数理解析研究所講究録
1009 巻 1997 年 162-180

162

Semigroup Properties by
Gaussian Estimatesby

W. Arendt
May 2, 1997

$0$Introduction

It is a most interesting phenomenon that, in many cases, solutions of parabolic
equations are given by kernels (so-called heat kernels) which can be estimated by
the Gaussian kernel. Investigations in this direction go back at least to the sixties.
We refer to the monographs Davies [Dal], Robinson [R] and Varopoulos, SaloffCoste, Coulhon [VSC] for the history and more information. Besides their intrinsic
interest there are many different motivations to establish Gaussian estimates: they
concern H\"older continuity of the solutions, Harnack’s, the construction ofstochastic processes among others.
$0\dot{\mathrm{f}}$$\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{i}\dot{\mathrm{t}}\mathrm{y}$But more recently, it turned out that Gaussian estimates imply very important andinteresting properties of the underlying semigroup. In this article we give an accounton such implications. They concern spectral properties, regularity of the Cauchy
problem and functional calculus. One of the most important results is that Gaussian
estimates for a holomorphic semigroup on
imply holomorphy of the semigroup in. Whereas the other parts of the paper are surveys, we include acomplete simple proof of this fact in Section 4 and 5 (which is somehow dispersedin the literature, otherwise).
$L^{2}(\Omega)$$L^{1}(\Omega)$

It is striking that Gaussian estimates can be proved by quadratic form methods
for elliptic operators with various boundary conditions under very mild regularity
assumptions on the coefficients and the underlying domain. We give an accounton such results in the second section adding some elementary special examples, as
illustrations. Thus, as a concrete result, this paper contains aexamples$0.\mathrm{f}$$\mathrm{v}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{e}_{k}\mathrm{t}\mathrm{y}$.

163of holomorphic semigroups on
generated by symmetric and non-symmetric
elliptic operators with boundary conditions.
$L^{1}(\Omega)$

The author is grateful to Professor Sawashima, Professor Takeo, Professor Okazawa
and Professor Miyajima for many stimulating discussions and a most enjoyable andfruitful stay in Japan.

Gaussian estimates for semigroups.

1Let

be an open set, $1\leq p<\infty$ . We consider
as a subspace ofidentifying a function $f\in L^{p}(\Omega)$ with its extension to
by . The
Gaussian semigroup on
is denoted by $G=(G(t))_{t}\geq 0$ where$\Omega\subset R^{d}$$L^{p}(\Omega)$$L^{p}(R^{d})$$R^{d}$$0$$L^{p}(R^{d})$$(G(t)f)(_{X)=}(4 \pi t)-d/2\int_{R^{d}}e-|x-y|2/4tf(y)dy$, $1\leq p<\infty$ . The generator of.. :.
with distributional domain$(x\in R^{d} , t>0 , f\in L^{p}(R^{d}))$$\Delta$on$L^{p}(R^{d})$$D(\Delta)=\{f\in Lp(R^{d}):\Delta f\in Lp(R^{d})\}$

Definition 1.1 A-semigroup on
there exist constants $c>0,$ $b>0$ such that
$C_{0}$$L^{2}(\Omega)sa..tisfieS\backslash .\cdot$(1.1)

is the$G$$\mathrm{L}\mathrm{a}\mathrm{p}\mathrm{l},\mathrm{a}\mathrm{C}\mathrm{i}\mathrm{a}\mathrm{n}$$\backslash$$|T(t)f|\leq cG(bt)|f|$

$a$

.

an

$\mathrm{G}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{s}\mathrm{i}$$\backslash \cdot..$..

$\cdot$$\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{e}:$i.f$(0<t\leq 1)$

for all $f\in L^{2}(\Omega)$ .

Here the inequality (1.1) is understood. onestimate (1.1), then it is easy to see [Ar, Sec. 4] that$\mathrm{a}.\mathrm{e}$(1.2)

for all$|T(t)f|\leq ce^{\omega t}G(bt)|f|$$f\in L^{2}(\Omega)$$\Omega.$

If.

$T$

satisfies the Gaussian$(t\geq 0)$

where $\omega=\ln c$ .

Next we explain how (1.2) is related to a representation of $T(t)$ by an integral
kernel. Let. Then$K\in L^{\infty}(\Omega\cross\Omega)$(1.3)$(B_{K}f)(X)= \int_{\Omega}K(x, y)f(y)dy$

defines an operator. Usually the following propositionis called the Dunford-Pettis criterion. We refer to [Ar Bu] for a short andelementary proof.
$B_{K}\in \mathcal{L}(L^{1}(\Omega), L\infty(\Omega))$

164
The mapping $K\mapsto B_{K}$ is an isometric.isomorphism from
. $M_{oreove}r_{J}$ this mapping is an order
onto$i.e.,$ $B_{K}\geq 0$ (i.e., $f\geq 0\Rightarrow B_{K}f\geq 0$ ) if and only if $K(x, y)\geq 0a.e$ .
$\mathrm{p}_{\mathrm{r}\mathrm{o}_{\mathrm{P}^{\mathrm{O}}}}\mathrm{S}\mathrm{i}\mathrm{t}i‘ \mathrm{o}\mathrm{n}1.2$$iSomor_{\mathrm{P}^{him_{i}}}s$$\mathcal{L}(L^{1}(\Omega), L\infty(\Omega))$$\Omega)$$\mathrm{L}\mathrm{e}\mathrm{t}\prime B\in \mathcal{L}(L^{p}(\Omega))$

(1.4)

,

$1\leq p\leq\infty$. We set$||B||c_{\langle L^{\infty}}L^{1},)= \sup\{||Bf||L^{\infty} :If)a unique kernel$||B||_{\mathcal{L}}\langle L^{1},L\infty$$<\infty$.

f\in L^{1_{\cap}}L^{p}, ||f||_{L^{1}}\leq 1\}$, it follows from the Dunford-Pettis
such that $Bf=B_{K}f$ for all$\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{i}_{0}\mathrm{n}$$K\in L^{\infty}(\Omega\cross\Omega)$(1.5)$L^{\infty}(\Omega\cross$$l||I\grave{\dot{C}}||l\mathrm{i}\infty \mathrm{t}^{\Omega \mathrm{X}}\Omega)=||BK||_{\mathcal{L}(}L^{1},L\infty)$that there exists
and$f\in L^{p}\cap L^{\infty}$.

is a seis represented by the kernel K. If
We say that$|B_{0}f|\leq
B|f|(f\inL^{\mathrm{p}}(\Omega))$, then it follows thatiscond operator such that$|K_{0}(x,y)|\leq
I\mathrm{f}(x,y)a.e$.
As
such that
represented by a kernel
an immediate consequence we obtain an alternative more concrete description ofGaussian estimates.
$B$$B_{0}\in \mathcal{L}(L^{p}(\Omega))$$B_{0}$$I\mathrm{f}_{0}\in L^{\infty}(\Omega\cross\Omega)$on-semigroup
Corollary 1.3 A
and only if $T(t)$ is represe. $nte.db.y$ a kernel$T$$C_{0}$

(1.6)

$d/2$
$\cdot-,-|I,.C.(t, x, y,)|.\leq ce^{\omega t}t^{-}$
$-.$

for

.

$a\dot{l}lt>0$, where

$c,$

$b>0$

are

$L^{2}(\Omega)$satisfies a Gaussian estimate if$K(t, \cdot,.\cdot)\in L^{\infty}(\Omega\cross\Omega)$econstrants.

x.

such that$\mathrm{p}(-b|x-y|2/t)\backslash \cdot$

satisfies a Gaussian esti..mate, it can be extended to $L^{p}(\Omega)(1\leq p\leq\infty)$ inIf
the following way.
$T$satisfying a Gaussian-semigroup on
Proposition 1.4 Let
be aon-semigroup
estimate. Then for $1\leq p<\infty$ there exists a uniquesuch thatonand a unique adjoint semigroup$T$$L^{2}(\Omega)$$C_{0}$$C_{0}$$a)T_{2}(t)=T(t)$$T_{p}$$L^{p}(\Omega)$$L^{\infty}(\Omega)$$\tau_{\infty}$$(t\geq 0)_{i}$$b)T_{p}(t)f=T_{q}(t)f$if$f\in L^{p}(\Omega)\cap L^{q}(\Omega),$

Here we say thataon-semigroupare called the extensions of$1\leq p,$ $q\leq\infty,$$t>0$ .

is an adjoint semigroup if there exists
such that $R(t)^{*}=S(t)$ . The semigroups$S=(S(t)_{t\geq 0}\subset \mathcal{L}(L^{\infty}(\Omega))$$C_{0}$$(R(t))_{t}\geq 0$$L^{1}(\Omega)$

$T$

to

$T_{p}$$L^{p}(\Omega)$, $1\leq p\leq\infty$ .

such that
Proof. It is clear from domination that there exists$f\inL^{2}\capL^{p}$$1\leqp<\infty$$T_{p}(t)=T(t)f$ for,satisfies the semigroupandproperty. Strong continuity demands an additional argument (see [Ar] or [AE]).
can be defined as follows. Also he adjoint $\tau*=(T(t)*)t\geq 0$
The semigroupascan be defined as before. Definesatisfies a Gaussian estimate and so
the adjoint semigroup of$T_{p}(t)\in \mathcal{L}(L^{p}(\Omega))$$T_{p}$$\dot{\mathrm{t}}$$\tau_{\infty}$$\tau_{1}*$$\tau_{\infty}$$\blacksquare$$\tau_{1}*$

165

2

Examples.

We describe elliptic operators associated to sesquilinear forms. Let
an open set,$\Omega\subset R^{d}$be$H^{1}( \Omega)=\{u\in L2(\Omega):D_{j}u=\frac{\partial u}{\partial x_{i}}\in L^{2}(\Omega), j=1\cdots d\}$the first Sobolev space with norm$||u||_{H^{1}} \langle\Omega)=(||u||_{L}2)2\langle\Omega+\Omega\int|\nabla u|^{2}\mathrm{d}\mathrm{x})\frac{1}{2}$Letbe a closed subspace of
which is dense in. Let $a:V\cross Varrow G$
be a continuous sesquilinear form. Assume thatis coercive, i.e.,$V$$H^{1}(\Omega)$$L^{2}(\Omega)$

$a$

(2.1)

${\rm Re} a(u, u)+\lambda_{0}(u|u)_{L(\Omega}2)\geq\alpha||u||_{H^{1}}^{2}$for all $u\in V$ and someby$\lambda_{0}\in R,$$\alpha>0$. Define the operator

$A$

..

..
$\cdot$$D(A)=\{u\in V:\exists v\in L^{2}(\Omega)$

for all$\varphi\in V$}, $Au=v$ .

such thatassociated with.

$a$$a(u,\varphi)=(v|\varphi)_{L^{2}}$$\sim$

Then $-A$ generates a holomorphic-semigrouponthe semigroup associated with a. In the following, we giveof such semigroups which admit a Gaussian estimate.
$(e^{-tA})_{t\geq 0}$$C_{0}$$L^{2}(\Omega)$a.series. We call itof examples

First of all, we notice that the Gaussian semigroup itself is associated with the form
on $V=H^{1}(R^{d})$ given by.

$a$$a(u, v)= \int_{R^{d}}\nabla u\cdot\overline{\nabla v}\mathrm{d}\mathrm{X}$

Example 2.1 (The Dirichlet Laplacian). Let
space,of all test functions inassociated withis denoted byThensatisfying$H^{1}(\Omega))$$C_{\mathrm{c}}^{\infty}(\Omega)$$a$$(e^{t\Delta_{\Omega}^{D}})_{t\geq}0$$-\Delta_{\Omega}^{D}$(the closure of the. The operator
is represented by a kernel$V=H_{0}^{1}(\Omega)$$a(u, v)= \int\nabla u\overline{\nabla v}$$k(t, \cdot, \cdot)$(2.2)$0\leq k(t, x,y)\leq(4\pi t)^{-d/}2e-|x-y|2/4t$$(t>0, x, y\in R^{d})$.

This is well-known (cf. [AB 1], [AB 2]).
This estimate
very
since iswithants $c=b=1$ , $\omega=0$ in (1.2).
$\mathrm{i}\dot{\mathrm{t}}$$(2.2).\mathrm{i}\mathrm{S}-$$\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\dot{\mathrm{t}}$$\mathrm{s},\mathrm{p}\mathrm{e}\mathrm{c}’ \mathrm{i}\mathrm{a}\mathrm{l}$$\mathrm{a}.$.

directdomination..b the Gauss kernel$\mathrm{y}$166
Example 2.2 (Elliptic operator with constant coefficients). Let
be a strictly positive definit matrix and define the operator. .,by$(c_{ij})_{i,j=}1\cdots d$

$C=$
$A$

on

$L^{2}(R^{d})$$A=- \sum_{=i,,j1}^{d}cijD_{i}Dj$

Thus

$A$

is associated with the form$\sum_{i,j=1}^{d}c_{j}\dot{.}\int\nabla u\overline{\nabla v}R^{d}\mathrm{d}\mathrm{x}.$kernel of$e^{-tA}$D.enote by$v_{1},$.

defined on$a$$\cdots v_{d}.>0$$V=H^{1}(R^{d})$the eigenvalues ofby$C$$a(u, v)=$. Then theis given by$k(t, x, y)=(4 \pi t)-d/2(v1\ldots v_{d})-\frac{1}{2}\exp(-(C-1(x-y)|(x-y))/4t)$.

So we have the estimate$0\leq k(t, x, y)\leq(4\pi t)-d/2(v1\ldots v_{d})^{-}1/2\exp(-b|x-y|2/4t)$where$b=( \min_{k=1\cdots d}vk)^{-1}$

As an illustration we next consider the simplest example of a non-symmetric operator. Here we can also determine the kernel explicitly.
Example 2.3 Let $Af=-f”+f’$ on $L^{2}(R)$ ; i.e.
is associated with the form
, $V=H^{1}(R)$ . Then $e^{-tA}=G(t)S(t)$ where
is
$A$

$G$

$a(u, v)= \int_{R}u’\overline{v}’\mathrm{d}\mathrm{X}+\int u’\overline{v}\mathrm{d}\mathrm{X}$the Gaussian semigroup and $(S(t)f)(x)=f(x-t)$ . Hencethe kernel$k(t, x,y)$$e^{-tA}$is represented by$=$$(4 \pi t)-\frac{1}{2}\exp(-(x-y-t)2/4t)$$=$$(4Tt)^{-\frac{1}{2}}\exp(-(x-y)^{2}/4t-t/4+(x-y)/2)$.

By the old Young’s trick we have$\frac{1}{2}(x-y)=\frac{1}{4}2((x-y)\cdot\epsilon^{\frac{1}{2}}t-\frac{1}{2})\cdot(\epsilon^{-\frac{1}{2}}t\frac{1}{2})\leq\frac{1}{4}\dagger.(x-y)^{2}t-1\epsilon+t\epsilon^{-1}\}$

Choosing$\epsilon=\frac{1}{2}$one obtains$0 \leq k(t, x, y)\leq e^{t/4}(4Tt)-\frac{1}{2}\exp(-(x-y)2/8t)$

Example 2.4 (the Neumann Laplacian). Let $V=H^{1}(\Omega)$ ,We denote the operator associated with

$a$

by

.
$a(u, v)– \int_{\Omega}\nabla u\overline{\nabla v}$$-\Delta_{\Omega}^{N}$a) Assume thathas a Lipschitz boundary. Then the semigroupsatisfies a Gaussian estimate, (see [Dal]) or Theorem 2.6 below). Here, in
general, we cannot take $\omega=0$ in the estimate (1.6). In fact, assume thatis bounded. Then$\Omega$$(e^{t\Delta_{\Omega)_{t\geq}}^{N}}0$$\Omega$$e^{t\Delta_{\Omega}^{N}}1_{\Omega}=1_{\Omega}(t\geq 0)$

whereas$\lim_{tarrow\infty}||G(t)1_{\Omega}||_{L^{2}}=0$

.

dx

167
b) Some regularity property ofFor example, if $d=1$ andcompact since$\Omega$is needed in order to have Gaussian estimates.
. Thenis not$\Omega=(0,1)\backslash \{\frac{1}{n} :$e^{t\Delta_{\Omega}^{N}}1_{(\frac{1}{n+1}},$$\frac{1}{n})=1_{(\frac{1}{n+1},\frac{1}{n})}$n\in N\}$$e^{t\Delta_{\Omega}^{N}}$for all $n\in N$

so that the eigenvalue 1 has infinite multiplicity (cf. Proposition 3.1).

After t-hese first examples we formulate some general results on elliptic operators.

,
Letellipticity condition$a_{ij}$(2.3)$b_{j},$$c_{j},$$c_{0}\in L^{\infty}(\Omega)$, $i,j\in\{1, \cdots, d\}$ . We assume throughout thefor all $\xi\in R^{d}x-a.e$ . , where $\mu>0$ .

$\sum_{i,j=1}^{d}.aij(X)\xi_{i}\xi j\geq\mu|\xi|^{2}$

Moreover, we assume throughout that the second order coefficients are real-valued(cf. Remark 2.7). The first order coefficients may be complex in some cases. As. Define
which contains
ofbefore we consider a closed subspace$a:V\cross Varrow C$ by$V$$a(u, v)$$=$

$H^{1}(\Omega)$$H_{0}^{1}(\Omega)$$\sum_{i,j=1}^{d}\int_{\Omega}aii^{D_{i}}u\overline{Djv}+\sum_{j=1}^{d}\int\Omega bjD_{j}u\overline{v}+$(2.4)$\sum_{j=1}^{d}\int_{\Omega}c_{j}u\overline{v}+\int c0\Omega u\overline{v}$.

the operator
is continuous, sesquilinear and coercive. We note by
Theni.e., $V=$
conditions;boundary
Dirichlet
weconsider
First
with
.
associated$A_{V}$$a$$V$$H_{0}^{1}(\Omega)$.

Theorem 2.5 Let, $j=1,$in$W^{1,\infty}(\Omega)$$a)$$a_{ij}\in W^{1,\infty}(\Omega)$$b)$the coefficients

Then the semigroup$\cdots,$$d$$i,j=1,$$b_{i},$$c_{j}$. Assume that the first order coefficientsareisconditions
following
the
that
one
Assume.
satisfiedof$V=H_{0}^{1}(\Omega)$$\cdots,$$d$$b_{j},$$c_{j}$orare real-valued.

$(e^{-tA_{V}})_{t\geq 0}$

satisfies a Gaussian estimate.

Notice that the semigroup is positive in the second case [Ou3]. For more general
the following holds.

Theorem 2.6 Assume that all coefficients are real andwe assume the following hypotheses:. For$1,$ $\cdots d)$$a)v\in V$$V$

implies$|v|$, $\inf\{|v|, 1\}\in V$ ,$b)v\in V$ , $u\in H^{1}(\Omega),$$|u|\leq v$

implies $u\in V$ ,$b_{j},$$c_{j}\in W^{1,\infty}(\Omega)$

$V$

$(j=$

168
$c)v\in V$

impliesIn the case, where
dary. Then$b_{j}v,$$c_{j}v\in H_{0}^{1}(\Omega)$$(j=1\cdot\forall\cdot d)$.

has Lipschitz bounwe assume in addition thatis a positive semigroup satisfying Gaussian estimates.
$\Omega$$V\neq H_{0}^{1}(\Omega)$$(e^{-tA_{V}})_{t\geq,\backslash }0$which can be
has a Lipschitz boundary
For example, assume that:
where
are closed. Choose
written as
then the conditions a), b),: If, this corresponds toc) are satisfied. In the case of the Laplacian and if.
and Neumann boundary conditions on
Dirichlet boundary conditions on$\partial\Omega--\Gamma$$\Omega$$\Gamma=\Gamma_{1}\cup\Gamma_{2}$$\Gamma_{1}\}^{-H^{1}()}\Omega$$\Gamma_{1},$$c_{j}=\{\varphi|_{\Omega} :$b_{j},$$V=\{u_{\Omega}$$\Gamma_{2}$$u\in C_{c}^{\infty}(R^{d}\backslash$\varphi\in C_{c}(R^{d}\backslash \Gamma_{2})\}-W^{\mathrm{r}},\infty(\Omega)$$\Gamma_{1}\cap\Gamma_{2}=\emptyset$$\Gamma_{2}$$\Gamma_{1}$Remark 2.7 Complex second order coefficients are a delicate matter. A counfor $d\geq 5$ is given by Auscher-Tchamitchian [AT]. However,terexample onandalso in the complex case, seeGaussian estimates can be proved on-Tchamitchian [AMT].
$R^{d}$$R^{2}$$R^{1}$$\mathrm{A}\mathrm{u}\mathrm{s}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{r}-\mathrm{M}_{\mathrm{C}\mathrm{I}}\mathrm{n}\mathrm{t}_{\mathrm{o}\mathrm{S}\mathrm{h}}$For symmetric elliptic operators (i.e., $b_{j}=c_{i}=0,$ $a_{ij}=a_{ji}$ ) and $V=H_{0}^{1}(\Omega)$, Theorem 2.5 and Theorem 2.6 are proved in [Dal]; for the generalor
non-symmetric case presented here see [AE] where also more general boundary
conditions (e.g., Robin’s boundary conditions) are given. All proofs are based on, Proposition 3.3] or [Dal]) which reduces the proof to an
Davies’ trick (seeby a similarity transfor-estimate of a semigroup obtained from-estimate , andmation. However, there are various ways to prove thethey are the reason of the different hypotheses in Theorem 2.5 and 2.6. The proof of
the first given in [AE] is the simplest one, based only on a non-symmetric BeurlingDeny criterion established by Ouhabaz ([Ou 1], [Ou 3]). It has the disadvantage
to need a regularity hypothesis on the second order coefficients but allows complex
first order coefficients. Davies [Dal] uses logarithmic Sobolev inequalites.
$H^{1}(\Omega)$$[\mathrm{A}\mathrm{E}$$e^{-tA}$$\mathcal{L}(L^{1}, L^{\infty})$$\mathcal{L}(L^{1}, L^{\infty})$

As.. a last exaniple we consider Schr\"odinger operators.
where
Example 2.8 (Schr\"odinger semigroups). Let $A=\Delta-V$ on$R^{d}arrowR$.
is in Kato’s class andis measurable such that:
$C_{0^{-}}semi-$agenerates
operatorisaself-adjoint(with suitable domain)
Then
on
group
. Then by [Si, Prop. .B. 6.7] satisfies a Gaussian estimate.
$L^{2}(R^{d})$$V$$V_{+}\in L_{l_{oC}}^{1}(R^{d})$$V_{-}$$whic\dot{h}$

$A$

$T$

$T$

$L^{2}(R^{d})$Remark 2.9 Here we restricted our attention to the most common class of po-theory of Schr\"odinger operators with more general
tentials. Concerning thepotentials we refer to the recent article by Okazawa [Ok].
$L^{p}$

Consequences of Gaussian estimates.

3

be open and let
Gaussian estimate. It follows thatLet$\Omega\subset IR^{d}$$T$satisfying a-semigroup on
be a
has many remarkable properties. Most of$C_{0}$$T$$L^{2}(\Omega)$

169
them concern the solutions of the corresponding Cauchy problem, some concern
spectral properties of the generator. By
we denote the extended semigroup on
and byits generator, $1\leq p\leq\infty$ . (By definition,
where
is the generator of.
$T_{p}$$L^{p}(\Omega)$$A_{\infty}=B_{1}^{*}$$A_{p}$$(T^{*})_{1})$$B_{1}$

Regularity of the semi.group.

3.1

In this subsection, we show that Gaussian estimates imply remarkable regularity
properties of the semigroup, like norm continuity or holomorphy.

Proposition 3.1 Assume that$0,1\leq p\leq\infty$$\Omega$.

is bounded. Then

$T_{p}(t)$is compact for all $t>$

Proof. a) It follows from the kernel representation thatis Hilbert-Schmidt,$t>0$ . For $2\leq p\leq\infty$ one can factorize $T_{p}(2t)=T_{2,\infty}(t)0\tau_{2}(t)\mathrm{o}j$ where:is the injection andisseen as an operator frominto
. Henceis compact.
.
b) For $1\leq p\leq 2$ , the result follows from a) by taking adjoints.
$T_{2}(t)$$j$$L^{\infty}(\Omega)arrow L^{2}(\Omega)$$L^{2}\not\in\Omega)$$T_{2,\infty}(t)$$T_{1}(t)$$T_{p}(2i)$$L^{\infty}(\Omega)$$\blacksquare$

It follows in particular thatis norm continuous on
for $1\leq p<\infty$ .
But it seems to be unknown whether
is automatically holomorphic ifisbounded. If
is unbounded, this is not the case. Indeed, it has been shown by, properly defined [LM], is the
Voigt (oral communication) that the operator
generator of aon $L^{2}(R)$ such that $|T(t)f|\leq G(t)|f|$ $(t\geq 0)$ .
-semigroupBut
is not holomorphic and not even eventually norm continuous. We are
grateful to a clarifying discussion with F. R\"abiger on this last point. However,
if for some $p_{0}\in[1, \infty)$ ,is holomorphic, then all the semigroups
are
$(1
\leq
p<\infty)$, and this holds for arbitrary open subsets
holomorphicofIn the preceding section, we saw a large class of examples which all are holomorphicon. Recall, that we assume thoughout this section thatsatisfies a
Gaussian estimate.
$(0, \infty)$$T_{p}$$\Omega$$T_{p}$$\Omega$$\triangle+ix$$T$$C_{0}$$T$$T_{p}$$T_{p_{0}}$$\Omega$$L^{2}(\Omega)$

Theorem 3.2 If$R^{d}$$T_{2}$$T_{2}$

is holomorphic, then$T_{p}$

is holomorphic for all $p\in[1, \infty)$ .

First of all, it should be said that in the case $1<p<\infty$ , Theorem 3.2 is valid, $1\leq p<\infty$ . This is a consequence offor every semigroup which extends to
Stein’s interpolation theorem (see [Dal, Thm. 1.4.2]). But, in general, the semi(see [Dal, Thm. 4.3.6] for an example).
group may fail to be holomorphic in
The point in Theorem 3.2 is the case $p=1$ , and, for this case, applied to elliptic operators, it has several predecessors. In fact, Stewart [Stl], [St2] provedorholomorphy of the semigroup generated by elliptic operators on the spaceand for various boundary
for bounded regularbyconditions. Using Stewart’s result, Amann [Am] proved holomorphy onskillful duality arguments. These results are valid also for operators which are notin divergence form. Then it was shown by Cannarsa and Vespri [CV] that a certain$L_{p}$$L_{1}$$C(\overline{\Omega})$$C_{0}(\Omega)--\{f\in C(\overline{\Omega}):f_{|\partial\Omega}=0\}$$\Omega$$L^{1}(\Omega)$

170class of more general elliptic operators generates an analytic semigroup in, it(see also the remark at the end of Section 4). For arbitrary open subsets of
was shown in [AB 1], [AB 2] that the Laplacian with Dirichlet boundary conditions
generates a holomorphic semigroup in.
$L^{1}(R^{d})$$R^{d}$$L^{1}(\Omega)$

Now, Theorem 3.2 implies in particular that the elliptic operators considered in
Theorem 2.5, 2.6 generate holomorphic semigroups in.
$L^{1}(\Omega)$

In view of Example 2.8, it also follows that Schr\"odinger semigroups are holomorphic, $(1 \leq p\leq\infty)$ . This had been shown before by Kato [K2]. In thein allgeneral version, Theorem 3.2 is obtained in [AE] and [H2]. We give a complete proofof Theorem 3.2 in Section 4.
$L^{p}(R^{d})$3.2Spectral properties.

In the following we denote by
the connected component of the resolvent
set
which contains a right half-plan. Then the following holds.
$\rho_{\infty}(A_{p})$$\rho(A_{p})$Theorem 3.3 One has$\rho_{\infty}(A_{\mathrm{P}})=\rho\infty(A_{2})$In $parti_{C}ular_{j}$$p\in[1, \infty]$

.

if

$A$

$(1\leq p\leq\infty)$

is self-adjoint then the spectrum.
$\sigma(A_{p})$is independent ofTheorem 3.3 has been proved in [Ar] after a proof for Schr\"odinger operators had
been given by Hempel and Voigt [HV]. It implies in particular, that the spectrumof the elliptic operators given in Theorem 2.5 and 2.6 is independent of $p\in[1, \infty]$ .
This is remarkable since the spectrum of (very simple) self-adjoint degenerate elliptic operators may strongly depend on(see [Ar]).
$p$

It is interesting that Theorem 3.3 is also true if, instead of the Gaussian kernel,
estimates by more general kernels are considered. This is done by Miyajima andIshikawa [MI] where the Gaussian semigroup is replaced by the semigroup generatedby. In Section 5 we consider higher order Gaussian estimates. For symmetric operators Davies [Da 2] obtains spectral-independence
as consequence of a functiOnal, calculus designed for symmetric operators satisfying
kernel estimates.
$-(I-\Delta)^{\alpha}$$( \frac{1}{2}<\alpha\leq 1)$$L^{p}$

However, the full independence in the non-symmetric case is still open.

Problem. Is the spectrum of
not connected)?

$A_{p}$

independent of $p\in[1, \infty]$ (also if$\rho(A_{2})$

is

171
We should mention, that spectral-independence is trivial whenis bounded.
Then the resolvent ofis compact (by Proposition 3.1), $1\leq p\leq\infty$ . Thus
consists of eigenvalues only, and since
maps
into
,
the point spectrum is the same for all operators $A_{p}(1\leq p\leq\infty)$ . Some otherpartial answers involving conditions on the domain are given recently by Hieber andSchrohe [HS].
$L^{p}$$\Omega$$A_{p}$$\sigma(A_{p})$3.3$T_{p}(t)$$H^{\infty}$$L^{p}(\Omega)$$L^{\infty}(\Omega)$-functional calculus and maximal regularity.

It has been discovered recently that Gaussian estimates imply the existence ofan-functional calculus (see Duong [Du], Duong-Robinson [DR], Hieber [H1]).
More precisely, if
has an-functional calculus then the same is valid for, $1<p<\infty$ . This is a very useful result sincecalculus iseasy to obtain on a Hilbert space; e.g. contractivity of the semigroup suffices. The
following result is due to Duong and Robinson who proved it for $\Omega=R^{d}$ or moregenerally, for homogeneous spaces(see, Theorem 5.5] for the general result).
$H^{\infty}$$H^{\infty}$$A_{2}$$A_{p}$$H^{\infty}\sim \mathrm{f}\mathrm{u}\mathrm{n}\mathrm{C}\mathrm{t}\mathrm{i}\mathrm{o}..\mathrm{n}\mathrm{a}1$$\Omega$$[\mathrm{A}\mathrm{E}$

Theorem 3.4 Assume $t’ hatT$ is contractive on
and holomorphic of angle. Thenhas an.
-functional calculus for all$L^{2}(\Omega)$$-A_{p}$$\theta\in(0,\frac{\pi}{2})$$H^{\infty}(\Sigma(\nu))$$\nu>\frac{\pi}{2}-\theta$

In view of the Dore-Venni theorem-functional calculus implies maximal regularity for the non-homogeneous Cauchy problem. Very recently, Hieber and[HP] gave a direct proof of maximal regularity assuming Gaussian estimates without
passing by the Dore-Venni theorem.
$H^{\infty}$$\mathrm{P}\mathrm{r}\ddot{\mathrm{u}}\theta$

Theorem 3.5 Let $1<p,$ $q<\infty$ and assume thathas Lebesgue measure .
Assume further thatis holomorphic. Let $\tau>0$ and $f\in L^{p}((0, \tau);L^{q}(\Omega))$ .
Consider the function. Then $u\in W^{1,p}((0, \tau);L^{q}(\Omega))\cap$; $D(A_{q}))$ , where $D(A_{q})$ is considered as a Banach space with the graphnorm.
$\partial\Omega$

$0$

$T$

$u(t)= \int_{0}^{t}\tau_{q}(t-s)f(s)\mathrm{d}\mathrm{S}$$L^{p}((\mathrm{o}, \mathcal{T})$

Notice that we assume thatsatisfies a Gaussian estimate throughout this section. It seems to be an open problem whether Theorem 3.5 is valid without thisassumption. In fact, it is if $p=2$ . Let us make this point more precise.
$T$

Remark 3.6 (maximal regularity). Let
be the generator of a-semigroup
$X$
$f\in
L^{p}((0,\mathcal{T});X)$
on a Banach space. Let $1<p<\infty$ andwhere $\tau>0$ .
Then
$B$

$C_{0}$$T$$u(t)= \int_{0}^{t}T(t-S)f(s)$.

ds

is the mild solution of the Cauchy problem$(CP)$One says that$B$$\dot{u}(t)$$=$$Bu(t)$$u(0)$

$=$

$0$

$\{$

$0\leq t<\tau$.

satisfies maximal regularity if$u\in W^{1,p}((0, \tau);X)\cap L^{p}((0, \tau);D(B))$

for all $f\in L^{p}((0, \tau);X)$ .

172
is
It turns out that this property is independent of $p\in(1, \infty)$ . Assume thatshowed in 1964 that $B$ is maximal regular if $X$
holomorphic. Then deSimon
space. This result does not hold for $X=L^{1}$ but it seems to be an
is aopen problem whether it holds for $X=L^{p}$ , $1<p<\infty$ . Particular cases are
known, though. For example, Lamberton [La] proved 1987 maximal regularity if-norm for all $1\leq r\leq\infty$ . Theorem 3.5 isthe semigroup is contractive for the
another particular solution of this open problem.
$S$$[\mathrm{d}\mathrm{S}]$$\mathrm{H}\mathrm{i}\mathrm{I}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{t}$$L^{r}$

Remark 3.7 Theorem 3.5 is proved in [ , Theorem 3.1] for $\Omega=R^{d}$ (and moregenerally, for t.o.pological spaces satisfying the doubling property). As in [ , Theorem 5.5] the more general result follows from this by a direct sum argument:. Con. Since
is a null set we haveLet
and apply the result [ , Theorem 3.1] to thethe operator
sider onas a-valued:considering
anddirect sum of.
function. Then the claim follows from the result on$\mathrm{H}\mathrm{P}$$\mathrm{A}\mathrm{E}$$L^{2}(R^{d})=L^{2}(\Omega)\oplus L^{2}(\Omega_{1})$$\partial\Omega$$\Omega_{1}=R^{d}\backslash \overline{\Omega}$$L^{2}(\Omega_{1})$$\mathrm{H}\mathrm{P}$$\Delta_{\Omega_{1}}$

$A$

$f$

$\triangle_{\Omega_{1}}$$(0, \infty)arrow L^{2}(\Omega)$$L^{2}(R^{d})$$lR^{d}$$\blacksquare$

Concerning elliptic operators of order 2 (as defined in Theorem 2.5, 2.6) Lamberton’s
direct result is more suitable than the general Theorem 3.5 based on Gaussian
estimates. In fact, the following theorem hold.
be an arbitrary open set. Let
Theorem 3.8 Let
coefficients satisfying the ellipticity condition (2.9) and let
be a closed subspace ofcomplex. Let,
Assume that$\Omega\subset R^{d}$$b_{j},$$1,$ $\cdots d)$$H_{0}^{1}(\Omega)$.

$V$$c_{0}\in L^{\infty}(\Omega)$(3.1)$u\in V$implies$\inf\{1, |u|\}$where

sign $u\{$

be real$a_{ij}\in L^{\infty}(\Omega)$$u(x)/\overline{u(x)}$
$0$

if
if

$(j=$

$c_{i}\in W^{1,\infty}(\Omega)$$H^{1}.(\Omega)$conta\’ining

sign $u\in V$$u(x)\neq 0$,$u(x)=0$ .

is continuous, sesquilinear andon $V\cross V$ by (2.10). Thenandthe operator associated with. Denote bycoercive with respect to$-A_{V}$
familyathere
exists. Thengenerated byby
the semigroup onof$(t\geq 0)$$1<p<\infty$ , such that $S_{2}(t)=S(t)$andon-semigroups$(f\in L^{p}\cap L^{q}, t\geq 0)$$S_{p}(t)f=S_{q}(t)f$thefor $1<p,$ $q<\infty$ Denote by
regularity.
maximalgenerator of. Thensatisfies

Define the form

$a$

$a$

$L^{2}(\Omega)$$S$$C_{0}$$A_{V}$$a$$L^{2}(\Omega)$$S_{p}$$L^{p}(\Omega),$.

$A_{p}$$A_{p}$$S_{p}$.

It follows from [Oul] or [Ou3,is holomorphic on
Proof. The semigroup$||e^{-\omegat}S(t)f||LP\leq||f||L^{p}$for all
Theorem 4.2 and Remark 4.3] thatand all $t\geq 0$ , $1\leq p\leq\infty$ and for some $w\in R$ . Now it follows from the above
mentioned theorem by Lamberton [La, Theorem 1] that $A_{p}-w$ satisfies maximal
satisfies maximal regularity.
regularity. Then also$S$

$L^{2}(\Omega)$$f\in L^{2}\cap L^{\infty}$$A_{p}$$\blacksquare$

We mention, however, that the above proof is restricted to second order elliptic
operators. On the other hand, Theorem 3.5 is also valid if other kernel estimates
are assumed and can be applied in particular to Example 5.2 below.

173

4Proof of holomorphy by Gaussian estimates.
In this section we give a complete proof of Theorem 3.2. First of all we need tochoose holomorphic representations of the kernel (cf. [Ar-Bu, Lemma 3.4]).
Let
be open and $F$ : $Darrow L^{\infty}(S)$ holomorphic whereis a measure space. Then there exists: $D\cross Sarrow G$ such that $f(z, \cdot)=$$F(z)a.e$ . andis holomorphic for all $x\in S$ .

Lemma 4.1

$D\subset$$\mathbb{C}$$(S, \Sigma, \mu)$$f$$f(\cdot, x)$

Proof.

Let

there exist$B=B(z_{0}, r)=\{z\in \mathbb{C} , |z-z_{\mathrm{O}}|<\mathrm{r}\}$$a_{n}\in L^{\infty}(S)$such that$\sum_{n=0}^{\infty}||a_{n}||_{\infty}r^{n}<\infty$

such that$\overline{B}\subset D$. Thenand $F(z)= \sum_{n=0}^{\infty}an(z-z_{0)^{n}}$Define $h:B\cross Sarrow G$ by $h(z, x)= \sum_{n=0}^{\infty}an(X)(z-z_{0)^{n}}$. Then$h(\cdot, x)$: $Barrow C$

is holomorphic for all $x\in S$ and $h(z, \cdot)=F(z)$ in
be two. Let$B_{j}\cross
Darrow
G$,such discs such thatand let:be functions$(z\in B_{j})$$(j=1,2)$ such that $h_{j}(z, \cdot)=F(z)$ inand such thatis holomorphic on
for all $x\in S$ . If
, then $h_{1}(z, x)=h_{2}(z, x)$ for
$x\in
S$
$z\in
B_{1}\cap
B_{2},$
allby the identity theorem. Now it suffices to cover $D$ bysuch discs.
$L^{\infty}(S).$$\overline{B_{1}}\subset D$$\overline{B_{2}}\subset D$$B_{1},$ $B_{2}$$h_{j}$$L^{\infty}(S)$$h_{j}(\cdot, x)$$B_{1}\cap B_{2}\neq\emptyset$$B_{i}$$\blacksquare$

The next criterion is very convenient to prove holomorphy of vector-valued functions.
A subset $W$ ofis called
if for all $x\in X$ there exists $\varphi\in W$
such that.
$X^{*}$$\mathrm{S}\mathrm{e}_{\mathrm{P}^{\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}}}\backslash$$\varphi(x)\neq 0$

Theorem 4.2 Let $D\subset G$ be open and : $Darrow X$ be locally bounded such thatis holomorphic for all
where
is separating. Thenisholomorphic.
$f$$W\subset X^{*}$$\varphi\in W$$\varphi \mathrm{o}f$$f$

In the case where $W$ is a norming subspace (i.e. if $||x||_{W}= \sup\{|\varphi(X)|$ :$W,$
is an equivalent norm), then this theorem can be found in [Kl, .
139]. The general case, proved in [AN], is much more convenient, but also the more
special hypothesis can be checked in the situation considered in the following proof.
Next we recall the following well-known version of the Phragmen-Lindel\"of theorem[Co,. $6.4.4$].
$\varphi\in$$||\varphi||\leq 1\}$$\mathrm{p}$$\mathrm{c}\mathrm{o}\mathrm{r}$Proposition 4.3 Letand let $D=\{re^{i\theta} : r>0,0<\theta<\gamma\}$ . Let
be continuous, holomorphic onsuch that $|h(z)|\leq\alpha\exp(\beta|z|)$ $(z\in$$M$$D)$$\beta>0$ .
where
If $|h(r)|\leq M\sim$.
for all $r>0$ , then. .
$|h(z)|\leq M$ for all $z\in D$ .
$\gamma\in(0, \frac{\pi}{2})$$k:\overline{D}arrow G$$D$$\alpha,$$\leq$$r’.\backslash |h(re^{i\gamma}.\cdot)|$

The following consequence is a simplified version of [Dal, Thm. 3.4.8] which suffices
for our purposes.

174
and let:$\Sigma(\gamma):=\{re^{\alpha} : r>0 , |\alpha|<\gamma\}$such that

Lemma 4.4 Letin$a)$$|k(z)|\leq c$$b)$$|k(t)|\leq ce^{-b/t}$

where

$c,$

$b>0$

$k$

$\gamma\in(0, \frac{\pi}{2})$$\overline{\Sigma(\gamma)}arrow C$

and$(z\in\Sigma(\gamma))$$(t>0)$ ,. Then for

one has$0<\theta_{1}<\gamma$$|k(Z)|\leq c\cdot\exp(-b_{1}/|z|)$where$b_{1}= \frac{\sin(\gamma-\theta 1)}{\sin\gamma}\cdot b$

Proof. Let
(i)

(ii)
(iii)

be continuous, holomorphic$(z\in\Sigma(\theta_{1}))$.

$g(z)=k(z^{-1}) \cdot\exp\{be(\frac{\pi}{2}-\gamma).

$|g(r)| \leq ce^{-b\cdot r}\exp\{b{\rm Re} e^{i(}\frac{\pi}{2}-\gamma).

Z\frac{1}{\sin\gamma}i\}$$(r>0)$ ;r\frac{1}{\sin\gamma}\}=c$$|g(re^{i\gamma})| \leq c\cdot\exp\{b{\rm Re}(e^{i()}\frac{\pi}{2}-\gamma re^{i}\gamma)\frac{1}{\sin\gamma}\}=c$$|g(z)| \leq c\exp\{\frac{b}{\sin\gamma}|z|\}$,$z\in\Sigma(\gamma)$. Then,.

It follows from Proposition 4.3 that $|g(z)|\leq c$ for all $z\in D$ . Replacing
we see that $|g(z)|\leq c$ for all. Hence$z\mapsto\overline{g(\overline{z})}$$z=re^{i\theta}$,$|\theta|\leq\gamma$

Lemma 4.5 Let$h(x-y)$by$z\in\Sigma(\gamma)$$|k(_{Z)|}$

for

$g$

$IC$

$=$

$|g(z^{-1} \rangle\exp(-be^{i}\frac{\pi}{2}-\gamma\rangle-1\frac{1}{\sin\gamma}\iota)Z|$$\leq$$c \cdot\exp(-b/r\cdot{\rm Re} e^{i(\frac{\pi}{2}}-\gamma-\theta)\frac{1}{\sin\gamma})$$=$$c\exp(-b/r\sin(\gamma+\theta)/\sin\gamma)$$\leq$$c\exp(-b\cdot\sin(\gamma-|\theta|)/(r\cdot\sin\gamma))$.
:$\blacksquare$$\Omega\cross\Omegaarrow C$. Then$(x, y\in\Omega)$measurable, $h\in L^{1}(R^{d})$ such that $|K(x, y)|\leq$$(B_{K}f)(X)= \int K(x, y)f(y)dy$

defines a bounded operator$B_{K}$

on$L^{\mathrm{p}}(\Omega)$

and$||B_{K}||C\mathrm{t}^{L^{\mathrm{p}})}\leq||h||_{L^{1}}(R^{d})$.

This is immediate from. inequality. For the proof of Theorem 3.2 we use thefollowing terminology. Let $S=(S(t))_{t}\geq 0$ be a-semigroup on a Banach space
X. We say thatis holomorphic of angle
if
has a holomorphic
extension (still denoted by) from the sector
intowhich is boundedon the sets
for all. Then the following iseasy to show (cf. , Thm. 2.5.2]):$\mathrm{Y}_{\mathrm{o}\mathrm{u}}\mathrm{n}\mathrm{g}_{\mathrm{S}}’$$C_{0}$$S$$\{re^{i\alpha} :$[\mathrm{P}$

(4.1)

$S$

$\theta\in(0, \frac{\pi}{2}]$$S$$\Sigma(\theta)$

0<r\leq 1 , |\alpha|\leq\theta-\epsilon\}$$\mathcal{L}(X)$$\epsilon>0$$(d)\Rightarrow(a)$$S(z_{1})S(Z_{2})=^{s(}Z_{1}+z_{2})$$(Z_{1}, Z_{2}\in\Sigma(\theta))$

;

175

(4.2)

$z \in\Sigma(\theta’)\lim_{0zarrow}S(Z)_{X=}X$

for all(4.3)

for all$(x\in X)$$\theta’\in(0, \theta)$

there exist $M\geq 0,$ $w\in R$ such that$\theta’<\theta$$||S(Z)||\leq Me^{|z|\omega}$$(z\in\Sigma(\theta’))$.

Proof of Theorem 3.2 Assume thatis a holomorphic-semigroup of angle$\theta\in(0, \pi/2]$
onwhich satisfies a Gaussian estimate. Let $0<\theta_{1}<\theta$ .
Choose. Replacing by
we can assumethat$||T(Z)||C(L^{2})\leq$
(4.4)
const
;
$|T(t)f|\leq$(4.5)
const $G(bt)|f|$ $(t\geq 0 , f\in L^{2}(\Omega))$ .
$T$$C_{0}$$L^{2}(\Omega)$$\theta_{1}<\gamma<\theta_{2}<\theta_{3}<\theta$$T$$(e^{-wt}\tau(t))_{t\geq 0}$$(z\in\Sigma(\theta_{3}))$

From this follows
(4.6)
(4.7)

$||T(t)||_{C\mathrm{t})}L^{1},L^{2}\leq$const$t^{-d/4}$$(t>0)$ ;$||T(t)||c(L^{2},L^{\infty})\leq$const$t^{-d/4}$$(t>0)$ .

(In fact, one sees from the kernel and Proposition 1.2 that$(4.6)$ and (4.7) follow fromconst
. Sinceis bounded on
andthe Riesz-Thorin theorem). Choose $\delta\in(0,1)$ such thatwhenever $t+is\in\Sigma(\theta_{2})$ . Let $z=t+is\in\Sigma(\theta_{2})$ . Then by (4.4), (4.6), (4.7),$||T(t)||c(L^{1},L^{\infty)}\leq$$t^{-d/2}$$T$$L^{1}$$L^{\infty},$$\delta t+is\in\Sigma(\theta_{3})$.

$||T(z)||c(L^{1},L\infty)\leq||T((1-\delta)t/2||_{\mathcal{L}\mathrm{t}^{L,L^{2})}}1||T(\delta t+iS)||c(L^{2})$const$t^{-d/2}=$

const$||\tau((1-\delta)t/2)||\mathcal{L}(L^{1},L^{\infty})\leq$$(\mathrm{R}\mathrm{e}\mathrm{z})-d/2$

It follows from Theorem 4.2 that the mapping $z\mapsto T(z)|L^{1}$ :is holomorphic. By Lemma 4.1 there exists:$K(\cdot,x,y)$$y\in\Omega,$such thatis holomorphic for all$(f\in L^{1}\cap L^{2})$ . Moreover,$(T(z)f)(x)= \int_{\Omega}I\mathrm{f}(Z, x, y)f(y)dy$$\Sigma(\theta_{2})arrow \mathcal{L}(L^{1}, L^{\infty})=$$IC$$L^{\infty}(\Omega\cross\Omega)$$K(z\cdot, \cdot)\in L^{\infty}(\Omega\cross\Omega)$

$x,$

(4.8)

$|IC(z, x, y)|\leq$const$\Sigma(\theta_{2})\cross\Omega\cross\Omegaarrow \mathbb{C}$$(\mathrm{R}\mathrm{e}\mathrm{Z})-d/2$

and$(z\in\Sigma(\theta_{2}))$and by (4.5),(4.9)$|I\mathrm{f}(t, x,y)|\leq$

const $t^{-d/2}\exp(-b|x-y|2/t)$$(t>0)$ .

Applying Lemma 4.4 to the function $z^{d/2}K(Z, x, y)$ we obtain a constant $c>0$
such that for$b_{1}=b \cdot\frac{\sin(\gamma-\theta 1)}{\sin\gamma}$(4.10)
for all$|K(z, x, y)|\leq c\cdot|z|^{-}d/2\exp(-b_{1}|x-y|^{2}/|z|)$$z\in\overline{\Sigma}(\theta_{1})$. It follows from Lemma 4.5 that$\sup||T(Z)||c(L^{p})<\infty,$.

and all$x,$ $y\in\Omega$$1\leq p<\infty$$z\in\Sigma(\theta_{1})$Thus there exist operatorssuch that. It follows from Theorem 4.2 that
$T_{p}(z)\in \mathcal{L}(L^{p})$$L^{p}\cap L^{2})(z\in\Sigma(\theta_{1}))$$T_{p}(z)f=T(z)f$$T_{p}(\cdot)$:$(f\in$$\Sigma(\theta_{1})arrow \mathcal{L}(L^{p}(\Omega))$176
is holomorphic. This finishes the proof.

$\blacksquare$

As application we obtain that the operators described in Theorem 2.5, 2.6 gene, $1\leq p<\infty$ . In the case of
rate holomorphic semigroups in all spaces
for real
Dirichlet boundary conditions, Vespri [V] proved holomorphy on-coefficients conditions. This is more general than our result based on Theorem
2.5, 2.6 where some regularity assumption on the first order coefficients is needed.
However, Vespri’s proof is not elementary at all, using estimates due to De Giorgi
technique. In addition some regularity assumptions on theand variants ofboundary of
are needed,, for his proof.
$L^{p}(\Omega)$$L^{1}(\Omega)$$L^{\infty}$$\mathrm{S}\mathrm{t}\mathrm{e}\mathrm{W}\mathrm{a}\mathrm{r}\dot{\mathrm{t}}’ \mathrm{s}$$\Omega$Higher order Gaussian estimates and holomor-

5.

$\mathrm{p}\mathrm{h}\mathrm{y}$So far, we considered merely Gaussian estimates of order 2. However, the proof of
Theorem 3.2 carries over to higher order with small alterations. We first give theprecise definition and examples for the higher-order-case, following Hieber [H2]. Let. We had considered $m=2$ , before. Let$m\in m^{r}\backslash \{1\}$Let$k_{t}(x)$$=$$t^{-d/m}\exp(-|X|m/(m-1).

$G(t)f$$=$$k_{t}*f$t^{-1}/\langle m-1))$$(f\in L^{2}(R^{d}))$.

be open.

$\Omega\subset R^{d}$on-semigroup
Definition 5.1 A
$w\in
R,$
order
if there exist constants$T$$C_{0}$$\mathrm{m}$$L^{2}(\Omega)Sat_{\dot{i}Sfies}$$(t>0)$ ,As in the case of order 2, this implies that there exist
such that$T(t)=T_{2}(t)$b)$T_{p}(t)f=T_{q}(t)f$-semigroups$T_{p}$$(f\in L^{p}\cap L^{q} , t>0)$

Let

for

$x\in R^{d}$$L^{p}(\Omega)$.

$\rho\in(0,1)$$|\alpha|\leq m$,$a_{\alpha}\in BUC^{\rho}(R^{d}, \mathbb{C})$. Suppose that there exists

for $|\alpha|=m$ and$\delta>0$such that${\rm Re} \sum_{|\alpha|=m}a_{\alpha}(x)(i.\xi)^{\alpha}<-\delta|\xi|^{2}$for allon

can be represented by a bounded integral kernel.

Example 5.2$L^{\infty}(R^{d}, \mathbb{C})$$C_{0}$$(t>0)$ ;a)$T$a Gaussian estimate of$M>0$ such that$b,$$|T(t)f|\leq Me^{wt}G(bt)|f|$

As before,$(t>0 , x\in R^{d})$,$\xi\in IR^{d}$. Define the elliptic operator

$A$

on

$D(A)=H^{m}(R^{d});Af= \sum_{\leq|\alpha|m}a_{\alpha}D^{\alpha}f$.

$L^{2}(R^{d})$by$a_{\alpha}\in$177
Then A generates a holomorphic
Gaussian estimate of order $m$ .

$C_{0}$-semigroup

$T$

on

which satisfies a$L^{2}(R^{d})$See [ , Theorem 9.4.2] for a proof.
$\mathrm{F}$

Theorem 5.3 Let
be a holomorphic-semigroup of angle
on$m\geq 2$ . Then the. Assume thataGaussianestimateordersatisfiesofextended, $1\leq p<\infty$ , are holomorphic of angle .
on-semigroup$T$$C_{0}$$\theta\in(0, \frac{\pi}{2}]$$T$$L^{2}(\Omega)$$C_{0}$.

$L^{2}(\Omega)$$T_{p}$$\theta$Theofcarries over if the following modifications are
made. From (4.5) one deduces here$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{o}\dot{\mathrm{f}}$$\mathrm{p}\mathrm{r}\mathrm{o}\dot{\mathrm{o}}\mathrm{f}$$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{e}\check{\mathrm{m}}3.2$

(5.1)

Since

$||T(t)||c(L^{1},L^{\infty})\leq$$T$is bounded onand$L^{1}$$L^{\infty}$

const $t^{-d/m}$

the Riesz-Thorin theorem gives(5.2)$||T(t)||_{C\mathrm{t})}L^{1},L^{2}\leq$

const $t^{-d/2m}$(5.3)$||T(t)||C(L^{2},L^{\infty})\leq$

const $t^{-d/2m}$

As before one deduces from this
(5.4)$||T(z)||C\mathrm{t}L^{1},L\infty)\leq$const$(\mathrm{R}\mathrm{e}\mathrm{z})^{-d/}m$$(z\in\Sigma(\theta_{2}))$.

Thus the kernel If $(z;x, y)$ satisfies(5.5)$|I\mathrm{f}(z, x, y).|\leq$

(5.6)

where

$|I\mathrm{f}(t, x, y)|\leq$$\beta=\frac{1}{m-1}$. Let$|z|^{-d/m}$$k(z)=z^{d/m}k\beta(z^{\frac{1}{\beta}}, X, y)$$|k(z)|\leq$

By Lemma 4.4 we obtainwhere$(z\in\Sigma(\theta_{2}))$,const $t^{-d/m}\exp(-b|x-y|m\beta. t-\beta)$$|k(z)|\leq$$\Sigma(\theta_{1}))$

const

const

. Then$(z\in\Sigma(\gamma))$andconst . $\exp(-b|X-y|^{\beta m.-}t)1$ .
const . $\exp(-b_{1}|X-y|^{\beta}m|Z|^{-1})(x,$. Now the conclusion follows as before.

$|k(z)|\leq$$b_{1}= \frac{\sin(\gamma-\theta_{1})}{\sin\gamma}b$$y\in\Omega$,$z\in$$\blacksquare$

Theorem 5.3 is due to Hieber [H2]. He uses the resolvent estimate characterizing
holomorphy, which is more complicated to prove, it seems.

178
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$\mathrm{N}.\mathrm{T}$Universit\"at Ulm

Mathematik V
Helmhoitzstr. 18
D- 89081 Ulm, Germany
arendt@mathematik.uni-ulm.de$\mathrm{A}\mathrm{b}\mathrm{t}\mathrm{e}\mathrm{i}\mathrm{l}\mathrm{u}\mathrm{n}\sim \mathrm{g}$