha,\beta,d)\) there holds
\[
u_n(t,0)\ge cn^{\frac{d}{p}}\int_{B(0,2\delta)}\frac{d\xi}{1+|n\xi|^\beta t^\alpha}\gtrsim n^{\frac{d}{p}-\beta}\int_{\frac{\delta}{n}}^{2\delta} r^{d-1-\beta} dr,
\]
where in the last step we introduced the spherical coordinates. This implies
\[
u_n(t,0)\gtrsim \begin{cases}
n^{\frac{d}{p}-\beta}, &\text{if $d>\beta$},\\
n^{\frac{d}{p}-\beta}\log(n), &\text{if $d=\beta$}.
\end{cases}
\]
In particular, \(u_n(t,0)\to \infty\), if (i) \(p<\frac{d}{\beta}\) and if (ii) \(p=1\) and \(d=\beta\). On the other hand, Young's inequality for convolutions~\cite[Theorem 1.2.10]{Graf04} and the fact that \(Z(t,\cdot)\) is a probability distribution function for all \(t>0\) (see~\cite{KempSiljZach16}) imply
\[
\|u_n(t,\cdot)\|_{L^p(\Rn)}\le\|Z(t,\cdot)\|_{L^1(\Rn)}\|u_0^n\|_{L^p(\Rn)}=\|u_0\|_{L^p(\Rn)}<\infty,
\]
which finishes the proof.
\end{proof}

\section{Harnack inequality}

\begin{proof}[Proof of Theorem~\ref{main_theorem}]

For the Harnack inequality, we need to consider the value
of the solution in two separate points, and for this purpose, we denote
\begin{equation*}
R_1:=R_1(y) = \frac{|x_1-y|^\beta}{2^\beta t_1^\alpha} \quad \text{and}\quad R_2:=R_2(y) =
\frac{|x_2-y|^\beta}{2^\beta t_2^\alpha}.
\end{equation*}

We split the integral defining the mild solution into two parts and have

\begin{align}\label{integral_split}
u(t_1,x_1) &= \int_{\Rn} Z(t_1, x_1-y) u_0(y) \, dy \notag\\
&= \int_{\{R_1 > 1 \}} Z(t_1, x_1-y) u_0(y) \, dy +  \int_{\{R_1 \le 1\}} Z(t_1, x_1-y)
u_0(y) \, dy     \notag\\
& =: I_1 + I_2.
\end{align}

We will use the asymptotics of \(Z\) given in Lemma~\ref{fund_sol_asympt}. We start with the integral \(I_1\) and consider only the case \(\beta=2\), since this case is more difficult. The case \(\beta\neq 2\) follows from the argument for \(\beta=2\), since for \(\beta\neq 2\) there is no exponential function in the asymptotics for large values of the similarity variable \(R_1\) and we only need the estimate for the ratio
\[
\frac{|x_1-y|}{|x_2-y|}
\]
given below.

By Lemma~\ref{fund_sol_asympt} we have
\begin{align*}
Z(t, x)
\sim t^{-\frac{\alpha d}2}R^{\frac{d(\alpha-1)}{2(2-\alpha)}}e^{-\sigma
R^{\frac1{2-\alpha}}},\quad R\ge 1,
\end{align*}
for a constant $\sigma=\alpha^{\frac\alpha{2-\alpha}}$. Observe that for
\begin{align*}
x_1, x_2 \in B_{r}(0),\quad  R_1 \ge 1 \quad\text{and}\quad t_1 \ge (2r)^{2/\alpha}
\end{align*}
there holds
\begin{align*}
R_2 = \frac{|x_2-y|^2}{4t_2^\alpha} &\ge \frac{(|x_1-y|-|x_1-x_2|)^2}{4t_2^\alpha} \\
&\ge \frac{(2t_1^{\alpha/2}-2r)^2}{4t_2^\alpha} \\
&\ge \frac14\left(\frac{t_1}{t_2}\right)^\alpha,
\end{align*}
which implies
\begin{align}\label{inclusion}

\{R_1 \ge 1 \}  \subset \left\{R_2 \ge \frac14\left(\frac{t_1}{t_2}\right)^\alpha \right\}.

\end{align}

 We obtain
\begin{align*}
I_1 &= \int_{\{R_1 > 1 \}} Z(t_1, x_1-y) u_0(y) \, dy \\
&\lesssim  t_1^{-\frac{\alpha d}2}\int_{{\{R_1 > 1\}}}
R_1^{\frac{d(\alpha-1)}{2(2-\alpha)}}e^{-\sigma R_1^{\frac1{2-\alpha}}} u_0(y) \, dy \\
&=  t_1^{-\frac{\alpha d}2}\int_{\{R_1 > 1 \}}
\left(\frac{R_1}{R_2}\right)^{\frac{d(\alpha-1)}{2(2-\alpha)}}e^{\sigma
\left(R_2^{\frac1{2-\alpha}}-
R_1^{\frac1{2-\alpha}}\right)}R_2^{\frac{d(\alpha-1)}{2(2-\alpha)}}e^{-\sigma
R_2^{\frac1{2-\alpha}}} u_0(y) \, dy.
\end{align*}
Next we estimate the exponential factor from above. If $R_2 \le R_1$, there is nothing to
prove, and for $R_2 \ge R_1$ we obtain for $\alpha \in (0,1]$ that
\begin{align*}
R_2^{\frac1{2-\alpha}}- R_1^{\frac1{2-\alpha}} &\le (R_2-R_1)^{\frac1{2-\alpha}} \\
&= \left(\frac{|x_2-y|^2}{4t_2^\alpha} -\frac{|x_1-y|^2}{4t_1^\alpha}
\right)^{\frac1{2-\alpha}} \\
&= \left(\frac{t_1^\alpha|x_2-y|^2-t_2^\alpha|x_1-y|^2}{4t_2^\alpha t_1^\alpha}
\right)^{\frac1{2-\alpha}} \\
&= \left(\frac{(t_1^\alpha-t_2^\alpha)|y|^2-2y \cdot (t_1^\alpha x_2-t_2^\alpha
x_1)+t_1^\alpha|x_2|^2-t_2^\alpha|x_1|^2}{4t_2^\alpha t_1^\alpha}
\right)^{\frac1{2-\alpha}} \\
\end{align*}

Since $ (2r)^{2/\alpha} \le t_1 < t_2 \le t_1 + (2r)^{2/\alpha}\le 2t_1 $ and
\(x_1,x_2\in B_r(0)\), we can estimate
\[
\frac{t_1^\alpha|x_2|^2-t_2^2|x_1|^2}{4t_2^\alpha
t_1^\alpha}\le\frac{r^2}{4t_2^\alpha}\le\frac{1}{16}.
\]
It remains to estimate the sum of the first two terms in the brackets. We estimate the
numerator of the second term as
\[
|y\cdot (t_1^\alpha x_2-t_2^\alpha x_1)|\le |y|(1+2^\alpha)r t_1^\alpha\le 3|y|r
t_1^\alpha.
\]
Denote \(r=|y|\) and study the parabola
\[
r\mapsto-a r^2+br
\]
with
\[
a=\frac{t_2^\alpha-t_1^\alpha}{4t_1^\alpha t_2^\alpha}\quad\text{and}\quad b=\frac{6r
}{4 t_2^\alpha}.
\]
Since the maximum is attained at \(r=\frac{b}{2a}\), the upper bound
for the sum of the first two terms is
\begin{equation}\label{est_blow_up}
\frac{b^2}{4a}=\frac{36 t_1^\alpha r^2}{16 t_2^\alpha
(t_2^\alpha-t_1^\alpha)}\le\frac{3r^2}{t_2^\alpha-t_1^\alpha}\le\frac{3
t_1^\alpha}{4(t_2^\alpha-t_1^\alpha)}=\left(\left(\frac{t_2}{t_1}\right)^\alpha-1\right)^{-1}.
\end{equation}

Next we estimate the power term. It is enough to estimate the ratio
\[
\frac{R_1}{R_2}=\Big(\frac{t_2}{t_1}
\Big)^\alpha\frac{|x_1-y|^2}{|x_2-y|^2}.
\]
Since the exponent is negative, we need a lower bound. In the set \(\{R_1\ge 1\}\) there
holds
\begin{equation}\label{triangle_ineq_est}
|x_2-y|\le |x_1-x_2|+|x_1-y|\le 2r+|x_1-y|\le t_1^{\alpha/2}+|x_1-y|\le\frac32|x_1-y|,
\end{equation}
which gives the desired lower bound.

Altogether we obtain
\begin{align}\label{I1_est1}
I_1 &\le   Ct_2^{-\frac{\alpha d}2}\int_{\{R_2 \ge \frac1{16}\}}
R_2^{\frac{d(\alpha-1)}{2(2-\alpha)}}e^{-\sigma R_2^{\frac1{2-\alpha}}} u_0(y) \, dy\notag \\
&\le  C\int_{\{R_2 \ge \frac1{16}\}} Z(t_2, x_2-y)u_0(y) \, dy,
\end{align}
where we used the inclusion~\eqref{inclusion} and estimated
\[
\Big(\frac{t_1}{t_2}\Big)^\alpha\ge \frac{t_1}{t_2}\ge\frac12.
\]

The proof shows that $C$ depends only on $d,
\alpha$ and $t_2/t_1$. Moreover, we see from~\eqref{est_blow_up} that the constant \(C\) given by the proof
blows up as \(t_2\to t_1\).

For $I_2$ we divide the treatment into two separate cases depending on the dimension. \\

\noindent {\bf Case: $d=\beta$.}

In this case we have from Lemma~\ref{fund_sol_asympt}
\begin{align*}
Z(t_1, x_1-y) \sim t_1^{-\alpha}\left(1-\log\left(\frac{|x_1-y|^\beta}{2^\beta t_1^\alpha}\right)
\right),\quad R_1\le 1
\end{align*}
Noting that \(c(d,\beta)=-(d\pi)^{d/2}\) in~\eqref{riesz_kernel} we can estimate

\begin{align}\label{I2_est1}
I_2 &= \int_{\{R_1 \le 1 \}} Z(t_1, x_1-y) u_0(y) \, dy \notag\\
&\le Ct_1^{-\alpha} \int_{\{R_1 \le 1 \}} \left[1-
\log\left(\frac{|x_1-y|^\beta}{2^\beta t_1^\alpha}\right) \right]u_0(y) \, dy\notag \\

&= Ct_1^{-\alpha} \int_{\Rn} \left[1+\beta\log 2+\alpha\log t_1-\beta\log|x_1-y|
\right]\chi_{B_{r_1}(x_1)}u_0(y) \, dy \notag\\
&= Ct_1^{-\alpha}[(1+\beta\log 2+\alpha \log t_1+\beta c(d,\beta)G) \star
\big(\chi_{B_{r_1}(x_1)}u_0\big)](x_1)\notag\\
&\le C\frac{[(c+\alpha \log t_1+\beta c(d,\beta)G) \star \big(\chi_{B_{r_1}(x_1)}u_0
\big)](x_1)}{[(c+\alpha \log t_2+\beta c(d,\beta) G) \star
\big(\chi_{B_{r_2}(x_2)}u_0\big)](x_2)}\int_{\{R_2 \le 1 \}} Z(t_2, x_2-y) u_0(y) \, dy.

\end{align}
In the last step we controlled the time factor $\left(\tfrac{t_2}{t_1}\right)^\alpha$
similarly as before and denoted \(c=1+\beta\log 2\). Combining~\eqref{integral_split},~\eqref{I1_est1} and~\eqref{I2_est1} gives the claim.\\

\noindent {\bf Case: $d>\beta$.}

Now we obtain
\begin{align*}
I_2 &= \int_{\{R_1 \le 1 \}} Z(t_1, x_1-y) u_0(y) \, dy \\
&\le Ct_1^{-\alpha} \int_{\{R_1 \le 1 \}} |x_1-y|^{\beta-d}u_0(y) \, dy \\
&= C\left(\frac{t_2}{t_1}\right)^\alpha\frac{\int_{\{R_1 \le 1 \}}  |x_1-y|^{\beta-d}u_0(y)
\, dy}{\int_{\{R_2 \le 1 \}} |x_2-y|^{\beta-d}u_0(y) \, dy} \int_{\{R_2 \le 1 \}} Z(t_2,
x_2-y)u_0(y) \, dy,

\end{align*}
where
\begin{align*}
t_i^{-\alpha}\int_{\{R_i \le 1 \}} |x_i-y|^{\beta-d}u_0(y) \, dy &= \int_{\Rn}
|x_i-y|^{\beta-d}\chi_{\{B_{r_i}(x_i)\}}u_0(y) \, dy \\
&= c(d,\beta)G*(u_0\chi_{\{B_{r_i}(x_i)\}})
\end{align*}
for $r_i = 2t_i^{\alpha/\beta}, i=1,2$. We obtain
\begin{align*}
I_2 \le C\left[\frac{[G \star (u_0\chi_{B_{r_1}(x_1)})](x_1)}{[G \star
(u_0\chi_{B_{r_2}(x_2)})](x_2)}\right] \int_{\{R_2 \le 1 \}} Z(t_2, x_2-y)u_0(y) \, dy.
\end{align*}

Combining the estimates for \(I_1\) and \(I_2\) we have obtained
\begin{align*}
u(t_1, x_1) \le C\left[1+\frac{[G \star (u_0\chi_{B_{r_1}(x_1)})](x_1)}{[G \star
(u_0\chi_{B_{r_2}(x_2)})](x_2)}\right] u(t_2, x_2),\quad d>\beta,
\end{align*}
for a constant $C$ depending on $d, \alpha,\beta$ and $t_2/t_1$.
\end{proof}

The Corollary~\ref{main_corollary} is now an easy consequence of the above proof.

\begin{proof}[Proof of Corollary~\ref{main_corollary}]

Since $r_2 \ge r_1$, by using the
assumption~\eqref{initial_harnack} we obtain
\begin{align*}
G \star (u_0\chi_{B_{r_2}(x_2)})(x_2)
&=\int_{\Rn} G(y)u_0(x_2-y)\chi_{B_{r_2}(x_2)}(x_2-y) \, dy \\
&=\int_{B_{r_2}(0)} G(y)u_0(x_2-y) \, dy \\
&\ge \inf_{B_{r_2}(x_2)}u_0\int_{B_{r_2}(0)} G(y) \, dy  \\
&\ge \frac1{H_0}\sup_{B_{r_1}(x_1)}u_0\int_{B_{r_1}(0)} G(y) \, dy \\
&\ge \frac1{H_0}\int_{B_{r_1}(0)} G(y)u_0(x_1-y) \, dy \\
&= \frac1{H_0}\int_{\Rn} G(y)(u_0\chi_{B_{r_1}(x_1)})(x_1-y) \, dy\\
&=G \star (u_0\chi_{B_{r_1}(x_1)})(x_1).
\end{align*}
Now the result follows from Theorem~\ref{main_theorem}.

\end{proof}

Finally, we turn into studying the one-dimensional case.

\begin{proof}[Proof of Theorem~\ref{harnack_local_in_time}]

By the definition of the mild solution we have
\[
{u}(t,x)=\int_{\R^d}Z(t,x-y){u}_0(y)\, d y

\]
for nonnegative \({u}_0$.

We evaluate \({u}\)
at the point \((t_1,x_1)\) and divide the integration into two
parts as before depending on whether \(R_1=\frac{|x_1-y|^\beta}{2^\beta t_1^\alpha}\le 1\) or
\(R_1\ge1\). The argument in the case \(R_1\le 1\) is the same for all \(\beta\), but in the case \(R_1\ge 1\) the argument is different depending on whether \(\beta=2\) or not.  We give here the proof in the case \(\beta\neq 2\). The case \(\beta=2\) can be handled similarly as in the proof of Theorem~\ref{main_theorem}.

By Lemma~\ref{fund_sol_asympt}
\[
Z(t_1,x_1-y)\sim t_1^{-\frac{\alpha d}{\beta}},\quad R_1\le 1,
\]
and
\[
Z(t_1,x_1-y)\sim
t_1^{\alpha}|x_1-y|^{-d-\beta},\quad
R_1\ge 1.
\]
Write
\begin{align}\label{integral_split2}
u(t_1,x_1) &= \int_{\Rn} Z(t_1, x_1-y) u_0(y) \, dy\notag \\
&= \int_{\{R_1 > 1 \}} Z(t_1, x_1-y) u_0(y) \, dy +  \int_{\{R_1 \le 1\}} Z(t_1, x_1-y)
u_0(y) \, dy     \notag\\
& =: I_1 + I_2
\end{align}
as in the proof of Theorem~\ref{main_theorem}.

\noindent {\bf The estimate for $I_1$.} The argument is essentially the same as in the proof of Theorem~\ref{main_theorem} for the power term. We will repeat the argument here with more details. Recall
\[
(2r)^{\alpha/\beta}\le t_1<t_2\le t_1+(2r)^{\beta/\alpha},
\]
which implies
\[
t_2\le t_1+(2r)^{\beta/\alpha}\le 2t_1.
\]
Then

\[
R_2 = \frac{|x_2-y|^\beta}{2^\beta t_2^\alpha} \ge \frac{(|x_1-y|-|x_1-x_2|)^\beta}{2^\beta t_2^\alpha} \ge \frac{(2t_1^{\alpha/\beta}-2r)^\beta}{2^\beta t_2^\alpha} \ge \left(\frac{t_1}{t_2}\right)^\alpha\ge\frac12,
\]

so
\[
\{R_1\ge 1\}\subset \{R_2\ge\tfrac12\}.
\]
Hence
\begin{align*}
I_1&\lesssim t_1^\alpha\int_{\{R_1\ge 1\}}|x_1-y|^{-d-\beta}u_0(y) dy\\
&\lesssim\left(\frac{t_1}{t_2}\right)^\alpha\int_{\{R_2\ge\tfrac12\}}\left(\frac{|x_2-y|}{|x_1-y|}\right)^{d+\beta} t_2^\alpha |x_2-y|^{-d-\beta}u_0(y) dy.
\end{align*}
Since the asymptotics of \(Z\) is sharp, it remains to have a uniform bound for the ratio
\[
\frac{|x_2-y|}{|x_1-y|}.
\]
But noting that \(t_1^{\alpha/\beta}\le\frac{|x_1-y|}{2}\) in the set \(\{R_1\ge 1\}\), we may estimate similarly as in~\eqref{triangle_ineq_est} to get
\[
|x_2-y|\le\frac32|x_1-y|,
\]
which gives the desired upper bound
\begin{equation}\label{I1_est2}
I_1\lesssim\int_{\{R_2\ge\tfrac12\}}Z(t_2,x_2-y)u_0(y) dy\le u(t_2,x_2).
\end{equation}

\noindent {\bf The estimate for $I_2$.} For \(R_1\le 1\) we have
\begin{align*}
|x_2-y|\le |x_1-y|+|x_1-x_2|\le 2t_2^{\alpha/\beta}+2r\le 3 t_2^{\alpha/\beta}\Rightarrow
R_2\le\left(\frac{3}{2}\right)^\beta
\end{align*}
and we may  estimate
\begin{equation}\label{I2_est2}
\begin{split}
I_2&=\int_{\{R_1\le 1\}}
Z(t_1,x_1-y)u_0(y)d y\\
&\lesssim
t_1^{-\frac{\alpha d}{\beta}}\int_{\{R_1\le 1\}}u_0(y)d
y\\
&\lesssim\left(\frac{t_2}{t_1}\right)^\frac{\alpha d}{\beta}\int_{\left\{R_2\le\left(\frac32\right)^\beta\right\}}Z(t_2,
x_2-y) u_0(y)d y\\
&\lesssim {u}(t_2,x_2).
\end{split}
\end{equation}
Combining~\eqref{integral_split2},~\eqref{I1_est2} and~\eqref{I2_est2} completes the proof.
\end{proof}

\section*{Acknowledgements}

J. S. was supported by the Academy of Finland grant 259363 and a V\"{a}is\"{a}l\"{a} foundation travel grant.

config: arxiv


text:
\section{Proof of inequality \eqref{remarkable}}\label{proof:remarkable_id}

Let us drop the index in the bracket $\langle -\rangle_t$ and simply denote $R\equiv R_{n}(t,\epsilon)$. We start by proving the identity

\begin{align}
-2\,\mathbb{E}\big\langle Q(\mathcal{L} - \mathbb{E}\langle \mathcal{L}\rangle)\big\rangle
&=\mathbb{E}\big\langle (Q - \mathbb{E}\langle Q \rangle)^2\big\rangle
+ \mathbb{E}\big\langle (Q-   \langle Q \rangle)^2\big\rangle\,.\label{47}
\end{align}

Using the definitions $Q \equiv \frac1n \bx\cdot \bX$ and \eqref{def_L} gives
\begin{align}
2\,\mathbb{E}\big\langle Q ({\cal L} -\mathbb{E}\langle {\cal L} \rangle) \big\rangle
	 =\, &  \mathbb{E} \Big [ \frac{1}{n}\big\langle Q \|\bx\|^2 \big\rangle - 2 \langle Q^2 \rangle - \frac{1}{n\sqrt{R}}\big\langle Q (\tilde \bZ \cdot \bx) \big\rangle \Big ] \nonumber \\
	&  -  \mathbb{E}\langle Q \rangle  \, \mathbb{E} \Big [ \frac{1}{n}\big\langle \|\bx\|^2 \big\rangle - 2 \langle Q \rangle - \frac{1}{n\sqrt{R}} \tilde \bZ \cdot \langle \bx\rangle \Big ]\,. \label{eq:QL:1}
\end{align}
The gaussian integration by part formula \eqref{GaussIPP} with Hamiltonian \eqref{Ht} yields
\begin{align*}
\frac{1}{n\sqrt{R}}\mathbb{E}\big\langle Q (\tilde \bZ \cdot \bx) \big\rangle
	& = \frac{1}{n}\mathbb{E}\big\langle Q \|\bx\|^2 \big\rangle - \frac{1}{n}\EE\big\langle Q (\bx \cdot \langle \bx \rangle) \big \rangle \overset{\rm N}{=}\frac1{n}\mathbb{E}\big\langle Q \|\bx\|^2 \big\rangle - \EE [\langle Q  \rangle^2]\,.
\end{align*}
Fort the last equality we used the Nishimori identity as follows
$$
\frac1n\EE\big\langle Q (\bx \cdot \langle \bx \rangle) \big \rangle=\frac1{n^2}\EE\big\langle (\bx\cdot \bX) (\bx \cdot \langle \bx \rangle) \big \rangle\overset{\rm N}{=} \frac1{n^2}\EE \big\langle (\bX\cdot \bx) (\bX \cdot \langle \bx \rangle) \big \rangle=\EE [\langle Q \rangle^2]\,.
$$
Note that we already proved \eqref{NishiTildeZ}, namely
$$
\frac{1}{n\sqrt{R}}\mathbb{E} \langle \tilde \bZ\cdot  \bx \rangle
=   \frac1n \mathbb{E}\big\langle \|\bx\|^2 \big\rangle - \EE \langle Q\rangle \, .
$$
Therefore \eqref{eq:QL:1} finally simplifies to
\begin{align*}
&2\,\mathbb{E}\big\langle Q ({\cal L} -\mathbb{E}\langle {\cal L} \rangle) \big\rangle = \mathbb{E}[\langle Q\rangle^2] - 2\,\mathbb{E}\langle Q^2\rangle +  \mathbb{E}[\langle Q\rangle]^2 \nn
&\qquad\qquad= -  \big ( \mathbb{E}\langle Q^2\rangle - \mathbb{E}[\langle Q\rangle]^2 \big ) -  \big ( \mathbb{E}\langle Q^2\rangle - \mathbb{E}[\langle Q\rangle^2] \big ).
\end{align*}
which is identity \eqref{47}.

This identity implies the inequality
\begin{align*}
2\big|\mathbb{E}\big\langle Q(\mathcal{L} - \mathbb{E}\langle \mathcal{L}\rangle)\big\rangle\big|&=2\big|\mathbb{E}\big\langle (Q-\mathbb{E}\langle Q \rangle)(\mathcal{L} - \mathbb{E}\langle \mathcal{L}\rangle)\big\rangle\big|\ge \mathbb{E}\big\langle (Q - \mathbb{E}\langle Q \rangle)^2\big\rangle
\end{align*}
and an application of the Cauchy-Schwarz inequality gives

\begin{align*}
2\big\{\mathbb{E}\big\langle (Q-\mathbb{E}\langle Q \rangle)^2\big\rangle\, \mathbb{E}\big\langle(\mathcal{L} - \mathbb{E}\langle \mathcal{L}\rangle)^2\big\rangle \big\}^{1/2}	\ge\mathbb{E}\big\langle (Q - \mathbb{E}\langle Q \rangle)^2\big\rangle\,.
\end{align*}
This ends the proof of \eqref{remarkable}.\\