COMMENTS ON THE TROTTER PRODUCT FORMULA ERROR-BOUND
ESTIMATES FOR NONSELF-ADJOINT SEMIGROUPS
VINCENT CACHIA, HAGEN NEIDHARDT, and VALENTIN A. ZAGREBNOV
Let A be a positive self-adjoint operator and let B be an m-accretive operator
which is A-small with a relative bound less than one. Let H = A + B , then H is
well-dened on dom(H ) = dom(A) and m-accretive. If B is a strictly m-accretive
operator obeying
dom((H )) dom(A ) \ dom((B )) 6= f0g for some 2 (0; 1]; (1)
then for the Trotter product formula we prove that
?
?tB=n ?tA=n n
e
e
?
? e?tH ^ e?tA=ne?tB=n n ? e?tH = O(ln n=n ) (2)
(and similar for H ) as n ! 1, uniformly in t 0. We also show that:
(a) the A-smallness of B guarantees the condition (1) for 2 (0; 1=2), i.e. the
estimate (2) holds for 2 (0; 1=2);
(b) if B is strictly m-sectorial, then there are sucient conditions ensuring the
relation (1) for = 1=2, that implies (2);
(c) if B is A-small, m-sectorial and such that dom(A = ) is a subset of the formdomain of B , then again (2) is valid for = 1=2.
1 2
35K22,
1 Introduction
In our recent paper [2] we have shown the following : Let A be a densely dened self-adjoint
operator A = A I in a separable Hilbert space H and let B a densely dened m-accretive
operator B in H such that
dom(A) dom(B )
(1.1)
kBf k akAf k; f 2 dom(A); for some a 2 (0; 1):
(1.2)
and
If the adjoint operator B satises conditions
dom(A) dom(B )
(1.3)
kB f k akAf k; f 2 dom(A); for some a 2 (0; 1);
(1.4)
and
then the Trotter product formula convergence rate is of the order O(ln(n)=n). It turns out
that the proof crucially depends on conditions (1.3) and (1.4) and, consequently, on the
domain dom(B ). The aim of the present comments is to show that error-bound estimates
can be found even when the conditions (1.3) and (1.4) are not satised.
To this end we rst construct in Section 2 an example of a self-adjoint operator A
and an m-accretive operator B such that conditions (1.1) and (1.2) are satised but
dom(A) \ dom(B ) = f0g;
(1.5)
which demonstrates that condition (1.3) is not a consequence of (1.1) and (1.2). In [2] it is
proved that under conditions (1.1) and (1.2) H = A + B and H = (A + B ) are well-dened
m-accretive operators, but H 6= A + B in the case of (1.5). This is the reason why the
techniques used in [2] should be rened if one would like to relax the conditions (1.3),(1.4)
and to indicate operator-norm error bounds a la [2] for the Trotter product formula.
Therefore, in Section 3 we look for a condition which is weaker than (1.3), but
which is enough to provide error bound estimates for the Trotter product formula. The
intermediate condition we propose in the present paper is :
dom((H ) ) dom(A) \ dom((B )) 6= f0g; for some 2 (0; 1]:
(1.6)
Notice that conditions (1.3) and (1.4) imply (1.6) for = 1. Assuming (1.1), (1.2) and (1.6)
for a strictly m-accretive operator B we prove in Section 3 that the estimates
?
?tB=n ?tA=n n
(1.7)
e
? e?tH L lnnn ; n = 3; 4; : : : ;
e
and
?
?tA=n ?tB=n n
(1.8)
e
? e?tH L0 lnnn ; n = 3; 4; : : : ;
e
hold uniformly in t 0 for the corresponding 2 (0; 1). Similarly, these estimates are valid
for adjoint semigroup generated by ?H , as well as for the symmetrized Trotter product
formulae.
In Section 4 we present some applications of the condition (1.6) to the case of
m-accretive and m-sectorial operators B . They are related to the famous Kato's squareroot problem [6, 7] and to the Miyazaki condition [10]. First, we prove that our minimal
conditions (1.1), (1.2) for the m-accretive B imply the intermediate condition (1.6) for any
2 (0; 1=2). Consequently,
one gets error-bound estimates (1.7),(1.8) for those 's with
appropriate L ; L0 . Second, we show that Miyazaki's condition (see [10, Theorem 1, (2.1)])
2
ensures (1.6) for = 1=2, if B is m-sectorial, verifying (1.1), (1.2). In Section 5 we show
that to prove (1.7), (1.8) for = 1=2 one can relax the above conditions to: (1.1), (1.2), B is
m-sectorial and dom(a) dom(b). Here a and b are quadratic forms associated to operators
A and B . In the last Section 6 we make some remarks and comments. Finally, we note that
below we essentially follow terminology of the Kato book [9].
2 Example
The aim of this section is to construct a pair of operators A and B , where A is a positive
self-adjoint obeying A I and B is m-accretive (i.e. a closed operator with k(B + )? k (Re)? , for Re > 0, see [9], Ch.V), such that in spite of conditions (1.1), (1.2) one has
(1.5).
We start by some general remarks. Recall that a closed operator K is m-accretive,
if
1
1
Re (Kf; f ) 0; f 2 dom(K );
(2.1)
and ran( + K ) = H for Re() > 0 , see [9] Ch.V, i.e. the operator I + K is boundedly
invertible. The Cayley transform of K , see [3], is dened by
T := (I ? K )(I + K )? :
(2.2)
If K is m-accretive, then T is obviously a contraction with dom(T ) = H. Moreover, one has
1
the following lemma.
Lemma 2.1 A contraction T is the Cayley transform of an m-accretive operator K if and
only if
ker(I + T ) = f0g:
(2.3)
Moreover, the m-accretive operator K corresponding to T is given by
dom(K ) = ran(I + T )
(2.4)
Kf = (I ? T )(I + T )? f; f 2 ran(I + T ):
(2.5)
and
1
The map T ?! K is bijective.
For the proof see e.g. [3].
Let R and S be non-negative self-adjoint operators such that
dom(R) \ dom(S ) = f0g:
(2.6)
T := (I + R)? (I + S )? :
(2.7)
We set
1
3
1
The operator T is a contraction. Let us verify for it the condition (2.3). If we choose
f = (I + S )g, g 2 dom(S ), then the condition (2.3) transforms into
(I + S )g + (I + R)? g = 0:
(2.8)
1
Since S 0 and (I + R)? 0 we get g = 0 which yields f = 0. Consequently, by Lemma
2.1 the contraction T dened by (2.7) is the Cayley transform of an m-accretive operator K
which is given by (2.4) and (2.5), and one obviously has :
1
dom(K ) = ran(I + T ) =
ran((I + R)? (I + R + (I + S )? )) = ran((I + R)? ) = dom(R):
1
1
(2.9)
1
The Cayley transform of the m-accretive operator K is T . Applying again Lemma 2.1 we
nd
dom(K ) = ran(I + T ) =
ran((I + S )? (I + S + (I + R)? )) = ran((I + S )? ) = dom(S ):
1
1
(2.10)
1
So we have constructed an m-accretive operator K such that dom(K ) = dom(R) and
dom(K ) = dom(S ). By condition (2.6) we nd
dom(K ) \ dom(K ) = f0g:
(2.11)
Now we are going to construct the pair A and B . To this end we set
B := aK; for some a 2 (0; 1);
and
p
A := I + K K:
(2.12)
(2.13)
The operator B is obviously m-accretive with
dom(B ) = dom(K );
(2.14)
and A is a self-adjoint operator obeying A I and
p
dom(A) = dom( K K ) = dom(jK j) = dom(K ):
(2.15)
Hence dom(A) dom(B ). Moreover, we have
p
kBf k = akKf k ak I + K Kf k = akAf k; f 2 dom(A):
Hence dom(A) dom(B ). Since B = aK , we nally obtain that
dom(B ) \ dom(A) = dom(K ) \ dom(K ) = f0g:
4
(2.16)
(2.17)
To nish the construction of our example it remains only to indicate unbounded
non-negative self-adjoint operators R and S with the property (2.6). Let H = L ([0; 1]) and
let R be one-dimensional Laplace operator on the interval [0; 1]:
2
dom(R) := ff 2 W ; ([0; 1]) : f (0) = f (1) = 0g;
(2.18)
(2.19)
(Rf )(x) := ? d f (x); f 2 dom(R):
dx
Then the operator R is self-adjoint and non-negative. Now let frng1
n be set of rational
numbers of the open interval (0; 1). To dene S we set
22
2
2
=1
q(x) =
1
X
n=1
cn ; 2 ( 1 ; 1); c > 0;
n
jx ? rnj
2
such that
1
X
(2.20)
cn < 1:
(2.21)
1 dx jx ? rnj
n
Z
1 Z
1
1
? X
X
X
1
1
2
cn jxj dx = 2cn jxj dx = 1 ? cn < +1:
?
n
n
n
(2.22)
n=1
Then q 2 L ([0; 1]). Indeed,
1
Z
1
0
1
X
q(x)dx =1
cn
Z
1
0
2
2
2
=1
2
0
=1
=1
We dene
dom(S ) := ff 2 L ([0; 1]) : q(x)f (x) 2 L ([0; 1])g;
(Sf )(x) := q(x)f (x); f 2 dom(S ):
2
2
(2.23)
(2.24)
The operator S is unbounded, non-negative and self-adjoint.
To verify (2.6) assume that
f 2 dom(R) \ dom(S ) W ; ;([0; 1]) \ dom(S ):
22
(2.25)
By virtue of (2.25) the function f () is continuous and such that
q(x)f (x) =
1
X
n=1
cn f (x) 2 L ([0; 1]):
jx ? rnj
2
(2.26)
Then we get
Z
1
0
jq(x)f (x)j dx cn
2
Z
2
0
5
1
1 jf (x)j dx
jx ? rnj 2
2
(2.27)
for any n = 1; 2; : : : . Since f () is continuous, function jf ()j is also continuous. Assume that
f (rn) 6= 0. Therefore, there is a suciently small neighbourhood (rn ? ; rn + ) (0; 1),
> 0, of rn and a > 0 such that jf (x)j for x 2 (rn ? ; rn + ). Hence
Z
Z rn +
1 dx = +1:
(2.28)
rn ? jx ? rn j But this is impossible by (2.26). Therefore, one gets f (rn) = 0, n = 1; 2; : : : . Since f () is
continuous this yields f 0. Hence dom(R) \ dom(S ) = f0g.
So, we have in the Hilbert space H = L ([0; 1]) the self-adjoint operator A (see
(2.9),(2.16),(2.17)) and the m-accretive operator B (see (2.9),(2.12),(2.17)) such that
1
0
jq(x)f (x)j dx cn
2
2
2
2
2
dom(A) W ; ([0; 1]);
(2.29)
dom(B ) \ W ; ([0; 1]) = f0g:
(2.30)
22
but
22
This nishes construction of desired example.
3 Intermediate condition
The example of the previous section shows that the conditions (1.1) and (1.2) do not guarantee that
dom(H ) dom(A) \ dom(B ) 6= f0g;
(3.1)
Hf := Af + Bf; f 2 dom(H ) = dom(A):
(3.2)
where
Notice that condition (3.1) is satised, for example, if we assume (as in [2]) that in addition
to (1.1) and (1.2) conditions (1.3) and (1.4) are satised. In this case condition (3.1) takes
the form
dom(H ) = dom(A) dom(A) \ dom(B ):
(3.3)
In view of Example of Section 2 the conditions (1.3) and (1.4) sound rather strong. It is
known that together with (1.1) and (1.2) they guarantee the error bounds (1.7), (1.8) with
= 1, see [2]. In this section we show that one can relax them to the intermediate condition
(1.6), which is applicable even in the case dom(A) \ dom(B ) = f0g, cf. (1.5).
To this end we assume that B is a strictly m-accretive operator (see [9],Ch.V), i.e.
there is > 0 such that
Re(Bf; f ) kf k ; f 2 dom(B ):
2
6
(3.4)
In particular, this yields that the left open half-plane C ? = f 2 C : Re( ) < 0g is contained
in the resolvent set %(B ) of B . Then our minimal conditions (1.1) and (1.2) imply that H
dened by (3.2) is also strictly m-accretive. It is known (see [7]-[9]) that m-accretivity of B
and H implies that B and H are also m-accretive and that fractional powers of all of these
operators: B , H , (B ), (H ) are well-dened for 2 (0; 1). The strict m-accretivity of
B guarantees invertibility of these operators. Now we recall some properties of H , see [1, 2]:
Proposition 3.1 Let A = A I and let B be a m-accretive operator such that conditions
(1.1) and (1.2) are satised. Then for H dened by (3.2) one has:
(a) The operator ?H is a boundedly invertible generator of contraction holomorphic semigroup fe?tH gt in the sector S! = ft 2 C : j arg tj < ! = arccos(a)g.
(b) For any > 0 and t > 0 the range ran(e?tH ) dom(H ), and there is d (H ) > 0
such that
(3.5)
kH e?tH k dt(H ) ; t > 0:
0
Next we generalize Lemma 5 of [1] :
Lemma 3.2 Let A I and let B be a strictly m-accretive operator obeying conditions (1.1)
and (1.2). If H is given by (3.2), then there is a constant M > 0 such that
? ?B ?A
e
e
? e?H A? M
1
(3.6)
for 0. If in addition the intermediate condition (1.6) is satised, then for any 2 (0; 1]
there is N1 () > 0 such that
? ? ?B ?A
H
e e
? e?H N () (3.7)
1
for 0.
Proof: The rst estimate (3.6) readily follows from the inequality
? ?B ?A
?H A?1 e
e
e
Z ?
?
sB
?
1 ?A ds
e
BA
e
+ I
?
0
Z ? e?A A?1 + 0
ds e?sH HA?1
and conditions (1.1), (1.2) which give M := kBA? k + 1 + kHA? k 2(a + 1).
To prove the second estimate (3.7) notice that for any strictly m-accretive operator
C the fractional powers C ?, 0 < < 1, are dened by
Z 1
sin
?
?(C + )? d;
C = see [7]-[9]. Hence (C ?) = (C )? , which allows us to pass from (3.7) to the norm-estimate
of the conjugate operator
1
1
1
0
?
?
(H ? e?B e?A ? e?H ) = e?A e?B ? e?H (H )? :
7
To this end we use the identity
(e?Ae?B ? e?H )(H )?
(3.8)
?
A
?
?
A
?
B
?
?
H
?
= (e ? I )(H ) ? e (I ? e )(H ) + (I ? e )(H )
= (e?A ? I )A? A(H )? + e?A(e?B ? I )(B )? (B ) (H )? ? (e?H ? I )(H )?:
Since for any strictly m-accretive operator C and 0 < < 1 one has
Z 1
sin
?
tC
?
?(C + )? (e?tC ? I ) d;
(e ? I )C = we get for > 0
(3.9)
1
0
(e?tC ? I )C ?
Z ?
?
sin
? C (C + )? C ? e?tC ? I d
= Z 1
?
sin
+ ?(C + )? e?tC ? I d:
1
1
0
1
Since the above identity is valid for any > 0, by the estimates k(e?tC ? I )C ? k t,
kC (C + )? k 1 and k(C + )? k 1= we obtain
?
sin
t
2
?
tC
?
?
?
k(e ? I )C k >
= sin (12 ? ) t :
inf 1 ? + (3.10)
for 2 (0; 1) and t 0. Applying the estimate (3.10) to operators A, B and H , by virtue
of identity (3.8) we nd the assertion (3.7) with constant
? ?
(H )? k + k(B ) (H )? k + 1 ; 2 (0; 1):
k
A
N () = 2(1sin
? )
Here kA (H )?k and k(B )(H )?k are bounded due to our condition (1.6). The estimate
(3.7) can be extended to = 1 by setting N ( = 1) := lim! N () = kA(H )? k +
kB (H )? k + 1.
2
Next we have to modify Lemma 6 of [1].
Lemma 3.3 Let A I and let B be a strictly m-accretive operator obeying conditions (1.1)
and (1.2). If the operator H dened by (3.2) satises the intermediate condition (1.6), then
for any 2 (0; 1] there is a constant N () > 0 such that
1
1
1
1
1
0
1
1
1
1
1
1
1
2
? ? ?B ?A
H
e e
? e?H A? N ()
for all 0.
Proof: We use the following representation
?
1
2
H ? e?B e?A ? e?H A? =
H ?(I ? e?B )(I ? e?A )A? +
H ?(e?B + e?A ? e?H ? I )A? :
1
1
1
8
1+
(3.11)
(3.12)
(3.13)
(3.14)
Let 2 (0; 1). By virtue of (3.10) one estimates (3.13) as follows:
kH ?(I ? e?B )(I ? e?A)A? k
(3.15)
?
B
?
?
A
?
k(I ? e )(H ) kk(I ? e )A k
? ?
k(I ? e?B )(B )? kk(B )(H )?k 2(1sin
? ) k(B ) (H ) k:
1
1
1
1+
To estimate (3.14) we use the representation
e?B + e?A ? e?H ? I =
(e?B ? I + B ) + (e?A ? I + A) ? (e?H ? I + H ):
Since
e?C ? I + C
=
Z 0
ds (I ? e?sC )C:
(3.17)
one gets for a strictly m-accretive operator C that
Z kH ?(e?C ? I + C )A?1 k Z 0
(3.16)
0
ds (I ? e?sC )(H )? kCA?1k
dsk(I ? e?sC )(C )?k k(C )(H )?k kCA? k:
1
By (3.10) and (3.16), (3.17) this yields for (3.14) the estimate
kH ?(e?B + e?A ? e?H ? I )A? k (3.18)
?
2 sin ?k(B )(H )?kkBA? k + kA(H )?k + kHA? k :
1 + (1 ? )
1
1+
1
1
1
Hence (3.15) and (3.18) verify assertion (3.11) with the constant
? ) (H )? k + 1 ?ak(B ) (H )? k + kA (H )? k + 1 + a :
k
(
B
N () := 2(1sin
? )
1+
1
2
Setting N (1) := lim! N () one easily gets the estimate (3.11) for = 1.
2
1
2
2
Theorem 3.4 Let A I and let B be a strictly m-accretive operator obeying conditions
(1.1) and (1.2). If the operator H dened by (3.2) satises (1.6), then there is a constant
L > 0 such that the estimates
? e?tH L lnnn ;
?
?tA=n ?tB =n n
e
? e?tH L lnnn ;
e
?
?tB=n ?tA=n n
e
e
hold for 2 (0; 1] and n = 3; 4; : : : uniformly in t 0.
9
(3.19)
(3.20)
Proof: Notice that (3.19) implies (3.20) and vice versa. Therefore, we start with (3.20). Let
= t=n and F ( ) = e?B e?A, i.e., F ( ) = e?Ae?B . Then we use the operator identity
F ( )n ? e?nH =
n?1
X
m=0
?
F ( ) n?m? F ( ) ? e?H e?mH ;
(
1)
(3.21)
and the representation
I ? F ( ) = (I ? e?A)(I + X ( ));
where X ( ) = (I ? e?A )? e?A(I ? e?B ). By Lemma 3.2 of [2] and Lemma 2.3 of [11] the
operator I + X ( ) is invertible and one gets
1
k(I ? F ( ))? uk 1 k(I ? e?A)? uk 1 (kuk + 1 A? u):
1?a
1?a
(3.22)
1
1
1
Since (H ) is invertible one can use (3.22) to estimate (3.21) as follows:
kF ( )n ? e?nH k kF ( )n? (I ? F ( ))kk(I ? F ( ))? (F ( ) ? e?H )k+
1
n?1
X
m=1
1
(3.23)
kF ( )n?m? (I ? F ( ))k k(I ? F ( ))? (F ( ) ? e?H )(H )? k k(H )e?mH k:
1
1
Then Theorem 2.4 of [2] gives
kF ( )s(I ? F ( ))k s +c 1 ; c > 0 and s = 0; 1; 2; ::::
(3.24)
>From the estimate (3.22) we get
k(I ? F ( ))?1 (F ( ) ? e?H )k 1
1?a
kF ( ) ? e?H k +
1 A? ?F ( ) ? e?H (3.25)
1
and
k(I ? F ( ))? (F ( ) ? e?H )(H )?k
(3.26)
?
1 ?1 a k(F ( ) ? e?H )(H )?k + 1 kA? F ( ) ? e?H (H )?k :
1
1
Therefore, by Lemmata 3.2 and 3.3 (taking the adjoint) one obtains the estimates
k(I ? F ( ))? (F ( ) ? e?H )k 1 ?1 a (2 + M );
k(I ? F ( ))? (F ( ) ? e?H )(H )?k 1 ? a (N () + N ())
1
1
1
10
2
(3.27)
(3.28)
for 2 (0; ] and 0. By Proposition 3.1 one gets
d
(H )
?
mH
k(H ) e
k (m ) :
Then using (3.23) and inequalities (3.24) - (3.29) we nd
kF ( )n ? e?nH k
(3.29)
n?1
X
n(1 c? a) (2 + M ) + n ?c m 1 ?1 a (N () + N ()) dm(H ) ;
m
for n = 3; 4; : : : . Since for 0 < < 1 and n 3 one has
1
(3.30)
2
=1
n?1
X
m=1
n?
X
1
2 ln n ;
1
? 1 + 1
m
=
(n ? m)m n m
m n?m
n
1
1
(3.31)
=1
the estimate (3.30) leads to (3.20) with
L := 1 2?c a f(2 + M )=2 + (N () + N ())d(H )g
for 2 (0; 1] and 0.
2
Corollary 3.5 Let A = A I and let B be a strictly m-accretive operator obeying (1.1)
and (1.2). If the operator H dened by (3.2) satises the intermediate condition (1.6), then
we have the following estimates
?
0 ln n
?tA=n ?tB=n n
?
tH
e
e
?
e
(3.32)
L ;
n
?
?tA= n ?tB=n ?tA= n n
?
tH
(3.33)
e
e
? e L00 lnnn
e
for 2 (0; 1] and n = 3; 4; ::: uniformly in t 0.
Proof: Let T ( ) = e?A e?B and = t=n. We use the identity
1
2
2
2
T ( )n ? e?tH =
e?A(F ( )n? ? e?tH )e?B + (e?A ? I )e?tH e?B + (e?tH H )(H ?(e?B ? I )):
1
Since F ( )n? ? e?tH = F ( )n? (I ? F ( )) + (F ( )n ? e?tH ), by Theorem 2.4 of [2] and by
(3.19) one gets the estimate
ke?A(F ( )n? ? e?tH )e?B k kF ( )n? ? e?tH k nc + L lnnn ; n 3:
(3.34)
Following the line of reasoning developed in proofs of Lemmata 3.2 and 3.3 we obtain
?A
(e
? I )e?tH e?B k(e?A ? I )A? kkAH ? kkHe?tH k n1 d1 ?(Ha) ; n 1; (3.35)
1
1
1
1
1
1
11
1
and
kH e?tH H ?(e?B ? I )k (3.36)
?
k(B )(H )? k:
kH e?tH kk(e?B ? I )(B )?kk(B ) (H )?k n1 d(H ) 2(1sin
? )
1
Thus, estimates (3.34)-(3.36) yield the rst assertion (3.32) with
(H ) + d (H ) 2 ? sin k(B )(H )?k
L0 = L + c + d1 ?
a (1 ? )
1
1
for 2 (0; 1) and 0. If = 1, then one has to replace f2 ? sin =(1 ? )g by 1.
The proof of the second assertion is similar.
2
1
4 Applications to accretive and sectorial generators
In this section we indicate sucient conditions, which give error-bound estimates (1.7), (1.8)
for some 0 < 1. Since for m-accretive generator B these estimates are guaranteed by
(1.6) (see Section 3), we scrutinize, in particular, conditions ensuring (1.6) for some 2 (0; 1)
.
4.1 We start by the following simple observation. Let J be a conjugation on H, i.e,
an antilinear operator on H such that
(Jf; Jg) = (g; f ); f; g 2 H:
(4.1)
JAJ = A and JBJ = B ;
(4.2)
If J obeys the condition
then conditions (1.1) and (1.2) imply conditions (1.3) and (1.4) with a = a. Therefore, the
assumptions of Theorem 3.5 of [2] are satised. That yields the error estimates (1.7), (1.8)
with = 1. As an example one can consider the Schrodinger operator. If A is the Laplace
operator on H = L (R n ) and B is a multiplication operator arising from some complex
potential V () such that Re V 0 and conditions (1.1) and (1.2) are satised, then denoting
by J the complex conjugation one easily veries that conditions (4.1) and (4.2) are valid.
4.2 The problem becomes more dicult if we loose the condition dom(A) dom(B ),
in particular, if we face up to dom(A) \ dom(B ) = f0g. Then it seems to be a delicate
problem to satisfy the intermediate condition (1.6). But it turns out that our minimal conditions (1.1) and (1.2) are sucient to imply (1.6) for any 0 < 1=2. This arises from
the following line of reasoning:
2
(i) Notice that for any m-accretive operators A and B satisfying (1.1) one has dom(A) dom(B ) for 0 1, see [8]. Since operator H is m-accretive and dom(H ) dom(A) (in fact, dom(H ) = dom(A), cf.(3.2)), we also obtain that dom(H ) dom(A ) for 0 1.
12
(ii) Since B and H are m-accretive operators, Theorem 1.1 of [6] yields dom(B ) =
dom((B ) ) and dom(H ) = dom((H ) ) for 0 < 1=2.
(iii) Summing up the statements in (i) and (ii) one immediately gets that dom((H )) =
dom(H ) dom(A ) dom(A ) \ dom(B ) = dom(A ) \ dom((B ) ) for 0 <
1=2. Hence, we obtain the intermediate condition (1.6) for 0 < 1=2.
Thus, taking into account Theorem 3.4 we have proved the following statement.
Theorem 4.1 Let A I and let B be a strictly m-accretive operator. If the conditions (1.1)
and (1.2) are satised, then estimates (1.7) and (1.8) hold for any 0 < < 1=2 uniformly
in t 2 [0; 1).
Therefore, to cover the case of the error bound with 0 < < 1=2 our minimal conditions
(1.1), (1.2) are sucient, in spite of the fact that dom(A) \ dom(B ) = f0g can happen.
4.3 Now we are going to demonstrate how we can ensure condition (1.6) in the
limit case = 1=2. The considerations below are mainly relied on the paper of Miyazaki
[10] which improves some results of Kato [6, 7] and Shimakura [14]. Again we assume our
minimal conditions (1.1), (1.2), and we suppose that B is strictly m-accretive and sectorial
(in short m-sectorial) with semi-angle B 2 [0; =2), i.e. with numerical range in the sector
SB := ft 2 C : j arg tj < B g. For simplicity, we set = 1 in (3.4). With B we associate a
sesquilinear form ~b dened by
~b[f; g] = (Bf; g); f; g 2 dom(~b) = dom(B ):
(4.3)
The form ~b is closable. Its closure is denote by b. This denition implies
dom(B ) dom(b):
(4.4)
One can associate with b its real part bR dened by
(4.5)
bR [f; g] = 21 fb[f; g] + b[g; f ]g; dom(bR ) = dom(b):
The form bR is symmetric and, by virtue of (3.4), it is semibounded from below with lower
bound = 1. Since B is m-accretive, the form bR is closed. By representation theorems (see
[9],Ch.VI) there is a non-negative self-adjoint operator BR I such that
bR [f; g] = (BR= f; BR= g); f; g 2 dom(BR= ) = dom(bR ):
(4.6)
The operator BR is called the real part of B . Similarly, we can introduce the imaginary part
bI of b dened by
bI [f; g] = 21i fb[f; g] ? b[g; f ]g; f; g 2 dom(bI ) = dom(b):
(4.7)
Since B is m-sectorial, the associated form b is also m-sectorial. It is known (see [9],Ch.VI)
that in this case there is a non-negative constant C = tan B such that
jbI [f; g]j CbR [f; f ] = bR [g; g] = ; f; g 2 dom(b) = dom(bR );
(4.8)
1 2
1 2
1 2
1 2
1 2
13
which is equivalent to
jbI [f; g]j C kBR= f kkBR= gk; f; g 2 dom(bI ) = dom(bR ) = dom(BR= ):
1 2
dition
1 2
1 2
(4.9)
Following Miyazaki [10] we assume that there is a constant M such that the con-
jbI [f; g]j M kBR= f kkBR= gk; f; g 2 dom(BR= ):
holds for some 2 [0; 1). Since BR I , one has
kBR= f k kBR= f k; f 2 dom(BR= ) dom(BR= );
2
2
2
1 2
1 2
(4.10)
2
(4.11)
1 2
which shows that the Miyazaki condition (4.10) is stronger than (4.9).
Let a be closed quadratic form associated with a positive self-adjoint operator A :
a[f; g] = (A = f; A = g); dom(a) = dom(A = ):
1 2
1 2
1 2
(4.12)
By condition (1.1) the domain
D = dom(a) \ dom(b)
(4.13)
is dense in H. Then the sum of the closed sectorial forms
h[f; g] = a[f; g] + b[f; g]; f; g 2 dom(h) = D
(4.14)
is a well-dened on D closed sectorial form by [9, Theorem VI-1.31]. By the representation
:
Theorem 2.1 of [9, Ch.VI] there is an m-sectorial operator H denoted by H = A + B (the
form-sum of A and B ) such that one has
h[f; g] = (Hf; g); f 2 dom(H ) dom(h); g 2 dom(h):
(4.15)
By virtue of conditions (1.1) and (1.2) the operator H coincides with that dened by the
algebraic sum A + B on dom(A), see (3.2).
Theorem 4.2 Let A I and let B be a strictly m-sectorial operator obeying conditions
(1.1) and (1.2). If in addition the operator B satises conditions (3.4), with = 1, and
(4.10), then there are constants L = and L0 = such that the estimates (1.7) and (1.8)
hold for = 1=2 uniformly in [0; 1).
Proof: Since by (3.4), (4.5) and (4.6) one has
=1 2
=1 2
Re(b[f; f ]) = kBR= f k kf k f 2 dom(BR= );
1 2
1 2
2
(4.16)
the sesquilinear form b is strongly coercive, see [10]. Then condition (4.10) and Theorem 2
of [10] imply
dom(B = ) = dom((B ) = ) = dom(BR= ) = dom(b):
1 2
1 2
14
1 2
(4.17)
Next, the real part hR of h (4.14) is given by
hR [f; g] = a[f; g] + bR [f; g]; f; g 2 dom(hR ) = dom(h):
(4.18)
The corresponding self-adjoint operator is denoted by HR. By the same line of reasoning
as above for the form b, the sesquilinear form h is strongly coercive. By denition (4.18) we
have BR HR . Then taking into account the Heinz inequality for self-adjoint operators we
nd
BR HR ; 2 [0; 1):
(4.19)
By virtue of (4.14) for the imaginary part of h we get :
hI [f; g] = 21i fh[f; g] ? h[g; f ]g = bI [f; g]; f; g 2 dom(hI ) = dom(h):
Taking into account (4.10) and (4.19) we nd that the Miyazaki condition
(4.20)
jhI [f; g]j M kHR= f kkHR= gk; f; g 2 dom(bI ); 2 [0; 1);
(4.21)
2
2
is valid for the form h. Then again by Theorem 2 of [10] one gets
dom(H = ) = dom((H ) = ) = dom(HR= ) = dom(h):
1 2
(4.22)
dom(h) = dom(a) \ dom(b);
(4.23)
1 2
1 2
By (4.13), (4.14) and [9, Theorem VI-1.31] we have
which together with (4.17) yields
dom((H ) = ) = dom(A = ) \ dom((B ) = ):
1 2
1 2
1 2
(4.24)
Hence, our condition (1.6) is satised for = 1=2. By virtue of Theorem 3.4 and Corollary
3.5 this gives the estimates (1.7) and (1.8) for = 1=2.
2
5 More about the case = 1=2
The aim of this section is to relax the Miyazaki condition (4.10), which is sucient to ensure
(1.6) and, hence, to apply our main Theorem 3.4 for the case = 1=2. Here we propose for
this case another strategy of the proof essentially based on Theorem 3 of [8]. We recall it
for the reader convenience:
Proposition 5.1 Let R, S be bounded accretive operators in Hilbert spaces H, H0,respectively,
and let T be a bounded operator from H to H0 . Then
kRTS k e ?)=2 kT k1? kRTS k ;
2 (1
15
0 1:
(5.1)
Below we shall prove the error-bound estimates (1.7) and (1.8) for = 1=2 under
assumptions avoiding our intermediate condition (1.6), i.e. involving only operators A
and B without any conditions on the adjoint operator B . We begin by setting F^ ( ) =
e?A= e?B e?A= and F ( ) = e?B e?A, for 0. Then we have
F^ ( )n = e?A= F ( )n? e?B e?A= :
(5.2)
2
2
2
1
2
Lemma 5.2 Let A I and let B be an m-accretive operator. If conditions (1.1) and (1.2)
are satised, then there is a constant c > 0 such that for n 1 one has
c
^ n
(5.3)
F ( ) I ? F^ ( ) ; > 0:
n
Proof: By (5.2) we get
F^ ( )n I ? F^ ( )
=
=
e?A= F ( )n? e?B e?A= I ? F^ ( )
e?A= F ( )n? (I ? F ( )) e?B e?A= ;
2
1
2
1
2
(5.4)
2
which yields the estimate
^ n
^
F ( ) I ? F ( ) F ( )n? (I ? F ( )) :
1
(5.5)
Then by [2, Theorem 2.4], we obtain (5.3).
2
Lemma 5.3 Let A I and let B be an m-accretive operator. If conditions (1.1) and (1.2)
are satised, then (I ? F^ ( )) is boundedly invertible for any > 0.
Proof: We use the representation
I ? F^ ( ) = I ? e?A + e?A= (I ? e?B )e?A= :
(5.6)
2
2
Setting
Y ( ) = e?A= (I ? e?B )e?A= (I ? e?A)? ;
(5.7)
I ? F^ ( ) = (I + Y ( ))(I ? e?A):
(5.8)
kAe?A= (I ? e?A)? k 1; 0;
(5.9)
k(I ? e?B )(A)? k a
(5.10)
2
2
1
we get
Since
2
1
and
1
we get kY ( )k a < 1. Hence, the operator I + Y ( ) is boundedly invertible. Since for
> 0 the operator (I ? e?A) is also boundedly invertible, the same is valid for (I ? F^ ( )),
see (5.8).
2
16
Lemma 5.4 Let A I and let B be an m-accretive operator. If conditions (1.1) and(1.2)
are satised, then there is a constant b > 0 such that for any > 0 one has
(I
khk + p1 kA? = hk ; h 2 H:
? F^ ( ))?1=2 h b
1 2
(5.11)
Proof: From (5.8) we nd the representation
?
^
I ? F ( ) (I + Y ( )) = (I ? e?A )? :
1
(5.12)
1
Since kF^ ( )k 1, the bounded operators I ? F^ ( ) and (I ? F^ ( ))? are accretive. Then we
can apply Proposition 5.1 to k(I ? F^ ( )) = (I ? e?A)? = k and T = I to obtain the estimate
1
1 2
(I
1 2
? F^ ( ))? = (I + Y ( ))f e2 = kI + Y ( )k = k(I ? e?A)? = f k; f 2 H:
1 2
1 2
8
1 2
(5.13)
Since kY ( )k a < 1, we get
(I
p
? F^ ( ))? = h e2 = 1 + a k(I ? e?A)? = (I + Y ( ))? hk; h 2 H:
1 2
8
1 2
1
(5.14)
Moreover, we have
(I + Y ( ))(I ? e?A) = = (I ? e?A) = (I + Z ( ));
1 2
1 2
(5.15)
where Z ( ) is the bounded operator
?A=
?A=
Z ( ) = (I ?e e?A) = (I ? e?B ) (I ?e e?A) = :
2
2
1 2
1 2
(5.16)
Notice that conditions on B imply that the operator Z ( ) is m-accretive, thus k(I +
Z ( ))? k 1. Then by (5.14), (5.15), and by the spectral representation for operator
(I ? e?A= )? = we get the inequality
1
2
1 2
(I
p
? F^ ( ))? = h e = 1 + a (I ? e?A= )? = h
!
r
p
2
e = 1 + a khk + kA? = hk :
1 2
2 8
2
2 8
p
(5.17)
1 2
p
Setting b := 2e2 = 1 + a, one obtains the estimate (5.11).
8
1 2
2
Lemma 5.5 For any m-sectorial operator C in a Hilbert space H with the real part CR , one
has the following estimate:
1=2 ?tC CR e
r
d (tC ) ; t > 0:
1
17
(5.18)
Proof: First we remark that
d ke?tC uk = ?2<e (e?tC u; Ce?tC u) = ?2kC = e?tC uk 0; h 2 H:
(5.19)
R
dt
Then since 0 ke?tC uk ? ke? t h C uk ke?tC u ? e? t h C uk for t; h > 0, one gets (cf.(3.5) :
h? (ke?tC uk ? ke? t h C uk) (5.20)
? dtd ke?tC uk = hlim
!
? ke?tC u ? e? t h C uk d e?tC u kCe?tC uk d (C ) kuk:
lim
h
h!
dt
t
1 2
2
( + )
2
( + )
1
( + )
0
1
1
( + )
0
By (5.19) and (5.20) we nd
2kCR= e?tC uk = ?2ke?tC uk d ke?tC uk 2 d (C ) kuk ;
dt
t
which leads to the estimate (5.18).
1 2
1
2
2
(5.21)
2
Lemma 5.6 Let A I be a self-adjoint operator and let B be an m-sectorial operator with
semi-angle B 2 [0; =2). If dom(A = ) dom(b), then we have
?=
A
(I ? e?B ) (1 + tan B )kBR= A? = k2 = d (B ) = ; 0:
(5.22)
Proof: For any f; g 2 H one has
1 2
1 2
1 2
? ?1=2
A (I
? e?B )f; g =
? ?1=2
A (I
? e?B )f; g =
Then we get
Z 0
Z 0
1 2
1 2
1
?
1 2
ds Be?sB f; A? = g :
1 2
ds b e?sB f; A? = g ;
1 2
(5.23)
where b[; ] is closed sesquilinear sectorial form corresponding to B . By denitions (4.5),
(4.7) of its real and imaginary parts, and by inequality (4.9) we get :
jb[u; v]j (1 + tan B )bR [u] = bR [v] = = (1 + tan B )kBR= ukkBR= vk:
1 2
1 2
1 2
1 2
(5.24)
Thus, for the integrand in (5.23) we obtain that
jb[e?sB f; A? = g]j (1 + tan B )kBR= e?sB f kkBR= A? = gk:
1 2
1 2
1 2
Together with inequality (5.18) this gives :
j
? ?1=2
A (I
? e?B )f; g j
1 2
r
Z =
?
1=2
(1 + tan B )kBR A gk ds d1(B ) kf k
s
0
1=2 ?1=2
1=2 1=2
(1 + tan B )kBR A k2d1(B ) kf kkgk;
1 2
which implies the annonced estimate (5.22). Here we used that by condition of lemma
dom(A = ) dom(b) = dom(BR= ) , cf.(4.17), and thus kBR= A? = k < 1.
2
1 2
1 2
1 2
18
1 2
Lemma 5.7 Let A I and let B be an m-sectorial operator obeying conditions (1.1) and
(1.2). If0 in addition the condition dom(A1=2 ) dom(b) is satised, then there are constants
M1=2 ; M1=2 > 0 such that the estimates
?1=2 ^
F ( )
A
and
?1=2 ^
F ( )
A
? e?H M
=
1 2
=
(5.25)
1 2
? e?H H ? M 0 = =
1
(5.26)
3 2
1 2
are valid for 0.
Proof: To prove (5.25) we use the representation
A? = (e?A= e?B e?A= ? e?H ) =
?e?A= A? = (I ? e?B )e?A ? A? = (I ? e?A) + A? = (I ? e?H ):
1 2
2
2
2
1 2
1 2
1 2
By spectral representation one readily gets the estimate kA? = (I ? e?A)k (2=e) = .
Taking into account Lemma 5.6 for estimations of A? = (I ? e?B ) and A? = (I ? e?H ), we
obtain (5.25) with
1 2
1 2
1 2
1 2
M = = (2=e) = + 2(1 + tan B )(kBR= A? = kd (B ) = + kHR= A? = kd (H ) = ):
1 2
1 2
1 2
1 2
1 2
1 2
1
1 2
1 2
1
To prove (5.26) we use the representation
A? = (F^ ( ) ? e?H )H ? = A? = (I ? e?A= )(I ? e?B )e?A= H ? +
A? = (I ? e?B )(I ? e?A= )H ? +
A? = (e?A + e?B ? e?H ? I )H ? :
1 2
1
1 2
2
2
1 2
2
1
1
1 2
1
(5.27)
(5.28)
(5.29)
Since by the spectral representation: kA? = (I ? e?A= )k (=e) = and k(I ? e?B )A? k kBA? k, we get for the term (5.27) the following estimate :
(5.30)
kA? = (I ? e?A= )(I ? e?B )e?A= H ? k p1e kBA? kkAH ? k = :
For the second term (5.28) we nd by Lemma 5.6 that
1 2
2
1 2
1
1
1 2
2
2
1
1
1
3 2
kA? = (I ? e?B )(I ? e?A= )A? AH ? k (1 + tan B )d (B ) = kBR= A? = k = kAH ? k:
1 2
2
1
1
1
1 2
1 2
1 2
3 2
1
(5.31)
Taking into account (3.16) and (3.17), we obtain representation for the third term (5.29):
A? =Z (e?A + e?B ? e?H ? I )H ? Z=
A? = (I ? e?sA)AH ? ds + A? = (I ? e?sB )BA? AH ? ds +
1 2
1
1 2
Z0 0
1
1 2
0
A? = (e?sH ? I ) ds:
1 2
19
1
1
Taking into account that kA? = (I ? e?sA)k (2s=e) = , and applying Lemma 5.6 to
A? = (I ? e?sB ) and to A? = (e?sH ? I ), one gets the estimate for the last term (5.29):
1 2
1 2
1 2
1 2
kA? = (e?Ar+ e?B ? e?H ? I )H ? k
(5.32)
32 = 2e kAH ? k + 2(1 + tan B ) d (B ) = kBR= A? = kkBA? kkAH ? k
+d (H ) = kHR= A? = k = :
1 2
1
3 2
1
1 2
1 2
1
1 2
1 2
1 2
1
1 2
1
1
3 2
Together with (1.1) and (1.2) this gives for (5.32) the estimate :
kA? = (e?A + e?B ? e?H ? I )H? k
(5.33)
=
=
3p2 e(1 ? a) + 2(1 + tan B ) d (B ) = kBR= A? = k 1 ?a a + d (H ) = kHR= A? = k = :
1 2
1
3 2
3 2
1 2
1 2
1
1 2
1 2
1
1 2
1 2
3 2
Applying conditions (1.1), (1.2) to (5.30), (5.31) one gets similar estimates, which together
with (5.33) give the proof of (5.26), if we set
B kB = A? = kd (B ) =
M 0 = := pe(1a? a) + 1 +1 tan
(5.34)
? a R
=
2
a
=
=
=
?
=
=
?
=
+ p
+ 2(1 + tan B ) d (B ) kBR A k 1 ? a + d (H ) kHR A k :
3 e(1 ? a)
1 2
1 2
3 2
1
1 2
1 2
1 2
1
1 2
1 2
1
1 2
1 2
1 2
2
Theorem 5.8 Let A I be a self-adjoint operator. Let B be an m-sectorial operator obeying
conditions (1.1) and (1.2) with associated closed sectorial form b. If in addition one has
dom(A = ) dom(b);
(5.35)
1 2
then there exists a constant L^ > 0 such that
?
?tA=2n ?tB=n ?tA=2n n
e
e
e
? e?tH L^ lnpnn
(5.36)
uniformly in t 2 [0; 1).
Proof: We begin by identity
F^ ( )n ? e?nH =
n?1
X
m=0
F^ ( )n?m? (F^ ( ) ? e?H )e?mH :
1
(5.37)
Since the operator (I ? F^ ( ))? = exists (Lemma 5.3), we nd the representation:
1 2
F^ ( )n ? e?nH =
n?1
X
m=0
F^ ( )n?m? (I ? F^ ( )) = (I ? F^ ( ))? = (F^ ( ) ? e?H )e?mH ; (5.38)
1
1 2
20
1 2
which leads to the estimate
kF^ ( )n ? e?nH k n?1
X
(5.39)
kF^ ( )n?m? (I ? F^ ( )) = kk(I ? F^ ( ))? = (F^ ( ) ? e?H )e?mH k:
1
m=0
1 2
1 2
Since kF^ ( )k 1, the operator I ? F^ ( ) is m-accretive, and we can use Proposition 5.1 for
R = I , T = F^ ( )n?m? and S = (I ? F^ ( )). This together with Lemma 5.2 gives :
n?m? (I ? F^ ( ))k =
kF^ ( )n?m? (I ? F^ ( )) = k e2= kF^ ( )n?m? k = kF^ ( )r
(5.40)
e2 = kF^ ( )n?m? (I ? F^ ( ))k = e2 = n ? mc ? 1 :
To estimate the term corresponding in (5.37), (5.39) to m = 0, we use again Proposition 5.1
as well as Lemmata 5.4 and 5.7. This gives :
kF^ ( )n? (F^ ( ) ? e?H )k
(5.41)
n
?
=
?
=
?
H
^
^
^
^
kF (r
) (I ? F( )) kk(I ? F ( )) (F ( ) ? e )k e2 = n ?c 1 b kF^ ( ) ? e?H k + p1 kA? = (F^ ( ) ? e?H )k
r
c b(2 + M ):
2=
e
=
n?1
For the terms with m > 0 we use Lemmata 5.4, 5.7 to get the estimate
k(I ? F^( ))? = (F^ ( ) ? e?H )H ? k
(5.42)
b k(F^ ( ) ? e?H )H ? k + p1 kA? = (F^ ( ) ? e?H )H ? k
b(M 0 + M 0 = ):
Notice that the estimate M 0 one nds in the same way as (5.26). Since kHe?mH k d (H )=m , we obtain from (5.42) that
k(I ? F^ ( ))? = (F^ ( ) ? e?H )e?mH k bd (H )(M 0 + M 0 = )=m:
(5.43)
By (5.39)-(5.43) one gets the estimate
kF^ ( )n ? e?nH k (5.44)
r
r
n?
X
e2 = n ?c 1 b(2 + M = ) + e2 = n ? mc ? 1 bd (H )(M 0 + M 0 = ) m1
m
r
e2 = n ?c 1 b(2 + M = ) + e2 = pcbd (H )(M 0 + M 0 = ) 2plnnn ;
where we have used (3.31). Thus, setting
p
L^ := e2 = cbf2(2 + M = ) + 4d (H )(M 0 + M 0 = )g
we obtain (5.36).
2
1
1
1 2
8
8
1
1
1 2
1
1 2
1
1 2
1 2
8
1
1 2
8
1 2
8
1 2
1 2
1
1
1 2
1
1 2
1
1 2
1
1 2
1
8
8
1 2
1
1 2
=1
8
1 2
8
8
1 2
21
1
1 2
1
1 2
Corollary 5.9 Let A I be a self-adjoint operator. Let B be an m-sectorial operator of
semi-angle B , obeying conditions (1.1) and (1.2), with associated closed sectorial form b. If
in addition the condition (5.35) is satised, then there are L > 0 and L^ 0 > 0 such that
?
?tB=n ?tA=n n
?
tH
(5.45)
e
? e L lnpnn ;
e
?
?tA=n ?tB=n n
?
tH
e
? e L0 lnp n ;
(5.46)
e
n
uniformly in t 2 [0; 1).
Proof: We use the identity (cf.(5.2)) :
F ( )n ? e?tH = e?B e?A= (F^ ( )n? ? e?tH )e?A= +
(5.47)
?
B
?
A=
?
tH
?
A=
+(e ? I )e
e e
+
(5.48)
?
A=
?
tH
?
A=
+(e
? I )e e
+
(5.49)
?
tH
?
A=
+e (e
? I ):
(5.50)
Since F^ ( )n? ? e?tH = F^ ( )n? (I ? F^ ( )) + (F^ ( )n ? e?tH ), by estimates (5.3) and (5.36)
we obtain an estimate for the rst term (5.47). For the second term (5.48) we have
2
1
2
2
2
2
2
2
1
1
k(e?B ? I )e?A= e?tH e?A= k (5.51)
d (H ) d (H )a :
k(e?B ? I )B ? kkBA? kke?A= kkAH ? kkHe?tH k 1 a
?a t
n(1 ? a)
2
2
1
1
2
1
1
1
For the third term (5.49) we nd
k(e?A= ? I )e?tH e?A= k d (H ) d (H ) :
k(e?A= ? I )A? kkAH ? kkHe?tH k 2(1
? a)t 2n(1 ? a)
2
(5.52)
2
2
1
1
1
1
To estimate the last term (5.50) we use the identity A = A? = = I . Then
1 2
ke?tH A = A? = (e?A= ? I )k 1 2
1 2
2
1 2
ke?tH HR= kkHR? = A = kkpA? = (e?A= ? I )k
kHRe?tH kkA = HR? = k =e
p1e d (H ) = kA = HR? = kn? = ;
(5.53)
1 2
1 2
1 2
1 2
1
1 2
1 2
1 2
2
1 2
1 2
1 2
where we used estimate (5.18) and kA? = (e?A= ? I )k (=e) = for the self-adjoint operator A I . Notice that kA = HR? = k < 1 because dom(A = ) = dom(a) = dom(h) =
dom(HR= ). Taking into account the estimate of (5.47) and (5.51), (5.52), (5.53), we get
1 2
1 2
2
1 2
1 2
1 2
1 2
F ( )n
1=2
pn) :
? e?tH n ?c 1 + L^ lnpnn + nd(1(H?)aa) + 2nd(1(H?)a) + p1e kA = HR? = k d (H
1
1
1 2
1 2
1
(5.54)
22
which veries the estimate (5.45).
In order to obtain (5.46) for T ( ) = e?Ae?B , we use the identity :
T ( )n ? e?tH = e?A= (F^ ( )n? ? e?tH )e?A= e?B +
+e?A= e?tH e?A= (e?B ? I ) +
+(e?A= ? I )e?tH e?A= +
+e?tH (e?A= ? I ):
2
1
(5.55)
(5.56)
(5.57)
(5.58)
2
2
2
2
2
2
The last two terms (5.57), (5.58) are identical with those in (5.49), (5.50). To estimate the
rst term one has to proceed exactly as in the proof of (5.47). For the second term (5.56)
we get the estimate :
ke?A= e?tH e?A= (e?B ? I )k ke?tH A = kkA? = (e?B ? I )k
kA = e?tH kkA? = (e?B ? I )k
=
pn) (1 + tan B )kBR= A? = k2d (B ) = ;
kA = HR? = k d (H
2
2
1 2
1 2
1 2
1 2
1 2
1 2
1
1 2
1 2
1 2
1
1 2
where we have used Lemma 5.6, the estimate of ke?tH A = k as it is done in (5.53), and the
fact that dom(A = ) = dom(HR= ) dom(BR= ).
2
1 2
1 2
1 2
1 2
6 Remarks
Finally, let us make the following remarks:
(i) Taking into account Theorem 5.5 of [12], our Theorem 3.4 as well as results of the
recent papers of [4] and [5], it seems natural to believe that the following conjecture
is true: Let A be a self-adjoint operator such that A I , let B be an m-accretive
operator and let H be generator of a holomorphic semigroup such that the condition
dom(H 0 ) dom(A0 ) \ dom(B 0 )
(6.1)
holds for some 0 2 (0; 1] and condition (1.6) is valid for some 2 (0; 1]. If one has
0 + > 1, then the error-bound
estimate for the Trotter product formula convergence
0 ?
is of the order O(1=n
).
+
1
(ii) Notice that if B is a self-adjoint operator, then one has 0 = . Hence 0 + > 1
yields 0 = > 1=2. This assumption is used in [12].
(iii) If 0 = = 1 and B = B , then the error-bound estimate for the Trotter product
formula should be O(1=n) instead of O(ln(n)=n) as it was proved in [11]. Indeed, this
optimal error-bound is announced in [4] and [5]. We recall that in [12] the optimal
error-bound O(1=n ? ) is proved for 0 = 2 (1=2; 1) and B = B under condition
(1.6).
2
1
23
(iv) The0 conjecture (i) implicitly contains another one : the smallness of B 0 with respect
A , which was assumed in [11] and [12], is not indispensable. This conjecture relies
on recent results obtained in [4] and [5], which show that in the case B = B and
0 = = 1 the smallness of B with respect to A used in [11] is indeed not necessary.
For the case 0 = 2 (1=2; 1) studied in [12] this problem remains open.
(v) In application to the present accretive situation, i.e., when B is maximal accretive, this
conjecture gives a strong hint that in [2] as well as in the present paper the smallness
assumption (1.1) and (1.2) can be replaced by condition (6.1). For the beginning it
would be useful to study the case 0 = = 1, and after that to make an extension to
the general case 0 + > 1.
Acknowledgement
H.N. thanks Universite de la Mediterranee and Centre de Physique Theorique-CNRS-Luminy
for hospitality and for nancial support extended to him.
References
[1] Cachia, V.; Zagrebnov, V.A.: Operator-norm convergence of the Trotter product formula for holomorphic semigroups, to appear in J. Operator Theory (2001).
[2] Cachia, V.; Neidhardt, H.; Zagrebnov, V.A.: Accretive perturbations and error estimates for the Trotter product formula, to appear in Integr. Equ. Oper. Theory (2001).
[3] Foias, C.: Sz.-Nagy, B.: Harmonic analysis of operators on Hilbert spaces. Akademiai
Kiado, Budapest, 1970.
[4] Ichinose, T.; Tamura, Hideo.: The norm convergence of the Trotter-Kato product formula with error bound. Preprint (2000).
[5] Ichinose, T., Tamura, Hideo; Tamura, Hiroshi; Zagrebnov, V.A.: Note on the paper
\The norm convergence of the Trotter-Kato product formula with error bound" by
Ichinose and Tamura. Preprint CPT-00/P.4057 (2000).
[6] Kato, T.: Fractional powers of dissipative operators. J. Math. Soc. Japan 13 (1961),
246-274.
[7] Kato, T.: Fractional powers of dissipative operators. J. Math. Soc. Japan 14 (1962),
242-248.
[8] Kato, T.: A Generalization of the Heinz Inequality. Poc. Japan Acad. 37 (1961), 305308.
[9] Kato, T.: Perturbation theory for linear operators. Grundlehren der math. Wissenschaften, Vol. 132, Springer (1980).
[10] Miyazaki, Y.: Domains of square roots of regularly accretive operators. Proc. Japan.
Acad. 67(A) (1991), 38-42.
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[11] Neidhardt, H.; Zagrebnov, V.A.: On error estimates for the Trotter-Kato product formula. Lett. Math.Phys. 44 (1998), 169-186.
[12] Neidhardt, H.; Zagrebnov, V.A.: Fractional powers of self-adjoint operators and TrotterKato product formula. Integr. Equ. Oper. Theory 35 (1999), 209-231.
[13] Pazy, A.: Semigroups of linear operators and applications to partial dierential equations. Springer-Verlag 1983.
[14] Shimakura, N.: Sur les domaines des puissances fractionnaires d'operateurs. Bull. Soc.
Math. France 96 (1968), 265-288.
Vincent Cachia
Centre de Physique Theorique, CNRS-Luminy-Case 907,
F-13288 Marseille Cedex 9, France
Email: [email protected]
Hagen Neidhardt
Weierstra-Institut fur Angewandte Analysis und Stochastik,
Mohrenstr. 39, D-10117 Berlin, Germany
Email: [email protected]
Valentin A. Zagrebnov
Universite de la Mediterranee (Aix-Marseille II) and
Centre de Physique Theorique, CNRS-Luminy-Case 907,
F-13288 Marseille Cedex 9, France
Email: [email protected]
AMS Classication: 47D03, 47B25, 35K22, 41A80
25