text:
\begin{document}

\title[Stochastic Heat Equation: germ Markov Property]
{The Stochastic Heat Equation Driven by a Gaussian Noise: germ
Markov Property}

\author[Raluca Balan]{Raluca Balan*}
\thanks{The first author was supported by a grant from the Natural Sciences and
Engineering Research Council of Canada. The second author was
supported by a post-doctoral fellowship at University of Ottawa.}
\address{Corresponding author. Department of Mathematics and Statistics,
University of Ottawa, 585, King Edward Avenue, Ottawa, Ontario K1N
6N5 Canada} \email{rbalan@uottawa.ca}
\urladdr{http://aix1.uottawa.ca/$\sim$rbalan}

\author{Doyoon Kim}
\address{Department of Mathematics,
University of Southern California, Los Angeles, 3620 Vermont Avenue,
KAP108, Los Angeles, CA 90089, USA} \email{doyoonki@usc.edu}
\urladdr{http://college.usc.edu/faculty/faculty1017406.html}

\subjclass[2000] {Primary 60H15; Secondary 60G60}

\keywords{Gaussian noise; stochastic integral; stochastic heat
equation; Reproducing Kernel Hilbert Space; germ Markov property}

\begin{abstract}
\noindent Let $u=\{u(t,x);t \in [0,T], x \in {\mathbb{R}}^{d}\}$ be
the process solution of the stochastic heat equation $u_{t}=\Delta
u+ \dot F, u(0,\cdot)=0$ driven by a Gaussian noise $\dot F$, which
is white in time and has spatial covariance induced by the kernel
$f$. In this paper we prove that the process $u$ is locally germ
Markov, if $f$ is the Bessel kernel of order $\alpha=2k,k \in
\bN_{+}$, or $f$ is the Riesz kernel of order $\alpha=4k,k \in
\bN_{+}$.
\end{abstract}

\maketitle

\section{Introduction}

This article is based on the theory of stochastic partial
differential equations (s.p.d.e.'s) initiated by John Walsh in 1986 \cite{walsh86}.
This theory relies on the construction of a stochastic integral with
respect to the so-called ``worthy martingale measures'' and focuses
mainly on equations driven by a space-time white noise.

Recently, there has been a considerable amount of interest in
s.p.d.e.'s driven by a noise term which is white in time, but is
``colored'' in space, in the sense that it has a spatial covariance
structure induced by a kernel $f$, which is the Fourier transform of
a tempered distribution $\mu$. This line of research was initiated
in \cite{mueller97}, \cite{dalang-frangos98} for the stochastic wave
equation in the case $d=2$, and then generalized in \cite{dalang99},
\cite{dalang-mueller03} to a larger class of s.p.d.e.'s in arbitrary
spatial dimensions. The new theory requires an extension of Walsh's
stochastic integral, to include the case when the integrand is a
(Schwartz) distribution; this type of extension is needed for
instance for the stochastic wave equation with $d \geq 3$. Under a
certain integrability condition imposed on the measure $\mu$, one
can prove that any second-order s.p.d.e. with constant coefficients
has a process solution, i.e. a solution which can be identified with
a multiparameter stochastic process $u$. As far as we know, the
literature to date does not contain any study of the germ Markov
property of this process solution. Such a study is of great
importance because it allows us to conclude that the behavior of the
process in any time-space region $A$ is independent of its behavior
outside the region, given the values of the process in a thin area
around the boundary $\partial A$ of the region.

There is an immense amount of literature dedicated to the study of
different types of Markov properties for multiparameter processes
(see
\cite{balan-ivanoff02} and the references
therein). In particular, the germ Markov property of multiparameter
Gaussian processes is an area which received a lot of attention in
the 1970's, which allows for the use of various tools imported from
Hilbert space analysis, like the method of the Reproducing Kernel
Hilbert Space (RKHS). An excellent all-time reference is the
monograph \cite{rozanov82}; earlier references include
\cite{pitt71}, \cite{kunsch79}, \cite{nualart-sanz79}.

In the 1990's, the problem of the germ Markov property for
multi-dimensional Gaussian processes appears again in the
literature, this time for processes which arise as solutions of
s.p.d.e.'s driven by a white noise: the case of elliptic equations
was thoroughly treated in \cite{donatimartin-nualart94}, while the
systematic study of this problem for the quasi-linear parabolic
equations (in the case $d=1$) is found in \cite{nualart-pardoux94}.
The case of hyperbolic equations turns out to be the most difficult
one: the only reference here seems to be \cite{dalang-hou97}, in
which the authors investigate very carefully the structure of the
germ $\sigma$-fields induced by the process solution of the
stochastic wave equation (in the case $d=2$) driven by a L\'{e}vy
noise without Gaussian component. (This type of noise induces a very
particular form for the process solution, which is used in a
fundamental way for deriving its germ Markov property.) Finally, the
more recent work \cite{pitt-robeva03} contains a detailed analysis
of the relationship between the germ and sharp $\sigma$-fields
(which are used for defining the sharp and germ Markov property,
respectively), in the case of the process solution associated to the
Bessel equation driven by a white noise (in the case $d=2$).

The goal of the present paper is to investigate the structure of the
RKHS associated to the process solution $u$ of the stochastic heat
equation driven by a spatial covariance kernel $f$, and to identify
some examples of functions $f$ for which the process $u$ possesses
the germ Markov property. As far as we know, this is the first
attempt to tackle a problem of this type in the literature. Our two
main results (Theorem \ref{main-theorem-Bessel} and Theorem
\ref{main-theorem-Riesz}) state that the process solution $u$ has
the germ Markov property, if the function $f$ is either the Bessel
kernel of order $\alpha=2k, k \in {\bf N}$ or the Riesz kernel of
order $\alpha=4k, k \in {\bf N}$. In order to prove these results,
we use some results from the $L_{p}$-theory of parabolic equations
with mixed norms, due to \cite{krylov99}, \cite{krylov01}. The germ
Markov property of the process solution $u$ in the case of the
Bessel or Riesz kernels with $\alpha>0$ arbitrary (or in the case of
other kernel functions) remains an open problem.

The paper is organized as follows. In Section 2, we introduce the
framework for the study of s.p.d.e.'s, including the construction of
the extended stochastic integral due to Dalang (as in
\cite{dalang99}). In Section 3, we include all the ingredients which
are necessary to formulate the result about the existence of the
process solution $u$, and we examine the structure of the Gaussian
space and the RKHS associated to this process. In Section 4, we give
the definition of the germ Markov property and we prove our two main
results, by applying a fundamental result of \cite{kunsch79}. The
two appendices contain some technical proofs.

\section{The Framework}
\subsection{Basic Notation}

We denote by $\cD(U)$ the space of all infinitely differentiable
functions on $\mathbb{R}^{n}$ whose support is compact and contained
in the open set $U$, and by $\cD'(U)$ the space of {\em
distributions}, i.e. continuous linear functionals on $\cD(U)$. We
denote by $\cS(\mathbb{R}^{n})$ the Schwartz space of all rapidly
decreasing functions on $\mathbb{R}^{n}$, and by
$\cS'(\mathbb{R}^{n})$ the space of {\em tempered distributions},
i.e. continuous linear functionals on $\cS(\mathbb{R}^{n})$. The
space $\cS'(\mathbb{R}^{n})$ can be viewed as a subspace of
$\cD'(\mathbb{R}^{n})$. An important subspace of
$\cS'(\mathbb{R}^{n})$ is the space $\cO_{C}'(\mathbb{R}^{n})$ of
all distributions with {\em rapid decrease} (see Chapter VII,
section 5, \cite{schwartz66}).

For an arbitrary function $\phi$ on $\mathbb{R}^{n}$, the
translation by $x$ is denoted with $\phi_{x}$, i.e.
$\phi_{x}(y)=\phi(x+y)$, and the reflection by $0$ is denoted with
$\tilde \phi$, i.e. $\tilde \phi(x)=\phi(-x)$. These notions have
obvious extensions to distributions. The Fourier transform of $\phi
\in L_{1}(\mathbb{R}^{n})$ is defined by $\cF\phi (\xi)=(2
\pi)^{-n/2}\int_{\mathbb{R}^{n}} \exp(- i \xi \cdot x) \phi(x)dx$.
The map $\cF:\cS(\mathbb{R}^{n}) \rightarrow \cS(\mathbb{R}^{n})$

can be extended to $L_{2}(\mathbb{R}^{n})$ and
$\cS'(\mathbb{R}^{n})$. For every subset $U \subset \bR^n$, we
denote
$$ \langle \varphi,\psi \rangle_{L_2(U)} = \int_{U} \varphi(x)
\overline{\psi(x)} dx, $$ whenever the integral is defined.

\subsection{The Gaussian Noise}

\noindent Let $T>0$ and $f$ be a locally integrable function on
$\mathbb{R}^{d}$. As in \cite{dalang99}, for every $\varphi, \psi
\in \cD((0,T) \times \mathbb{R}^{d})$ we define
\begin{equation}
\label{def-inner-product} J_{f}(\varphi, \psi)=\langle \varphi,\psi
\rangle_{0}=\int_{0}^{T} \int_{\bR^{d}} \int_{\bR^{d}}
\varphi(t,x)f(x-y) \psi(t,y)dy dx dt.
\end{equation}

 The Gaussian
 noise mentioned in the Introduction will be a zero-mean Gaussian
 process with covariance $J_{f}$. The existence of this process is based on the
following Bochner-Schwartz type result (see Theorem 2, p. 157,
\cite{gelfand-vilenkin64}):

\begin{lemma}
\label{3-equivalences} The bi-functional $J_{f}$ is
nonnegative-definite if and only if there exists a tempered measure
$\mu$ on $\mathbb{R}^{d}$ such that
$$f(x)=\int_{\mathbb{R}^{d}}e^{-i \xi \cdot x}\mu(d \xi), \ \ \ \forall
x \in R^{d}.$$
\end{lemma}

In this case, for every $\varphi, \psi \in \cD((0,T) \times
\mathbb{R}^{d})$ $$ \langle \varphi,\psi \rangle_{0}=\int_0^T
\int_{\bR^d} \cF\varphi(t,\xi)
\overline{\cF\psi(t,\xi)}\mu(d\xi)dt,$$ where $\cF \varphi(t,\cdot)$
denotes the Fourier transform of $\varphi (t,\cdot)$.

The basic example of kernel functions is the white noise kernel:
$f(x)=\delta(x)$, $\mu(d\xi)=d\xi$.

More interesting examples of kernel functions $f$ arise when
$\mu(d\xi)=g(|\xi|^{2})d\xi$, for a certain function $g$ on
$(0,\infty)$. In this case, $f$ becomes the inverse Fourier
transform of the tempered distribution $\xi \mapsto g(|\xi|^{2})$
and we can define the operator $g(-\Delta):\cS(\mathbb{R}^{d})
\rightarrow \cS'(\mathbb{R}^{d})$ by
$\cF[g(-\Delta)\phi](\xi)=g(|\xi|^{2})\cF \phi(\xi)$, or
equivalently

$$
g(-\Delta)\phi=\phi *f.
$$

(see p. 149-152, \cite{folland95}, or p. 117-132, \cite{stein70}).
Here are some typical examples:

\begin{example}
\label{riesz} {\rm If $\mu(d\xi)=|\xi|^{-\alpha}d\xi$, then $f$ is
the Riesz kernel of order $\alpha$:
$$f(x)=R_{\alpha}(x):=
\gamma_{\alpha}|x|^{-d+\alpha}, \ \ \ 0<\alpha<d,$$
 where $\gamma_{\alpha}$ is an appropriate constant.
In this case, for every $\phi \in \cS(\mathbb{R}^{d})$ we have
$(-\Delta)^{-\alpha/2}\phi=\phi*R_{\alpha}$.}
\end{example}

\begin{example}
\label{bessel}{\rm If $\mu(d\xi)=(1+|\xi|^{2})^{-\alpha/2}d\xi$,
then $f$ is the Bessel kernel of order $\alpha$:
$$f(x)=B_{\alpha}(x):=
c_{\alpha}\int_{0}^{\infty}\tau^{(\alpha-d)/2-1}
e^{-\tau-|x|^{2}/(4\tau)} d\tau, \ \ \ \alpha>0,$$ where
$c_{\alpha}$ is an appropriate constant. In this case, for every
$\phi \in \cS(\mathbb{R}^{d})$ we have
$(1-\Delta)^{-\alpha/2}\phi=\phi*B_{\alpha}$.}
\end{example}

\begin{example} {\rm If $\mu(d\xi)=e^{-4\pi^2 \alpha|\xi|^{2}}$, then
$f$ is the heat kernel
$$f(x)=G_{\alpha}(x):= (4\pi \alpha)^{-d/2}
e^{-|x|^{2}/(4\alpha)}, \ \ \ \alpha>0.$$ In this case, for every
$\phi \in \cS(\mathbb{R}^{d})$ we have
$e^{\alpha\Delta}\phi=\phi*G_{\alpha}$.}
\end{example}

In what follows, we let  $F=\{F(\varphi);\varphi \in \cD((0,T)
\times \mathbb{R}^{d})\}$ be a zero-mean Gaussian process with
covariance $J_{f}$, i.e. $\forall \varphi,\psi \in \cD((0,T) \times
\mathbb{R}^{d})$
\begin{equation}
\label{covariance} E(F(\varphi) F(\psi))=\langle \varphi,\psi
\rangle_{0}.
\end{equation}
The Gaussian space $H^{F}$ of the process $F$ is defined as the
closed linear subspace of $L_{2}(\Omega)$, generated by the
variables $\{F(\varphi),\varphi \in \cD((0,T) \times
\mathbb{R}^{d})\}$.

\subsection{The Stochastic Integral}

In this subsection, we summarize the construction of the generalized
stochastic integral $M(\varphi)=\int_{0}^{T}\int_{\mathbb{R}^{d}}
\varphi(t,x)M(dt,dx)$ with respect to the martingale measure $M$
induced by the noise $F$, due to Dalang (see \cite{dalang99}). For
our purposes, it is enough to consider only the case of
deterministic integrands. Here is the construction procedure:

\vspace{2mm}

{\em Step 0.} For each $\varphi \in \cD((0,T) \times R^{d})$, we set
$M(\varphi)=F(\varphi) \in H^{F}$.

\vspace{2mm}

 {\em Step 1.} Let ${\cB}_{b}(\mathbb{R}^{d})$ be the class of all bounded Borel subsets
 of $\mathbb{R}^{d}$ and $\cE^{(d)}$ be the class
of all linear combinations of functions $1_{[0,t] \times A}$, $t \in
[0,T], A \in {\cB}_{b}(\mathbb{R}^{d})$. We endow $\cE^{(d)}$ with
the inner product $\langle \cdot,\cdot \rangle_{0}$ given by formula
(\ref{def-inner-product}) and we denote by $\| \cdot \|_{0}$ the
corresponding norm.

For each $t \in (0,T), A \in {\cB}_{b}(\mathbb{R}^{d})$, there
exists a sequence $\{\varphi_{n}\}_{n} \subset \cD((0,T) \times
\mathbb{R}^{d})$ such that $\varphi_{n} \to 1_{[0,t] \times A}$ and
${\rm supp} \varphi_{n} \subseteq K$ for all $n$ (see
\cite{dalang-frangos98}, p. 190).

By the bounded convergence theorem, $\| \varphi_{n}-1_{[0,t] \times
A}\|_{+} \to 0$, where $\| \cdot \|_{+}$ is the norm defined in Step
2 below. Hence $\| \varphi_{n}-1_{[0,t] \times A}\|_{0} \to 0$ and
$\bE( M(\varphi_{m})-M(\varphi_{n}))^2= \| \varphi_{m}-\varphi_{n}
\|_{0} \to 0$ as $m,n \rightarrow \infty$, i.e. the sequence
$\{M(\varphi_{n})\}_{n}$ is Cauchy in $L_{2}(\Omega)$. A standard
argument shows that its limit does not depend on
$\{\varphi_{n}\}_{n}$. We set $M_{t}(A)=M(1_{[0,t] \times
A})=_{L_{2}(\Omega)}\lim_{n}M(\varphi_{n}) \in H^{F}$.
 (Note that the
process $M=\{M_{t}(A); t \in [0,T],A \in {\cB}(\mathbb{R}^{d}) \}$
is a worthy martingale measure, in the sense of \cite{walsh86}.) We
extend $M$ by linearity to $\cE^{(d)}$. A limiting argument and
relation (\ref{covariance}) shows that, for every $\varphi,\psi \in
{\cE}^{(d)}$
\begin{equation}
\label{covariance-1} \bE(M(\varphi)M(\psi))=\langle \varphi,\psi
\rangle_{0}.
\end{equation}

\vspace{2mm}

 {\em Step 2.} Let $\cP_{+}^{(d)}=\{\varphi:[0,T] \times
\mathbb{R}^{d} \rightarrow \mathbb{R} \ {\rm measurable} ; \|
\varphi \|_{+}<\infty\}$, where $\| \varphi \|_{+}^{2}:=
 \int_{0}^{T} \int_{\mathbb{R}^{d}} \int_{\mathbb{R}^{d}}
|\varphi(t,x)|f(x-y) |\varphi(t,y)|dy dx dt$. We endow
$\cP_{+}^{(d)}$ with the inner product $\langle \cdot, \cdot
\rangle_{0}$ given by formula (\ref{def-inner-product}) and we
denote by $\| \cdot \|_{0}$ the corresponding norm; note that $\|
\cdot \|_{0} \leq \| \cdot \|_{+} $. An argument similar to that
used in the proof of Proposition 2.3, \cite{walsh86}, shows that
$\cE^{(d)}$ is dense in $\cP_{+}^{(d)}$ with respect to $\| \cdot
\|_{+}$, and hence for each $\varphi \in \cP_{+}^{(d)}$,  there
exists a sequence $\{ \varphi_{n} \}_{n} \subset \cE^{(d)}$ such
that $\| \varphi_{n}-\varphi
\|_{+} \to 0$. As in Step 1,

we set $M(\varphi)=_{L_{2}(\Omega)}\lim_{n}M(\varphi_{n}) \in
H^{F}$. Note that relation (\ref{covariance-1}) holds for every
$\varphi,\psi \in {\cP}_{+}^{(d)}$.

\vspace{2mm}

{\em Step 3.}  Let $\cP_{0}^{(d)}$ be the completion of $\cE^{(d)}$
with respect to $< \cdot, \cdot>_{0}$. The space $\cP_{0}^{(d)}$ is
the largest space of integrands $\varphi$ for which we can define
the stochastic integral $M(\varphi)$. According to
\cite{dalang99},p. 9, the space $\cP_{0}^{(d)}$ has the following
alternative definition.

 Let $\overline{{\cP}}^{(d)}=\{\varphi:[0,T]
\to {\cS}'(\mathbb{R}^{d}) ; \ {\cF}\varphi(t, \cdot) \
\mbox{function} \forall t \in [0,T], \ (t,\xi) \mapsto
{\cF}\varphi(t,\xi) \linebreak \mbox{ measurable}, \| \varphi
\|_{0}<\infty\}$, where $\| \varphi \|_{0}^{2}=\int_{0}^{T}
\int_{\mathbb{R}^{d}} |{\cF}\varphi(t,\xi)|^{2}\mu(d\xi)dt$. Let
${\cE}_{0}^{(d)}=\{\varphi \in {\cP}_{+}^{(d)}; \varphi(t, \cdot)
\in {\cS}(\mathbb{R}^{d}), \ \forall t \in [0,T]\}$ and
$\cP_{0}^{(d)}$ be the closure of $\cE_{0}^{(d)}$ in
$\overline{\cP}^{(d)}$ with respect to $\| \cdot \|_{0}$.
Note that
$$\langle \varphi,\psi \rangle_{0}=\int_{0}^{T} \int_{\mathbb{R}^{d}}{\cF}
\varphi(t,\xi) \overline{{\cF} \psi(t,\xi)}\mu(d\xi)dt, \  \ \
\varphi,\psi \in {\cP}_{0}^{(d)}.$$

\noindent For each $\varphi \in {\cP}_{0}^{(d)}$, there exists a
sequence $\{ \varphi_{n} \}_{n} \subset {\cE}_{0}^{(d)}$ such that
$\| \varphi_{n}-\varphi \|_{0} \rightarrow 0$. As in Step 1,

we set $M(\varphi)=_{L_{2}(\Omega)}\lim_{n}M(\varphi_{n}) \in
H^{F}$. Note that relation (\ref{covariance-1}) holds for every
$\varphi,\psi \in {\cP}_{0}^{(d)}$, and hence
\begin{equation}
\label{M-HF} \varphi \mapsto M(\varphi) \ \mbox{is an isometry
between} \ {\cP}_{0}^{(d)} \ \mbox{and} \

H^{F}.
\end{equation}

\vspace{3mm}

In summary, the previous construction is based on the diagram
\begin{eqnarray*}
{\cD}((0,T) \times \mathbb{R}^{d}) \subset {\cE}_{0}^{(d)} \subset &
{\cP}_{+}^{(d)} & \subset {\cP}_{0}^{(d)}
\subset \overline{\cP}^{(d)} \\
& \cup & \\
& {\cE}^{(d)} &
\end{eqnarray*} and the following
$3$ approximation techniques:
\begin{itemize}
\item indicator functions can be approximated by functions in
${\cD}((0,T) \times \mathbb{R}^{d})$;

\item functions in ${\cP}_{+}^{(d)}$ can be approximated by
indicator functions;

\item {\em distributions} in ${\cP}_{0}^{(d)}$ can be
approximated by ``smooth'' functions in ${\cP}_{+}^{(d)}$.
\end{itemize}

\noindent In particular, we conclude that $\cD((0,T) \times
\mathbb{R}^{d})$ is dense in ${\cP}_{0}^{(d)}$ with respect to $\|
\cdot \|_{0}$.

\subsection{Alternative Characterization of the Space
$\cP_{0}^{(d)}$}

As in \cite{ferrante-sanzsole06}, for every $\varphi,\psi \in
{\cD}(\bR^d)$, define
$$\langle \varphi,\psi
\rangle_{0,x}:=\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\varphi(x)f(x-y)\psi(y)
dydx= \int_{\bR^d} \cF \varphi(\xi) \overline{\cF \psi(\xi) } \, \mu
(d\xi),$$ and let $\cP_{0,x}^{(d)}$ be the completion of
$\cD(\bR^{d})$ with respect to $\langle \cdot, \cdot \rangle_{0,x}$.
Note that
\begin{equation}
\label{product0-produxt0x} \langle \varphi,\psi
\rangle_{0}=\int_{0}^{T} \langle \varphi(t,\cdot),\psi(t,\cdot)
\rangle_{0,x}dt, \ \ \ \forall \varphi,\psi \in {\cP}_{0}^{(d)}.
\end{equation}

\begin{remark}
\label{P0d-in-L2}

{\rm Note that for every $\varphi \in \cP_{0}^{(d)}$, $\varphi(t,
\cdot) \in \cP_{0,x}^{(d)}$ for a.e. $t \in [0,T]$. Since
$\cE^{(d)}$ is dense in $\cP^{(d)}$, one can prove that the map $t
\mapsto \varphi(t, \cdot)$ is strongly measurable from $[0,T]$ to
$\cP_{0,x}^{(d)}$ (in the sense of Definition on p.649,
\cite{evans98}). Using (\ref{product0-produxt0x}), we conclude that
${\cP}_{0}^{(d)} \subset L_{2}((0,T), {\cP}_{0,x}^{(d)})$ and $\|
\varphi \|_{0}=\| \varphi \|_{L_{2}((0,T),\cP_{0,x}^{(d)})}$,
$\forall \varphi \in \cP_{0}^{(d)}$.}
\end{remark}

\section{The Process Solution}
\subsection{The Equation and its Solution}

We consider the stochastic heat equation with vanishing initial
conditions, written {\em formally} as:

\begin{equation}
\label{heat-eq}u_{t}-\Delta u =  \dot F, \ \ \ \mbox{in} \ (0,T)
\times \mathbb{R}^{d}, \ \ \  u(0,\cdot)= 0.
\end{equation}

\noindent The solution of this equation is defined formally as
follows. Suppose for the moment that $\dot F$ is a random variable
with values in $\cS'(\mathbb{R}^{d+1})$. For every fixed $\omega \in
\Omega$,
 let $\{u(\varphi); \varphi \in \cD((0,T) \times \mathbb{R}^{d})\}$
be the distribution solution of (\ref{heat-eq}). It is known that
\begin{equation}
\label{analogy} u(\varphi)=(G
* \dot F)(\varphi)= \dot F(\varphi
* \tilde G)
\end{equation}
where $G$ is the fundamental solution of the heat equation:
$$G(t,x)=\left\{
\begin{array}{ll} (4 \pi t)^{-d/2} \exp\left( -\frac{|x|^{2}}{4t}\right) & \mbox{if $t>0, x \in R^{d}$} \\
0 & \mbox{if $t \leq 0, x \in R^{d}$}
\end{array} \right.$$
(Since $G \in {\cO}_{C}'(\mathbb{R}^{d+1})$, we have $\varphi *
\tilde G \in {\cS}(\mathbb{R}^{d+1})$ for every $\varphi \in
{\cD}(\mathbb{R}^{d+1})$ and the convolution $G * \dot F$ is well
defined; see Theorem XI, Chapter VII, \cite{schwartz66}.)

\vspace{2mm}

Going back to our framework, we have the following lemma.

\begin{lemma}
\label{psi*Gtilde} If
\begin{equation}
\label{cond-mu} \int_{R^{d}} \frac{1}{1+|\xi|^{2}} \mu(d \xi)<
\infty,
\end{equation}
then: (a) $(G_{tx})^{\verb2~2} \in {\cP}_{+}^{(d)}$ for every $(t,x)
\in [0,T] \times \mathbb{R}^{d}$; (b) $\varphi * \tilde G \in
\cP_0^{(d)}$ for every $\varphi \in \cD((0,T) \times \bR^d)$.
\end{lemma}

\begin{proof} (a) Set $g_{t,x}=
(G_{tx})^{\verb2~2}$. Note that $g_{tx}$ is measurable and $\|
g_{tx} \|_{+}=\| g_{tx} \|_{0}$, since $g_{tx} \geq 0$. We have $\cF
g_{t,x}(s,\xi) = c_d 1_{t>s} \exp ( - \mathrm{i} \xi x- (t-s)
|\xi|^2)$, where $c_d$ is an appropriate constant. Hence
$$\| g_{tx} \|_{0}^{2}=
\int_0^T \int_{\bR^d} | \cF g_{t,x}(s,\xi) |^2 \, \mu (d \xi) \, ds
\leq N(T) \int_{\bR^d} \frac{1}{1+|\xi|^2} \, \mu ( d \xi) < \infty,
$$
where $N(T)$ is a constant depending on $T$.

(b) We apply Remark 4, \cite{dalang99} to the function $\psi=\varphi
* \tilde G$. For this we need to check that: (i) $t \mapsto {\cF}
\psi(t,\xi)$ is continuous $\forall \xi \in R^{d}$; (ii) there
exists a nonnegative function $k(t, \xi)$ which is square-integrable
with respect to $dt \times \mu(d \xi)$ such that $|{\cF}\psi(t,\xi)|
\leq k(t, \xi)$, $\forall t \in [0,T], \xi \in R^{d}$.

(i) is clearly satisfied, and (ii) will follow from (\ref{cond-mu})
once we prove that
\begin{equation} \label{boundedness}
|\cF \psi(t,\xi)| \le \frac{N}{1+|\xi|^{2}}:=k(t,\xi), \ \ \ \forall
t \in [0,T],\ \forall \xi \in \mathbb{R}^{d},
\end{equation}
where $N$ is a constant. Since $(1+|\xi|^{2})|\cF \psi(t,x)|=|{\cF
\phi}(t,\xi)| \leq \| \phi(t,\cdot) \|_{L_1(\mathbb{R}^d)}$ where
$\phi=(1-\Delta)\psi$, relation (\ref{boundedness}) follows if we
prove that $\| \phi(t,\cdot) \|_{L_1(\mathbb{R}^d)} \le N$. Note
that
$$
- \phi_t - \Delta \phi = (1- \Delta) \varphi \quad \text{in} \quad
(0,T) \times \mathbb{R}^d, \quad \phi(T,x) = 0,
$$
since $\psi$ is the unique solution of: $- \psi_t - \Delta \psi =
\varphi \ \text{in} \ (0,T) \times \mathbb{R}^d, \ \psi(T,x) = 0$.
Thus $\phi(s,y) = \int_s^T \int_{\mathbb{R}^d} G(t-s,x-y)
(1-\Delta)\varphi(t,x) \, dx \, dt$. From this we calculate $\|
\phi(s,\cdot) \|_{L_1(\mathbb{R}^d)}$, which turns out to be bounded
(note that $(1-\Delta)\varphi$ has a compact support in $(0,T)
\times \bR^d$).
\end{proof}

A consequence of Lemma \ref{psi*Gtilde}.(b) is the fact that
$M(\varphi
* \tilde G)$ is well-defined for every
$\varphi \in {\cD}((0,T) \times \mathbb{R}^{d})$. By analogy with
(\ref{analogy}) (and using a slight abuse of terminology), we
introduce the following definition:

\begin{definition}
{\rm The process $\{u(\varphi); \varphi \in {\cD}((0,T) \times
\mathbb{R}^{d})\}$ defined by
$$u(\varphi):  = M(\varphi * \tilde G)=
\int_{0}^{T} \int_{\mathbb{R}^{d}} \left( \int_{\mathbb{R}_{+}}
\int_{\mathbb{R}^{d}}\varphi(t+s,x+y)G(s,y)dyds \right) M(dt,dx)$$
is called the {\bf distribution-valued solution} of the stochastic
heat equation (\ref{heat-eq}), with vanishing initial conditions.}
\end{definition}

\begin{theorem}[\cite{dalang99}, \cite{dalang-mueller03}]
\label{dalang-existence-theorem} Let $\{u(\varphi); \varphi \in
{\cD}((0,T) \times \mathbb{R}^{d})\}$ be the distribution-valued
solution of the stochastic heat equation (\ref{heat-eq}). In order
that there exists a jointly measurable process $X=\{X(t,x);t \in
[0,T],x \in \mathbb{R}^{d}\}$ such that
$$u(\varphi)=\int_{0}^{T} \int_{\mathbb{R}^{d}} X(t,x)\varphi(t,x)
dxdt \ \ \ \forall \varphi \in {\cD}((0,T) \times \mathbb{R}^{d}) \
\ \ a.s.$$ it is necessary and sufficient that (\ref{cond-mu})
holds. In this case, $X$ is a modification of the process
$u=\{u(t,x);t \in [0,T],x \in \mathbb{R}^{d}\}$ defined by
\begin{equation}
\label{solution-process} u(t,x):=M((G_{tx})^{\verb2~2})=\int_{0}^{T}
\int_{\mathbb{R}^{d}} G(t-s,x-y)M(ds,dy).
\end{equation}
\end{theorem}

\begin{definition}
{\rm The process $u=\{u(t,x);t \in [0,T], x \in \mathbb{R}^{d}\}$
defined by (\ref{solution-process}) is called the {\bf process
solution} of the stochastic heat equation (\ref{heat-eq}), with
vanishing initial conditions.}
\end{definition}

\noindent Note that the process $u$ is a zero-mean Gaussian process.
In the present paper, we are interested in examining the germ Markov
property of the process $u$.

\subsection{The Gaussian space}

The Gaussian space $H^{u}$ of the process $u$ is defined as the
closed linear subspace of $L_{2}(\Omega)$, generated by the
variables $\{u(t,x), t \in [0,T], x \in \bR^{d}\}$.

The next result shows that this space coincides with the space
$H^{F}$.

\begin{lemma}
\label{Hu=HF} We have $H^{u}=H^{F}$.
\end{lemma}

\begin{proof} a) First, we prove that $H^{u} \subseteq H^{F}$. Recall
that $u(t,x)=M((G_{tx})^{\verb2~2})$ and $(G_{tx})^{\verb2~2} \in
{\cP}_{+}^{(d)} \subset {\cP}_{0}^{(d)}$, by Lemma
\ref{psi*Gtilde}.(a). Since ${\cD}((0,T) \times \mathbb{R}^{d})$ is
dense in ${\cP}_{0}^{(d)}$ with respect to $\| \cdot \|_{0}$, there
exists a sequence $\{\varphi_{n}\}_{n \geq 1} \subseteq {\cD}((0,T)
\times \mathbb{R}^{d})$ such that
$\|\varphi_{n}-(G_{tx})^{\verb2~2}\|_{0} \to 0$. Hence
$\bE(M(\varphi_{n})-u(t,x))^{2} \to 0$. But $M(\varphi_{n}) \in
H^{F}$ for all $n$ and therefore $u(t,x) \in H^{F}$.

b) Let $H_{*}^{u}$ be the closed linear subspace of $L_{2}(\Omega)$,
generated by the variables $\{u(\eta), \eta \in \cD((0,T) \times
\bR^{d})\}$. To prove that $H^F \subseteq H_{*}^{u}$, let $\varphi
\in \cD((0,T) \times \mathbb{R}^{d})$ be arbitrary and
$\eta=-\varphi_{t}-\Delta \varphi \in \cD((0,T) \times
\mathbb{R}^{d})$. Then $\varphi=\eta
* \tilde G$ and $F(\varphi)=M(\varphi)=M(\eta * \tilde G)=u(\eta) \in H_{*}^{u}$.

c) Finally, we prove that $H_{*}^{u} \subseteq H^{u}$. Let $\eta \in
{\cD}((0,T) \times \mathbb{R}^{d})$ be arbitrary. Note that
$u(\eta)=M(\varphi)$ where $\varphi=\eta* \tilde G$. Let $K$ be a
compact set such that $\text{supp} \, \eta \subset [0,T] \times K$.

For each $n \geq 1$, let $\{ Q_m^{(n)}, m \in \bZ \}$ be a partition
of $(0,T) \times \bR^d$ such that each $Q_m^{(n)}=R_{m}^{(n)} \times
S_{m}^{(n)}$, where $R_{m}^{(n)}$ is an interval in $[0,T]$ and
$S_{m}^{(n)}$ is a cube in $\bR^d$. Suppose that $\max_{m \in \bZ}
|Q_m^{(n)} | \to 0$ as $n \to \infty$. For each $Q_m^{(n)}$, we
choose $(t_m^{(n)}, x_m^{(n)}) \in Q_m^{(n)}$ such that $t_m^{(n)}
\ge t$ for all $t \in R_m^{(n)}$. We consider the Riemann sum:
$$\varphi_n(s,y) = \sum_{m \in I_n} |Q_m^{(n)}|  G(t_m^{(n)}-s,x_m^{(n)}-y)
\eta(t_m^{(n)},x_m^{(n)}),$$ where $I_n=\{m \in \bZ ; t_m^{(n)} > s,
x_m^{(n)} \in K\}$.

Clearly $\varphi_n(s,y) \rightarrow \varphi (s,y)$ for every
$(s,y)$. We claim that
\begin{equation}
\label{conv-norm0} \|\varphi_{n}-\varphi \|_0 \rightarrow 0.
\end{equation}

\noindent From here, it follows that
$\bE(M(\varphi_n)-M(\varphi))^{2} \to 0$. This concludes that proof,
since $\varphi_{n}=\sum_{m \in I_n}a_m^{(n)}
(G_{t_m^{(n)},x_m^{(n)}})^{\verb2~2}$ and hence
$M(\varphi_n)=\sum_{m \in I_n}a_m^{(n)} u(t_m^{(n)},x_m^{(n)}) \in
H^{u}$, where $a_{m}^{(n)}=|Q_m^{(n)}| \ \eta(t_m^{(n)},x_m^{(n)})$.
The proof of (\ref{conv-norm0}) is given in Appendix A.
\end{proof}

\subsection{The RKHS}

Let ${\cH}^{u}=\{h_{Y}; h_{Y}(t,x)=\bE (Yu(t,x)),Y \in H^{u}\}$ be
the RKHS of the process $u$, endowed with the inner product $\langle
h_{Y},h_{Z} \rangle_{{\cH}^{u}}:=\bE(YZ), \ Y,Z \in H^{u}$. Note
that any function $h \in \cH^{u}$ is continuous on $[0,T] \times
\mathbb{R}^{d}$ and satisfies $h(0,\cdot)=0$. Moreover, $\cH^u$ is
the closure of $\{R((t,x), \cdot \ );(t,x) \in [0,T] \times \bR^d\}$
with respect to $\| \cdot \|_{\cH^u}$, where
$R((t,x),(s,y))=\bE(u(t,x)u(s,y))$.

By Lemma \ref{Hu=HF} and (\ref{M-HF}), it follows that
$${\cH}^{u}=\{h(t,x)=\bE (M(\varphi)u(t,x)); \varphi \in
{\cP}_{0}^{(d)}\}.$$ Moreover, if $h(t,x)=\bE (M(\varphi)u(t,x))$,
$g(t,x)=\bE (M(\eta)u(t,x))$ with $\varphi,\eta \in \cP_0^{(d)}$,
then $\langle h,g \rangle_{{\cH}^{u}}=\langle \varphi,\eta
\rangle_{0}$.

Let $g_{t,x}= (G_{tx})^{\verb2~2}$. Using the fact that
$u(t,x)=M(g_{tx})$ and (\ref{product0-produxt0x}), we get:
\begin{equation}
\label{formula-h} h(t,x) = \bE (M(\varphi) u(t,x))=\langle \varphi,
g_{tx} \rangle_{0}=

\int_0^t \int_{\bR^d} \cF \varphi (s,\xi) \overline{ \cF
g_{t,x}(s,\xi)} \, \mu(d\xi) \, ds.
\end{equation}

\begin{lemma}
\label{key-calculation} Let  $h(t,x) = \bE (M(\varphi) u(t,x)),
\varphi \in \cP_0^{(d)}$. For $\eta \in \cD((0,T) \times \bR^{d})$,
let $\phi$ be a solution to the equation $-\phi_t-\Delta \phi=\eta$
in $(0,T) \times \bR^d$, $\phi(T,\cdot)=0$. Then
$$
\langle h, \eta \rangle_{L_{2}((0,T) \times \bR^d)} =\langle
\varphi, \phi \rangle_{0}.
$$
 In particular, for any
$\phi \in \cD((0,T) \times \bR^d)$, we have
$$\langle h, -\phi_{t}-\Delta \phi \rangle_{L_{2}((0,T) \times \bR^{d})}=\langle
\varphi,\phi \rangle_{0}.$$
\end{lemma}

\begin{proof}
Using (\ref{formula-h}) and applying Fubini's theorem (since
$\varphi \in \cP_{0}^{(d)}$, $g_{tx} \in \cP_{+}^{(d)}$, and $\eta$
has compact support), we get
\begin{multline}\label{interm-calcul1}
\langle h, \eta \rangle_{L_{2}((0,T) \times \bR^d)}
= \int_0^T \langle h(s,\cdot), \eta(s,\cdot) \rangle_{L_2(\bR^d)} \, ds\\

= \int_0^T \int_{\bR^d} \cF \varphi (r,\xi) \int_r^T \int_{\bR^d}
\overline{\cF g_{s,x}(r,\xi) \, \eta(s,x)} \, dx \, ds \, \mu(d\xi)
\, dr.
\end{multline}

Note that

for every $r \in (0,T)$ and $y \in \bR^d$
$$
\phi(r,y) = \int_r^T \int_{\bR^d} G(s-r,x-y) \eta(s,x) \, dx \,ds.

$$

Hence

\begin{equation}
\label{interm-calcul2} \cF \phi(r,\xi)= \int_r^T \int_{\bR^d} \cF
g_{s,x}(r,\xi) \eta(s,x) \, dx \, ds

\end{equation}

\noindent From (\ref{interm-calcul1}) and (\ref{interm-calcul2}), we
conclude that
$$
\langle h, \eta \rangle_{L_{2}((0,T) \times \bR^d)}

= \langle \varphi, \phi \rangle_0

$$
(recall that

$\phi \in \cP_0^{(d)}$ by Lemma \ref{psi*Gtilde}). This finishes the
proof.

\end{proof}

\section{The Germ Markov Property}
\subsection{The Definition}

Let $S \subseteq [0,T] \times \mathbb{R}^{d}$ be an arbitrary set.
Let ${\cF}_{S}^{u}$ be the $\sigma$-field generated by the variables
$\{u(t,x);(t,x) \in S\}$, $K^{u}_{S}$ be the closed linear subspace
of $L_{2}(\Omega)$ generated by the variables $\{u(t,x);(t,x) \in
S\}$, and ${\cK}_{S}^{u}$ be the closed linear subspace of
${\cH}^{u}$ generated by the functions $\{R((t,x), \cdot \ ); (t,x)
\in S\}$. Let

$${\cG}_{S}^{u}
=\bigcap_{O \ {\rm open}; \ O \supset S}{\cF}_{O}^{u}, \ \ \
H_{S}^{u} =\bigcap_{O \ {\rm open}; \ O \supset S}{K}_{O}^{u}, \ \ \
{\cH}_{S}^{u} =\bigcap_{O \ {\rm open}; \ O \supset
S}{\cK}_{O}^{u}.$$

\begin{definition}
{\rm The process $u=\{u(t,x); (t,x) \in [0,T] \times \bR^d\}$ is
{\bf locally germ Markov} if for every relatively compact open set
$A \subset [0,T] \times \mathbb{R}^{d}$, ${\cG}_{\overline{A}}^{u}$
and ${\cG}_{\overline{A^{c}}}^{u}$ are conditionally independent
given ${\cG}_{\partial A}^{u}$, where $\partial A=\overline A \cap
\overline{A^{c}}$. }
\end{definition}

Based on the fact that ${\cG}_{S}^{u}=\sigma(H_{S}^{u})$,

 one can prove that $u$ is locally
germ Markov if and only if for every open set $A$,
$H^{u}=H_{\overline{A}}^{u} \oplus (H_{\overline{A^{c}}}^{u} \ominus
H_{\partial A}^{u})$ (see Lemma 1.3, \cite{kunsch79}), or
equivalently, using the isometry between $H^{u}$ and $\cH^{u}$
\begin{equation}
\label{direct-product} {\cH}^{u}={\cH}_{\overline{A}}^{u} \oplus
({\cH}_{\overline{A^{c}}}^{u} \ominus {\cH}_{\partial A}^{u}).
\end{equation}
(Here $\oplus, \ominus$ denote the usual operations on Hilbert
subspaces: if $H$ is a Hilbert space, $S$ is a closed subspace and
$S^{\perp}$ is its orthogonal complement, then we write $H=S \oplus
S^{\perp}$ and $S^{\perp}=H \ominus S$.)

On the other hand, $h(t,x)=\langle h,R((t,x), \cdot \ )
\rangle_{{\cH}^{u}}$ for every $h \in {\cH}^{u}$. Hence
\begin{equation}
\label{support-orthog} {\rm supp} \ h \subseteq B^{c} \ \ \ \mbox{if
and only if} \ \ \ h \in ({\cH}_{B}^{u})^{\perp}.
\end{equation}

\noindent Based on (\ref{direct-product}) and
(\ref{support-orthog}), we have the following fundamental result.

\begin{theorem}[Theorem 5.1, \cite{kunsch79}]
\label{kunsch-theorem} The Gaussian process $u$ is locally germ
Markov if and only if the following two conditions hold:

(i) If $h,g \in {\cH}^{u}$ are such that $({\rm supp} \ h) \cap
({\rm supp} \ g)=\emptyset$ and ${\rm supp} \ h$ is compact,
 then $<h,g>_{{\cH}^{u}}=0$.

(ii)  If $\zeta=h+g \in {\cH}^{u}$, where $h,g$ are such that $({\rm
supp} \ h) \cap ({\rm supp} \ g)=\emptyset$ and ${\rm supp} \ h$ is
compact, then $h,g \in {\cH}^{u}$.
\end{theorem}

In the next two subsections, we will suppose that $f$ is the Bessel
kernel of order $\alpha> 0$ (Example \ref{bessel}), respectively the
Riesz kernel of order $0<\alpha<d$ (Example \ref{riesz}). In both
cases we must have $\alpha>d-2$, in order to have condition
(\ref{cond-mu}) satisfied. Our goal is to prove that conditions (i)
and (ii) of Theorem \ref{kunsch-theorem} hold. For this, we will
assume that $\alpha=2k, k \in \bN_+$ in the case of the the Bessel
kernel, respectively $\alpha=4k, k \in \bN_+$ in the case of the
Riesz kernel (and $d=4k+1$, since $d-2<\alpha<d$).

\subsection{The Case of the Bessel Kernel}

In this subsection we will assume that $f$ is the Bessel kernel,
i.e. $f=B_{\alpha}$ with $\alpha>\max\{0,d-2\}$.

 In this case,
$\cP_{0,x}^{(d)}
=H_{2}^{-\alpha/2}(\bR^{d})$, where $$H_{2}^{\gamma}(\bR^d)
=\{\varphi \in \cS'(\bR^d); \cF \varphi \ \mbox{is a
function},\|\varphi \|_{\gamma,2}^2=\int_{\bR^d}|\cF \varphi
(\xi)|^{2} (1+|\xi|^{2})^{\gamma}d\xi<\infty\}$$ denotes the
fractional Sobolev space of index $\gamma \in \bR$ (see e.g.
\cite{folland95}, p.191-192).

By Remark \ref{P0d-in-L2}, $\cP_0^{(d)} \subset L_2((0,T),
H_2^{-\alpha/2}(\bR^d))$.

On the other hand, we have the denseness of $\cE_0^{(d)}$ in
$L_2((0,T),H_2^{-\alpha/2}(\bR^d))$ (for a proof, one may see the
proof of Theorem 3.10 in \cite{krylov99}). Thus $\cP_0^{(d)} =
L_2((0,T),H_2^{-\alpha/2}(\bR^d))$ and
$$\| \varphi \|_{0}=\| \varphi
\|_{L_{2}((0,T),H_{2}^{-\alpha/2}(\bR^{d}))}, \ \ \ \forall \varphi
\in \cP_{0}^{(d)}.$$

\noindent For each $t \in [0,T]$, let $\varphi_{1}(t,
\cdot):=(1-\Delta)^{-\alpha/2}\varphi (t, \cdot)$ and note that the
map $\varphi \mapsto \varphi_{1}$ is an isometry between $L_2((0,T),
H_2^{-\alpha/2}(\bR^d))$ and $L_2((0,T), H_2^{\alpha/2}(\bR^d))$.

Since

${\cF} \varphi_{1}(t,\xi)=(1+|\xi|^{2})^{-\alpha/2} \cF
\varphi(t,\xi)$, it is not difficult to see that: $\forall \varphi
\in L_2((0,T), H_2^{-\alpha/2}(\bR^d))$
\begin{equation}
\label{product-varphi1} \langle \varphi_{1}, \phi
\rangle_{L_{2}((0,T) \times \bR^{d})}=\langle \varphi, \phi
\rangle_{0}, \ \ \ \forall \phi \in {\cD}((0,T) \times \bR^{d}).
\end{equation}

To investigate the relationship between $h$ and $\varphi_1$,

we need some general results from the $L_p$-theory of parabolic
equations.

Recall that $H_2^{0}(\bR^d)=L_2(\bR^d)$, and $H_2^{\gamma}(\bR^d)
\subset H_2^{\gamma'}(\bR^d)$ if $\gamma>\gamma'$. For every
$\gamma,\beta \in \bR$, $(1-\Delta)^{\gamma/2}$ is a unitary
isomorphism between $H_2^{\beta}(\bR^d)$ and
$H_{2}^{\gamma-\beta}(\bR^d)$. In particular,
$(1-\Delta)^{\gamma/2}$ is a unitary isomorphism between
$H_2^{\gamma}(\bR^d)$ and $L_{2}(\bR^d)$. For every $v \in
H_2^{\gamma}(\bR^d)$, $\phi \in \cD(\bR^d)$, we denote
$$(v,\phi)=\int_{\bR^d}
[(1-\Delta)^{\gamma/2}v](x)\overline{[(1-\Delta)^{-\gamma/2}\phi](x)}dx.$$
Note that, if $v \in H_2^{\gamma}(\bR^d)$ and $\gamma \geq 0$, then
$(v,\phi)=\langle v,\phi \rangle_{L_{2}(\bR^d)}$ for all $\phi \in
\cD(\bR^d)$.

\begin{definition}
{\rm If $t \mapsto v(t, \cdot)$ is a function from $[0,T]$ to
$H_2^{\gamma}(\bR^d)$, with $\gamma \in \bR$, we say that $v$ is a
{\bf solution} of
$$d v = ( \Delta v + g ) d t \quad \text{in} \quad (0,T) \times
\bR^d, \quad v(0,\cdot) = 0$$ if for any $t \in (0,T)$ and for any
$\psi \in \cD(\bR^d)$, we have}
\begin{equation}\label{Krylov-eq}
(v(t,\cdot), \psi) = \int_0^t (v(s,\cdot), \Delta \psi)ds +\int_0^t
(g(s,\cdot), \psi) ds.
\end{equation}

We write $v \in \cH_{2,0}^{\gamma}(T)$ if $v_{xx} \in L_2((0,T),
H_2^{\gamma-2}(\bR^d))$, $v(0,\cdot) = 0$, and there exists $g \in
L_2((0,T), H_2^{\gamma-2}(\bR^d))$ satisfying \eqref{Krylov-eq}. By
$\|v\|_{\cH_{2,0}^{\gamma}(T)}$ we mean
$$
\|v\|_{\cH_{2,0}^{\gamma}(T)} = \|v_{xx}\|_{L_2((0,T),
H_2^{\gamma-2}(\bR^d))} + \|g\|_{L_2((0,T), H_2^{\gamma-2}(\bR^d))}.
$$

\end{definition}

\begin{remark}
{\rm It is known that (\ref{Krylov-eq}) implies that: $\forall t \in
(0,T)$, $\forall \phi \in \cD((0,T) \times \bR^d)$
$$(v(t,\cdot),
\phi(t,\cdot))=\int_0^t (v(s,\cdot), (\partial/\partial t+\Delta)
\phi(s,\cdot))ds +\int_0^t (g(s,\cdot), \phi(s,\cdot)) ds$$ (see
e.g. Proposition 10, \cite{ferrante-sanzsole06} for a stochastic
version of this result). In particular, (\ref{Krylov-eq}) implies
that: $\forall \phi \in \cD((0,T) \times \bR^d)$
\begin{equation}
\label{Krylov-eq2} \int_0^T (v(s,\cdot), (-\partial/\partial
t-\Delta) \phi(s,\cdot))ds =\int_0^T (g(s,\cdot), \phi(s,\cdot)) ds
\end{equation}
(by taking $t=T$ and using the fact that $\phi(T, \cdot)=0$).}
\end{remark}

\begin{theorem}[\cite{krylov99}, \cite{krylov01}]
\label{theoremA} Given $g \in L_2((0,T), H_2^{\gamma}(\bR^d))$,
$\gamma \in \bR$, there exists a unique solution

$v \in \cH_{2,0}^{\gamma+2}(T)$ to the equation

$$d v = ( \Delta v + g ) d t \quad \text{in} \quad (0,T) \times \bR^d,
\quad v(0,\cdot) = 0.$$

Moreover, there exists a constant $N$ (independent of $v$) such that
$$

\| v \|_{\cH_{2,0}^{\gamma+2}(T)} \le N \| g\|_{L_2((0,T),
L_2(\bR^d))}.
$$

\end{theorem}

We now return to our framework.

\begin{theorem}\label{h-unique-alpha+2}
Let $h(t,x) = \bE (M(\varphi) u(t,x))$, where $u$ is defined in
\eqref{solution-process} and $\varphi \in \cP_0^{(d)}$. Set
$\varphi_1=(1-\Delta)^{-\alpha/2}\varphi$. Then $h$ is the unique
solution in

$\cH_{2,0}^{\alpha/2+2}(T)$ to the equation

\begin{equation}
\label{Krylov-equation} d h = (\Delta h + \varphi_1) \, dt \ \ \ in
\ (0,T) \times \bR^{d}, \ \ \
 \quad h(0,\cdot) = 0.
 \end{equation}
\end{theorem}

\begin{proof}

Recall that $\varphi_1 \in L_{2}((0,T), H_2^{\alpha/2}(\bR^d))$.
Thus by Theorem \ref{theoremA} there exists a unique solution $v \in
\cH_{2,0}^{\alpha/2+2}(T)$ to the equation \eqref{Krylov-equation}.

We are now proving that $h=v$. This will follow once we prove that
$$
\langle h,\eta \rangle_{L_2((0,T) \times \bR^d)} = \langle v, \eta
\rangle_{L_2((0,T)\times\bR^d)}, \quad \forall{\eta \in \cD((0,T)
\times \bR^d)}.
$$
Let $\eta \in \cD((0,T) \times \bR^d)$ be arbitrary and $\phi$ be
the unique solution of
$$-\phi_t-\Delta \phi=\eta \quad \mbox{in} \ (0,T) \times \bR^{d}, \quad \phi(T,\cdot) = 0.$$
By Lemma \ref{key-calculation} and (\ref{product-varphi1})
$$
\langle h,\eta \rangle_{L_2((0,T) \times \bR^d)} =\langle \varphi,
\phi \rangle_{0} =\langle \varphi_1, \phi
\rangle_{L_2((0,T)\times\bR^d)}.
$$
On the other hand, by \eqref{Krylov-eq2} we have
$$
\langle v, \eta \rangle_{L_2((0,T)\times\bR^d)} = \langle \varphi_1,
\phi \rangle_{L_2((0,T) \times \bR^d)}
$$
This concludes the proof.

\end{proof}

\begin{corollary}\label{corollary20070928}
If $f$ is the Bessel kernel of order $\alpha$, then
$$
\cH^u = \cH_{2,0}^{\alpha/2+2}(T)
$$
and the norms in the two spaces are equivalent.
\end{corollary}

\begin{proof}
From the argument at the beginning of subsection 3.3, for every $h
\in \cH^u$, we have $h(t,x) = \bE M(\varphi)u(t,x)$, where $u$ is
defined in \eqref{solution-process} and $\varphi \in \cP_0^{(d)}$.
Then by Theorem \ref{h-unique-alpha+2}, $h \in
\cH_{2,0}^{\alpha/2+2}(T)$.

For $v \in \cH_{2,0}^{\alpha/2+2}(T)$, there is a $\varphi_1 \in
L_2((0,T), H_2^{\alpha/2}(\bR^d))$ satisfying \eqref{Krylov-eq} with
$\varphi_1$ in place of $g$. Then
$$
\varphi:= (1-\Delta)^{\alpha/2}\varphi_1 \in
L_2((0,T),H_2^{-\alpha/2}(\bR^d)) = \cP_0^{(d)}.
$$
Set $h(t,x) = \bE M(\varphi)u(t,x) \in \cH^u$, then by Theorem
\ref{h-unique-alpha+2} it follows that $v=h$, so $v \in \cH^u$.

Now notice that
$$
\| h \|_{\cH^u} = \| \varphi \|_{0} = \| \varphi \|_{L_2((0,T),
H_2^{-\alpha/2}(\bR^d))} = \| (1-\Delta)^{-\alpha/2} \varphi
\|_{L_2((0,T), H_2^{\alpha/2}(\bR^d))}
$$
$$
= \| \varphi_1 \|_{L_2((0,T), H_2^{\alpha/2}(\bR^d))} \simeq \| h
\|_{\cH_{2,0}^{\alpha/2+2}(T)},
$$
where $\simeq$ indicates the equivalence of the norms, which follows
from the definition of the norm of $\cH_{2,0}^{\alpha/2+2}(T)$ and
the estimate in Theorem \ref{theoremA}. This finishes the proof.
\end{proof}

\begin{remark}\label{remark20070928}
{\rm Under the conditions of Theorem \ref{h-unique-alpha+2}, we can
also say that $h$ is the unique solution in $W_{2}^{1,2}((0,T)
\times \bR^{d})$ of:
$$h_{t} = \Delta h + \varphi_1 \ \ \  {\rm in} \ \ \ (0,T) \times
\bR^{d}, \quad h(0,\cdot) = 0.$$ Here $W_2^{1,2}((0,T) \times
\bR^d)$ is the space of all measurable functions $v:(0,T) \times
\bR^d \rightarrow \bR$, such that the weak derivatives $v_t$,
$v_{x_i}$, $v_{x_i x_j}$ exist and are in $L_2((0,T) \times
\bR^d)$.}
\end{remark}

We are now ready to prove the main result of this subsection.

\begin{theorem}
\label{main-theorem-Bessel} Suppose that $f$ is the Bessel kernel of
order $\alpha=2k,k \in \bN_{+}$ such that $\alpha>d-2$. Then the
process solution $u$ of the stochastic heat equation (\ref{heat-eq})
with vanishing initial conditions is locally germ Markov.
\end{theorem}

\begin{proof}
We need to verify conditions (i) and (ii) of Theorem
\ref{kunsch-theorem}.

We first verify condition (i). Let $h,g \in {\cH}^{u}$ are such that
$({\rm supp} \ h) \cap ({\rm supp} \ g)=\emptyset$ and ${\rm supp} \
h$ is compact. We have to prove that $\langle h,g
\rangle_{\cH^u}=0$. We know that there exist $\varphi, \eta \in
\cP_0^{(d)}$ such that $h(t,x) = \bE (M(\varphi) u(t,x))$ and
$g(t,x) = \bE (M(\eta) u(t,x))$. Then by Theorem
\ref{h-unique-alpha+2} and Corollary \ref{corollary20070928} (also
recall the definition of the norm of $\cH_{2,0}^{\gamma}(T)$),
$$
\langle h, g \rangle_{\cH^u} = \langle h, g
\rangle_{\cH_{2,0}^{k+2}(T)} = \langle h_{xx}, g_{xx}
\rangle_{L_2((0,T), H_2^k(\bR^d))} + \langle \varphi_1, \eta_1
\rangle_{L_2((0,T), H_2^k(\bR^d))},
$$
where $\varphi_1 = (1-\Delta)^k\varphi$ and $\eta_1 =
(1-\Delta)^k\eta$. Note that (see also Remark \ref{remark20070928})
$$
\text{supp}\, h_{xx} \subset \text{supp}\, h, \quad \text{supp}\,
\varphi_1 \subset \text{supp}\, h.
$$
Similar inclusions hold for $g$ and $\eta_1$. Thus it is clear that
$$
\langle h_{xx}, g_{xx} \rangle_{L_2((0,T), H_2^k(\bR^d))} = \langle
\varphi_1, \eta_1 \rangle_{L_2((0,T), H_2^k(\bR^d))} = 0,
$$
from which we arrive at $\langle h, g \rangle_{\cH^u} = 0$. This
finishes the proof of the condition (i).

To prove the condition (ii), let $\zeta=h+g \in {\cH}^{u}$, where
$h$ and $g$ are such that $({\rm supp} \ h) \cap ({\rm supp} \
g)=\emptyset$ and ${\rm supp} \ h$ is compact. We have to prove that
$h \in \cH^u$.

Let $\chi$ be an infinitely differentiable function such that
$\chi=1$ on ${\rm supp} \ h$ and $\chi=0$ on an open set containing
${\rm supp} \ g$. By Corollary \ref{corollary20070928}, it follows
that $\zeta \in \cH_{2,0}^{k+2}(T)$, and hence $h = \chi \zeta \in
\cH_{2,0}^{k+2}(T)= \cH^u$ . The theorem is proved.
\end{proof}

\subsection{The Case of the Riesz Kernel}

In this subsection, we will assume that $f$ is the Riesz kernel,
i.e. $f=R_{\alpha}$ with $\max\{d-2,0\}< \alpha = 4k<d$, $k \in
\bN_+$.

According to \cite{samko76}, for any $0<\beta<d/2$ we can define the
Riesz potential
$$
I^{\beta}\varphi(y):=\varphi
*R_{\beta}(y)=\frac{1}{\gamma_{n}(\beta)}
\int_{\bR^d}\frac{\varphi(y)}{|x-y|^{d-\beta}}dy \quad \mbox{for
all} \ \varphi \in L_{2}(\bR^{d}).$$ Let $q=q_{d,\beta}>2$ be such
that $1/q=1/2-\beta/d$.

The space $I^{\beta}(L_{2}(R^d))$ of all Riesz potentials has the
following properties (see \cite{samko76}):
\begin{enumerate}
\item $I^{\beta}(L_{2}(R^d)) \subset L_{q}(\bR^d)$ and there exists a
constant $N>0$ such that $$\|I^{\beta}\varphi \|_{L_{q}(\bR^d)} \leq
N \|\varphi \|_{L_{2}(\bR^d)}, \quad \mbox{for all} \ \varphi \in
L_{2}(\bR^d).$$

\item For every $f \in I^{\beta}(L_{2}(R^d))$, we define the Riesz
derivative $\bD^{\beta}f$ as \linebreak $\lim_{\varepsilon \to
0}\bD_{\varepsilon}^{\beta}f$ in $L_{2}(\bR^d)$, where
$$(\bD_{\varepsilon}^{\beta}f)(x)=
\frac{1}{c_{d,l}(\beta)}\int_{|y|>\varepsilon}
\frac{(\Delta_y^{l}f)(x)}{|y|^{d+\beta}}dy \quad (\mbox{is
independent of} \ l),$$ $\Delta_{y}^{l}f=(I-\tau_y)^{l}f$ and
$(\tau_y f)(x)=f(x-y)$. Then
$$
\cF (\bD^{\beta}f) (\xi)={\cF}f(\xi)|\xi|^{\beta}, \ \ \ \forall f
\in \cS(\bR^d).
$$

\item $I^{\beta}(L_{2}(R^d))=\{f \in L_{q}(\bR^d); \bD^{\beta}f \
\mbox{exists and is in} \ L_{2}(\bR^d)\}$.

\item $\bD^{\beta}$ is the left inverse of $I^{\beta}$ in
$L_{2}(\bR^d)$, i.e. $\bD^{\beta}(I^{\beta}\varphi)=\varphi$ for all
$\varphi \in
L_{2}(\bR^d)$.

\item $\cD(\bR^d)$ is dense in $I^{\beta}(L_{2}(R^d))$ with respect to
 the norm $\|f\|_{L_{q}(\bR^d)}+\linebreak \|\bD^{\beta}f
 \|_{L_{2}(\bR^d)}$.

\end{enumerate}

In what follows, we will use these properties with
$\beta=\alpha/2=2k$, where $d-2<4k<d$. We let $q=q_{d,4k}>2$ be such
that $1/q=1/2-2k/d$.

\begin{proposition}
\label{Fourier-varphi1-Riesz} Let $2 < d/(2k)$ and $1/q = 1/2 -
2k/d$. If $f \in I^{2k}(L_2(\bR^d))$, where $k \in \bN_+$, then
$$
\cF \left( \bD^{2k} f \right) = |\xi|^{2k} \cF f,
$$
where $\cF f$ is an element in $\cS'(\bR^d)$.
\end{proposition}

\begin{proof}
First of all, $|\xi|^{2k} \cF f$ is well-defined as an element of
$\cS'$ since $|\xi|^{2k} = ( \xi_1^2 + \cdots + \xi_d^2 )^{k}$ is
infinitely differentiable.

Due to the fact that $\cD(\bR^d)$ is dense in $I^{2k}(L_2(\bR^d))$,
there exists a sequence $\{ f_n \} \subset \cD(\bR^d)$ such that
$$
\| f_n - f \|_{L_q(\bR^d)} + \| \bD^{2k} f_n - \bD^{2k} f
\|_{L_2(\bR^d)} \to 0.
$$
Note that $\cF \left( \bD^{2k} f_n \right) = |\xi|^{2k} \cF f_n \in
\cS(\bR^d)$. Also note that
$$
\left( \cF \left( \bD^{2k} f_n \right), \phi \right) = \left(
|\xi|^{2k} \cF f_n, \phi \right) = \left( \cF f_n, |\xi|^{2k} \phi
\right)
$$
$$
= \left( f_n, \cF^{-1} \left(|\xi|^{2k} \phi \right) \right) \to
\left( f, \cF^{-1} \left(|\xi|^{2k} \phi \right) \right) = \left(
|\xi|^{2k} \cF f, \phi \right),
$$
where the convergence is possible because $f_n \to f$ in
$L_q(\bR^d)$. This along with the fact that $\bD^{2k} f_n \to
\bD^{2k} f$ in $L_2(\bR^d)$ implies that $$ \cF \left( \bD^{2k} f
\right) = |\xi|^{2k} \cF f.
$$
\end{proof}

\begin{corollary} \label{claim_2006_12_19}
Let $2 < d/(2k)$ and $1/q = 1/2 - 2k/d$. If $f \in
I^{2k}(L_2(\bR^d))$, where $k \in \bN_+$, then
$$
\bD^{2k} f = (-\Delta)^k f.
$$
\end{corollary}

\begin{proof} Note that
$$
\left( \cF \left( \bD^{2k} f \right), \phi \right) = \left(
|\xi|^{2k} \cF f, \phi \right) = \left( \cF f, |\xi|^{2k} \phi
\right).
$$
On the other hand,
$$
\left( \cF \left( (-\Delta)^{k} f \right), \phi \right) = \left(
(-\Delta)^{k} f, \cF^{-1}\phi \right) = \left( f, (-\Delta)^{k}
\cF^{-1}\phi \right)
$$
$$
= \left( f, \cF^{-1} \left( |\xi|^{2k} \phi \right) \right) = \left(
\cF f, |\xi|^{2k} \phi \right).
$$

\end{proof}

\begin{lemma}\label{lemma20070927_01}
Let $2 < d/2k$. There exists a linear operator $J:\cP_0^{(d)} \to
L_2((0,T) \times \bR^d)$ such that
 $J$ is one-to-one and onto, and satisfies:
$$
J \varphi (t,x) = I^{2k} \left(\varphi(t,\cdot) \right) (x)
$$
for any $\varphi \in \cD((0,T) \times \bR^d)$.
\end{lemma}

\begin{proof}
For all $\varphi \in \cD((0,T) \times \bR^d)$, set
$$
J(\varphi) = I^{2k} \left( \varphi(t,\cdot) \right)(x) = c_{d,k}
\int_{\bR^d} \frac{\varphi(t,y)}{|x-y|^{d-2k}} \, dy,
$$
where $c_{d,k}$ is an appropriate constant. Denote $\varphi_0 :=
J(\varphi)$. We see that $\varphi_0(t,x)$ is measurable in $(t,x)
\in (0,T) \times \bR^d$. Since $\varphi(t,\cdot) \in \cS(\bR^d)$ for
each fixed $t \in (0,T)$, we have
$$
\cF \varphi_0(t,\cdot) (\xi) = \cF \varphi(t,\xi) |\xi|^{-2k}
$$
in the sense that
$$
\int_{\bR^d} \varphi_0(t,x) \overline{\phi(x)} \, dx = \int_{\bR^d}
\cF \varphi(t,\xi) |\xi|^{-2k} \overline{\cF \phi(\xi)} \, d \xi,
$$
where $\phi \in \cS(\bR^d)$. Also notice that, for a.e. $t \in
(0,T)$,
$$
\cF \varphi(t,\xi) |\xi|^{-2k} \in L_2(\bR^d)
$$
because $\varphi \in \cP_0^{(d)}$ (recall that $\alpha = 4k$), i.e.,
$$
\int_0^T\int_{\bR^d} |\cF \varphi(t,\xi) |\xi|^{-2k}|^2 \, d \xi \,
dt < \infty.
$$
Thus, for a.e. $t \in (0,T)$, as a function of $x \in \bR^d$,
$\varphi_0(t,x) \in L_2(\bR^d)$ and
$$
\cF \varphi_0 (t,\cdot)(\xi) = \cF \varphi(t,\xi) |\xi|^{-2k}.
$$
Thus
$$
\|J(\varphi)\|_{L_2((0,T) \times \bR^d)}^2 = \int_0^T \int_{\bR^d} |
\varphi_0(t,x) |^2 \, dx \, dt
$$
$$
= \int_0^T \int_{\bR^d} | \cF \varphi(t,\xi)|^2 |\xi|^{-4k} \, d \xi
\, dt = \| \varphi \|_0^2.
$$
From this and the fact that $\cD((0,T) \times \bR^d)$ is dense in
$\cP_0^{(d)}$, we extend $J$ to all elements in $\cP_0^{(d)}$.

It is easy to see that $J$ is linear as well as one-to-one. To prove
the fact that $J$ is onto, take $\varphi_0 \in L_2((0,T) \times
\bR^d)$. Also take a sequence $\{{\varphi_0}_n\} \subset \cD((0,T)
\times \bR^d)$ such that ${\varphi_0}_n \to \varphi_0$ in $L_2((0,T)
\times \bR^d)$. Let $\varphi_n = (-\Delta)^{k} {\varphi_0}_n$. Then
$$
\cF \varphi_n (t, \xi) = |\xi|^{2k} \cF {\varphi_0}_n (t, \xi)
$$
and
$$
\int_0^T \int_{\bR^d} |\cF \varphi_n(t,\xi)|^2 |\xi|^{-4k} \, d \xi
\, dt = \int_0^T \int_{\bR^d} |\cF {\varphi_0}_n |^2 \, d \xi \, dt.
$$
Thus $\{\varphi_n\}$ is a Cauchy sequence in $\cP_0^{(d)}$, so there
is a $\varphi \in \cP_0^{(d)}$ such that $\varphi_n \to \varphi$ in
$\cP_0^{(d)}$. To prove $J(\varphi) = \varphi_0$, we only need to
prove that
$$
I^{2k}(\varphi_n) = {\varphi_0}_n,
$$
which follows from
$$
\cF \left( I^{2k} (\varphi_n) \right) = |\xi|^{-2k} \cF \varphi_n.
$$
\end{proof}

From the above result it follows that:

\begin{corollary}\label{corollary20070927_01}
For $\varphi$, $\eta \in \cP_0^{(d)}$, let $\varphi_0 := J(\varphi)$
and $\eta_0 : = J(\eta)$. Then
$$
\langle \varphi, \eta \rangle_0 = \langle \varphi_0, \eta_0
\rangle_{L_2((0,T) \times \bR^d)}.
$$
\end{corollary}

Now we set
$$
\varphi_1(t,x) = I^{2k} \left( \varphi_0(t,\cdot) \right) (x) =
c_{d,k} \int_{\bR^d} \frac{\varphi_0(t,y)}{|x-y|^{d-2k}} \, dy,
$$
where $\varphi_0 \in L_2((0,T)\times\bR^d)$. Notice that $\varphi_1$
is a measurable function of $(t,x) \in (0,T) \times \bR^d$. Since
$\varphi_0(t,x) \in L_2(\bR^d)$ for a.e. $t \in (0,T)$,
$$
\varphi_1(t,\cdot) \in L_q(\bR^d), \quad \quad \| \varphi_1(t,\cdot)
\|_{L_q(\bR^d)} \le N \| \varphi_0(t,\cdot) \|_{L_2(\bR^d)},
$$
for a.e. $t \in (0,T)$, where $1/q = 1/2 - 2k/d$. This implies that
$$
\varphi_1 \in L_{2,q}((0,T) \times \bR^d),
$$
$$
\| \varphi_1 \|_{L_{2,q}((0,T) \times \bR^d)} = \left(\int_0^T \|
\varphi_1(t,\cdot) \|_{L_q(\bR^d)}^2 \, dt \right)^{1/2} \le N \|
\varphi_0 \|_{L_2((0,T) \times \bR^d)}.
$$

From the $L_p$-theory of parabolic equations with mixed norms, there
exists a unique function $w \in W_{2,q}^{1,2}((0,T) \times \bR^d)$
satisfying
\begin{equation}                            \label{eq01}
w_t - \Delta w = \varphi_1, \qquad w(0,\cdot) = 0.
\end{equation}

\begin{lemma}
\label{lemma-w-equal-h} Let $h(t,x) = \bE M(\varphi) u(t,x)$, where
$u$ is defined in \eqref{solution-process}. Let $\varphi_0 =
J(\varphi)$, $\varphi_1$ be a function defined as above, and $w$ be
the solution to \eqref{eq01}. Then
$$
w = h.
$$
\end{lemma}

\begin{proof}
For $\eta \in \cD((0,T) \times \bR^d)$, find a function $\phi$, a
unique solution to
$$
- \phi_t - \Delta \phi = \eta, \qquad \phi(T,\cdot) = 0.
$$
Then by Lemma \ref{key-calculation}
$$
\langle h, \eta \rangle_{L_2((0,T)\times\bR^d)} = \langle \varphi,
\phi \rangle_0
$$
and
$$
\langle w, \eta \rangle_{L_2((0,T)\times\bR^d)} = \langle w_t -
\Delta w, \phi \rangle_{L_2((0,T)\times\bR^d)} = \langle \varphi_1,
\phi \rangle_{L_2((0,T)\times\bR^d)}.
$$
Find a sequence $\{ {\varphi_0}_n \} \subset \cD((0,T) \times
\bR^d)$ such that ${\varphi_0}_n \to \varphi_0$ in $L_2((0,T) \times
\bR^d)$. Let $\psi_n = I^{2k} ({\varphi_0}_n)$ and $w_n$ be the
solution to $(\partial/\partial t - \Delta) w_n = \psi_n$ and
$w_n(0,\cdot) = 0$. Then
$$
\langle w, \eta \rangle_{L_2((0,T)\times\bR^d)} = \lim_{n \to
\infty} \langle w_n, \eta \rangle_{L_2((0,T)\times\bR^d)} = \lim_{n
\to \infty} \langle \psi_n, \phi \rangle_{L_2((0,T)\times\bR^d)}
$$
$$
= \lim_{n \to \infty} \langle I^{2k}({\varphi_0}_n), \phi
\rangle_{L_2((0,T)\times\bR^d)} = \lim_{n \to \infty} \int_0^T
\int_{\bR^d} \cF {\varphi_0}_n(t,\xi) \overline{\cF \phi(t, \xi)}
|\xi|^{-2k} \, d \xi \, dt.
$$
On the other hand,
$$
\langle h, \eta \rangle_{L_2((0,T)\times\bR^d)} = \langle \varphi,
\phi \rangle_0 = \lim_{n \to \infty} \int_0^T \int_{\bR^d} \cF
\varphi_n(t,\xi) \overline{\cF \phi(t,\xi)} |\xi|^{-4k} \, d \xi \,
dt
$$
$$
= \lim_{n \to \infty} \int_0^T \int_{\bR^d} \cF {\varphi_0}_n(t,\xi)
\overline{\cF \phi(t,\xi)} |\xi|^{-2k} \, d \xi \, dt,
$$
where $\varphi_n = (-\Delta)^{2k} {\varphi_0}_n$. This finishes the
proof.
\end{proof}

Now we prove the main result of this subsection.

\begin{theorem}\label{main-theorem-Riesz}
Suppose that $f$ is the Riesz kernel of order $\alpha=4k,k \in
\bN_{+}$ and $d=4k+1$. Then the process solution $u$ of the
stochastic heat equation (\ref{heat-eq}) with vanishing initial
conditions is locally germ Markov.
\end{theorem}

\begin{proof}
Again we need to verify conditions (i) and (ii) in Theorem
\ref{kunsch-theorem}.

We first verify condition (i).

Let $h(t,x) = \bE M(\varphi) u(t,x)$ and $g(t,x) = \bE M(\eta)
u(t,x)$, where $\varphi, \eta \in \cP_0^{(d)}$, such that
$\text{supp} \, h \cap \text{supp} \, g = \emptyset$. We need to
show that $\langle \varphi, \eta \rangle_0 = 0$. Let $\varphi_0 =
J(\varphi)$ and $\eta_0 = J(\eta)$, where $J$ is the operator
defined in Lemma \ref{lemma20070927_01}. Also let $\varphi_1 =
I^{2k}\left(\varphi_0(t,\cdot)\right)$ and $\eta_1 =
I^{2k}\left(\eta_0(t,\cdot)\right)$. For a.e. $t \in (0, T)$, we
have
$$\varphi_0(t,x)=\bD^{2k} \varphi_1(t,x) =(-\Delta)^k \varphi_1,
$$
where the second equality follows from Corollary
\ref{claim_2006_12_19}. Let $\phi(t,x) \in \cD((0,T) \times \bR^d)$
such that $\phi(t,x) = 0$ on $\text{supp} \, \varphi_1$. Then for
a.e. $t \in (0,T)$,
$$
\int_{\bR^d} \varphi_0(t,x) \phi(t,x) \, dx = \int_{\bR^d}
\varphi_1(t,x) (-\Delta)^k \phi(t,x) \, dx.
$$
Thus
$$
\int_0^T \int_{\bR^d} \varphi_0(t,x) \phi(t,x) \, dx \, dt =
\int_0^T \int_{\bR^d} \varphi_1(t,x) (-\Delta)^k \phi(t,x) \, dx \,
dt = 0.
$$
This shows that $\text{supp} \, \varphi_0 \subset \text{supp} \,
\varphi_1$. Similarly, we have $\text{supp} \, \eta_0 \subset
\text{supp} \, \eta_1$. Hence we have $\langle \varphi_0, \eta_0
\rangle_{L_2((0,T)\times\bR^d)} = 0$. This and Corollary
\ref{corollary20070927_01} prove that $\langle \varphi, \eta
\rangle_0 = 0$.

We now verify condition (ii).

Assume that $\zeta = h + g$, where $\zeta(t,x) = \bE M(\nu) u(t,x)$,
$\text{supp} \, h \cap \text{supp} \, g = \emptyset$, and
$\text{supp} \, h$ is compact. We have to prove that $h \in \cH^u$.

Let $\chi$ be an infinitely differentiable function defined on
$[0,T] \times \bR^d$ such that $0 \le \chi \le 1$, $\chi = 1$ on
$\text{supp} \, \mu$, and $\chi = 0$ in an (relative) open set
containing $\text{supp} \, \nu$. Then $\chi \zeta = h$ and $h \in
W_{2,q}^{1,2}((0,T) \times \bR^d)$. Let $(\partial/\partial t -
\Delta) \zeta = \nu_1$, where $\nu_1 = I^{2k}(\nu_0)$ and $\nu_0 =
J(\nu)$. Now we set
$$
\varphi_1 := \chi \nu_1, \quad \eta_1 := (1-\chi) \nu_1, \quad
\varphi_0 := \chi \nu_0.
$$
Since $\varphi_0 \in L_2((0,T) \times \bR^d)$ there exists a
sequence $\{ {\varphi_0}_n \} \subset \cD((0,T) \times \bR^d)$ such
that ${\varphi_0}_n \to \varphi_0$ in $L_2((0,T) \times \bR^d)$. Let
$\varphi_n = (-\Delta)^{k} {\varphi_0}_n$. Then $\varphi_n \in
\cD((0,T) \times \bR^d)$ and $\{\varphi_n\}$ is a Cauchy sequence in
$\cP_0^{(d)}$ due to the fact that
$$
\cF \varphi_n(t,\xi) = |\xi|^{2k} \cF {\varphi_0}_n(t,\xi).
$$
Let $\varphi$ be the limit in $\cP_0^{(d)}$ of $\varphi_n$. Then
$J(\varphi) = \varphi_0$. Now let $\hat{\varphi}_1 =
I^{2k}(\varphi_0)$ and $\hat{h}(t,x) = \bE M(\varphi) u(t,x)$.

We now prove that $h=\hat h$, which will imply that $h \in \cH^u$.
For this, it suffices to prove that $\varphi_1 = \hat{\varphi}_1$.
By Lemma \ref{lemma-w-equal-h}, we have $(\partial/\partial t -
\Delta) \hat{h} = \hat{\varphi}_1$. Notice that $\nu_1 = \varphi_1 +
\eta_1$ and $\text{supp} \, \varphi_1 \cap \text{supp} \, \eta_1 =
\emptyset$. Thus for a.e. $t \in (0,T)$, we have
$$
\text{supp}_x \, \varphi_1(t,\cdot) \cap \text{supp}_x \, \eta_1(t,
\cdot) = \emptyset.
$$
Thus by Lemma \ref{lemma01} it follows that, for a.e. $t \in (0,T)$,
$$
\bD^{2k} \varphi_1(t,\cdot) = \varphi_0(t,\cdot).
$$
We also have
$$
\bD^{2k} \hat{\varphi}_1 (t,\cdot) = \varphi_0(t,\cdot)
$$
for a.e. $t \in (0,T)$. Since $\varphi_1(t,\cdot) -
\hat{\varphi}_1(t,\cdot) \in I^{2k}(L_2(\bR^d))$ for a.e. $t \in
(0,T)$, there exists a function $f_t(x)$ such that
$$
\varphi_1(t,\cdot) - \hat{\varphi}_1(t,\cdot) = I^{2k}(f_t)
$$
for a.e. $t \in (0,T)$. From the fact that $\bD^{2k}$ is the left
inverse of the $I^{2k}$, we know that
$$
0 = \bD^{2k} \left( \varphi_1(t,\cdot) - \hat{\varphi}_1(t,\cdot)
\right) = \bD^{2k} I^{2k}(f_t) = f_t.
$$
We also know that $I^{2k}(0) = 0$. Hence $\varphi_1(t,\cdot) =
\hat{\varphi}_1(t,\cdot)$ for a.e. $t \in (0,T)$. Therefore,
$\varphi_1 = \hat{\varphi}_1$ as elements in $L_{2,q}((0,T) \times
\bR^d)$. The theorem is proved.
\end{proof}

\appendix

\section{Proof of Relation (\ref{conv-norm0})}
\noindent We need to prove that
$$\int_0^T \int_{\bR^d}
|\cF \varphi_n(s,\xi)-\cF \varphi(s,\xi) |^{2}\mu(d\xi)ds \to 0, \ \
\ {\rm as} \ n \to \infty.$$

\noindent Note that $\cF \varphi_n(s,\xi) = \sum_{m \in I_n} c_d
|Q^{(n)}_m| \exp \{- \mathrm{i} \xi \cdot x^{(n)}_m - (t^{(n)}_m-s)
|\xi|^2\} \eta(t^{(n)}_m,x^{(n)}_m)$ converges pointwise to $$\cF
\varphi(s,\xi)= \int_s^T \int_{\bR^d} c_d \exp \{ - \mathrm{i} \xi
\cdot x - (t-s) |\xi|^2 \} \eta(t,x) \, dx \, dt.$$

\noindent In order to apply the dominated convergence theorem, we
need to prove that there exists a function $\Psi$, which is
square-integrable with respect to $ds \times \mu(d\xi)$ such that
$|\cF \varphi_n(s,\xi)| \le \Psi(s,\xi)$. Let $N$ be a constant such
that $|\eta| \le N$. Then
$$|\cF \varphi_n(s,\xi)| \le \sum_{m \in I_n} c_d |Q^n_m| |\eta(t^{(n)}_m,x^{(n)}_m)|
\exp \{- (t^{(n)}_m-s) |\xi|^2\}$$
$$\le \sum_{m \in I_n} c_d N
|Q^{(n)}_m| e^{- (t^{(n)}_m-s) |\xi|^2} \le \int_s^T \int_{K'} c_d N
e^{-(t-s)|\xi|^2} \, dx \, dt$$
$$\le c_d N |K'| (T-s) 1_{|\xi|< 1} + c_d N |K'|
\frac{1-e^{(s-T)|\xi|^2}}{|\xi|^2} 1_{|\xi| \ge 1}: = \Psi(s,\xi),$$
where $K'$ is a compact set containing $K$, and the third inequality
above is due to the fact that $1_{Q^{(n)}_m}(t,x) e^{-(t_m^{(n)}-s)
|\xi|^2 } \le 1_{Q^{(n)}_m}(t,x) e^{-(t-s) |\xi|^2 }$. Clearly
$\Psi$ is square-integrable with respect to $ds \times \mu (d \xi)$.
This concludes the proof of (\ref{conv-norm0}).

\section{An Auxiliary Lemma}
The following technical result was used in the proof of Theorem
\ref{main-theorem-Riesz} for the verification of condition (ii).

\begin{lemma}
\label{lemma01} Let $2 < d/ 2k$ and $1/q = 1/2 - 2k/d$. Assume that
$\kappa \in I^{2k}(L_2(\bR^d))$ and $\kappa = \mu + \nu$, where
$\mu$, $\nu \in L_q(\bR^d)$, $\text{supp} \, \mu \cap \text{supp} \,
\nu = \emptyset$, and $\text{supp} \, \mu$ is compact. Then
$$
\mu \in I^{2k}(L_2(\bR^d)), \qquad \bD^{2k} \mu = \chi
\bD^{2k}\kappa,
$$
where $\chi$ is an infinitely differentiable function such that $0
\le \chi \le 1$, $\chi = 1$ on $\text{supp} \, \mu$, and $\chi = 0$
in an open set containing $\text{supp} \, \nu$.
\end{lemma}

\begin{proof}
Since $\kappa \in I^{2k}(L_2(\bR^d))$, we have
$$
\bD_{\varepsilon}^{2k} \kappa := \frac{1}{c(d,l,2k)} \int_{|z| >
\varepsilon} \frac{\left( \Delta_z^{l} \kappa
\right)(x)}{|z|^{d+2k}} \, dz \to \bD^{2k} \kappa \quad \text{in}
\quad L_2(\bR^d).
$$
Note that $\bD_{\varepsilon}^{2k} \kappa \in L_2(\bR^d)$ (see
Proposition 2.4 in \cite{almeida-samko06}). Let
$$
\Lambda_{\varepsilon}^{2k} \kappa := \frac{1}{c(d,l,2k)}
\int_{\varepsilon < |z| \le \varepsilon_0} \frac{\left( \Delta_z^{l}
\kappa \right)(x)}{|z|^{d+2k}} \, dz, \quad \varepsilon <
\varepsilon_0.
$$
Since $\Lambda_{\varepsilon}^{2k} \kappa = \bD_{\varepsilon}^{2k}
\kappa - \bD_{\varepsilon_0}^{2k} \kappa$, we have
$\Lambda_{\varepsilon}^{2k} \kappa \in L_2(\bR^d)$. Moreover,
$$
\Lambda_{\varepsilon_1}^{2k} \kappa - \Lambda_{\varepsilon_2}^{2k}
\kappa = \int_{\varepsilon_1 < |z| \le \varepsilon_2} \frac{\left(
\Delta_z^{l} \kappa \right)(x)}{|z|^{d+2k}} \, dz =
\bD_{\varepsilon_1}^{2k} \kappa - \bD_{\varepsilon_2}^{2k} \kappa.
$$
Form this and the fact that $\bD_{\varepsilon}^{2k} \kappa$ is
Cauchy in $L_2(\bR^d)$, it follows that
$\{\Lambda_{\varepsilon}^{2k} \kappa\}$ is a Cauchy sequence in
$L_2(\bR^d)$. Consider $\bD^{2k}_{\varepsilon} (\chi \kappa)$, which
is well-defined because $\chi \kappa \in L_q(\bR^d)$. Note that
$$
\bD^{2k}_{\varepsilon} (\chi \kappa) = \Lambda^{2k}_{\varepsilon}
(\chi \kappa) + \int_{|z| \ge \varepsilon_0} \frac{\left(
\Delta_z^{l} \chi \kappa \right)(x)}{|z|^{d+2k}} \, dz
$$
and
\begin{equation}
\label{technical-eq-appendixC} \Lambda^{2k}_{\varepsilon} (\chi
\kappa) = \chi \Lambda^{2k}_{\varepsilon} \kappa
\end{equation}
for a sufficiently small $\varepsilon_0 > 0$. The proof of
\eqref{technical-eq-appendixC} is stated below. Also note that
$$
\Lambda^{2k}_{\varepsilon} (\chi \kappa) \in L_2(\bR^d), \quad
\int_{|z| \ge \varepsilon_0} \frac{\left( \Delta_z^{l} \chi \kappa
\right)(x)}{|z|^{d+2k}} \, dz \in L_2(\bR^d),
$$
where the latter follows from the fact that $\chi \kappa \in
L_2(\bR^d)$ ($\chi$ has a compact support and $2 \le q$). In
addition, $\Lambda^{2k}_{\varepsilon} \kappa$ approaches a function
in $L_2(\bR^d)$ as $\varepsilon \searrow 0$. Thus
$\bD^{2k}_{\varepsilon} (\chi \kappa)$ converges a function in
$L_2(\bR^d)$. This shows that $\mu = \chi \kappa \in
I^{2k}(L_2(\bR^d))$. To prove $\bD^{2k} \mu = \chi \bD^{2k}\kappa$,
we can just use Corollary \ref{claim_2006_12_19}.

Let us prove \eqref{technical-eq-appendixC} as follows. For each $ x
\in \text{supp} \, \mu$, find $\delta_x > 0$ such that $B(x,
\delta_x) \cap \text{supp} \, \nu = \emptyset$. Consider
$$
\{ B(x, \delta_x/5) : x \in \text{supp} \, \mu \}.
$$
Since $\text{supp} \, \mu$ is compact, we have a finite subset $\{
x_i : i \in I \} \subset \text{supp} \, \mu$ such that
$$
\bigcup_{i \in I} B(x_i, \delta_{x_i}/5) \supset \text{supp} \, \mu.
$$
Let
$$
\cO_1 := \bigcup_{i \in I} B(x_i, \delta_{x_i}/4), \quad \cO_2 :=
\bigcup_{i \in I} B(x_i, \delta_{x_i}/3), \quad \cO_3 := \bigcup_{i
\in I} B(x_i, \delta_{x_i}/2).
$$
Then
$$
\cO_i \supset \text{supp} \, \mu, \quad i = 1, 2, 3
$$
and
$$
\cO_i \cap \text{supp} \, \nu = \emptyset, \quad i = 1, 2, 3.
$$
We may assume that $\chi = 1$ on $\cO_2$ and $\chi = 0$ on
$\cO_3^c$. We also assume that $\varepsilon_0 < \delta/(20 l)$,
where $\delta = \min_{i \in I} \delta_{x_i}$. By definition we have
$$
\Lambda_{\varepsilon}^{2k} \left( \chi \kappa  \right) (x)=
\frac{1}{c} \int_{\varepsilon < |z| \le \varepsilon_0} \frac{\left(
\Delta_z^{l} (\chi \kappa) \right)(x)}{|z|^{d+2k}} \, dz
$$
$$
= \frac{1}{c} \int_{\varepsilon < |z| \le \varepsilon_0} \frac{
\sum_{k=0}^l (-1)^k c_l^k \chi(x-kz) \kappa(x-kz)}{|z|^{d+2k}} \, dz
$$
Let $x \in \cO_1$, i.e. $x \in B(x_i, \delta_{x_i}/4)$. Then
$$
|x_i - (x - k z)| \le |x_i - x| + |k z| \le \delta_{x_i}/4 + l
\delta/(20 l)
$$
$$
\le \delta_{x_i}/4 + \delta_{x_i}/12 = \delta_{x_i}/3.
$$
Thus $x-kz \in \cO_2$. This shows that $\chi(x-kz) = 1$ and
$$
\Lambda_{\varepsilon}^{2k} \left( \chi \kappa  \right) (x)= \chi(x)
\Lambda_{\varepsilon}^{2k} \kappa (x), \quad x \in \cO_1.
$$
Now let $x \in \cO_1^c$. Then for every $x_i$, $i \in I$,
$$
|x_i - (x - k z)| \ge |x_i - x| - l|z|
$$
$$
\ge \delta_{x_i}/4 - l \delta/(20 l) \ge \delta_{x_i}/4 -
\delta_{x_i}/20 = \delta_{x_i}/5.
$$
This implies that $x-kz \notin \text{supp} \, \mu$. In that case we
have
$$
\chi(x-kz) \kappa(x-kz) = 0
$$
because $\kappa(x-kz) = 0$ in case $\chi(x-kz) > 0$. Thus
$\Lambda_{\varepsilon}^{2k} \left( \chi \kappa  \right) (x) = 0$. In
addition, $\chi(x) \Lambda_{\varepsilon}^{2k} \kappa (x) = 0$
(recall that $x \in \cO_1^c$ and $x-kz \notin \text{supp} \, \mu$)
because $\kappa(x-kz) = 0$ in case $\chi(x) > 0$. Indeed, if $x \in
\cO_3 \setminus \cO_1$, then there exists $x_j$, $j \in I$, such
that $x \in B(x_j, \delta_{x_j}/2)$. Then
$$
|x_j - (x-kz)| \le |x_j - x| + l |z| < \delta_{x_j}.
$$
Thus by the choice of $\delta_{x_j}$ it follows that $x-kz \notin
\text{supp} \, \nu$, that is, $\kappa(x-kz) = 0$.
\end{proof}

\par\bigskip\noindent
{\bf Acknowledgment.} The authors are grateful to an anonymous
referee who carefully read the article and made some suggestions for
future research.

\bibliographystyle{amsplain}


TITLE: $ o(ab)= o(ba)$, What about when $ab$ has infinite order?
QUESTION [1 upvotes]: I have been able to prove this when I say simply assume $o(ab)= n$, but I have been pondering the case of where $o(ab)$ is infinite?
Is this something that needs to be considered? And if not, why?
My rationale is that if $o(ab)$ is infinite, then no comparison can really be made with infinity. i.e. It can't really be said that two infinite things are equal. Is that correct?

REPLY [4 votes]: The order is always the same, since $ab$ and $ba$ are conjugated elements through the automorphisms:
$$\phi_a : x \to a^{-1} x a,$$
$$\phi_b : x \to b^{-1} x b,$$
hence if $o(ab)$ is finite, $o(ba)|o(ab)$, so $o(ba)$ is finite, and since $o(ba)|o(ab)$, $o(ab)=o(ba)$.
If $o(ab)$ is infinite, $o(ba)$ is infinite, too.

REPLY [3 votes]: Suppose $ab$ has infinite order but $ba$ had finite order $n$. Then you can use your previous working to prove that $ab$ also had order $n$, a contradiction.
Hence, if $ab$ has infinite order then so does $ba$.


TITLE: On the generators of free $R$-modules.
QUESTION [0 upvotes]: Let $R$ be a ring and $M$ be an $R$-module. Let $(F(M),\eta)$ be the free $R$-module over $M$. By that I mean that, for every $R$-module $N$ and every function $f\colon M\to N$ there exists exactly one $R$-linear mapping $\hat{f}\colon F(M)\to N$ such that $f=\hat{f}\circ \eta$. Now let $X$ be a set of generators for $M$. Is it possible to prove that $\eta(X)$ is a set of generators for $F(M)$, using just the universal property of $F(M)$? I don't want to use any realisation of the free $R$-module F(M), such as the direct sum of $|M|$ copies of $R$, I would like to use only universality. Is that possible?

REPLY [2 votes]: The generators of $M$ have nothing to do with generators of $F(M)$. In fact, if $X \subsetneq M$ is a proper subset of $M$, then $\eta(X)$ does not generate $F(M)$. The point here is that the free module structure $F(M)$ does not in any way depend on module structure on $M$, it only sees $M$ as a set.


TITLE: A question about time in Special and General Relativity.
QUESTION [0 upvotes]: I apologize if this question is considered too mathematically imprecise. My understanding of Special and
General Relativity comes from reading books which attempt to explain them to non-expertsin these fields.
Let C denote a clock (such as an atomic clock) fixed at the earth's surface. It seems that the rate at
which another clock B runs can be "slowed down" to almost any extent one chooses (when compared with A's
rate) if either (1) B moves at a speed sufficiently close to that of light-with respect to A-or (2) B is
at rest with respect to A but is located in a sufficiently strong gravitational field. I have never found
mention of any "scenario" in which B's rate could be "speeded up" to any desired extent (when compared
with A's rate). Are there any such "scenarios"? If not, is there some fact or law which implies that their
existence is impossible?

REPLY [3 votes]: It is possible, but not in special relativity (because of the answer given by Youloush). Use a 2-dimensional space-time diffeomorphic to $\mathbb{R}^2$ and define a pseudo-riemannian metric (using coordinates for the tangent space from the flat case):
$g_{t,x}((a,b), (c,d))=(1+x^4)\cdot a\cdot c-b\cdot d$
The coordinate $t$ is identical to the proper time of an observer resting at point $0$. Now consider the proper time of an observer moving with constant velocity (with respect to our coordinates) from $(0,0)$ to $(2,2)$ and back to $(4,2)$ (he starts at the speed of light, but we do not have to worry, we would get the same result when we would start a little bit slower). The proper time elapsed is:
$2\cdot\int_0^2 \sqrt{(1+t^4)-1}\mathrm{d}t=2\cdot\int_0^2 t^2\mathrm{d}t=\frac{16}{3}$
This number is larger than $4$ (the proper time elapsed for the resting observer).
However, you can generalize the result from special relativity: Geodesics are those paths (locally) extremising the elapsed proper time. In the most typical cases you actually maximise the elapsed proper time (for example in flat space-time, where the geodesics are given by movements at constant velocity, look at the example above: you move far away (to 2), where the metric is very different, to get the effect, if you would stay within $(-1,1)$, the effect would not happen and it would be like in the flat case: dilatation). Then on every non-geodesic path you get a smaller proper time. However, in general it is not the case.


text:
\begin{document}

\maketitle

\section{Introduction}

The genus field of a number field $K$ is defined to be the maximal extension $K^*$ of $K$
that is unramified at all finite primes and is a compositum of the form $Kk^*$
where $k^*$ is absolutely abelian.  The genus number is defined as $g_K=[K^*:K]$.
It follows immediately from class field theory that $g_K$ divides the narrow class number $h_K^+$.
See~\cite{mcgown.tucker} for a slightly more detailed introduction
and~\cite{ishida} for a comprehensive account of genus fields.

One may consider the density
of genus number one fields among all number fields of a fixed degree and signature,
ordered by their discriminants.  It essentially follows from a classical theorem of Gauss
that $0\%$ of quadratic fields have genus number one.  On the other hand,
McGown and Tucker proved that a positive proportion (roughly $96.23\%$)
of cubic fields have genus number one.  Due to subtleties that arise in the quartic case,
we will temporarily put degree four fields aside and consider this
problem in the quintic case.

Let $\mathcal{F}$ denote the collection of all quintic fields $K$,
and $\mathcal{G}$ denote the collection of all quintic fields with $g_K=1$.
For $i=0,1,2$,
write $\mathcal{F}^{(i)}$ and $\mathcal{G}^{(i)}$ to denote the subsets of $\mathcal{F}$
and $\mathcal{G}$, respectively, consisting of fields with precisely~$i$ complex embeddings.
Define $N^{(i)}(X)=\#\{K\in\mathcal{F}^{(i)}:|\Disc(K)|\leq X\}$ and
$N_g^{(i)}(X)=\#\{K\in\mathcal{G}^{(i)}:|\Disc(K)|\leq X\}$.
Bhargava proved \cite{bhargava.quintic} that
$N^{(i)}(X)\sim C^{(i)}X$
where
\[
  C^{(i)}=\prod_p(1+p^{-2}-p^{-4}-p^{-5})\cdot
  \begin{cases}
  \frac{1}{240} & \text{if $r=0$}\\  \frac{1}{24} & \text{if $r=1$} \\  \frac{1}{16} & \text{if $r=2$}\,.
  \end{cases}
\]

Our main result concerning counting genus number one fields is the following:
\begin{theorem}\label{T:1}
\[
  N_g^{(i)}(X)=\left(C^{(i)}\frac{506874}{506875}\prod_{p\equiv 1\pmod{5}} \frac{p^4+p^3+2p^2+2p}{p^4+p^3+2p^2+2p+1} \right)X + O(X^{1-\frac{1}{400} + \eps})
  \,.
\]

\end{theorem}
\begin{corollary}
  The proportion of quintic fields with $i$ complex embeddings having genus number one equals
  \[
  \frac{506874}{506875}
  \prod_{p\equiv 1\pmod{5}} \frac{p^4+p^3+2p^2+2p}{p^4+p^3+2p^2+2p+1}
  \,.
  \]
\end{corollary}
Although the number above (which is approximately $0.999935$) is quite close to $1$, this implies
that a positive proportion of quintic fields have class number divisible by $5$.
(Note that since $g_K$ divides the narrow class number and $g_K$ is a power of $5$,
we also have that $g_K$ divides the class number.)
In fact, our methods yield the following stronger result:
\begin{theorem}\label{T:2}
Given any $k\geq 0$, there is a positive proportion of quintic fields with
genus number equal to $5^k$, and hence with class number divisible by $5^k$.
\end{theorem}
We also prove the following:
\begin{theorem}\label{T:3}
The average of the genus number taken over all quintic fields of a given signature is constant.
\end{theorem}

Numerically, this constant is $1.00026\dots$; see \eqref{eq:average_genus} for an exact expression.

Late in the preparation of this paper, we became aware of the work of Kim (see~\cite{kim}) on
 statistical questions concerning the genus number in cyclic and dihedral extensions of prime degree, in which he says ``it seems very difficult to compute genus numbers of [$S_5$] fields''.
All of the results in this paper hold, with essentially identical proofs, when restricted to $S_5$-quintic fields only.
As such, our results establish that a variety of statistics can indeed be computed for genus numbers of such fields.

Finally, we also show the following:
\begin{theorem}\label{T:4}
A positive proportion of quintic fields with genus number one fail to be norm-Euclidean.
\end{theorem}

\section{Counting quintic fields with specified local completions}

Our proofs will apply two results on counting quintic fields. The first, essentially due to Ellenberg, Pierce, and Wood
\cite[Theorem 5.1]{ellenberg.pierce.wood}, and building on Bhargava's work
\cite{bhargava.quintic},
is a result counting quintic fields
with a finite specified set of local conditions.

By a {\itshape local condition} $\mathcal{S}_p$ on a quintic field $K$, at a prime $p$, we mean
a demand that $K\otimes \Q_p\in\Sigma_p$ for some
fixed subset $\Sigma_p$ of the quintic \'etale algebras over $\Q_p$.
By a set of local conditions $\mathcal{S} \pmod{e}=(\mathcal{S}_p)_{p \mid e}$, we mean a choice of local condition
$\mathcal{S}_p$ for each prime $p \mid e$.

To avoid a technical complication, we assume, for each $p \mid e$ other than $p = 5$, that $\mathcal{S}_p$ corresponds to a
`splitting type' -- i.e., that $\Sigma_p$ consists of those algebras $K_{v_1} \times \dots \times K_{v_g}$ for which $g$, each
$e(K_{v_i}|\Q_p)$, and each $f(K_{v_i}|\Q_p)$ take a prescribed value. We write $e = e_1 e_2$ or $e = 5e_1 e_2$, where $e_1$ and $e_2$ are the products of those
primes $p \neq 5$ for which $\Sigma_p$ consists of unramified or ramified algebras, respectively.

\begin{theorem}[Ellenberg--Pierce--Wood \cite{ellenberg.pierce.wood}]

\label{T:counting}
Let $N^{(i)}(X,\mathcal{S})$ be the number of quintic fields $K$ with $|\Disc(K)|<X$
having $i$ complex embeddings, and
satisfying a set of local conditions $\mathcal{S} \pmod{e}$ with the restriction described above. Then, we have
\begin{align*}
  N^{(i)}(X;\mathcal{S})=C^{(i)}(\mathcal{S})
  X+O\left(e_1^{1/2} e_2 X^{79/80 + \eps} + X^{199/200 + \eps} \right),
\end{align*}
where
\begin{align*}
C^{(i)}(\mathcal{S})&=C^{(i)}\prod_p C_p(\mathcal{S}_p), \\
  C_p(\mathcal{S}_p)
  &=
  m(p)^{-1}\sum_{F\in\Sigma_p}\frac{1}{\Disc_p(F)}\frac{1}{\#\Aut(F)}
  \,,
  \\
  m(p)
  &=
  1+p^{-1}+2p^{-2}+2p^{-3}+p^{-4}
  \,.
\end{align*}
\end{theorem}
The function $m(p)$ appearing above is the total mass of all quintic \'etale extensions of $\Q_p$.
Indeed, Bhargava
(see~\cite{bhargava.mass})
proves a generalization of Serre's mass formula (see~\cite{serre})
which, specialized to the quintic case, yields
\[
  \sum_{[K:\Q_p]=5\text{ \'etale}}
  \frac{1}{\Disc_p(K)}\frac{1}{\#\Aut(K)}=m(p)
  \,.
\]
For the condition that $p$ not totally ramify, i.e.,
$$\Sigma_p=\{[F:\Q_p]=5 \text{ \'etale}\mid e(F/\Q_p)<5\}\,,$$
we have
\[
\sum_{F\in\Sigma_p}\frac{1}{\Disc_p(F)}\frac{1}{\#\Aut(F)} = 1+p^{-1}+2p^{-2}+2p^{-3}
\,.
\]

\begin{proof} The proof is essentially identical to that of \cite[Theorem 5.1]{ellenberg.pierce.wood}.
The explanation given there is quite thorough, and we will indicate only what needs to be changed.

As in \cite{ellenberg.pierce.wood}, the proof applies an inclusion-exclusion argument to points in the lattice
$V_{\Z} = \Z^{40}$ studied in \cite{bhargava.quintic}. For each squarefree integer $q$ coprime to $5e$, let
$W_{q, e} \subset V_\Z$ denote the set of elements corresponding to quintic rings that are:
nonmaximal at each prime dividing $q$; maximal and of the desired splitting type at primes dividing $5e$.

Write $e_1$ for the product of primes $p \mid e$, excluding $5$, for which $\mathcal{S}_p$ prescribes
an unramified splitting type at $p$, and write $e_2$ for the analogous product where
$\mathcal{S}_p$ prescribes
a ramified splitting type.

Then, $W_{q, e}$ is defined by congruence conditions modulo $q^2 e_1 e_2^2 5^k$, for a positive integer $k$ which
to the authors' knowledge has not been computed. This contrasts to the situation in \cite{ellenberg.pierce.wood},
where no special conditions modulo $5$ were imposed, and where the conditions modulo $e$ were all unramified.
In the ramified case, the splitting type $(1^5)$ is defined $\pmod{e}$, as described in
\cite[Section 12]{bhargava.quintic}, but the maximality condition is no longer automatic and this is defined only $\pmod{e^2}$.

The remainder of the analysis remains essentially unchanged. The error term of \cite[(5.3)]{ellenberg.pierce.wood} includes an error term $q^2 \delta_P e$,
and $e$ must be replaced by $e_1 e_2^2$ as described above. (The $5^k$ term contributes to the implied constant, and we may
ignore this contribution.) We now have $\delta_P \ll e_2^{-1}$, reflecting the fact that the fields being counted are rare.
Finally, in invoking \cite[(27)]{bhargava.quintic}, it must be assumed that $q^2 e_1 e_2^2 \ll X^{1/40}$.

The ensuing analysis then remains valid, with identical or improved bounds on the error terms $E_1, E_2, E_3, E_4$. The restriction
that $q^2 e_1 e_2^2 \ll X^{1/40}$ proves to be the bottleneck, and choosing $Q = X^{1/80} e_1^{-1/2} e_2^{-1}$ completes the proof.
\end{proof}

\begin{remark} For a general set of local conditions $\mathcal{S} \pmod{e}$, analogous results hold with an undetermined
$e$-dependence in the error term. As the proof shows, this dependence can be computed in terms
of a modulus for which $W_{q, e}$ can be defined by congruence
conditions.
\end{remark}

We will also apply the following complementary `tail estimate', which was proved by Bhargava, Cojocaru, and the second author \cite{bhargava.cojocaru.thorne}:
\begin{theorem}\label{thm:tail}
For any $Y > 1$, define
\begin{equation}
N_5(X, Y) :=
\# \left\{ K \textnormal{ quintic} : \ |\Disc(K)| \leq X, \ q^2 \mid \Disc(K) \textnormal{ for some squarefree } q > Y \right\}.
\end{equation}
Then
\[
N_5(X, Y) \ll_\varepsilon X^{\frac{39}{40} + \varepsilon} + \frac{X}{Y^{1 - \varepsilon}}.
\]
\end{theorem}

\section{The genus theory of quintic fields}

For $p\neq 5$, write $k(p)$ to denote the unique quintic subfield of $\Q(\zeta_p)$ if such a field exists and $\Q$ otherwise,
and write $k(5)$ to denote the unique quintic subfield of $\Q(\zeta_{25})$.
Let $K$ be a non-cyclic quintic field.  Ishida proves that for $p\neq 5$,
$Kk(p)/K$ is a nontrivial unramified extension if and only if $p$ is totally ramified in $K$ and
$p\equiv 1\pmod{5}$.
Moreover, he shows that $Kk(5)/K$ is a nontrivial unramified extension if and only if $5$ is totally ramified and
$N(\gamma)^{4}\equiv 1\pmod{25}$ for all $\gamma\in\OK$ coprime to $5$.
We will refer to this latter congruence condition as condition~$(\star)$.
Ultimately, Ishida proves the following result (see~\cite[Chapter 5]{ishida}).

\begin{theorem}[Ishida]\label{T:ishida}
Let $t$ denote the number of primes $p$ such that $p$ is totally ramified in $K$ and $p\equiv 1\pmod{5}$,
and add $+1$ to $t$ if $5$ is totally ramified in $K$ and condition $(\star)$ is satisfied.
Then we have:
\[
   g_K=\begin{cases}
   5^{t-1} & \text{if $K$ is cyclic}\\
   5^t & \text{if $K$ is not cyclic}
   \end{cases}
\]
\end{theorem}

We wish to reformulate condition $(\star)$ in a manner suitable for our calculations.
Suppose $5$ is totally ramified in $K$.
Let $\mathfrak{p}$ be the unique prime in $K$ above $5$.
It is plain that the congruence condition above involving the norm can be checked in the quintic extension
$K_\mathfrak{p}/\Q_5$ of local fields.  It is equivalent to requiring that
$N(u)^4\equiv 1\pmod{25}$ for all $u\in\mathcal{O}_{\mathfrak{p}}^\times$.
Since every element of $\mathcal{O}_{\mathfrak{p}}^\times$ is a $4$-th root of unity
times an element of the principal units $\mathcal{U}^{(1)}=1+\mathfrak{p}$, it suffices
to check the congruence for all $u\in\mathcal{U}^{(1)}$.

Since $5$ is totally ramified in $K_\mathfrak{p}$ it is possible to choose a generating polynomial
$f(x)=x^5 + a_4x^4 + a_3x^3 +a_2x^2 +a_1x+a_0$ that is Eisenstein at the prime $5$.
A simple calculation (given on~pages 57--59 of~\cite{ishida})
shows that the condition
$N(u)^4\equiv 1\pmod{25}$ for all $u\in\mathcal{U}^{(1)}$
is equivalent to
\[
  a_1 \equiv a_2 \equiv a_3 \equiv a_4+a_0 \equiv 0 \pmod{5^2}
  \,.
\]
There are $25$ quintic ramified extensions of $\Q_5$.  See~\cite{jones.roberts, amano}.  For each of these, we use the generating polynomial given in the Jones--Roberts database to check condition ($\star)$.
We find that precisely $5$ of these $25$ extensions satisfy the condition.
Generating polynomials for these extensions are given by
$x^5-5x^4+5(1+5a)$ for $0\leq a \leq 4$.
Moreover, these are precisely the Galois extensions.
We mention in passing that all of these extensions have discriminant $5^8$.
We have thus proved the following two results.

\begin{lemma}\label{5totram}
Suppose $K$ is a non-cyclic quintic field where $5$ is totally ramified.
Let $k(5)$ be the unique quintic subfield of $\Q(\zeta_{25})$.
We have that $Kk(5)/K$ is unramified if and only if $K\otimes_{\Q} \Q_5$ is a Galois extension of $\Q_5$.

\end{lemma}

\begin{lemma}\label{5mass}
The mass of all totally ramified quintic Galois extensions of $\Q_5$ is equal to
\[
  \sum_{[K:\Q_5]=5 \text{ Galois}}
    \frac{1}{\Disc_5(K)}\frac{1}{\#\Aut(K)}
  =\frac{1}{5^8}
  \,.
\]
\end{lemma}

\section{Proofs of Theorems~\ref{T:1},~\ref{T:2},~\ref{T:3},~\ref{T:4}.}

\begin{proof}[Proof of Theorem~\ref{T:1}]
Theorem \ref{T:ishida} and Lemma \ref{5totram}
establish that $\mathcal{G}$, the set of non-cyclic quintic fields $K$
with genus number $g_K = 1$, consists precisely of those $K$ satisfying the following local conditions:
\begin{itemize}
\item No prime $p \equiv 1 \pmod{5}$ is totally ramified in $K$.
\item Either $5$ is not totally ramified in $K$, or $K\otimes \Q_5$ is not a Galois field extension of $\Q_5$.
\end{itemize}

We count these fields (and hence prove Theorem~\ref{T:1}) by means of an inclusion-exclusion sieve, adapting the approach of
Belabas, Bhargava, and Pomerance in \cite{belabas.bhargava.pomerance}. Throughout this section, we ignore the cyclic
quintic fields; by \cite{CoDiOl3} there are $\sim c X^{1/4}$ of them with discriminant bounded by $X$, and hence they
do not contribute to any of our asymptotics. Alternatively, one may exclude all of the non-$S_5$ quintic fields, of which there
are $\ll X^{39/40 + \epsilon}$ by \cite{shankar.tsimerman.s5} and \cite{bhargava.cojocaru.thorne}.

For each quintic field $K$, let $f = f(K)$ denote the product of
the primes $p \equiv 0, 1 \pmod{5}$ that are totally ramified in $K$, with the
additional condition for $p = 5$ that $K\otimes \Q_5$ be a Galois field extension of $\Q_5$.
Then $\mathcal{G}^{(i)}$ is the set of all $K$ in $\mathcal{F}^{(i)}$
for which $f(K) = 1$.
Denote by $N_{g}^{(i)}(X)$ the counting function for $\mathcal{G}^{(i)}$.

Let $T$ denote the collection of all squarefree $f$ whose prime divisors $p$ all satisfy
$p \equiv 0, 1\pmod{5}$. For each $f \in T$, write $\mathcal{S}_f$
for the set of local conditions specifying that
$f \mid f(K)$.
Then it follows, for an arbitrary parameter $Y$, that
\begin{align}
N_{g}^{(i)}(X) & = \sum_{f\in T} \mu(f) N^{(i)}(X;\mathcal{S}_f) \label{eq:ng1} \\
& = \sum_{\substack{f\in T \\ f \leq Y}} \mu(f) N^{(i)}(X;\mathcal{S}_f) + \nonumber
 \sum_{\substack{f\in T \\ f > Y}} \mu(f) N^{(i)}(X;\mathcal{S}_f).
 \end{align}
The sum over $f > Y$ is handled by the tail estimate of Theorem \ref{thm:tail}. Each field $K$
is counted with multiplicity at most
$d(\Disc(K)) = O(X^{\eps})$ in this sum (where $d(n)$ is the number of positive divisors of $n$), so that
we have
\begin{equation}\label{eq:tail}
 \sum_{\substack{f\in T \\ f > Y}} \mu(f) N^{(i)}(X;\mathcal{S}_f)
 \ll X^{39/40 + \eps} + X^{1 + \eps}/Y.
 \end{equation}

Define $m^*(p)$ to be $m(p)$ for $p \neq 5$, with $m^*(5) = 5^4 m(5)$.
For each $f \leq Y$, upon
applying Theorem~\ref{T:counting} with Lemma~\ref{5mass} we obtain
\begin{align*}
N^{(i)}(X;\mathcal{S}_f)&=\left( C^{(i)} \prod_{p|f}m^*(p)^{-1}p^{-4} \right) \cdot X + O(X^{199/200 + \eps})

\,,
\end{align*}
provided that $Y \leq X^{\frac{199}{200} - \frac{79}{80}} = X^{\frac{3}{400}}$,
so that
\begin{equation}\label{eq:avg1}
\sum_{\substack{f\in T \\ f \leq Y}} \mu(f) N^{(i)}(X;\mathcal{S}_f)
=
C^{(i)}
X
\sum_{\substack{f\in T \\ f \leq Y}} \left( \prod_{p|f}-m^*(p)^{-1}p^{-4} \right) + O\left(Y X^{199/200 + \eps}\right).
\end{equation}
In the main term we extend the sum over $f \leq Y$ to all $f \in T$, at the expense of an error term
$\ll X Y^{-3 + \eps}$. We thus have that
\begin{align*}
\sum_{\substack{f\in T \\ f \leq Y}} \mu(f) N^{(i)}(X;\mathcal{S}_f)
& =
C^{(i)}
 X
\sum_{\substack{f\in T}} \left( \prod_{p|f}-m^*(p)^{-1}p^{-4} \right) + O\left( Y X^{199/200 + \eps} + X Y^{-3 + \eps} \right).

\end{align*}

Choosing $Y = X^{\frac{1}{400}}$, we see that the error terms above and in \eqref{eq:tail} are all $O(X^{1 -\frac{1}{400} + \eps})$.
We therefore have that
\begin{align*}
N_{g}^{(i)}(X)& =
C^{(i)}
X
\sum_{\substack{f\in T}} \left( \prod_{p|f}-m^*(p)^{-1}p^{-4} \right) + O\left( X^{1 - \frac{1}{400} + \eps} \right)
\\
& =
C^{(i)}
\left(1 - \frac{1}{5^{8}m(5)} \right) X
\prod_{ p \equiv 1 \pmod{5}} \left( 1-m(p)^{-1}p^{-4} \right) + O\left( X^{1 - \frac{1}{400} + \eps} \right),
\end{align*}
which is what we wanted to prove.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{T:2}]

Let $U$ denote the set of the first $k$ primes satisfying $p\equiv 1\pmod{5}$,
and write $T(U)$ for the set of squarefree $f$ whose prime factors $p$ all satisfy
$p \equiv 0, 1 \pmod{5}$ and $p \not \in U$.

We modify the proof of Theorem \ref{T:1} by replacing $T$ with $T(U)$, and by adding the
condition throughout that each $p \in U$ be totally ramified in each field being counted.

The proof is otherwise identical, and we obtain that the number of fields with $g_k = 5^k$ is bounded below by
\begin{equation*}
C^{(i)}
\left(1 - \frac{1}{5^{8}m(5)} \right)
\left( \prod_{\substack{ p \equiv 1 \pmod{5} \\ p \not \in U}} \left( 1-m(p)^{-1}p^{-4} \right) \right) \cdot
\left( \prod_{p \in U} m(p)^{-1} p^{-4} \right) X
+ O_k\left( X^{1 - \frac{1}{400} + \eps} \right).
\end{equation*}
\end{proof}

\begin{proof}[Proof of Theorem~\ref{T:3}]
By construction we have $g_K = 5^{\omega(f(K))}$, and hence also
\[
g_K = \sum_{f \mid f(K)} 4^{\omega(f)},
\]
so that in analogy with \eqref{eq:ng1} we have
\[
\sum_{\substack{K\in\mathcal{G}^{(i)} \\|\Disc(K)|\leq X}} g_K =
\sum_{f \in T} 4^{\omega(f)} N^{(i)}(X; S_f).
\]
This sum is evaluated exactly as in the proof of Theorem \ref{T:1}, with $- m^*(p)$ replaced by $4 m^*(p)$ at every occurrence.
All of the error terms satisfy identical bounds since $4^{\omega(f)} \ll X^{\epsilon}$
for $f \leq X$. We therefore conclude that
\begin{equation}\label{eq:average_genus}
\sum_{\substack{K\in\mathcal{G}^{(i)} \\|\Disc(K)|\leq X\}}} g_K =
C^{(i)}
\left(1 + \frac{4}{5^{8}m(5)} \right) X
\prod_{ p \equiv 1 \pmod{5}} \left( 1+4m(p)^{-1}p^{-4} \right) + O\left( X^{1 - \frac{1}{400} + \eps} \right),
\end{equation}
as desired.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{T:4}]
Let $\mathcal{S}$ denote the local conditions that
$2$ is inert, $5$ is inert, $7$ is totally ramified,
and no prime $p\equiv 1\pmod{5}$ is totally ramified.
Observe that any quintic field $K$ satisfying the conditions $\mathcal{S}$ must be genus
number one.
By way of contradiction, suppose $K$ is norm-Euclidean and satisfies the conditions $\mathcal{S}$.
Let $\mathfrak{p}$ denote the unique prime lying over $7$.
Then there exists $\alpha\in\OK$
such that
$4\equiv \alpha\pmod{\mathfrak{p}}$
with
 $|N(\alpha)|<|N(\mathfrak{p})|=7$.
It follows that $2\equiv 4^5\equiv N(\alpha)\pmod{7}$ and therefore
$N(\alpha)\in\{2,-5\}$.  We are forced to conclude that either
$2$ or $5$ is not inert, a contradiction.
The result now follows from techniques similar to the proof of Theorem~\ref{T:1}.
\end{proof}

\section*{Acknowledgments}

We would like to thank Melanie Matchett Wood for helpful comments.

This work was supported by a grant from the Simons Foundation (No. 586594, F.T.)
and an internal research grant through California State University, Chico.
\nocite{}
\bibliographystyle{alpha}
\bibliography{counting}

\vspace{1ex}
\footnotesize{
\noindent
Kevin J. McGown\\
Department of Mathematics and Statistics\\
California State University, Chico\\

Chico, CA 95929\\
\emph{E-mail address:} {\tt kmcgown@csuchico.edu}\\[1ex]

\noindent
Frank Thorne\\
Department of Mathematics\\
University of South Carolina\\

Columbia, SC 29208\\
\emph{E-mail address:} {\tt thorne@math.sc.edu}\\[1ex]

\noindent
Amanda Tucker\\
Department of Mathematics\\
University of Rochester\\
Rochester, NY 14627\\
\emph{E-mail address:} {\tt amanda.tucker@rochester.edu}
}

\end{document}

config: arxiv


TITLE: Do two sets generate the same subgroup/normal subgroup?
QUESTION [1 upvotes]: Suppose I have a group $G$, given as finitely many generators and finitely many relations, say. Furthermore, suppose I have two finite sets $A,B$ of elements of the group. I would like to know if the subgroups generated by $A$ and $B$ are the same, and more generally if the normal subgroups generated by $A$ and $B$ (set of all elements formed by products of generators and their conjugates) are the same.
Perhaps I can find a way of writing elements of $A$ in terms of elements of $B$ (and their conjugates) and vice versa, in which case I'm done. Alternatively, if I can come up with a homomorphism to some simpler group (like the abelianization) under which the images of $A$ and $B$ clearly generate different groups, I have my answer. But both of these rely on my ingenuity; is there a more systematic way of working it out?

An example (sorry I can't come up with something simpler which isn't obvious)!:
$$G=\langle a_1,b_1,a_2,b_2 \, | \, [a_1,b_1][a_2,b_2]=1 \rangle \\
A = \{ [a_1,b_1]b_1 b_2^{-1}, a_1 a_2 \} \\
B= \{ [a_1,b_1] b_2^{-1} b_1, a_1 [a_1,b_1] a_2 [a_1,b_1]^{-1}\}
$$
Do $A$ and $B$ generate the same normal subgroup of $G$?

REPLY [2 votes]: Virtually all questions involving groups defined by finite presentations are theoretically undecidable in general, so you cannot hope to find a general procedure for solving these sorts of problems.
The answer to your specific question is no. I adjoined the generators of $A$ to the relations of $G$ to get the quotient of $G$ by the normal closure of $A$, computed the $3$-quotient to class $3$, which had order $19683$, and found that the image of the element $[a_1.b_1]b_2^{-1}b_1$ for $B$ was nontrivial. (I did that in a CAS of course!)


TITLE: Cantor set is compact?
QUESTION [7 upvotes]: Cantor set
See the link, I am referring to cantor set on the real line. I wish to show that it is compact. I am doing this by pointing following arguments. I am not sure if this is enough.

Cantor set is bounded by definition in the region $[0,1]$
Cantor set is the union of closed intervals, and hence it is a closed set.
Since the Cantor set is both bounded and closed it is compact by Heine-Borel Theorem.

REPLY [1 votes]: This question is tagged solution-verification, which indicates that the asker is seeking a critique of their argument.  As neither of the previous answers have addressed this, and as there is an important (and common!) error in the argument, I am answering in order to fill that gap.
The argument presented in the question is incorrect, though the spirit of the argument is good.  The error is in the second line, where the asker writes:

Cantor set is the union of closed intervals, and hence it is a closed set.

There are at least two errors in this statement:
The union of closed sets needn't be closed
It is not true that the union of closed sets is closed.  For example, it is possible to write the open unit interval as the union of closed sets:
$$ (0,1) = \bigcup_{n=1}^{\infty} \left[ \frac{1}{n}, 1-\frac{1}{n} \right]. $$
For all integers $n$, the inequalities
$$ 0 < \frac{1}{n} \qquad\text{and}\qquad 1-\frac{1}{n} < 1, $$
which implies that neither $0$ nor $1$ is in any of the intervals in the union (therefore neither $0$ nor $1$ is in the union itself).  However, for any real number $x \in (0,1)$, there exists some $n$ so large that
$$ \frac{1}{n} \le x \le 1 - \frac{1}{n}; $$
if $x < 1/2$, take $n$ to be larger than $1/x$.  Otherwise, take $n$ to be larger than $1/((1/2)-x)$.
In short, an arbitrary union of closed sets needn't be closed.  Indeed, the correct statements are:

arbitrary unions and finite intersections of open sets are open
finite unions and arbitrary intersections of closed sets are closed

The first statement will be useful, below.
The Cantor set is not a union of closed intervals
The Cantor set is not a union of intervals.  In fact, the Cantor set doesn't contain any intervals.  For contradiction, suppose that there is some interval $[a,b]$ contained in the Cantor set.  While there are more elegant arguments, it might be simplest to observe that if $m,n\in\mathbb{Z}$ (with $0 < n < 3^m$, then the Cantor set cannot contain both of the points
$$\frac{n}{3^m}-\frac{1}{2\cdot 3^m}, \qquad\text{and}\qquad
\frac{n}{3^m}+\frac{1}{2\cdot 3^m}. \tag{1}$$
This is because the point $n/3^m$ is the endpoint of an interval that is removed in the $m$-th step of the construction.  The interval removed is either
$$ \left( \frac{n-1}{3^m}, \frac{n}{3^m} \right)
\qquad\text{or}\qquad
\left( \frac{n}{3^m}, \frac{n+1}{3^m} \right). $$
As one of these intervals is removed in the construction of the Cantor set, at least one of the two points listed in (1) must be removed.
To show that the Cantor set contains no intervals, choose $m$ and $n$ such that both of the points given in (1) are in the interval $(a,b)$ (note that this shows not only that $[a,b]$ is not contained in the Cantor set, but also that $(a,b)$ is not contained in the Cantor set).  This is possible, as the points are dense in $[0,1]$.
In short, the Cantor set contains no intervals, and cannot be written as the union of intervals.
Fixing the argument presented
Cyriac Antony gave a good argument, using the fact that arbitrary intersections of closed sets are closed.  Summarizing that argument, take $C_0 = [0,1]$, and construct $C_{n+1}$ by removing the open middle third from each interval contained in $C_n$.  Then each $C_n$ is closed and the Cantor set is given by
$$ \bigcap_{n=0}^{\infty} C_n, $$
which is a countable intersection of closed sets, therefore closed.
An alternative is to look at the sets which are removed: the core idea is that the Cantor set can be thought of as the complement of the sets which are removed at each stage of the construction.
For each $n$, define
$$ U_n = \bigcup_{j=1}^{\frac{3^n-1}{2}} \left(\frac{2j-1}{3^n}, \frac{2j}{3^n} \right). $$
Observe that $U_n$ is the set of all intervals of length $1/3^n$ where the left endpoint is an odd multiple of $1/3^n$.  All of these intervals are removed in the construction of the Cantor set (there is redundancy here, but who cares?).  Then define
$$ U = \bigcup_{n=1}^{\infty} U_n, $$
which is a countable union of open sets, and therefore open itself.
Observe that the Cantor set is precisely $[0,1] \setminus U$, and the complement of an open set is closed by definition (or, as this might make the point a little more clearly, the Cantor set is given by
$$ \mathbb{R} \setminus ( (-\infty,0) \cup U \cup (1, \infty), $$
which gives the Cantor set as the complement of an open set in $\mathbb{R}$, so that one doesn't need to muck about with subspace topologies at all).
In any event, the Cantor set is the complement of an open set, therefore closed.


TITLE: Proving bases of representation for a specific base, image question part D. (please give hint, no answer)
QUESTION [0 upvotes]: I recently went on CEMC waterloo's website and stumbled across this question. Part a, b and c were pretty easy and using the fact from part c, the two facts in part d can be proven. But I am stuck on knowing how those two facts would help me to prove the original question in d, which is that every positive integer has a base $\phi$ representation. Maybe the two facts imply directly that every real number has a base $\phi$ representation and positive integers are a subset of real numbers, so it's already proven? I am very stuck on how to use the two facts, and I don't want answers but just hints. I need hints on how to incorporate the two facts into my proof. This is a grade 12 level math contest question, so I would assume not too many advanced number theory techniques would be involved in doing this. However, looking at this, I am thinking base of representation theorem, but I still have no luck approaching it. If anyone can give me one hint, it would be much appreciated.

REPLY [1 votes]: $\phi^{n+2}=\phi^{n+1}+\phi^n$ means that in base-$\phi$, $100_\phi=011_\phi$.
Assume that we know the base-$\phi$ represenation of $n$. Then the base-$\phi$ represenation of $n+1$ is just adding $1$ on $d_0$. But, what if already $d_0=1$? Can we shift down $d_0$ using the fact above?


TITLE: Quantum Field Theory: completing the "A Bridge between Mathematicians and Physicists" series
QUESTION [7 upvotes]: I decided to read the series "A Bridge between Mathematicians and Physicists" written by Eberhard Zeidler. But when I read the preface of the first book I realized that at first this series should be composed of six volumes, namely:

Quantum Field Theory I: Basics in Mathematics and Physics
Quantum Field Theory II: Quantum Electrodynamics
Quantum Field Theory III: Gauge Theory
Quantum Field Theory IV: Quantum Mathematics
Quantum Field Theory V: The Physics of the Standard Model
Quantum Field Theory VI: Quantum Gravity and String Theory

I have only the first three books. Searching for the last three I discovered that Eberhard Zeidler died and therefore the last three volumes will never be published.
For this reason I ask: could someone please indicate me books that cover the subject of the last three books and that focus a lot on the mathematical part?
Thank you for your attention.

REPLY [4 votes]: The volume number 4 has a vague title. 2 volumes of "Quantum fields and strings: a course for mathematicians" probably cover whatever that term is supposed to mean (there is also a partial QFT-mathematics dictionary in the books, very nice). The books are outdated by this point, but they should be enough to get you started (then you should be able to read physical mathematics papers without too much pain).