q$.
{\bf Remark 2:} Instead of using the results of [L90] one might approach
Theorem 1.1 a la Gross [G76],i.e., trying to reduce the problem to an integral
over infinitesimal time steps,
and estimating the semigroup over these time steps by employing
logarithmic Sobolev inequalities. It is clear that one would get some
smoothing estimates but can one obtain them in the sharp form?
That there are some obstructions to reach this goal by this method
can be seen as follows.
Eqs.~(1.5) and (1.7) determine the Gaussian function $u_0$
that yields the norm
of the magnetic heat kernel as an operator from $L^p$ to $L^q$. With this
Gaussian we may write
%
$$
C_0(t;p,q) = {\|u_0(t)\|_q\over \|u_0\|_p} =
{\|u_0(s)\|_r\over \|u_0\|_p}\, {\|u_0(t)\|_q\over \|u_0(s)\|_r}\ ,
$$
%
for some $r$ and $s$ with $p\le r\le q$, $0\le s\le t$. Hence it is obvious that
%
$$
C_0(t;p,q) \le C_0(s;p,r)\,C_0(t-s,r,q)\ .
$$
%
Clearly, the proposed method can only work, if there exists a number
$r$ with $p 0$ and $1\le p~~0$ is continuous. Then the estimate
%
$$
C(t;p,q) = \|e^{tH(B)}\|_{L^p\to L^q} \le C_0(t;p,q)
$$
%
(with $C_0(t;p,q)$ given by Eq.~(1.8)) holds for
\smallskip\noindent
{\bf a}) all $t>0$, if $p\le 2$, $q\ge 2$,
\smallskip\noindent
{\bf b}) $t>0$ and $2~~0$ is continuous. Then the magnetic heat
kernel satisfies
the bound
%
$$
|e^{tH}(x,y)| \leq {B_0 \over 4 \pi \sinh({B_0 t \over 2})} e^{-{(x-y)^2
\over 2t}} \ .
\eqno(1.14)
$$
%
}
\medskip
{\bf Remark:} The Gaussian decay on the right side of (1.14) is the one of
the heat
kernel, which is considerably weaker than the decay of the Mehler kernel.
Is it true that $|e^{tH}(x,y)|$ is bounded by the Mehler kernel $|G(t,x,y)|$?
The truth of this estimate would reveal a robust dependence of the magnetic
heat kernel on the magnetic field. This is an open problem.
\medskip
{\bf Acknowledgment:} We would like to thank Eric Carlen, Laszlo Erd\"os, and Vitali Vugalter
for many helpful discussions.
\bigskip\bigskip\bigskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent{\bf 2. PROOF OF THEOREM 1.1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\medskip
\noindent
Let $G_t$ be the integral operator with the Mehler kernel $G(t,x,y)$.
We want to determine
%
$$
\sup_u {\| G_t u\|_q \over \|u\|_p} = C_0(t;p,q),
$$
%
which is either finite or infinite.
By a result of E.~Lieb (see [L90], Theorem~4.1) it is sufficient to take
the supremum
over centered Gaussian functions of the form
%
$
\exp( - x\cdot J_\ve x )
$,
%
where $J_\ve$ is a complex-valued matrix, symmetric with
respect to the scalar product in ${\Bbb R}^2$, and with a strictly positive
real part.
Even for this case the calculation is still somewhat tedious.
Instead it is convenient to rework the proof of Theorem~4.1 in [L90].
Let $t>0$ be arbitrary.
We approximate $G_t$ by the operator $G_t^\ve$ with kernel
%
$$
G_t^\ve (x,y) = {\rm e}^{-\ve x^2}\,G(t,x,y),
$$
%
which for $\ve>0$ is a non-degenerate, centered Gaussian kernel.
According to [L90], Theorem~3.4, there is a unique (up to a multiplicative
constant)
centered Gaussian function $u_\ve$ which is the maximum of
$\| G_t^\ve u\|_q/\|u\|_p$ over all $u\in L^p({\Bbb R}^2)$.
The function $u_\ve$ is of the form
%
$
u_\ve(x) = \exp( - x\cdot J_\ve x )
$,
%
where $J_\ve$ is a (possibly complex-valued) matrix which is symmetric with
respect to the scalar product in ${\Bbb R}^2$ and has a strictly positive
real part.
But since the integral operator $G_t^\ve$ commutes with
rotations,
the unique maximum $u_\epsilon$ must also be rotationally invariant.
Hence $J_\ve= \alpha_\ve {\bf 1} + {\rm i}\beta_\ve {\bf 1}$,
where $\alpha_\ve>0$, and $\beta_\ve$ is real.
Since the integrals of Gaussian functions can be evaluated explicitly,
we can evaluate the quotient $\|G_t^\ve u\|_q/\|u\|_p$ for
$u(x)=\exp[-(\alpha + \rmi \beta)\,x^2]$
and maximize this expression over all $\alpha$ and $\beta$.
The maximum is obtained for $(\alpha,\beta)=(\alpha_\ve,\beta_\ve)$,
with $\alpha_\ve>0$ and $\beta_\ve = 0$.
Since $\exp(-\ve x^2)\le 1$ we find for any Gaussian function $u$
%
$$
\|G_t^\ve u\|_q \le \|G_tu\|_q
$$
%
and hence
%
$$
C^\ve={\|G_t^\ve u_\ve\|_q\over \|u_\ve\|_p} \le C_0(t;p,q),
$$
%
where $u_\ve = \exp(-\alpha_\ve x^2)$ is the unique Gaussian maximizer for
$G_t^\ve$.
By an explicit calculation one sees that
%
$$
\lim_{\ve \to 0} C^\ve \equiv C^0 = {\| G_t u_0 \|_q \over \|u_0\|_p},
$$
%
with $u_0(x) = \exp(-\alpha x^2)$, $\alpha = \lim_{\ve\to 0} \alpha_\ve$.
Of course, we have $C^0\le C_0(t;p,q)$.
Finally, for any Gaussian function $u$, the limit
%
$$
\lim_{\ve \to 0 } {\|G_t^\ve u\|_q\over \|u\|_p} = {\| G_tu\|_q\over \|u\|_p}
$$
%
exists by Lebesgue's dominated convergence, and from
%
$$
{\|G_t^\ve u\|_q\over \|u\|_p} \le C^\ve \quad \hbox{for all $\ve > 0$}
$$
%
we find immediately that
%
$$
{\| G_t u\|_q\over \|u\|_p} \le \lim_{\ve \to 0} C^\ve = C^0
\quad\hbox{for all Gaussian functions $u$}.
$$
%
Hence, also $\sup \|G_t u\|_q/\|u\|_p = C_0(t;p,q) \le C^0$.
This proves $C^0 = C_0(t;p,q)$.
A little calculation easily gives the explicit value of this constant.
\vfill\eject
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent{\bf 3. A DIFFERENTIAL INEQUALITY}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\medskip
\noindent In the following we assume that the magnetic field $B$ is given by a
differentiable
vector field $A$ as in Eq.~(1.1) and satisfies
$B(x) \ge B_0 > 0$, for all $x \in {\Bbb R}^2$. In particular we choose the
vector
potential to be in $L^4_{\rm loc}(\Bbb R^2)$ and then by the
Leinfelder--Simader Theorem
[LS81] the formal expression (1.1) defines a selfadjoint operator on some
domain
$\Dom(H)$ with
$C^{\infty}_0(\Bbb R^2)$ as a core.
If we set $u(s)=e^{Hs}u_0$
then $u(s)$ is a solution of
%
$$
{d\over ds}\,u = H\,u={1 \over 2}(\nabla + \rmi A)^2\,u \ .
\eqno(3.1)
$$
%
If $u_0 \in \Dom(H) \cap L^1(\Bbb R^2) \cap L^{\infty}(\Bbb R^2)$ then we
have that
$u(s)\in \Dom(H) \cap L^1(\Bbb R^2) \cap L^{\infty}(\Bbb R^2)$.
This follows from the diamagnetic inequality (1.2)
for each $s\in (0,t]$ and the explicit form of the heat kernel.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\medskip
\noindent
{\bf Theorem 3.1: }
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent
{\it Let $r:(0,t]\to {\Bbb R}$ be a twice differentiable function with
$r(s) \geq 2$ and $\dot r(s) \ge 0$ for all $s\in (0,t]$.
Then
%
$$
{d\over ds}\,\ln \|u(s)\|_{r(s)} \le - L(r(s),\dot r(s)) \ ,
\eqno(3.2)
$$
%
for all $s\in (0,t]$, where
%
$$
L(r, \dot r) = {\dot r\over r^2}\,\left\{
2+\ln{4\pi a(r,\dot r)\over B_0 r}
\right\} + {a(r,\dot r) B_0\over r} \ .
\eqno(3.3)
$$
%
and
%
$$
a(r,\dot r) = {1\over B_0 r}\bigl( \sqrt{\dot r^2 + B_0^2 r^2 (r-1)}
- \dot r \bigr).
$$
%
Equality holds in Eq~(3.2) for $B(x) = B_0$ and
%
$$
u(s,x) = \exp\left( - {B_0\over 4 a\bigl(r(s),\dot r(s)\bigr)} \,x^2 \right) \ .
$$
%
}
\medskip\noindent
{\it Proof\/}: Pick $u_0 \in \Dom(H) \cap L^1(\Bbb R^2) \cap L^{\infty}(\Bbb R^2)$.
Since ${d\over d\tau}\,\ln k\|u(\tau)\|_{r(\tau)}$ does not depend on $k>0$, we
assume during the following calculation, without loss of generality, that
at the time
$\tau=s>0$ the solution is normalized such that
$\|u(s)\|_{r(s)} = 1$. For the derivative at $\tau=s$ we obtain therefore
%
$$
\eqalign{{d\over ds}\,\ln \|u(s)\|_{r(s)}
= & \,{\dot r(s)\over r(s)^2}\int | u(s,x) |^{r(s)}
\ln | u(s,x) |^{r(s)}\,d^2\!x \cr
& + {1 \over 2}\int |u(s,x)|^{(r(s)-2)}\, {d\over ds}|u(s,x)|^2 \,d^2\!x \ .\cr}
\eqno(3.4)
$$
%
The integrals have to be taken over ${\Bbb R}^2$.
The formal computation can be easily justified by an approximation argument.
For simplicity, the arguments $s$ and $x$ in the integrand on the right side
will be omitted from now on.
Using (3.1) we obtain, after a partial integration
%
$$
\eqalign{{1\over 2}\int |u|^{(r-2)}\,{d\over ds} |u|^2
= & -{1 \over 2}(r-2)\int|u|^{(r-2)}\,(\nabla |u|)^2\cr
& -{1\over 2}\int |u|^{(r-2)}\,|(\nabla + \rmi A)\,u|^2 \ .}
\eqno(3.5)
$$
%
The integration by parts can be justified as follows. Since $u \in
\Dom(H)$, by the Leinfelder--Simader Theorem [LS81] we can pick a sequence
$u_n \in C^{\infty}_0
({\Bbb R}^2)$ such that $u_n \to u, Hu_n \to Hu$ in $L^2$. Inspecting
the proof of the Leinfelder--Simader Theorem one sees that the sequence
$u_n$ can be chosen to converge to $u$ in $L^1$ and to have a uniform bound
on the $L^{\infty}$ norm.
Thus $u_n$ converges to $u$ in $L^p$ for all $1\leq p <\infty$.
In particular, all the following computations can be justified in the same
fashion and we can assume without restriction that
$u \in C^{\infty}_0({\Bbb R}^2)$.
If we set $u=f+\rmi g$ then $|u|= \sqrt {f^2+g^2}$.
We find
%
$$
|(\nabla + \rmi A)\,u|^2= (\nabla |u|)^2 + |A+\nabla S|^2\,|u|^2 = X^2+Y^2
\eqno(3.6)
$$
%
where we have introduced two
real vector fields $X$ and $Y$ over ${\Bbb R}^2$,
%
$$
X = \nabla^\bot |u| =
\Bigl({\partial\over \partial x_2},-{\partial\over \partial x_1}\Bigr)|u|,
$$
%
$$
Y = (A+\nabla S)\,|u|.
$$
%
Here the symbol $\nabla S$ denotes the expression
%
$$
{f \nabla g - g \nabla f \over f^2+g^2} ,
$$
%
which is defined wherever $f^2+g^2 > 0$.
For any $c>0$ we may estimate
%
$$
c^2X^2 + Y^2 \ge 2c \, X \cdot Y
\eqno(3.7)
$$
%
with equality if and only if $cX=Y$.
If $B=B_0$ is a constant magnetic field we can choose a Gaussian
function
%
$$
u=N\,\exp\left\{-{B_0\over 4c}\,x^2\right\}
\eqno(3.8)
$$
%
such that equality holds everywhere in (3.7), because
(with $A$ as in Eq.~(1.2))
%
$$
cX = c\,\nabla^\bot u = {B_0\over 2}\,(-x_2,x_1)\,u = Au = Y \ .
$$
%
Let us now insert (3.6) and (3.7) into (3.5).
Since $r-1>0$ we can add and subtract a positive constant $c^2$ with
$0