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Diamagnetic inequality, domination of semigroups,  
random Schr"odinger operators, density of states. 
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% Diamagnetic Inequality for Semigroup Differences  
% 
%  by Dirk Hundertmark and Barry Simon  
% 
% Original Version: December 29, 2002 
% made 2 changes due to referees comments 
% on  February 24, 2003 
% 
% accepted for publication in Crelles Journal
% (aka Journal f\"ur die reine und angewandte Mathematik)
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\begin{document} 
\title[Diamagnetic Inequality]{A Diamagnetic Inequality for Semigroup Differences} 
%\author[D.~Hundertmark and B.~Simon]{Dirk Hundertmark$^1$ and Barry Simon$^{1,2}$} 
\author[D.~Hundertmark]{Dirk Hundertmark} 
\address{Department of Mathematics, University of Illinois at Urbana-Champaign,  
                1409 W.~Green Street, Urbana, IL 61801.}%  
\email{dirk@math.uiuc.edu}% 
\author[B.~Simon]{Barry Simon$^{1}$} 
\address{Department of Mathematics 253-37, California Institute of Technology,     
Pasadena, CA 91125.}% 
\email{bsimon@caltech.edu} 
 
\thanks{$^1$ Supported in part by NSF grants DMS-9707661 and DMS-0140592. } 
\thanks{\copyright 2003 by the authors. Faithful reproduction of this article,  
        in its entirety, by any means is permitted for non-commercial purposes} 
\date{February 24, 2003}  
 
 
\begin{abstract}  
The diamagnetic inequality for the magnetic \mbox{Schr\"odinger} semigroup is  
extended to the difference of the semigroups of magnetic Schr\"odinger operators with  
Neumann and Dirichlet boundary conditions on arbitrary open domains and rather  
general magnetic vector potentials $A$ and potentials $V$. In particular, this bound  
renders moot all the technical issues in the recent proofs of the independence of the  
boundary conditions for the integrated density of states for magnetic Schr\"odinger operators:  
Independence of the boundary conditions for the free case, that is, for  
vanishing potentials and vector potentials, immediately implies independence of the boundary 
conditions of the integrated density of states for a large class of magnetic  
Schr\"odinger operators.  
\end{abstract} 
 
\maketitle 
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\section{Introduction}\label{Dia:sec1} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
Let $H(A,V)= (-i\nabla -A)^2+V$ be the magnetic Schr\"odinger operator with electric potential $V$  
and magnetic vector potential $A$. The diamagnetic inequality of Simon \cite{Simon76,Simon77}  
says that, under some rather general conditions on $A$ and $V$, the bound  
\begin{equation}\label{eq:sec1-dia} 
\abs{(\e^{-tH(A,V)}f)(x)} \le (\e^{-tH(0,V)}\abs{f})(x)  
\end{equation}  
holds for all $t\ge 0$ and almost all $x\in\R^d$. In some sense this inequality shows that the magnetic  
operator is dominated by the non-magnetic Schr\"odinger operator. The importance of this inequality  
was already noted in \cite{AHS,CSS}. Its strength lies in the fact that it is valid for rather  
arbitrary vector potentials $A$ and a large class of potentials $V$; see \cite{HSU,Simon79b} and  
\cite{Ouhabaz} for further developments.  
 
 
Our interest in the diamagnetic inequalities comes from the recent studies of the integrated density  
of states (IDS) of magnetic Schr\"odinger operators, especially the proofs of independence of the  
boundary conditions \cite{DIT,Nakamura,HLMW2}. The integrated density of states is a fundamental  
quantity in the theory and applications of random Schr\"odinger operators \cite{CL,Kirsch,LGP,PF}.  
It is defined as follows: Let $\Lambda\subset \R^d$ be an open (and bounded) set, for example,  
$\Lambda=\Lambda_L= (-L,L)^d$, and consider $H_\Lambda(A,V)$, the restriction of $H(A,V)$ to  
$L^2(\Lambda)$. Of course, one has to consider boundary conditions in order to get a self-adjoint operator.  
This is most conveniently done with the help of quadratic forms; see Section \ref{Dia:sec2}.  
Since $\Lambda$ is bounded, the spectrum of $H_\Lambda(A,V)$ has a good chance of being  
discrete (at least with Dirichlet boundary conditions), in which case the eigenvalue distribution  
function  
\begin{displaymath} 
N_\Lambda(\lambda)= \tfrac{1}{\abs{\Lambda}} \#\{\text{eigenvalues of }H_\Lambda(A,V)\le \lambda  \} 
\end{displaymath} 
is well-defined. The IDS is then given by the macroscopic, or infinite-volume, limit  
\begin{equation}\label{eq:ml} 
N(\lambda)= \lim_{\Lambda\to \R^d}N_\Lambda(\lambda) ,  
\end{equation} 
where $\Lambda\to \R^d$ is usually interpreted as $\Lambda=\Lambda_L=(-L,L)^d$, $L\to\infty$,  
or, more generally, in the sense of Fisher \cite{Fisher,Ruelle}. We interpret the limit in  
\eqref{eq:ml} not in a pointwise sense, but as vague convergence of the  
corresponding measures. For example, \eqref{eq:ml} holds at all continuity points of N,  
in particular, for almost all $\lambda$.   
Basic questions are whether this limit exists at all, is independent of the chosen  
boundary conditions, and, in case the potential $V$ is random, that is, given by  
realizations of some random field, is independent of the realizations of the random potential.  
 
For vanishing magnetic vector potentials, these questions had been solved some time ago  
\cite{AS,CL,DK,Kirsch,KM1,Nakao,Pastur71,Pastur80,PF}. For non-zero magnetic vector potentials, the  
existence and non-randomness of the IDS are well-known \cite{BHL1,Matsumoto,Ueki}. However,  
uniqueness has only recently been studied. Bounded potentials were considered in \cite{Nakamura},  
extended to non-negative potentials and arbitrary vector potentials  
$A\in L^2_{\mathrm{loc}}(\R^d,\R^d)$ in \cite{DIT}, and \cite{HLMW2} used the method of \cite{PF}  
and the a priori input from \cite{DIT} to extend this to some random potentials which are  
unbounded from below. All these results have somewhat technical and complicated proofs.  
 
The main point of this paper is to show that independence of the boundary conditions of the IDS  
is a very natural property. In the free case, that is, for vanishing $A$ and $V$, it was already  
known to Weyl \cite{Weyl}, see also \cite{ReSi4}, Chapter XIII.15,  
and we will show that the  
interacting system, for rather general vector potentials $A$ and potentials $V\ge 0$,  
inherits this property from the free case.  
Thus it is a \emph{geometrical property} of (the sequence of) the domains $\Lambda\to \R^d$.  
 
To study the IDS for different boundary conditions, it is convenient, following Avron-Simon  
\cite{AS}, to look at their Laplace transforms. Let $N_\Lambda^\sharp$ be the finite-volume  
IDS for $\sharp = N,D$,  Neumann, respectively Dirichlet boundary conditions and  
$\calL_\Lambda^\sharp(t)= \int \e^{-t\lambda} dN_\Lambda^\sharp(\lambda)$ be its  
Laplace transform. By the usual Dirichlet-Neumann bracketing,  
\begin{equation}\label{eq:sec1-D-N}  
N^D_\Lambda(\lambda)\le N^N_\Lambda(\lambda)  
\end{equation}  
as long as $N^N_\Lambda$ is well-defined and consequently,  
\begin{displaymath}  
0\le \calL^D_\Lambda(t)\le\calL^N_\Lambda(t) . 
\end{displaymath}  
Using the uniqueness theorem for Laplace transforms, it is enough to show that,  
for all fixed $t>0$,  
\begin{equation}\label{eq:sec1-independence}  
\lim_{\Lambda\to \R^d} \big( \calL^N_\Lambda(t) - \calL^D_\Lambda(t) \big) = 0 .  
\end{equation}  
 
Let $\calL^\sharp_{\Lambda,\mathrm{free}}$ be the Laplace transforms of the free IDS, that is,  
for vanishing electric and vector potentials. {}From Weyl's result, we know that  
\begin{equation} \label{eq:sec1-Weyl} 
\calL^N_{\Lambda,\mathrm{free}}(t) - \calL^D_{\Lambda,\mathrm{free}}(t) \le   
\frac{C(t)}{\abs{\Lambda}} \big(\abs{\partial\Lambda}+1\big) 
\end{equation} 
for a large class of domains $\Lambda$. Of course, here one has to interpret  
$\abs{\partial\Lambda}$ correctly, which is not a problem for domains with nice enough boundaries;  
see also Remark \ref{rem:KM1} below, and one should keep in mind that, in our notation,  
the finite-volume IDS contains the factor $1/\Lambda$. If one had a bound of the form  
\begin{equation}\label{eq:sec1-diadifference} 
\calL^N_\Lambda(t) - \calL^D_\Lambda(t) \le  
\calL^N_{\Lambda,\mathrm{free}} - \calL^D_{\Lambda,\mathrm{free}}, 
\end{equation}  
then \xref{eq:sec1-Weyl} would lead to the desired result \xref{eq:sec1-independence}, since the  
difference on the left-hand side is non-negative by Dirichlet-Neumann bracketing.  
 
By the functional calculus,  
$\calL^\sharp_\lambda(t)= \tfrac{1}{\abs{\Lambda}}\tr[\e^{-tH_\Lambda^\sharp(A,V)}]$.  
Assume for the moment that the diamagnetic inequality \xref{eq:sec1-dia} also holds for the  
operators restricted to $L^2(\Lambda)$, at least for Dirichlet and Neumann boundary conditions.  
Then some standard estimates on integral kernels \cite{SimonFIandQPh} imply that  
\begin{displaymath} 
\tr[\e^{-tH_\Lambda^\sharp(A,V)}] \le \tr[\e^{-tH_\Lambda^\sharp(0,V)}]  
\le \tr[\e^{-tH_\Lambda^\sharp(0,0)}] 
\end{displaymath}  
for $\sharp=D$ or $N$. The last inequality holds as long as $V\ge 0$. In particular,  
$\calL^\sharp_\Lambda(t)\le \calL^\sharp_{\Lambda,\mathrm{free}}(t)$. {}From this point of view, the  
inequality \xref{eq:sec1-diadifference} looks rather strange, since it is in general not true that  
$a-b\le \abs{a}-\abs{b}$. However, as for the usual diamagnetic inequality \xref{eq:sec1-dia}, there  
is a simple probabilistic heuristic for the inequality \xref{eq:sec1-diadifference}: Let  
$b$ be the usual Brownian motion process in $\R^d$ and  $T=\inf\{s\ge 0: b_s \not\in \Lambda\}$, the  
hitting time of the complement of $\Lambda$. The Feynman-Kac-It\^{o} formula for the Dirichlet semigroup  
$\e^{-tH_\Lambda^D(A,V)}$ then reads  
\begin{equation}\label{eq:sec1-FKI} 
(\e^{-tH_\Lambda^D(A,V)}f)(x)= \E^x[\e^{iI^t(b)-\int_0^t V(b_s)\, ds}\ind_{\{T<t\}}(b) f(b_t)] . 
\end{equation} 
Here $\E^x$ stand for integrating with respect to Brownian motion starting at $x$,  
$\ind_A$ is the indicator function of the set $A$, and $I^t$ is the line integral of $A$ along  
Brownian paths. Since Brownian paths are not of finite variation, this needs some interpretation;  
see, e.g., \cite{SimonFIandQPh}, Chapters 14 and 15, \cite{Roepstorff},  Chapter 6, and \cite{BHL2}.  
Usually one also has to impose some additional conditions on the divergence of  
$A$ for this probabilistic approach. 
In any case, since both $I^t(b)$ and $ \int_0^t V(b_s)\, ds$ are real-valued random variables,  
the Feynman-Kac-It\^{o} formula together with the triangle inequality immediately implies  
\xref{eq:sec1-dia}.  
So from a probabilistic point of view, the diamagnetic inequality is, at least for Dirichlet  
boundary conditions, just the triangle inequality for integrals. This point of view of the  
diamagnetic inequalities goes back to Nelson; see \cite{SimonFIandQPh}, Chapter 15.  
 
Assume for the moment that a Feynman-Kac-It\^{o} type formula also holds for the Neumann semigroup,  
that is,  
\begin{equation}\label{eq:sec1-FKINeumann} 
(\e^{-tH_\Lambda^N(A,V)}f)(x) 
= \wti{\E}^x[\e^{iI^t(\tilde{b})-\int_0^t V(\tilde{b}_s)\, ds} f(\tilde{b}_t)], 
\end{equation} 
where $\tilde{b}$ is reflected Brownian motion. The magnetic phase $I^t$ and $\int_0^t V(b_s)\, ds$  
depend only locally on the paths. Moreover, one has $b_s=\tilde{b_s}$ for $s\le t$ and paths in  
$\{T\le t\}$, since, at least morally, reflected Brownian motion up to the stopping time $T$  
has not reached the boundary and hence is not reflected, yet. Accordingly, we can extend the  
integration in \xref{eq:sec1-FKI} to all reflected paths $\tilde{b}$ and, subtracting it from  
\xref{eq:sec1-FKINeumann}, one gets  
\begin{align}\label{eq:sec1-difference1} 
\abs{(\e^{-tH_\Lambda^N(A,V)}- \e^{-tH_\Lambda^D(A,V)})f(x)}  
&=  
\abs{  
\wti{\E}_x[\e^{iI^t(\tilde{b})-\int_0^t V(\tilde{b}_s)\, ds}(1-\ind_{\{T<t\}}(\tilde{b})) f(\tilde{b}_t)]  
} 
\nonumber\\ 
&\le 
\wti{\E}_x[\e^{-\int_0^t V(\tilde{b}_s)\, ds}(1-\ind_{\{T<t\}}(\tilde{b})) \abs{f(\tilde{b}_t)}] 
\nonumber \\ 
&= (\e^{-tH_\Lambda^N(0,V)}- \e^{-tH_\Lambda^D(0,V)})\abs{f}(x),  
\end{align} 
since $1-\ind_{\{T<t\}}= \ind_{\{T\ge t\}}\ge 0$. Arguing the same way, one also sees  
\begin{equation}\label{eq:sec1-difference2} 
\abs{(\e^{-tH_\Lambda^N(0,V)}- \e^{-tH_\Lambda^D(0,V)})f(x)}  
\le  
(\e^{-tH_\Lambda^N(0,0)}- \e^{-tH_\Lambda^D(0,0)})\abs{f}(x)  
\end{equation} 
for all non-negative potentials $V$. Taking traces, \xref{eq:sec1-difference1} and  
\xref{eq:sec1-difference2} imply \xref{eq:sec1-diadifference}, at least on a formal level.  
Hence, independence of the boundary conditions of the integrated states should follow from the  
free case.  
 
The catch in the above argument is that we do not know whether a Feynman-Kac-It\^{o} type formula holds  
for the Neumann case. Although this is conceivably the case, it will pose strong restrictions on the  
domain $\Lambda$. It is well-known that Neumann boundary conditions are a tricky business and even the  
free Neumann operator on bounded domains can have some surprising spectral properties, like non-trivial  essential spectrum or even non-trivial ``scattering" asymptotics \cite{DS,HSS,Simon92}. In particular,  
reflected Brownian motion will exist only for nice enough domains  
\cite{Tanaka,Williams,WZ}.  
Moreover, some restrictions on $\Lambda$ will already have to be imposed in order that the  
finite-volume IDS, $N_\Lambda^N$, is well  
defined, see Remark \ref{rem:dia-compact}, but we do not want to impose these restrictions from the  
beginning in the proof of the diamagnetic inequalities.  
So instead, we use the probabilistic heuristic as a guiding principle for the right kind  
of inequality to be proven, that is, \xref{eq:sec1-difference1} and \xref{eq:sec1-difference2},  
and will give an analytic proof for them. Moreover, it is conceivable that a probabilistic approach  
will also need some regularity assumptions on $A$ near the boundary of $\Lambda$, whereas the analytic  
approach we propose will only need $\Lambda$ open and $A\in L^2_{\mathrm{loc}}(\Lambda,\R^d)$, a  
condition which is already very convenient for the definition of the operators $H_\Lambda^N(A,V)$  
and $H_\Lambda^D(A,V)$ via quadratic forms, see Section \ref{Dia:sec2}, and thus poses no further  
restriction on the class of vector potentials to be considered. As a bonus, we obtain a new  
proof of the normal diamagnetic inequalities which is ``form theoretic" rather than the operator  
theoretic proof of Simon \cite{Simon77,Simon79b}.  
 
The main result of this paper is given by the following two theorems: 
\begin{theorem} \label{thm:dia}  
Let $\Lambda\subset \R^d$ be open, $A\in L_{\mathrm{loc}}^2(\Lambda,\R^d)$, and $V=V_+-V_-$  
with $V_\pm\ge 0$ and $V_+\in L^1_{\mathrm{loc}}(\Lambda)$, $V_-$ relatively form bounded  
w.r.t.\ $H^N_\Lambda(0,0)$.  
Then, for all $f\in L^2(\Lambda)$ and almost all $x\in\Lambda$,  
\begin{displaymath}  
\abs{\e^{-t H^N_\Lambda(A,V)} f (x)}  
\le  
\e^{-t H^N_\Lambda(0,V)} \abs{f}(x)  
\le  
\e^{-t H^N_\Lambda(0,-V_-)} \abs{f} (x) . 
\end{displaymath}  
\end{theorem}  
 
\begin{remarks} \itemthm  Given the discussion above, this result is certainly to be expected,  
but, maybe somewhat surprisingly, no regularity assumptions on $A$ and $V_+$ are made close to  
the boundary of $\Lambda$. Also, since $\Lambda$ is an arbitrary open set, its boundary can  
be quite wild. Following the approach in \cite{Simon79a}, a weaker version of the first inequality  
is proven in \cite{HLMW1} under the condition that $A$ and the positive part of the potential  
$V_+$ are restrictions of a vector potential in $L^2_{\mathrm{loc}}(\R^d,\R^d)$ and a potential in  
$L^1_{\mathrm{loc}}(\R^d)$.\\[0.2em] 
\itemthm Choosing for $f$ an approximate delta-function, Theorem \ref{thm:dia} implies the bound  
$\abs{\e^{-t H^N_\Lambda(A,V)}(x,y)}\le \e^{-t H^N_\Lambda(0,V)}(x,y)$ for almost all  
$(x,y)\in\Lambda\times\Lambda$ for the integral kernel of the Neumann semigroup  
(as long as it exists). \\[0.2em] 
\itemthm Of course, the analogous result for Dirichlet boundary conditions also holds.   
\end{remarks} 
 
The second theorem  shows that the probabilistic heuristic, suggesting a diamagnetic inequality  
for the \emph{difference} of the Neumann and Dirichlet semigroup, is, indeed, correct. 
\begin{theorem}\label{thm:diadifference}  
Let $\Lambda\subset \R^d$ be open, $A\in L_{\mathrm{loc}}^2(\Lambda,\R^d)$, and $V=V_+-V_-$  
with $V_\pm\ge 0$ and $V_+\in L^1_{\mathrm{loc}}(\Lambda)$, $V_-$ relatively form bounded  
w.r.t.\ $H^N_\Lambda(0,0)$. Then 
\begin{align*} 
\abs{(\e^{-t H^N_\Lambda(A,V)}-\e^{-t H^D_\Lambda(A,V)} )f (x)}  
&\le  
(\e^{-t H^N_\Lambda(0,V)}-\e^{-t H^D_\Lambda(0,V)} ) \abs{f} (x) \\ 
&\le  
(\e^{-t H^N_\Lambda(0,-V_-)}-\e^{-t H^D_\Lambda(0,-V_-)} ) \abs{f} (x)  
\end{align*} 
for all $f\in L^2(\Lambda)$ and almost all $x\in\Lambda$.  
\end{theorem} 
 
Recall that an open set $\Lambda$ has the extension property if, with $W^{1,p}(\Lambda)$ denoting the  
usual Sobolev space, there exists a bounded extension operator $E: W^{1,p}(\Lambda)\to W^{1,p}(\R^d)$  
for all $1\le p<\infty$. If $\Lambda$ has the extension property, then it is known that the free  
Neumann semigroup $\e^{-tH_\Lambda^N(0,0)}$ is an integral operator, see \cite{Davies}, Theorem 2.4.4.  
If, moreover, $\Lambda$ is also bounded, then $\e^{-tH_\Lambda^N(0,0)}$ is a trace class operator for  
all $t>0$; see Remarks \ref{rem:dia-compact} and \ref{rem:dia-rooms-and-passages} below.  
An immediate corollary of the two theorems above is given by  
\begin{corollary} \label{cor:dia-maincoro}  
Let $\Lambda$ be a bounded set having the extension property. \\[0.1em] 
\itemthm \label{cor:dia-Neumann} For any $t>0$, $\e^{-tH^N_\Lambda(A,V)}$ is a trace class  
operator for arbitrary vector potentials $A\in L_{\mathrm{loc}}^2(\Lambda,\R^d)$, and  
potentials $V= V_+ -V_-$ with $V_\pm\ge 0$, $V_+\in L^1_{\mathrm{loc}}(\Lambda)$, and  
$V_-$ form small with respect to $H^N_\Lambda(0,0)$. Moreover,  
\begin{displaymath} 
\tr_{L^2(\Lambda)}\big[\e^{-tH^N_\Lambda(A,V)}\big] 
\le  
\tr_{L^2(\Lambda)}\big[\e^{-tH^N_\Lambda(0,V)}\big]  
\le 
\tr_{L^2(\Lambda)}\big[\e^{-tH^N_\Lambda(0,-V_-)}\big]  
\end{displaymath} 
for all $t>0$. \\ 
\itemthm \label{cor:dia-difference} The bounds  
\begin{align*}  
0\le \tr_{L^2(\Lambda)}\big[ \e^{-t H^N_\Lambda(A,V)}-\e^{-t H^D_\Lambda(A,V)} \big]  
& \le 
\tr_{L^2(\Lambda)}\big[ \e^{-t H^N_\Lambda(0,V)}-\e^{-t H^D_\Lambda(0,V)} \big] \\ 
& \le 
\tr_{L^2(\Lambda)}\big[ \e^{-t H^N_\Lambda(0,-V_-)}-\e^{-t H^D_\Lambda(0,-V_-)} \big]  
\end{align*} 
hold for all $A\in L^2_{\mathrm{loc}}(\Lambda,\R^d)$.  
\end{corollary} 
The proof of this corollary is given at the end of Section \ref{Dia:sec3}.   
%The second inequality in the corollary is well-known from statistical mechanics.... 
 
\begin{remarks} \itemthm For the free Dirichlet semigroup $\e^{-tH_\Lambda^D(0,0)}$ to be  
trace class for $t>0$, it is enough that $\Lambda$ is bounded,  
see \cite{ReSi4}, Theorem XIII.76. The analog of Corollary \ref{cor:dia-Neumann} holds for  
the Dirichlet semigroup assuming only boundedness of $\Lambda$.\\[0.2em] 
\itemthm \label{rem:dia-compact} Any bounded convex domain and every bounded domain  
with a piecewise smooth boundary has the extension property. In general, a domain which  
is ``minimally smooth" will have the extension property; see \cite{Stein}, page 189, for  
the precise conditions. The discussion in \cite{Davies}, Chapters 1 and 2, also shows that  
if $\Lambda$ is bounded and has the extension property, then the spectrum of  
$H_\Lambda^N(0,0)$ is discrete, since $H_\Lambda^N(0,0)$ can be compared with the  
Dirichlet Laplacian on a large enough ball containing $\Lambda$. For that reason, one  
knows also that the eigenvalues of $H_\Lambda^N(0,0)$ go to infinity fast enough such that  
$\e^{-tH_\Lambda^N(0,0)}$ is a trace class operator. In particular, for these domains, the  
spectrum of $H_\Lambda^N(0,0)$ and of $H_\Lambda^N(A,V)$ is discrete. Thus the finite-volume  
IDS for Neumann boundary conditions, $N_\Lambda^N$, is well-defined. \\[0.2em]  
\itemthm\label{rem:dia-rooms-and-passages} By the semigroup property, $\e^{-tH_\Lambda^N(0,0)}$  
is a trace class operator for all $t>0$ as soon as it is in some von Neumann-Schatten  
ideal $\calI_p$ for all small enough $t>0$ and some finite $p$.  
It seems conceivable that, by mimicking the rooms-and-passages construction  
in \cite{HSS}, one could construct a bounded open set $\Lambda$, necessarily not having the  
extension property, such that for some $t_0>0$, $\e^{-t_0H_\Lambda^N(0,0)}$ is compact  
but not in any von Neumann-Schatten ideal for finite $p$.\\[0.2em] 
\itemthm \label{rem:KM1}   
Corollary \ref{cor:dia-maincoro} shows that for non-negative (or, more generally,  
bounded below) potentials and arbitrary vector potentials, the IDS is independent of  
the boundary conditions as soon as this is true for the free case. If the sequence  
of domains is given by boxes,  
$\Lambda=\{ -L_j/2<x_j<L_j/2, j= 1,\ldots,d \}$, this again can be seen rather naturally  
by probabilistic methods.  
With the help of the method of images, one can give an explicit expression for $\e^{-tH_\Lambda^N(0,0)}$  
in terms of a Feynman-Kac formula, see \cite{BR}, Example 6.3.11, and \cite{KM1}.  
Using this, one can easily show that for any $\veps>0$ and with  
$\partial\Lambda_\veps= \{x\in\Lambda\vert\,\dist(x,\Lambda^c)\le \veps\}$,  
\bdm 
\tr\left[ \e^{-t H^N_\Lambda(0,0)}-\e^{-t H^D_\Lambda(0,0)} \right]  
\le  
C_{t}\left( \abs{\partial\Lambda_\veps}+\e^{-\veps^2/(2t)}\abs{\Lambda} \right) , 
\edm 
as was already noticed in \cite{Pastur71}. Thus a suitably modified version of  
\xref{eq:sec1-Weyl} holds. Hence, for sequences $\Lambda\to\R^d$ of boxes (for which one  
always has $\abs{\partial\Lambda_\veps}/\abs{\Lambda}\to 0$ for fixed $\veps>0$),  
the IDS is independent of the boundary conditions used in its definition  
for arbitrary vector potentials $A$ and non-negative potentials $V$. \\[0.2em] 
\itemthm Using the methods of \cite{DK,KM1} or \cite{PF}, one can extend this result to  
certain not necessarily non-negative potentials $V$. The most general result in this  
direction is in \cite{HLMW2}, which uses the a priori information that the IDS is  
independent of the boundary conditions for non-negative potentials, or equivalently,  
potentials which are bounded from below.  
In particular, it is shown that for $\R^d$-ergodic random potentials, the IDS is  
independent of the boundary conditions as soon as  
$\E[(V_\omega(0))_-^p]<\infty$ for some $p>d+1$. A straightforward modification of the  
result in \cite{KM2}, see also \cite{Kirsch}, Theorem 1 in Section 5, shows that for differentiable  
vector potentials $A$, the random magnetic Schr\"odinger operator $(-i\nabla+A)^2 +V_\omega$ is  
almost surely self-adjoint on $\calC^\infty_0(\R^d)$ as long as $p>d$.  
This leaves open the question of uniqueness of the IDS for $d<p\le d+1$.  
We are convinced that uniqueness holds as soon as one has essential self-adjointness,  
that is, one should have uniqueness for $p> d$.  
\end{remarks} 
 
 
The organization of the paper is as follows: In Section \ref{Dia:sec2} we give  
the construction of the magnetic Schr\"odinger operator on arbitrary open domains with  
Neumann and Dirichlet boundary conditions and gather some technical tools.  
Theorems \ref{thm:dia} and \ref{thm:diadifference} are proven in Section \ref{Dia:sec3}.  
In the probabilistic heuristics given above, the main observation was that the diamagnetic  
inequality for the difference of semigroups follows from the Feynman-Kac-It\^{o} representation  
\xref{eq:sec1-FKINeumann}. In our analytic proof, this is replaced by the diamagnetic inequality  
for the Neumann semigroup: Modulo some approximation arguments, Theorem \ref{thm:diadifference}  
is a corollary of Theorem \ref{thm:dia}; we first prove a result similar to that in  
Theorem~\ref{thm:diadifference} for the difference of two Neumann semigroups whose generators  
differ by a positive potential, see Lemma \ref{lem:difference}, and then use an approximation  
argument to recover Dirichlet from Neumann boundary conditions. The main tool for the proof  
of Theorem \ref{thm:dia} is given in Lemma~\ref{lem:kato-bilinear}.  
For the convenience of the reader, we present the approximation theorem needed in the proof  
of Theorem \ref{thm:diadifference} in the appendix.  
 
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\section{Some Preliminaries}\label{Dia:sec2} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
We begin with some general preliminaries. Even without magnetic vector potentials,  
restricting the Laplacian to an open subset poses the problem of introducing the right  
boundary conditions for which the restriction is realized by a self-adjoint operator. This  
is most conveniently done with the help of quadratic forms. It turns out that introducing a  
magnetic vector potential poses no real difficulty, except some notational effort.  
 
Let $\Lambda\subset \R^d$ be an open set. For $1\le p\le \infty$, we let  
$L^p(\Lambda)=L^p(\Lambda,\C)$ be the usual Lebesgue spaces of complex-valued functions  
whose $p^\mathrm{th}$ power is integrable, and $L^p_{\mathrm{loc}}(\Lambda)$ be the space of  
(measurable) functions $f\!:\Lambda\to\C$ for which $f\id_K\in L^p(\Lambda)$ for all compact  
subsets $K\subset \Lambda$. Here $\id_K$ denotes the indicator function of the set $K$.  
Furthermore, for $\K$ being either $\R$ or $\C$, the spaces $L^p(\Lambda,\K^d)$, respectively  
$L^p_{\mathrm{loc}}(\Lambda,\K^d)$, are the Lebesgue spaces of functions $f\!:\Lambda\to \K^d$  
whose coordinate functions $f_j\!:\lambda\to \K$ are in $L^p(\Lambda)$, respectively  
$L^p_{\mathrm{loc}}(\Lambda)$. For non-negative functions $f$ and $g$,  
we write $f\lesssim g$ if, for some constant $C>0$, one has $f\le Cg$.  
 
 
We write $\calC^\infty_0(\Lambda)$ for the infinitely often differentiable functions with  
compact support in $\Lambda$. Like the usual gradient, for a magnetic vector potential  
$A\in L^2_{\mathrm{loc}}(\Lambda)$, the magnetic gradient $D_{A}= \nabla -iA$ is a closable  
operator on $\calC^\infty_0(\Lambda)$; more precisely, its components  
$D_{A,j}=\tfrac{\partial}{\partial x_j}- iA_j$ are closable since they are anti-symmetric  
operators with respect to the standard scalar product  
$\langle f,g \rangle= \int_\Lambda \ol{f}g\, dx$ on $L^2(\Lambda)$, that is,  
$\langle \varphi, D_{A,j}\psi \rangle = \langle -D_{A,j}\varphi,\psi\rangle $ for all  
$\varphi,\psi\in\calD(D_{A})$ and $j=1,\ldots,d$. By slight abuse of  
notation, we will use $D_{A,j}$ to denote this closure.  
The magnetic Sobolev space $H^1_{A,0}(\Lambda)$ ($=W^{1,2}_{A,0}(\Lambda)$) is then given  
by $H^1_{A,0}(\Lambda):= \cap_{j=1}^d\calD(D_{A,j})$, which, for $A=0$, reduces to the  
well-known Sobolev space $H^1_{0}(\Lambda)= W^{1,2}_0(\Lambda)$. The operator  
$H^D_\Lambda(A,0)$ is defined as the unique operator associated with the closed,  
non-negative symmetric quadratic form  
\begin{equation} 
h^D_\Lambda(A,0)(\varphi,\psi) \defeq 
\sum_{j=1}^d \langle D_{A,j}\varphi, D_{A,j}\psi \rangle  
\qquad 
\end{equation} 
with form domain $\calD(h^D_\Lambda(A,0))\defeq H^1_{A,0}(\Lambda)$. This gives  
the well-known construction for the Dirichlet Laplacian  
$H^D_\Lambda(0,0)=-\Delta^D_\Lambda$ for vanishing magnetic vector potentials.  
 
For the Neumann boundary conditions, we need the maximal  
magnetic gradient, which is the magnetic analog of the maximal gradient.  
Since $D_{A,j}$ is an anti-symmetric operator, a natural closed extension is  
given by the negative of its adjoint, $D_{A,\mathrm{max},j}= - (D_{A,j})^*$,  
for  $j= 1,\ldots,d$.  That is,  
\begin{equation}\label{eq:maximalDA} 
\int_\Lambda \overline{\varphi}D_{A,\mathrm{max},j} f\, dx \defeq  
- \int_\Lambda \overline{D_{A,j}\varphi} f\, dx  
\end{equation} 
for all $\varphi \in \calC^\infty_0(\Lambda)\subset\calD(D_{A})$. The domain of  
$D_{A,\mathrm{max}}$ is given by   
$H^1_A(\Lambda)\equiv\calD(D_{A,\mathrm{max}})\defeq \cap_{j=1}^d \calD(D_{A,\mathrm{max},j})$  
with  
\begin{align*} 
\calD(D_{A,\mathrm{max},j}) = \calD(D_{A,j}^*)  
&= 
\{f\in L^2(\Lambda)\!\mid  
 \text{ there exists } \eta_f\in L^2(\Lambda) \text{ such that } \\ 
& \phantom{= =\{ }  
  \langle -D_{A,j} \varphi , f \rangle = \langle \varphi, \eta_f \rangle  
\text{ for all } \varphi\in \calC^\infty_0(\Lambda)\} \\ 
&= \{ f\in L^2(\Lambda)\!\mid  \partial_j f\in L^1_{\mathrm{loc}}(\Lambda),   
      A_jf\in L^1_{\mathrm{loc}}(\Lambda) \\ 
&      \phantom{= =\{f\in L^2(\Lambda)\vert\,} 
      \text{ with } \partial_j f - i A_j f \in L^2(\Lambda) 
   \},  
\end{align*} 
where the partial derivative $\partial_j f$ is in the weak sense.  
$D_{0,\mathrm{max}}$ is the well-known maximal (or distributional) gradient  
$\nabla_{\mathrm{max}}$.  
Since $A\in L^2_{\mathrm{loc}}(\Lambda,\R^d)$, by assumption, we have  
$Af\in L^1_\mathrm{loc}(\Lambda,\R^d)$ for all $f\in L^2(\Lambda)$.  
Thus, $H^1_A(\Lambda)=  
\{f\in L^2(\Lambda)\cap W^{1,1}_{\mathrm{loc}}(\Lambda)\!\!\mid   
D_{A,\mathrm{max}}f\in L^2(\Lambda,\C^d)\}\subset L^2(\Lambda)\cap W^{1,1}_{\mathrm{loc}}(\Lambda)$.  
Here $W^{1,1}_{\mathrm{loc}}(\Lambda) 
= \{f\in L^1_{\mathrm{loc}}(\Lambda)\!\!\mid  
\nabla_{\mathrm{max}}f\in L^1_{\mathrm{loc}}(\Lambda)\}$.  
Thus, in general, one only knows that $\nabla_{\max}f\in L^1_{\mathrm{loc}}(\Lambda)$ for  
$f\in \calD(D_{A,\mathrm{max}})$, which is the main reason for some of the technical  
difficulties with magnetic Schr\"odinger operators.  
 
The Neumann quadratic form is given by the closed, non-negative, symmetric form  
\begin{equation} 
h^N_\Lambda(A,0)(\varphi,\psi) \defeq 
\sum_{j=1}^d \langle D_{A,\mathrm{max},j}(\varphi), D_{A,\mathrm{max},j}(\psi) \rangle  
\end{equation} 
with domain $\calD(h^N_\Lambda(A,0))\defeq H^1_A(\Lambda)$. Again, it uniquely defines  
the Neumann magnetic Laplacian $H^N_\Lambda(A,0)$, which reduces to $-\Delta^N_\Lambda$ for $A=0$.  
\begin{remark}%\label{rem:D-N}  
Let $\sharp$ be a  boundary condition which leads to a  
quadratic form which is sandwiched between the Dirichlet and Neumann forms, that is,  
for which  
$\calD(h^D_\Lambda(A,0))\subset\calD(h^\sharp_\Lambda(A,0))\subset\calD(h^N_\Lambda(A,0))$  
and $h^N_\Lambda(A,0)[\varphi,\varphi]\le h^\sharp_\Lambda(A,0))[\varphi,\varphi]$ for all  
$\varphi\in \calD(h^\sharp_\Lambda[A,0])$ together with  
$h^\sharp_\Lambda(A,0)[\varphi,\varphi]\le h^D_\Lambda(A,0))[\varphi,\varphi]$ for all  
$\varphi\in \calD(h^D_\Lambda[A,0])$ holds. For such a boundary condition,   
the min-max principle gives the bound  
$N^D_\Lambda(\lambda)\le N^\sharp_\Lambda(\lambda)\le N^N_\Lambda(\lambda)$  
for the finite volume IDS $N^\sharp$ and consequently  
$\calL^D_\Lambda(t)\le \calL^\sharp_\Lambda(t)\le \calL^N_\Lambda(t)$ for the Laplace transforms.  
Thus, for this type of boundary conditions, the finite-volume IDS $N_\Lambda^\sharp $ is  
well-defined as soon as this is the case for Neumann boundary conditions and its infinite  
volume limit will not depend on the chosen boundary condition as soon as the infinite volume  
limits constructed with the help of the Neumann and Dirichlet boundary conditions  are the same.  
However, one should note that there are self-adjoint extensions of $(-i\nabla+A)^2$ on finite  
domains, which have associated quadratic forms not comparable to the Dirichlet or Neumann form  
even for  vanishing magnetic vector potential.  
\end{remark} 
To include an electric potential $V\in L^1_{\mathrm{loc}}(\Lambda)$, first assume  
that it is  non-negative. Then, by setting  
$v[\varphi,\psi]= \langle \sqrt{V}\varphi,\sqrt{V}\psi\rangle$, $V$ can be seen  
as a closed quadratic form with domain $\calQ(V)= \calD(\sqrt{V})$ which includes  
$\calC^\infty_0(\Lambda)$. Hence, $\calQ(V)\cap \calD(h^{\sharp}_\Lambda(A,0))$ is dense  
in $L^2(\Lambda)$ and   
\bdm 
h^{\sharp}_\Lambda[A,V](\varphi,\psi) := h^{\sharp}_\Lambda[A,0](\varphi,\psi) 
+ \langle \sqrt{V}\varphi, \sqrt{V}\psi \rangle  
\edm 
on $\calD(h^{\sharp}_\Lambda[A,V]):= \calD(h^{\sharp}_\Lambda[A,0])\cap \calD(\sqrt{V})$  
are closed, densely defined quadratic forms, generating the magnetic Schr\"odinger operators $H^{\sharp}_\Lambda(A,V)$ with Dirichlet, $\sharp= D$, or Neumann, $\sharp= N$,  
boundary conditions. 
 
If the potential is not necessarily non-negative, write $V$ as $V= V_+ -V_-$ with  
$V_\pm := \sup(0,\pm V)$. Assuming $V_+\in L^1_{\mathrm{loc}}(\Lambda)$ and $V_-$  
form small with respect to $h^{\sharp}_\Lambda[A,V_+]$,  
$\langle \varphi, V_-\varphi\rangle \le  
a_\sharp h^\sharp[A,V_+](\varphi,\varphi)+ b\langle \varphi,\varphi\rangle$  
for some $0\le a_\sharp <1$ and $0\le b<\infty$, the KLMN theorem, see, e.g.,  
\cite{ReSi2}, shows that the form sum  
$h^{\sharp}_\Lambda[A,V](\varphi,\psi):= h^{\sharp}[A,V_+](\varphi,\psi)-  
\langle \sqrt{V_-}\varphi,\sqrt{V_-}\psi\rangle$ defines a closed quadratic from which is  
bounded from below. It again uniquely defines a self-adjoint magnetic Schr\"odinger operator  
$H^{\sharp}_\Lambda(A,V)$.  
 
\noindent  
\begin{remarks} \itemthm Since $h^D_\Lambda(A,V_+)$ is the restriction of $h^N_\Lambda(A,V_+)$  
to $H^1_{A,0}(\Lambda)\cap\calQ(V_+)$, any  potential $V_-$ which is form small with respect to  
the Neumann form is also form small with respect to $h^D_\Lambda[A,V_+]$. \\[0.2em] 
\itemthm As a consequence of the diamagnetic inequality in Theorem \ref{thm:maindia}, any  
potential which is form small with respect to $h^\sharp[0,V_+]$ with relative bound $a_\sharp$,  
for $\sharp=D$ or $N$, is also form small with respect to $h^\sharp[A,V_+]$ with relative bound  
$\tilde{a}_\sharp\le a_\sharp$ for any magnetic vector potential  
$A\in L^2_{\mathrm{loc}}(\Lambda,\R^d)$. \\[0.2em] 
\itemthm The conditions we impose on the potential are not the weakest possible, but are  
general enough to cover all cases of interest for the applications we have in mind. It is  
possible to study a larger class of positive perturbations. Without magnetic vector potentials,  
this has been extensively studied in \cite{Stollmann1,SV,Voigt1,Voigt2}. For example, the  
condition $V_+\in L^1_{\mathrm{loc}}(\Lambda)$ can be replaced by the assumption of regularity,  
that is, $\calQ{V_+}\cap \calD(h^\sharp_\Lambda(A,0))$ is dense in $L^2(\Lambda)$. This allows  
for somewhat strong local singularities in $V_+$, \cite{SV}.  
More general magnetic vector potentials, relaxing the condition  
$A^2\in L^1_{\mathrm{loc}}(\R^d)$ somewhat in the spirit of \cite{Voigt1, Voigt2},  
were discussed in \cite{LM}.  
\end{remarks} 
 
Our main tool for the diamagnetic inequality is Lemma \ref{lem:kato-bilinear} below.  
We collect some preparatory technical tools first.  
\begin{lemma}\label{lem:weak1}  
\itemthm \label{lem:weak1i} If $f,g\in W^{1,1}_{\mathrm{loc}}(\Lambda)$ with  
$f\nabla_{\mathrm{max}} g$ and $g \nabla_{\mathrm{max}} f \in L^1_{\mathrm{loc}}(\Lambda)$,  
then $fg\in W^{1,1}_{\mathrm{loc}}(\Lambda)$ and the product rule  
\bdm  
\nabla_{\mathrm{max}}(fg)= f\nabla_{\mathrm{max}} g + g\nabla_{\mathrm{max}} f  
\edm 
holds.\\ 
\itemthm\label{lem:weak1ii}  
If  $u\in W^{1,1}_{\mathrm{loc}}(\Lambda)$ is real valued with $u\ge \vare >0$, then  
$\tfrac{1}{u} \in W^{1,1}_{\mathrm{loc}}(\Lambda)$ and  
\bdm 
\nabla_{\mathrm{max}}\, \frac{1}{u} = -\frac{\nabla_{\mathrm{max}} u}{u^2} . 
\edm 
\end{lemma} 
\begin{remark} Part (i) of this lemma already appeared in \cite{Manavi}, Hilfssatz 14.1.  
The usual formulation of this result requires $f\in W^{1,p}_{\mathrm{loc}}(\Lambda)$ and  
$g\in W^{1,p'}_{\mathrm{loc}}(\Lambda)$ with $1/p + 1/p' =1$. It will be very convenient  
to have it formulated in this generality for the proof of Lemma \ref{lem:weaktwo} below.  
\end{remark} 
\begin{proof} Since the conclusions of the lemma are local, it is enough to fix some  
compact subset $K\subset\Lambda$ and to prove the result for almost all $x\in K$.  
Let $j_\delta$ be an approximate delta function with  
$\supp(j_\delta)\in B_\delta= \{\norm{x}<\delta \}$. We extend all functions to  
$\R^d$ by setting them to zero on $\Lambda^c$. Thus, for any  
$f\in L^1_{\mathrm{loc}}(\Lambda)$, the convolution $f_\delta =j_\delta *f$ is  
well-defined. Moreover, for small enough $\delta$, $u_\delta(x)\ge \veps$ for all  
$x\in K$. \\ 
To prove part (ii), note that for the smoothed function $u_\delta$, the  
conclusion of the lemma is immediate. Now take $\delta\to 0$ along a subsequence chosen  
such that the individual terms converge almost everywhere and use the definition of the  
weak derivative. \\  
To prove part (i), first assume that $f,g$ are bounded.  
Then $\norm{f_\delta}_\infty\le \norm{f}_\infty$ and the same for $g$.  
Again choose a subsequence $\delta_n\to 0$ such that individual terms converge  
almost everywhere in $\Lambda$. Writing $\nabla$ for $\nabla_{\mathrm{max}}$ and  
using dominated convergence,  
$\norm{f_{\delta_n} \nabla g_{\delta_n}- f\nabla g}_{L^1(\Lambda)}\le  
\norm{f}_\infty \norm{\nabla g_{\delta_n}- \nabla g}  
+\norm{(f_{\delta_n} -f)\nabla g}_{L^1(\Lambda)}\to 0$  
as ${\delta_n}\to 0$. Similarly,  
$g_{\delta_n}\nabla f_{\delta_n} \to g\nabla f$ and $f_{\delta_n}g_{\delta_n} \to fg$ as $\delta_n\to 0$.  
Since $\nabla (f_\delta g_\delta)= g_\delta\nabla f_\delta+ f_\delta\nabla g_\delta$,  
by the usual product rule, we see that $fg\in W^{1,1}(\Lambda)$ and  
$\nabla (fg)= g\nabla f+ f\nabla g$ in the limit $\delta_n\to 0$.  
Now assume $f$ and $g$ real-valued, but not necessarily bounded. Put  
$f_n= \ind_{\{\abs{f}\le n\}}f$ and similarly for $g$. By the usual arguments in the theory of  
Sobolev spaces, see, e.g., \cite{GT}, Lemma 7.6, we have  
$f_n, g_n \in W^{1,1}(\Lambda)$ with $\nabla f_n= \ind_{\{\abs{f}\le n\}}\nabla f$  
and $\nabla g_n= \ind_{\{\abs{g}\le n\}}\nabla g$. Thus $f_n g_n\in W^{1,1}(\Lambda)$ with  
$\nabla (f_n g_n)= g_n\ind_{\{\abs{f}\le n\}}\nabla f+ f_n\ind_{\{\abs{g}\le n\}}\nabla g$.  
Letting $n\to \infty$, using dominated convergence, gives the claim. For $f$ and $g$  
complex-valued the result follows by writing them as a sum of their real and imaginary parts. 
\end{proof} 
 
Following Kato \cite{Kato}, for each $\vare>0$ and $u:\Lambda\to \C$ measurable, let  
$\abs{u}_\vare:= \sqrt{\abs{u}^2+\vare^2} \ge \vare $ and $s_\vare:= u/\abs{u}_\vare$.  
Furthermore, set $s(x)\defeq u(x)/\abs{u(x)}$ where $u(x)\not = 0$ and $s(x)=0$ whenever  
$u(x)=0$. Then $s_\vare(x)\to s(x)$ for all $x$ as $\vare\to 0$.  
\begin{lemma}\label{lem:weaktwo} Let $u\in W^{1,1}_{\mathrm{loc}}(\Lambda)$. Then\\ 
\itemthm \label{lem:weaktwoi}  
    $\abs{u}, u_\vare\in W^{1,1}_{\mathrm{loc}}(\Lambda)$ for all $\vare >0$  
    with $\nabla_{\mathrm{max}} u_\vare= \re(\ol{s_\vare}\, \nabla_{\mathrm{max}} u)$ and  
    $\nabla_{\mathrm{max}} \abs{u} = \re(\ol{s}\, \nabla_{\mathrm{max}} u)$. \\ 
\itemthm \label{lem:weaktwoii}   
    For all $\vare >0$, $s_\vare \in W^{1,1}_{\mathrm{loc}}(\Lambda))$ with  
    $$ 
    \nabla_{\mathrm{max}} s_\vare =  
    \frac{\nabla_{\mathrm{max}} u - s_\vare\nabla_{\mathrm{max}} u_\vare }{u_\vare} . 
    $$ 
\itemthm \label{lem:weaktwoiii}   
    If $\varphi\in W^{1,1}_{\mathrm{loc}}(\Lambda)$ with $\abs{\varphi}\lesssim 1+\abs{u}$,  
    then $s_\vare \varphi \in W^{1,1}_{\mathrm{loc}}(\Lambda)$ for all $\vare>0$ and  
    \bdm 
    \nabla_{\mathrm{max}} (s_\vare \varphi)  
    =  
    \varphi \,\frac{\nabla_{\mathrm{max}} u - s_\vare\nabla_{\mathrm{max}} u_\vare }{u_\vare}  
    + s_\vare \nabla_{\mathrm{max}}\varphi.  
    \edm 
Moreover, assume $A\in L^2_{\mathrm{loc}}(\Lambda, \R^d)$ and $u\in H^1_A(\Lambda)$  
    {\rm (}$=\calD(D_{A,\mathrm{max}})${\rm )}. Then \\[0.3em]  
\itemthm \label{lem:weaktwoiv}  
    $\abs{u}, u_\vare\in H^1_A(\Lambda)$ for all $\vare>0$ with  
    $\abs{\nabla_{\mathrm{max}}u_\vare}\le \abs{D_{A,\mathrm{max}}u}$ and  
    $\abs{\nabla_{\mathrm{max}}\abs{u}}\le \abs{D_{A,\mathrm{max}}u}$. \\ 
\itemthm \label{lem:weaktwov}  
    If $\varphi\in H^{1}_A(\Lambda)$ with $\abs{\varphi}\lesssim 1+\abs{u}$,  
    we have $s_\vare \varphi \in H^1_A(\Lambda)$ for all $\vare>0$ and  
    \bdm 
    D_{A,\mathrm{max}} (s_\vare \varphi)  
    =  
    \varphi \,\frac{D_{A,\mathrm{max}} u - s_\vare\nabla_{\mathrm{max}} u_\vare }{u_\vare}  
    + s_\vare \nabla_{\mathrm{max}}\varphi.  
    \edm 
\end{lemma}  
\begin{proof} For simplicity, we will again write $\nabla$ for $\nabla_{\mathrm{max}}$ and  
$D_A$ for $D_{A,\mathrm{max}}$. The basic strategy of the proof of (i) is  
well-known, see \cite{Kato,ReSi2}. The map $\R^2\ni(s,t)\to \sqrt{s^2+t^2+\vare^2}$ is  
differentiable and the chain rule for Sobolev spaces, \cite{GT}, Lemma 7.5, shows that  
$u_\vare \in W^{1,1}_{\mathrm{loc}}(\Lambda)$. Using the same reduction argument  
as in the proof of Lemma \ref{lem:weak1ii}, we can assume that $u$ is bounded.  
Then $\ol{u}\nabla_{\mathrm{max}}u\in L^1_{\mathrm{loc}}(\Lambda)$, and since  
$u_\vare^2= \ol{u}u+\vare^2$, Lemma \ref{lem:weak1i} implies  
that $u_\vare^2\in W^{1,1}_{\mathrm{loc}}(\Lambda)$ with  
$\nabla u_\vare^2= 2u_\vare\nabla u_\vare= 2\re(\ol{u}\nabla u)$.  
Hence $\nabla u_\vare= \re(\ol{s_\vare}\nabla u)$ for all $\vare>0$.  
The result for $\abs{u}$ follows by taking the limit $\vare\to 0$, using the  
definition of the weak derivative and dominated convergence. \\ 
Once part (i) of the lemma  holds, (iv) is an immediate consequence, since  
$H^1_A(\Lambda)\subset W^{1,1}_{\mathrm{loc}}(\Lambda)$ and  
$\re(\ol{s_\vare}D_A u )= \re(\ol{s_\vare}\nabla u )$ for all $\vare > 0$. \\ 
By Lemma \ref{lem:weak1ii} and part (i) above, we have  
$\tfrac{1}{u_\vare}\in W^{1,1}_{\mathrm{loc}}(\Lambda)$.  
Since $u \tfrac{\nabla u_\vare}{u_\vare^2}$ and $\tfrac{\nabla u}{u_\vare}$ are locally  
integrable, Lemma \ref{lem:weak1i} shows  
$s_\vare = u/u_\vare\in W^{1,1}_{\mathrm{loc}}(\Lambda)$ and  
\bdm 
\nabla s_\vare = \nabla \Big(u \frac{1}{u_\vare} \Big) 
= \frac{\nabla u}{u_\vare}- u \,\frac{\nabla u_\vare}{u_\vare^2}  
= \frac{\nabla u -s_\vare \nabla u_\vare}{u_\vare} 
\edm 
which proves (ii). \\ 
(iii): Since $s_\vare$ is bounded, $s_\vare \nabla \varphi\in L^1_{\mathrm{loc}}(\Lambda)$.  
By assumption, $\varphi/u_\vare$ is also bounded. Hence, with part (ii),  
$\varphi \nabla s_\vare\in L^1_{\mathrm{loc}}(\Lambda)$ and consequently, using  
Lemma \ref{lem:weak1i} again, we see that $s_\vare\varphi\in W^{1,1}_{\mathrm{loc}}(\Lambda)$  
with  
\bdm 
\nabla (s_\vare\varphi)=  
\varphi \, \frac{\nabla u-s_\vare\nabla u_\vare}{u_\vare} + s_\vare \nabla \varphi . 
\edm 
Part (v) follows from this, by adding $-i\tfrac{\varphi}{u_\vare}Au$ to  
the above equation.  
\end{proof} 
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\section{The Diamagnetic Inequalities}\label{Dia:sec3} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\begin{lemma} \label{lem:kato-bilinear} 
Let $A\in L^2_{\mathrm{loc}}(\Lambda,\R^d)$ and $\varphi\ge 0$,  
$\varphi\in \calD(\nabla_{\mathrm{max}})$, $u\in\calD(D_{A,\mathrm{max}})$  
with $\varphi \lesssim 1+|u| $. Then $s_\vare \varphi\in \calD(D_{A,\mathrm{max}})$ and the  
pointwise inequality  
\bdm 
\re(\overline{D_{A,\mathrm{max}}(s_\vare \varphi)}\cdot D_{A,\mathrm{max}} (u))  
\ge \abs{s_\vare} \nabla_{\mathrm{max}} \varphi \cdot \nabla_{\mathrm{max}} \abs{u}   
\edm 
holds for all $\vare>0$.  
\end{lemma} 
\begin{remark} 
A similar inequality to the one given in Lemma \ref{lem:kato-bilinear}, more precisely its $\vare=0$  
limit, has already been used in \cite{LS}, proof of Lemma 6, for $\Lambda=\R^d$ and in \cite{DIT} to  
prove a diamagnetic inequality for the Dirichlet operator on open subsets $\Lambda\subset\R^d$.  
There it is enough to consider $\varphi,u\in \calC^\infty_0$, which simplifies some calculations.  
\end{remark} 
\begin{proof} 
That $s_\vare\varphi\in \calD(D_{A,\mathrm{max}})$ follows from Lemma \ref{lem:weaktwoiv}.  
We claim  
\begin{equation}\label{eq:kato} 
\re (\overline{D_{A,\mathrm{max}}(s_\vare \varphi)}\cdot D_{A,\mathrm{max}}(u)) =  
\varphi \,\frac{|D_{A,\mathrm{max}} u|^2- |\nabla_{\mathrm{max}} u_\vare|^2}{u_\vare}  
+ \abs{s_\vare} \nabla_{\mathrm{max}} \varphi\cdot\nabla_{\mathrm{max}} \abs{u}   
\end{equation} 
from which the inequality in the lemma follows, since $\varphi\ge 0$ and  
$|\nabla_{\mathrm{max}} u_\vare|\le |D_{A,\mathrm{max}} u|$ by Lemma \ref{lem:weaktwoiv}.  
Using Lemma \ref{lem:weaktwov} gives  
\begin{align*} 
\re(\overline{D_{A,\mathrm{max}}(s_\vare \varphi)}\cdot D_{A,\mathrm{max}}(u) )  
&=  
\varphi \,\frac{|D_{A,\mathrm{max}} u|^2 -\nabla_{\mathrm{max}}  
                u_\vare\, \re(\ol{s_\vare}D_{A,\mathrm{max}} u)}{u_\vare} \\ 
&\phantom{=}\,      
+ \nabla_{\mathrm{max}}\varphi\, \re(\ol{s_\vare}D_{A,\mathrm{max}} u)  \\ 
&=  
\varphi \,\frac{|D_{A,\mathrm{max}} u|^2 -|\nabla_{\mathrm{max}} u_\vare |^2}{u_\vare} 
+ \nabla_{\mathrm{max}}\varphi\, \nabla_{max} u_\vare 
\end{align*} 
where we also used Lemma \ref{lem:weaktwoi}.  
{}From the above equality, \xref{eq:kato} follows, since  
\bdm 
\nabla_{\mathrm{max}} u_\vare = \frac{1}{u_\vare} \nabla_{\mathrm{max}} u_\vare^2  
= \frac{1}{u_\vare} \nabla_{\mathrm{max}} \abs{u}^2  
= \frac{\abs{u}}{u_\vare} \nabla_{\mathrm{max}} \abs{u}  
= \abs{s_\vare} \nabla \abs{u}.  
\edm  
\end{proof} 
 
\begin{theorem}[Slight generalization of Theorem \ref{thm:dia}] \label{thm:maindia} 
Let $\Lambda\subset \R^d$ be open, $A\in L_{\mathrm{loc}}^2(\Lambda,\R^d)$,  
and $0\le V\in L^1_{\mathrm{loc}}(\Lambda)$. Then, for all $t\ge 0$ and $f\in L^2(\Lambda)$,  
\bdm 
\abs{\e^{-t H^N_\Lambda(A,V)} f }  
\le  
\e^{-t H^N_\Lambda(0,V)} \abs{f}  
\edm 
and  
\bdm  
\abs{\e^{-t H^N_\Lambda(0,V)} f }  
\le  
\e^{-t H^N_\Lambda(0,0)} \abs{f} .  
\edm 
Moreover, if $V=V_+-V_-$ with $V_\pm\ge 0$ and $V_+\in L^1_{\mathrm{loc}}(\Lambda)$, then  
$V_-$ relatively form bounded w.r.t.\ $H^N_\Lambda(0,0)$ with  
relative bound $a<1$ implies that  
$V_-$ is relatively form bounded w.r.t.\ $H^N_\Lambda(A, 0)$ with relative  
bound $\tilde{a}\le a<1$. Moreover, the two bounds  
\bdm  
\abs{\e^{-t H^N_\Lambda(A,V)} f }  
\le  
\e^{-t H^N_\Lambda(0,V)} \abs{f}  
\edm 
and 
\bdm 
\abs{\e^{-t H^N_\Lambda(0,V)} f } 
\le 
\e^{-t H^N_\Lambda(0,\wti{V})} \abs{f} 
\edm 
hold for all $f\in L^2(\Lambda)$ and all potentials  
$\wti{V}\le V$ for which $\wti{V}_-$ is form small with respect to $H^N_\Lambda(0,0)$.  
\end{theorem} 
\begin{remarks} \itemthm The same result holds for the Dirichlet semigroup, that is,  
$\abs{\e^{-t H^D_\Lambda(A,V)} f } \le \e^{-t H^D_\Lambda(0,V)} \abs{f}   
\le \e^{-t H^D_\Lambda(0,\wti{V})} \abs{f}$. For example, see \cite{BHL2} where this is  
proven under slightly more restrictive conditions on the vector potential $A$ using the  
Feynman-Kac-It\^{o} formula. With \xref{eq:slim} below, the Dirichlet case is an immediate  
corollary of the result for Neumann boundary conditions. \\[0.2em] 
\itemthm What we call diamagnetic inequality is often rephrased by semigroup 
people saying the magnetic semigroup is dominated by the non-magnetic one. 
In this direction one can find, e.g., in \cite{LM,Manavi}, more general (as far 
as the singularities of $V$ and $A$ are concerned) diamagnetic inequalities for the Dirichlet 
semigroup.\\[0.2em]
\itemthm The other proofs of the diamagnetic inequality mainly use the $\vare =0$ limit  
of Lemma \ref{lem:kato-bilinear} or similar bounds; see \cite{LS,DIT,LM,Ouhabaz}.  
This forces one to assume $\varphi\lesssim \abs{u}$.  
We deviate from this line of reasoning by trying to take the limit $\vare\to 0$ as late  
as possible in the proof.  
\end{remarks} 
 
\begin{proof} 
We start with the following well-known remark: Let $A,B$ be self-adjoint operators which  
are bounded from below.  
Then the bounds  
\begin{equation}\label{eq:abstract-dia-semigroup} 
\abs{\e^{-tB}f} \le \e^{-tA}\abs{f} 
\end{equation}  
for all $f$ and $t\ge 0$, and  
\begin{equation}\label{eq:abstract-dia-resolvent} 
\abs{(B+E)^{-n}f}\le (A+E)^{-n}\abs{f} 
\end{equation} 
for all $f$, $n\in\N$, and $E>0$ large enough, are equivalent. This follows from the representations 
\bdm 
\e^{-tB} = \slim_{n\to\infty}\big(n/t\big)\big(B+n/t\big)^{-n} 
\edm 
and 
\bdm 
(B+E)^{-\alpha} = c_\alpha \int_0^\infty t^{\alpha-1} \e^{-t(B+E)}\, dt 
\edm 
with $c_\alpha^{-1}= \int_0^\infty t^{\alpha-1}\e^{-t}\, dt$. In particular, once  
\xref{eq:abstract-dia-resolvent} holds for $n\in\N$, it holds for all $n\in (0,\infty)$.  
As a final preliminary, we note that for $V=V_+ -V_-$ with  
$0\le V_+\in L^1_{\mathrm{loc}}$ and $V_-$ form small with respect to $h^N_\Lambda(0,V_+)$,  
the first Beurling-Deny criterion implies that the resolvent  
$(H^N_\Lambda(0,V)+E)^{-1}$ is positivity preserving for large enough $E>0$. Similarly,  
the second Beurling-Deny criterion shows that, for non-negative  
$V\in L^1_{\mathrm{loc}}(\Lambda)$, $(H^N_\Lambda(0,V)+E)^{-1}$ maps bounded functions to  
bounded functions with  
$\norm{(H^N_\Lambda(0,V)+E)^{-1}f}_\infty\le \tfrac{1}{E}\norm{f}_\infty $; see, e.g.,  
\cite{ReSi4}, Theorems XIII.$50$, XIII.$51$, and problem $99$. If $V$ is merely bounded  
from below, the analogous result holds for $E>-\inf V$.  
 
In the following we will always assume that $A\in L^2_{\mathrm{loc}}(\Lambda, \R^d)$.  
For the moment, we also assume that $V=0$. Note that $\ol{s}_\vare u = \abs{s_\vare}\abs{u}$.  
Let $E>0$. With Lemma~\ref{lem:kato-bilinear}, we get  
\begin{align} 
  \langle \abs{s_\vare}\nabla_{\mathrm{max}} \varphi, \nabla_{\mathrm{max}} \abs{u} \rangle  
  +E \langle \abs{s_\vare}\varphi, u  \rangle  
&\le  
  \re \big( (h^N_\Lambda[A,0]+E)[s_\vare\varphi,u] \big) \label{eq:dia-resolvent01}\\ 
&\le 
  \abs{(h^N_\Lambda[A,0]+E)[s_\vare\varphi,u]} \label{eq:dia-resolvent02} 
\end{align} 
for any bounded, non-negative $\varphi\in\calD(\nabla_{\mathrm{max}})$ and all $\vare >0$.  
Now write $u= (H^N_\Lambda(A,0)+E)^{-1}f$ for some $f\in L^2(\Lambda)$. Then  
\xref{eq:dia-resolvent02} becomes  
\begin{equation} \label{eq:dia-resolvent03} 
\langle \abs{s_\vare}\nabla_{\mathrm{max}} \varphi, \nabla_{\mathrm{max}} \abs{u} \rangle  
  +E \langle \abs{s_\vare}\varphi, u  \rangle  
\le 
  \abs{\langle s_\vare\varphi, f \rangle} 
\le 
  \langle \varphi, \abs{f}\rangle . 
\end{equation} 
Since $\nabla\abs{u}=0$ on $\{u=0\}$ by Lemma \ref{lem:weaktwo}.i, we have that  
$\abs{s_\vare}\nabla\abs{u}$ tends to $\nabla\abs{u}$ in $L^2(\Lambda)$ as $\vare\to 0$.  
So taking $\vare \to 0$ in \xref{eq:dia-resolvent03} gives  
\begin{equation}\label{eq:dia-resolvent04} 
(h^N_\Lambda(0,0)+E)[\varphi, \abs{u}]  
\le 
\langle \varphi, \abs{f}\rangle  
\end{equation} 
Setting $\varphi = (H^N_\Lambda(0,0)+E)^{-1}\psi$ for $\psi\in L^2(\Lambda)$ non-negative  
and bounded preserves non-negativity and boundedness of $\varphi$ by the second remark at  
the beginning of the proof. Thus \xref{eq:dia-resolvent04} gives  
\begin{align*} 
\langle \psi, \abs{(H^N_\Lambda(A,0)+E)^{-1}f} \rangle  
&\le \langle  (H^N_\Lambda(0,0)+E)^{-1}\psi, \abs{f}\rangle \\ 
&= \langle  \psi, (H^N_\Lambda(0,0)+E)^{-1}\abs{f}\rangle  
\end{align*} 
for all bounded $0\le \psi\in L^2(\Lambda)$ and all $f\in L^2(\Lambda)$, which is equivalent to  
\bdm 
\abs{(H^N_\Lambda(A,0)+E)^{-1}f} \le (H^N_\Lambda(0,0)+E)^{-1}\abs{f}  
\edm 
for all $E>0$. Iterating the last inequality, we get  
\bdm 
\abs{(H^N_\Lambda(A,0)+E)^{-n}f} \le (H^N_\Lambda(0,0)+E)^{-n}\abs{f}  
\edm 
for all $n\in\N$, which is the diamagnetic inequality for resolvents. By the first remark at  
the beginning of the proof, the inequality   
\begin{equation}\label{eq:dia-resolvent2} 
\abs{\e^{-tH^N_\Lambda(A,0)}f} \le \e^{-tH^N_\Lambda(0,0)}\abs{f} \quad\text{for all }  
t\ge 0, f\in L^2(\Lambda)  
\end{equation} 
follows.  
 
Now let $V\ge 0$ and add  
$\langle s_\vare \varphi, Vu \rangle= \langle \abs{s_\vare} \varphi, V\abs{u}\rangle \ge 0$  
to both sides of \xref{eq:dia-resolvent01}.  
Setting $u= (H^N_\Lambda(A,V)+E)^{-1}f$, we get   
\begin{align*} 
\langle \abs{s_\vare}\nabla_{\mathrm{max}} \varphi, \nabla_{\mathrm{max}} \abs{u} \rangle  
  +\langle \abs{s_\vare}\varphi, (V+E)u  \rangle  
&\le 
  \abs{(h^N_\Lambda[A,V]+E)[s_\vare,u]} \\ 
&\le 
  \abs{\langle s_\vare\varphi, f \rangle} 
\le 
  \langle \varphi, \abs{f}\rangle  
\end{align*} 
for all $0\le \varphi\in\calD(\nabla)\cap \calQ(V)$. Again, if $\psi$ is non-negative and  
bounded, so is $\varphi=(H^N_\Lambda(0,V)+E)^{-1}\psi$. In turn, we can do the limit $\vare\to 0$  
to get  
\bdm  
\langle \psi, \abs{(H^N_\Lambda(A,V)+E)^{-1}f}\rangle  
\le \langle \psi , (H^N_\Lambda(0,V)+E)^{-1}\abs{f}\rangle  
\edm  
which leads to  
\bdm  
\abs{(H^N_\Lambda(A,V)+E)^{-n}f} \le (H^N_\Lambda(0,V)+E)^{-n}\abs{f}  
\edm  
for all $n\in\N$ and $f\in L^2(\Lambda)$, and  
\begin{equation}\label{eq:dia-resolvent-positive-V}  
\abs{\e^{-tH^N_\Lambda(A,V)}f} \le \e^{-tH^N_\Lambda(0,V)}\abs{f}  
\text{ for all } t\ge 0, f\in L^2(\Lambda) . 
\end{equation} 
Now consider the special case $A=0$ and  add  
$\langle s_\vare \varphi, Vu \rangle= \langle \abs{s_\vare} \varphi, V\abs{u}\rangle \ge 0$  
only to the right-hand side of \xref{eq:dia-resolvent01}. Making the obvious changes  
in the above argument leads to the bound  
\bdm  
\abs{(H^N_\Lambda(0,V)+E)^{-n}f} \le (H^N_\Lambda(0,0)+E)^{-n}\abs{f}  
\edm  
for all $n\in\N$, or equivalently,  
\bdm 
\abs{\e^{-tH^N_\Lambda(0,V)}f} \le \e^{-tH^N_\Lambda(0,0)}\abs{f}  
\edm 
for all $f\in L^2(\Lambda)$. Combining this with \xref{eq:dia-resolvent-positive-V}  
proves  
\bdm 
\abs{\e^{-tH^N_\Lambda(A,V)}f} \le \e^{-tH^N_\Lambda(0,0)}\abs{f}  
\edm 
or equivalently,  
\bdm 
\abs{(H^N_\Lambda(A,V)+E)^{-\alpha}f} \le (H^N_\Lambda(0,0)+E)^{-\alpha}\abs{f}  
\edm 
for all $\alpha >0$, non-negative $V\in L^1_{\mathrm{loc}}(\Lambda)$, and $f\in L^2(\Lambda)$.  
This proves the first and second inequalities in the theorem. Of course, by adding a  
constant, one sees that the results remain true if $V$ is merely bounded from below.  
 
The fact that $V_-$ relatively $H^N_\Lambda(0,0)$-form bounded with bound $a$ implies  
that it is relatively $H^n_\lambda(A,0)$-form bounded with bound $\tilde{a}\le a$ follows  
immediately from the diamagnetic inequality for the resolvent and the formula  
\bdm 
a =\lim_{E\to\infty}\norm{(H^N_\Lambda+E)^{-1/2}V_-(H^N_\Lambda+E)^{-1/2}} 
\edm 
for the relative form bound $a$, see, e.g., \cite{CFKS}.  
 
Now assume that $V=V_+-V_-$ with $V_-$ not necessarily bounded but form small  
with respect to  $H^N_\Lambda(A,0)$ and the same for $\wti{V}\le V$.  
For $n\in\N$ set $V_n= \max(V,-n)$ and $\wti{V}_n= \max(\wti{V},-n)$ .  
Using what we proved so far, we know that  
\bdm 
\abs{\e^{-tH^N_\Lambda(A,V_n)}f} \le \e^{-tH^N_\Lambda(0,V_n)}\abs{f} .  
\edm 
Furthermore, we have  
\bdm 
\e^{-tH^N_\Lambda(0,V_n)}\abs{f}  
\le \e^{-tH^N_\Lambda(0,\wti{V}_n)}\abs{f} 
\edm 
which follows similarly to the reasoning in the proof of  
\xref{eq:dia-resolvent-positive-V} by adding $V_n$ to the right and $\wti{V}_n$  
to the left-hand side of \xref{eq:dia-resolvent01} using $\wti{V}_n\le V_n$  
for all $n\in\N$.  
Letting $n\to \infty$, the strong convergence guaranteed, for example, by  
Theorem \ref{thm:monconv}, finishes the proof of the theorem.  
\end{proof} 
 
The next simple lemma is our main observation in extending the usual  
diamagnetic inequality to an inequality for the semigroup differences.  
\begin{lemma}\label{lem:difference}  
Let $V=V_+ -V_-$ with $V_+\in L^1_{\mathrm{loc}}(\Lambda)$, $V_-$ form  
small with respect to $H^N_\Lambda(0,0)$ and $U$ be any non-negative bounded  
function. Then  
\bdm 
\abs{(\e^{-tH^N_\Lambda(A,V)}-\e^{-tH^N_\Lambda(A,V+U)})f}  
\le 
(\e^{-tH^N_\Lambda(0,V)}-\e^{-tH^N_\Lambda(0,V+U)})\abs{f} 
\edm  
for all $f\in L^2(\Lambda)$. Furthermore, if $\wti{V}\le V$ is such that  
$\wti{V}_-$ is form small with respect to $H^N_\Lambda(0,0)$, then   
\bdm 
\abs{(\e^{-tH^N_\Lambda(0,V)}-\e^{-tH^N_\Lambda(0,V+U)})f} 
\le 
(\e^{-tH^N_\Lambda(0,\wti{V})}-\e^{-tH^N_\Lambda(0,\wti{V}+U)})\abs{f} 
\edm 
for all $f\in L^2(\Lambda)$.  
\end{lemma} 
\begin{proof} 
Let $C,D$ be non-negative operators with $D-C$ bounded. Then  
\begin{align*} 
\e^{-tC}- \e^{-tD}  
&= 
\int_0^t \frac{d}{ds}\big( \e^{-sC}\e^{-(t-s)D} \big)\, ds \\ 
&= 
\int_0^t  \e^{-sC}(D-C)\e^{-(t-s)D} \, ds 
\end{align*} 
which is Duhamel's formula. With $D= H^N_\Lambda(A,V)$ and $C= H^N_\Lambda(A,V)+U$,  
we get, for all $f\in L^2(\Lambda)$,  
\begin{align*} 
\abs{(\e^{-tH^N_\Lambda(A,V)}- \e^{-tH^N_\Lambda(A,V+U)})f}  
&\le  
\int_0^t  \abs{ \e^{-sH^N_\Lambda(A,V)}U\e^{-(t-s)H^N_\Lambda (A,V+U)} f} \, ds \\ 
&\le  
\int_0^t  \e^{-sH^N_\Lambda(0,V)}U\e^{-(t-s)H^N_\Lambda(0,V+U)}\abs{f} \, ds \\ 
&=  
(\e^{-tH^N_\Lambda(0,V)}- \e^{-tH^N_\Lambda(0,V+U)} )\abs{f}  
\end{align*} 
by the triangle inequality,  a repeated use of the diamagnetic inequality  
for $\e^{-tH^N_\Lambda(A,V)}$ and $\e^{-tH^N_\Lambda(A,V+U)}$, the positivity  
of $U$, and Duhamel's formula. This proves the first inequality. The second one follows  
along the same line, using again Duhamel's formula and the bound  
$\abs{\e^{-tH^N_\Lambda(0,V)}f}\le \e^{-tH^N_\Lambda(0,\wti{V})}\abs{f} $  
from Theorem~\ref{thm:maindia}. 
\end{proof} 
\begin{remark}  
Alternatively, one can use the Trotter product formula in place of Duhamel's formula.  
A third possibility is to use resolvent methods directly: by the remark in the beginning  
of the proof of Theorem \ref{thm:maindia}, the bound in Lemma \ref{lem:difference} is equivalent to  
\begin{align}  \label{rem:resolvent} 
\abs{\big((&H^N_\Lambda(A,V)+ E)^{-n} - (H^N_\Lambda(A,V+U)+E)^{-n}\big)f} \nonumber \\ 
&\le  
\big((H^N_\Lambda(0,V)+E)^{-n} - (H^N_\Lambda(0,V+U)+E)^{-n}\big)\abs{f} \\  
&\le  
\big((H^N_\Lambda(0,\wti{V})+E)^{-n} - (H^N_\Lambda(0,\wti{V}+U)+E)^{-n}\big)\abs{f} 
\nonumber 
\end{align} 
for all $E>-\inf\sigma ((H^N_\Lambda(0,\wti{V}) $ and $n\in\N$. For $n\!=\!1$, the bound  
\eqref{rem:resolvent} can be seen by inspection, using the resolvent identity and the diamagentic  
inequality for the resolvents. The case $n\ge 2$ then follows from the $n\!=\!1$ case by writing  
the difference as a telescoping sum, using the triangle inequality, the bound for $n\!=\!1$, and  
the diamagnetic inequality in each term of the sum.  
\end{remark} 
With this we can finally prove our main result. 
\begin{theorem}[Slight generalization of  Theorem \ref{thm:diadifference}] 
\label{thm:maindiadifference}  
Let $\Lambda\subset \R^d$ be open, $A\in L_{\mathrm{loc}}^2(\Lambda,\R^d)$, and $V=V_+-V_-$  
with $V_\pm\ge 0$ and $V_+\in L^1_{\mathrm{loc}}(\Lambda)$, $V_-$ relatively form bounded  
w.r.t.\ $H^N_\Lambda(0,0)$. Then 
\bdm 
\abs{(\e^{-t H^N_\Lambda(A,V)}-\e^{-t H^D_\Lambda(A,V)} )f }  
\le  
(\e^{-t H^N_\Lambda(0,V)}-\e^{-t H^D_\Lambda(0,V)} ) \abs{f}   
\edm 
for all $f\in L^2(\Lambda)$. Furthermore, if $\wti{V}\le V$ is such that  
$\wti{V}_-$ is form small with respect to $H^N_\Lambda(0,0)$ then   
\bdm 
\abs{(\e^{-t H^N_\Lambda(0,V)}-\e^{-t H^D_\Lambda(0,V)} )f }  
\le 
(\e^{-t H^N_\Lambda(0,\wti{V})}-\e^{-t H^D_\Lambda(0,\wti{V})} ) \abs{f}  
\edm  
for all $f\in L^2(\Lambda)$. 
\end{theorem} 
\begin{proof} 
Given Lemma \ref{lem:difference}, the proof of this theorem reduces to the existence  
of a non-negative function $U\in L^1_{\mathrm{loc}}(\Lambda)$ such that,  
setting $U_n\defeq \inf(U,n)$,  
\begin{equation}\label{eq:slim} 
\slim_{r\searrow 0}\,\slim_{n\to\infty}\,  
\e^{-tH^N_\Lambda(A,V+r U_{n})} 
=  
\e^{-tH^D_\Lambda(A,V)} . 
\end{equation}  
Let $\{\Lambda_n\}_{n\in\N}$ be a sequence of open sets $\Lambda_n\subset \Lambda$ such that the  
closures $\overline{\Lambda}_n $ are compact sets, $\overline{\Lambda}_n\subset\Lambda_{n+1}$ and  
$\Lambda = \cup_{n}\Lambda_n$. Choose smooth functions $0\le \psi_n \le 1$ with  
$\psi_n(x)=1$ for $x\in\Lambda_{n}$, $\psi_n(x)=0$ for $x\in\Lambda_{n+1}$, and set 
\begin{equation} \label{eq:def-U} 
U(x)\defeq \sum_{n\in\N} |\nabla\psi_n(x)|^2 . 
\end{equation}  
For each $x\in\Lambda$, only one term in the above sum can contribute, so, in fact,   
$U\in L^\infty_{\mathrm{loc}}(\Lambda)$. We claim that $U$ lets us do \xref{eq:slim}.  
To see this, it is enough to show that  
\begin{equation} \label{eq:slim1} 
\slim_{n\to\infty} \,\e^{-tH^N_\Lambda(A,V+r U_{n})}  
= 
\e^{-tH^N_\Lambda(A,V+r U)} 
\end{equation}  
for fixed $r>0$ and 
\begin{equation} \label{eq:slim2} 
\slim_{r\searrow 0} \,\e^{-tH^N_\Lambda(A,V+r U)} 
= 
\e^{-tH^D_\Lambda(A,V)},  
\end{equation}  
the second being the slightly trickier task. Let $h^N_\Lambda(A,V)$ be the quadratic form  
corresponding to $H^N_\Lambda(A,V)$. Then for any non-negative, locally integrable potential $U$,  
$H^N_\Lambda(A,V+U)$ is defined as the operator corresponding to the closed quadratic form  
\bdm 
h^N_\Lambda(A, V+U)[\varphi,\varphi] =  
h^N_\Lambda(A, V)[\varphi,\varphi] + \langle \sqrt{U}\varphi, \sqrt{U}\varphi\rangle  
\edm 
with domain $\calD(h^N_\Lambda(A, V+U))= \calD(h^N_\Lambda(A, V))\cap \calD(\sqrt{U})$.  
 
Since $U_n$ converges pointwise monotonically to $U_n$, we can use Lebesgue's monotone convergence  
theorem to see that $h^N_\Lambda(A, V+rU_n)$ converges monotonically to  
$h^N_\Lambda(A, V+rU)$. By Theorem \ref{thm:isotone}, the corresponding operators converge in strong  
resolvent sense. In particular, for any fixed  $r>0$, \xref{eq:slim1} holds.  
 
The proof of \xref{eq:slim2} is more complicated: Since  
$\cup_{r>0}\calD(h^N_\Lambda(A, V+rU))=h^N_\Lambda(A, V+U)=  
\calD(h^N_\Lambda(A, V))\cap \calD(\sqrt{U})$, we see that  
$\lim_{r\to 0}h^N_\Lambda(A, V+rU)[\varphi,\varphi]= h^N_\Lambda(A, V)[\varphi,\varphi]$  
for all $\varphi\in \cup_{r>0}\calD(h^N_\Lambda(A, V+rU))$. In order to apply  
Theorem \ref{thm:antitone}, it remains to show that  
$\overline{\calD(h^N_\Lambda(A, V+U))}^{\norm{.}_{h^D_\Lambda(A,V)}} 
= \calD(h^D_\Lambda(A, V)) = H^1_{0,A}(\Lambda)\cap \calQ(V)$.  
Since $\calC^\infty_0(\Lambda)$ is dense in $\calQ(V_+)$ with respect to $\norm{.}_{V_+}$ and  
$\calD(h^N_\Lambda(A, V+U))= \calD(h^N_\Lambda(A, 0))\cap\calQ(V_+)\cap\calQ(U) 
\subset \calD(h^N_\Lambda(A,U))$,  
this is the case if   
\begin{equation}\label{eq:dense} 
\calC^\infty_0(\Lambda) \text{ is dense in }  
\calD(h^N_\Lambda(A, U)) \text{ in the norm } \norm{. }_{h^D_\Lambda(A,0)}.  
\end{equation} 
We show \xref{eq:dense} in three steps: \\ 
\textit{Step 1:} $\calD(h^N_\Lambda(A,U))\cap L^\infty(\Lambda)$ is dense in  
                 $\calD(h^N_\Lambda(A,U))$.  
We follow \cite{LS}, proof of Lemma $2$, quite closely, but with some slight  
simplifications, since we cut smoothly in the range of functions:  
Given $n\in\N$, $n\ge 2$, let $\varphi_n\in \calC^1([0,\infty))$  
with $0\le\varphi_n\le 1$, $\varphi_n(t)=1$ for all $0\le t\le n-1$,  
$\varphi_n(t)= (n-1/2)/t$ for $t\ge n$ and  
$\sup_{t\in [n-1,n]}\abs{\varphi'(t)}\le \abs{\varphi'(n)}\le 1/n$. For such a sequence  
of functions, we have the bounds $(1-\varphi_n(t))\le \chi_{[n-1,\infty)}(t)$ and  
$t\abs{\varphi_n'(t)}\le \chi_{[n-1,\infty)}(t)$.  
Given $u\in \calD(h^N_\Lambda(A,U))$, set $u_n:= \varphi_n(u_1)u$, where  
$u_1= \sqrt{\abs{u}^2+1}$. Then, as distributions,   
\bdm 
\nabla u_n = \varphi_n(u_1)\nabla u + u \varphi'_n(u_1)\nabla u_1.  
\edm 
Hence  
\begin{equation} \label{eq:step1} 
D_A u -D_A u_n = (1-\varphi_n(u_1)) D_A u -u \varphi'_n(u_1)\nabla u_1.  
\end{equation} 
By Lemma \ref{lem:weaktwoiv}, we know $\nabla u_1 \in L^2(\Lambda)$. Note that   
$\abs{u\varphi'_n(u_1)}\le u_1 \abs{\varphi'_n(u_1)}\le \chi_{[n-1,\infty)}(u_1)$.  
Thus, by \xref{eq:step1}, $\norm{D_A u -D_A u_n}$ goes to zero as  $n\to\infty$. \\[0.35em] 
% 
\textit{Step 2:} $\calD(h^N_\Lambda(A,U))\cap L^\infty_{\mathrm{comp}}(\Lambda)$ is dense in  
                 $\calD(h^N_\Lambda(A,U))$.  
By Step 1 we can assume $u\in \calD(h^N_\Lambda(A,U))\cap L^\infty_{\mathrm{comp}}(\Lambda)$.  
Recall that $U=\sum_n \abs{\nabla\psi_n}^2$. Put $u_n\defeq \psi_n u$ and note that $u_n$  
has compact support in $\Lambda$. As distributions,  
\begin{equation} \label{eq:step2} 
D_A u - D_A u_n = (1-\psi_n)D_A u - u\nabla\psi_n . 
\end{equation} 
We have $\sum_n u\nabla\psi_n \in L^2(\Lambda)$ since $u\in\calQ(U)$. In particular,  
$u\nabla\psi_n$ goes to zero in $L^2(\Lambda)$. Hence, by \xref{eq:step2},  
$\norm{D_A u -D_A u_n}$ goes to zero as  $n\to\infty$. \\[0.35em] 
% 
\textit{Step 3:} $\calC^\infty_0(\Lambda)$ is dense in $\calD(h^N_\Lambda(A,U))$.  
Without loss of generality we can assume  
$u\in \calD(h^N_\Lambda(A,U))\cap L^\infty_{\mathrm{comp}}(\Lambda)$.  
Since $A\in L^2_{\mathrm{loc}}(\Lambda, \R^d)$, we know that  
\bdm 
A u \in L^2(\Lambda) \text{ and } \nabla u = D_Au +iAu \in L^2(\Lambda) . 
\edm 
Let $(j_n)_{n\in\N}$ be a sequence of approximate delta functions with  
$\supp(j_n)\subset B_{1/n}(0)$. Choose $n$ so large that  
$u_n\defeq j_n*u\in \calC^\infty_0(\Lambda)$. By standard mollifier arguments,  
$\norm{\nabla u- \nabla u_n}$ and $\norm{A u-Au_n}$ go to zero as $n\to\infty$. Since    
\bdm 
D_A u- D_A u_n = \nabla u - \nabla u_n -iA u +iA u_n , 
\edm 
this finishes the proof of \xref{eq:dense}.  
\end{proof} 
\begin{remark} The proof of \xref{eq:dense} is motivated by the proof of the Feynman-Kac  
formula for the Dirichlet Laplacian on arbitrary open subsets in \cite{Simon78b}.  
That one can extend this line of thought to certain magnetic vector potentials was  
already noticed in \cite{BHL2}. The observation that the ideas in \cite{Simon78b} even  
allow one to recover Dirichlet from Neumann boundary conditions for arbitrary vector  
potentials in $L^2_{\mathrm{loc}}(\Lambda, \R^d)$ seems to be new.  
\end{remark} 
 
\begin{proof}[Proof of Corollary \ref{cor:dia-maincoro}:] Part (i): By Remark  
\ref{rem:dia-compact}, we know that the free semigroup $\e^{-tH^N_\Lambda(0,0)}$ is a  
trace class operator, and standard perturbation arguments show that  
$\e^{-tH^N_\Lambda(0,-V_-)}$ and $\e^{-tH^N_\Lambda(0,V)}$ are trace class operators,  
see, e.g., \cite{ReSi4}, proof of Theorem XIII.76.  
 
 
In the following, let $\tr$ stand for the trace on $L^2(\Lambda)$ and  
$\norm{C}_{HS}= \tr[C^*C]^{1/2}$ be the Hilbert-Schmidt norm.  
Since $\e^{-tH^N_\Lambda(A,V)}$ is a non-negative operator, its trace norm is equal to  
its trace and, by the semigroup property and self-adjointness, we have  
\bdm 
\tr[\e^{-2tH^N_\Lambda(A,V)}] = \tr[(\e^{-tH^N_\Lambda(A,V)})^*\e^{-tH^N_\Lambda(A,V)}]  
= 
\big\Vert\e^{-tH^N_\Lambda(A,V)}\big\Vert_{HS}^2 . 
\edm 
{}From the diamagnetic inequality and \cite{Davies}, Theorem 2.4.4,  
we know that $\e^{-tH^N_\Lambda(A,V)}$ is an integral operator whose kernel obeys  
the bounds   
\bdm 
\abs{\e^{-tH^N_\Lambda(A,V)}(x,y)}\le \e^{-tH^N_\Lambda(0,V)}(x,y)  
\le \e^{-tH^N_\Lambda(0,-V_-)}(x,y) \text{ for almost all }x,y\in\Lambda . 
\edm  
Thus, using \cite{ReSi1}, Theorem VI.23, we conclude  
\begin{align*} 
\tr\big[\e^{-2tH^N_\Lambda(A,V)}\big]  
&= 
\big\Vert\e^{-tH^N_\Lambda(A,V)}\big\Vert_{HS}^2  
= \iint_{\Lambda\times\Lambda} \abs{\e^{-tH^N_\Lambda(A,V)}(x,y)}^2\, dxdy \\ 
&\le  \big\Vert\e^{-tH^N_\Lambda(0,V)}\big\Vert_{HS}^2  
  = \tr\big[\e^{-2tH^N_\Lambda(0,V)}\big] <\infty  
\end{align*} 
by the bound on the integral kernels. In an analogous way, one sees that  
$\tr[\e^{-tH^N_\Lambda(0,V)}] \le \tr[\e^{-tH^N_\Lambda(0,-V_-)}] $. \\[0.35em] 
Part (ii): For any pair $C,D$ of Hilbert-Schmidt operators on $L^2(\Lambda)$, an  
easy extension of the proof of Theorem VI.23 in \cite{ReSi1} shows the formula  
\begin{equation}\label{eq:dia-HSformula} 
\tr[CD]= \iint_{\Lambda\times\Lambda} C(x,y)D(y,x)\, dxdy . 
\end{equation} 
We know, by our assumptions on $\Lambda$ and part (i) of the corollary, that  
$\e^{-tH^N_\Lambda(A,V)}- \e^{-tH^D_\Lambda(A,V)}$ is a trace class operator  
(\emph{both} $\e^{-tH^N_\Lambda(A,V)}$ and $\e^{-tH^D_\Lambda(A,V)}$ are trace class!)  
and thus, in particular, Hilbert-Schmidt operators.  
Moreover, the difference of the kernels obeys, by the diamagnetic inequality for the  
differences, Theorem \ref{thm:maindiadifference},  
\begin{align}\label{eq:dia-kerneldifference} 
\abs{\e^{-tH^N_\Lambda(A,V)}(x,y)- \e^{-tH^D_\Lambda(A,V)}(x,y)}  
&\le \e^{-tH^N_\Lambda(0,V)}(x,y)- \e^{-tH^D_\Lambda(0,V)}(x,y) \nonumber \\ 
&\le \e^{-tH^N_\Lambda(0,-V_-)}(x,y)- \e^{-tH^D_\Lambda(0,-V_-)}(x,y)  
\end{align}  
for almost all $x,y\in\Lambda$. Factorizing  
\begin{align*} 
\e^{-2tH^N_\Lambda(A,V)}- \e^{-2tH^D_\Lambda(A,V)}  
=  
& \e^{-tH^N_\Lambda(A,V)} \big( \e^{-tH^N_\Lambda(A,V)}- \e^{-tH^D_\Lambda(A,V)} \big) \\ 
& + \big( \e^{-tH^N_\Lambda(A,V)}- \e^{-tH^D_\Lambda(A,V)} \big) \e^{-tH^D_\Lambda(A,V)} , 
\end{align*} 
using \xref{eq:dia-HSformula}, the diamagnetic inequalities for the respective  
kernels, and reasoning similarly as in part (i) gives the claim in Corollary  
\ref{cor:dia-difference}.  
\end{proof} 
\begin{remarks} \itemthm Of course, with the obvious notational changes, one can extend the  
above proof to cover the case $\wti{V}\le V$. \\ 
\itemthm It might be that, for some cleverly constructed bounded set $\Lambda$, the Neumann semigroup  
$\e^{-tH^N_\Lambda(0,0)}$ is compact but not trace class or Hilbert-Schmidt for small times  
$t$, see Remark \ref{rem:dia-rooms-and-passages}. Nevertheless, the diamagnetic inequality  
and the Dodds, Fremlin, and Pitt theorem \cite{AB,DF,Pitt} ensures that  
$\e^{-tH^N_\Lambda(A,0)}$ is also compact for arbitrary vector potentials  
$A\in L^2_{\mathrm{loc}}(\Lambda.\R^d)$.  
See also \cite{Leinfelder} for a nice extension of this basic compactness criterion.  
\end{remarks} 
 
% 
%\appendix  
\setcounter{section}{0} 
\renewcommand{\thesection}{\Alph{section}} 
\section*{Appendix: Monotone Convergence of Forms and Strong Resolvent Convergence} \label{Dia:Appendix}  
\renewcommand{\theequation}{A.\arabic{equation}} 
\renewcommand{\thetheorem}{A.\arabic{theorem}} 
\setcounter{theorem}{0} 
\setcounter{equation}{0} 
For the convenience of the reader we present here the basic convergence theorem we need in  
the proof of Theorem~\ref{thm:maindiadifference}   
\begin{theorem}\label{thm:monconv} 
Let $\calH$ be a Hilbert space, $s_n$, $u_n$, and $h$ non-negative, closed quadratic forms, and 
$S_n$, $U_n$, and $H$ be the corresponding self-adjoint operators. \\ 
\itemthm \label{thm:isotone}  
  If $u_n \!\le\! u_{n+1} \!\le\! h$ and $h(\varphi,\varphi)=\sup_n u_n(\varphi,\varphi)$,  
  that is,  
  \bdm 
    \Big\{ 
      \varphi\in\bigcap_{n\in\N}\calQ(u_n)\!: 
      \sup_{n\in\N} u_n(\varphi,\varphi) <\infty   
    \Big\} 
    = 
    \calQ(h),  
  \edm 
  \bdm 
    u_n(\varphi,\varphi) \rightarrow h(\varphi,\varphi) 
    \text{ as ~} n\to\infty,\quad \forall \varphi \in \calQ(h),  
  \edm 
  then $U_n \to H$ as $n\to\infty$ in the strong resolvent sense. \\ 
\itemthm \label{thm:antitone} 
  If $s_n \!\ge\! s_{n+1} \!\ge\! h$ and  
  \bdm 
    \overline{\bigcup_{n\in\N} \calQ(s_n)}^{\| \cdot \|_h}  
    = 
    \calQ(h), 
  \edm 
  \bdm 
    s_n(\varphi,\varphi) \longrightarrow h(\varphi,\varphi)  
    \text{ as } n\to\infty, \quad\forall \varphi \in  
    \bigcup_{n\in\N} \calQ(s_n)  , 
  \edm 
   then $S_n \to H$ as $n\to\infty$ in the strong resolvent sense. 
%\end{enumerate} 
\end{theorem} 
 
\begin{remarks} 
\itemthm   
    As usual, see, e.g., \cite{ReSi1}, Theorem VIII.20, strong resolvent convergence  
    of $U_n$ (resp.\ $S_n$) implies strong convergence of   
    $f(U_n)$ (resp.\ $f(S_n)$) for any continuous bounded function $f$. \\[0.2em] 
\itemthm  
    This theorem is taken from Simon \cite{Simon78a}, see also  
    \cite{ReSi1}, Theorems S.14 and S.16, where an even stronger result was proven:  
    The quadratic forms need not be densely defined, as long as one interprets strong  
    resolvent convergence correctly. \\[0.2em] 
\itemthm  
    A partly alternate proof of this result can be found in \cite{Weidmann84}. \\[0.2em] 
\itemthm  
    In general, (generalized) strong convergence of operators is equivalent to the  
    so-called $\Gamma$-convergence of the corresponding quadratic forms;  
    see \cite{DalMaso}.  
\end{remarks} 
 
\smallskip 
\noindent 
\textbf{Acknowledgements:} It is a pleasure to thank Alexander Elgart for  
discussions and Hendrik Vogt for pointing us to \cite{Manavi}.  
D.H.\ also thanks Ari Laptev and Kjell-Ove Widman for their hospitality at the  
Mittag-Leffler institute.  
 
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
% End of appendix for Diamagnetic Inequality 
\renewcommand{\thesection}{\arabic{chapter}.\arabic{section}} 
\renewcommand{\theequation}{\arabic{chapter}.\arabic{section}.\arabic{equation}} 
\renewcommand{\thetheorem}{\arabic{chapter}.\arabic{section}.\arabic{theorem}} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
 
 
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