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Landau's Type Inequalities
JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS ARTICLE NO.
202, 280]301 Ž1996.
0317
Landau’s Type Inequalities John M. Rassias Pedagogical Department, E. E. National Uni¨ ersity of Athens, 4, Agamemnonos Str., Aghia Paraske¨ i, Attikis 15342, Greece Submitted by A. M. Fink Received May 8, 1995
Let X be a complex Banach space, and let t ª T Ž t . Ž5 T Ž t .5 F 1, t G 0. be a strongly continuous contraction semigroup Žon X . with infinitesimal generator A. This paper proves that 5 Ax 5 4 F
1024 3
5 x 5 3 5 A4 x 5 ,
5 A2 x 5 4 F
10 4 9
5 x 5 2 5 A4 x 5 2 ,
5 A3 x 5 4 F 192 5 x 5 5 A4 x 5 3 hold for every x g DŽ A4 .. Inequalities are established also for uniformly bounded strongly continuous semigroups, groups, and cosine functions. Q 1996 Academic Press, Inc.
1. INTRODUCTION Edmund Landau w6x initiated the following extremum problem: The sharp inequality between the supremum-norms of derivatives of twice differentiable functions f such that 5 f 95 2 F 4 5 f 5 5 f 0 5
Ž q.
holds with norm referring to the space C w0, `x. Then R. R. Kallman and G.-C. Rota w3x found the more general result that inequality 5 Ax 5 2 F 4 5 x 5 5 A2 x 5
Ž 1.
holds for every x g DŽ A , and A the infinitesimal generator Ži.e., the strong right derivative of T at zero. of t ª T Ž t . Ž t G 0.: a semigroup of linear contractions on a complex Banach space X 2.
280 0022-247Xr96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
LANDAU’S TYPE INEQUALITIES
281
Z. Ditzian w1x achieved the better inequality 5 Ax 5 2 F 2 5 x 5 5 A2 x 5
Ž 2.
for every x g DŽ A2 ., where A is the infinitesimal generator of a group t ª T Ž t . Ž5 T Ž t .5 s 1, t g R. of linear isometries on X. Moreover H. Kraljevic´ and S. Kurepa w4x established the even shaper inequality 5 Ax 5 2 F
4 3
5 x 5 5 A2 x 5
Ž 3.
for every x g DŽ A2 ., and A the infinitesimal generator Ži.e., the strong right second derivative of T at zero. of t ª T Ž t . Ž t G 0.: a strongly continuous cosine function of linear contractions on X. Therefore the best Landau’s type constant is 43 Žfor cosine functions.. The above-mentioned inequalities Ž1. ] Ž3. were extended by H. Kraljevic´ and J. Pecaric ˇ ´ w5x so that new Landau’s type inequalities hold. In particular, they proved that 5 Ax 5 3 F
243 8
5 x 5 2 5 A3 x 5 ,
5 A2 x 5 3 F 24 5 x 5 5 A3 x 5 2 .
Ž 19 .
hold for every x g DŽ A3 ., where A is the infinitesimal generator of a strongly continuous contraction semigroup on X, Besides they obtained the analogous but better inequalities 5 Ax 5 3 F
9 8
5 x 5 2 5 A3 x 5 ,
5 A2 x 5 3 F 3 5 x 5 5 A3 x 5 2
Ž 29 .
which hold for every x g DŽ A3 ., where A is the infinitesimal generator of a strongly continuous contraction group on X. Moreover they got the set of analogous inequalities 5 Ax 5 3 F
81 40
5 x 5 2 5 A3 x 5 ,
5 A2 x 5 3 F
72 25
5 x 5 5 A3 x 5 2
Ž 39 .
for every x g DŽ A3 ., where A is the infinitesimal generator of a strongly continuous cosine function on X. In this paper, we extend the above inequalities Ž19. ] Ž39. so that other Landau’s inequalities hold for every x g DŽ A4 ., where A is infinitesimal generator of a uniformly bounded continuous semigroup Žresp. group, or cosine function..
282
JOHN M. RASSIAS
2. SEMIGROUPS Let t ª T Ž t . be uniformly bounded Ž5 T Ž t .5 F M - `, t G 0. strongly continuous semigroup of linear operators on X with infinitesimal generator A, such that T Ž0. s I Ž[ Identity. in B Ž X . [ the Banach algebra of bounded linear operators on X, lim t x 0 T Ž t . x s x, for every x, and Ax s lim
T Ž t. y I t
t x0
Ž s T 9 Ž 0. x .
x
Ž 4.
for every x in a linear subspace DŽ A. Ž[ Domain of A., dense in X w2x. For every x g DŽ A., we have the formula T Ž t. x s x q
t
H0 T Ž u . Ax du.
Ž 5.
Using integration by parts, we get the formula t
ž
u
H0 H0
T¨ A2 x d¨ du s
/
t
H0 Ž t y u . TuA x du. 2
Ž 6.
Employing Ž6. and iterating Ž5., we find for every x g DŽ A2 . that T Ž t . x s x q tAx q
t
H0 Ž t y u . TuA x du. 2
Ž 59 .
Similarly iterating Ž59., we obtain for every x g DŽ A4 . that T Ž t . x s x q tAx q
t2 2
A2 x q
t3 6
1
A3 x q
t
3
H Ž t y u . TuA x du. 6 0 4
Ž 50 .
THEOREM 1. Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t .5 F M - `, t G 0. strongly continuous semigroup of linear operators on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 Ax 5 F M
Ž Ž ts . 2 q Ž sr . 2 q Ž rt . 2 . q s 2 Ž rt y sr y st . tsr Ž t y s . Ž s y r . q
ts q sr q rt tsr
5 x5 q M
tsr 24
5 A4 x 5 ,
Ž 7.
LANDAU’S TYPE INEQUALITIES
5 A2 x 5 F 2 M
tqsqr Ž tr 2 q sr 2 q t 2 s q t 2 r . q s Ž rt y s 2 . 5 x5 q tsr Ž t y s . Ž s y r . tsr
qM 5 A3 x 5 F 6 M
283
ts q sr q rt 12
5 A4 x 5 ,
Ž 79 .
1 tqsqr Ž ts q sr q rt . y s 2 5 x5 q M 5 A4 x 5 , Ž 70 . q tsr Ž t y s . Ž s y r . tsr 4
hold for e¨ ery x g DŽ A4 . and for e¨ ery t, s, r g Rqs Ž0, `., 0 - t - s - r. THEOREM 2. Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t .5 F M - `, t G 0. strongly continuous semigroup of linear operators on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 32 5 Ax 5 4 F Mg 1 Ž m1 , m 2 . 5 x 5 3 5 A4 x 5 , Ž 8. 81 5 A2 x 5 4 F 5 A3 x 5 4 F
4 9
M 2 g 2 Ž m 1 , m 2 . 5 x 5 2 5 A4 x 5 2 ,
Ž 89 .
8
M 3 g 3 Ž m1 , m 2 . 5 x 5 5 A4 x 5 3 , Ž 80 . 9 hold for e¨ ery x g DŽ A4 ., and for some m1 , m 2 g Rq, m 2 ) m1 ) 1, where g 1 Ž m1 , m 2 . s Ž m1 m 2 . =
2 m12 q Ž m1 m 2 . q m22 . q m12 Ž m 2 y m1 m 2 y m1 . Ž M
m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . q
g 2 Ž m1 , m 2 . s Ž m1 q m1 m 2 q m 2 . = M
m1 q m1 m 2 q m 2 m1 m 2
3
,
2
Ž m22 q m1 m22 q m1 q m 2 . q m1 Ž m 2 y m12 . m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . q
g 3 Ž m1 , m 2 . s Ž 1 q m1 q m 2 .
3
M
1 q m1 q m 2 m1 m 2
2
,
1 Ž m1 q m1 m 2 q m 2 . y m12 q . m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . m1 m 2
284
JOHN M. RASSIAS
THEOREM 3. Let t ª T Ž t . be a strongly continuous contraction Ž5 T Ž t .5 F 1, t G 0. semigroup of linear operators on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 Ax 5 4 F 5 A2 x 5 4 F
1024
5 x 5 3 5 A4 x 5 ,
Ž 9.
5 x 5 2 5 A4 x 5 2 ,
Ž 99 .
3 10 4 9
5 A3 x 5 4 F 192 5 x 5 5 A4 x 5 3 ,
Ž 90 .
hold for e¨ ery x g DŽ A4 .. Proof of Theorem 1. In fact, formula Ž50 . yields the system
H0 Ž t y u . T Ž u . A x du¦ s 6 sAx q 3s A x q s A x s 6T Ž s . x y 6 x y H Ž s y u . T Ž u . A x du¥ t
6 tAx q 3t 2A2 x q t 3A3 x s 6T Ž t . x y 6 x y 2
2
3
3
3
4
3
4
0
6 rAx q 3r 2A2 x q r 3A3 x s 6T Ž r . x y 6 x y
H0 Ž r y u . T Ž u . A x du.§ r
3
4
Ž 10 . The coefficient determinant D of system Ž10. is D s 18tsr Ž t y s . Ž s y r . Ž r y t . .
Ž 11 .
It is clear that D is positive because of the hypothesis 0 - t - s - r. Therefore there is a unique solution of system Ž10. of the form 2
2
2
Ž sr . Ž r y s . T Ž t . x y Ž tr . Ž r y t . T Ž s . x q Ž ts . Ž s y t . T Ž r . x Ax s tsr Ž t y s . Ž s y r . Ž r y t . y
ts q sr q rt tsr
x y
r
4
H0 K Ž t , s, r ; u . T Ž u . A x du, 1
Ž 12 .
LANDAU’S TYPE INEQUALITIES
A2 x s 2
y Ž sr . Ž r 2 y s 2 . T Ž t . x q Ž tr . Ž r 2 y t 2 . T Ž s . x tsr Ž t y s . Ž s y r . Ž r y t . y q
A3 x s 6
285
Ž ts . Ž s 2 y t 2 . T Ž r . x tsr Ž t y s . Ž s y r . Ž r y t . tqsqr
x q
tsr
r
H0 K
2
Ž t , s, r ; u . T Ž u . A4 x du,
Ž 129.
Ž sr . Ž r y s . T Ž t . x y Ž tr . Ž r y t . T Ž s . x q Ž ts . Ž s y t . T Ž r . x y tsr Ž t y s . Ž s y r . Ž r y t . 1 tsr
x y
r
4
H0 K Ž t , s, r ; u . T Ž u . A x du, 3
Ž 120 .
where
¡ Ž sr . Ž r y s . Ž t y u. 2
3
2
y Ž tr . Ž r y t . Ž s y u .
3
6 tsr Ž t y s . Ž s y r . Ž r y t . 2
K1
s~
3
Ž ts . Ž s y t . Ž r y u . q , 6 tsr Ž t y s . Ž s y r . Ž r y t . 2
3
2
0FuFt
y Ž tr . Ž r y t . Ž s y u . q Ž ts . Ž s y t . Ž r y u .
3
6 tsr Ž t y s . Ž s y r . Ž r y t . 2
t F u F s,
,
3
Ž ts . Ž s y t . Ž r y u . , 6 tsr Ž t y s . Ž s y r . Ž r y t .
¢ ¡ Ž sr . Ž r
2
sFuFr
3
y s 2 . Ž t y u . y Ž tr . Ž r 2 y t 2 . Ž s y u .
3
3tsr Ž t y s . Ž s y r . Ž r y t . 3
K2
s~
Ž ts . Ž s 2 y t 2 . Ž r y u . q , 3tsr Ž t y s . Ž s y r . Ž r y t . 3
y Ž tr . Ž r 2 y t 2 . Ž s y u . q Ž ts . Ž s 2 y t 2 . Ž r y u . 3tsr Ž t y s . Ž s y r . Ž r y t .
¢
0FuFt 3
,
t F u F s,
3
Ž ts . Ž s 2 y t 2 . Ž r y u . , 3tsr Ž t y s . Ž s y r . Ž r y t .
sFuFr
286
JOHN M. RASSIAS
¡ Ž sr . Ž r y s . Ž t y u.
3
y Ž tr . Ž r y t . Ž s y u .
3
tsr Ž t y s . Ž s y r . Ž r y t . 3
K3 s
~
q
Ž ts . Ž s y t . Ž r y u . , tsr Ž t y s . Ž s y r . Ž r y t .
0FuFt
3
y Ž tr . Ž r y t . Ž s y u . q Ž ts . Ž s y t . Ž r y u .
3
,
tsr Ž t y s . Ž s y r . Ž r y t .
¢
t F u F s,
3
Ž ts . Ž s y t . Ž r y u . , tsr Ž t y s . Ž s y r . Ž r y t .
s F u F r.
It is obvious that K i s K i Ž t, s, r; u. G 0, i s 1, 2, 3, for every u g w0, r x Ž0 - t - s - r ., and that the following equalities r
H0 K
1
du s
tsr 24
r
,
H0 K
2
du s
ts q sr q rt 12
r
,
H0 K
3
du s
tqsqr 4
,
Ž 13 . hold. Note that Ž12. ] Ž120 . hold because the identities 2
2
¦
2
Ž sr . Ž r y s . y Ž tr . Ž r y t . q Ž ts . Ž s y t . s Ž t y s . Ž s y r . Ž r y t . Ž ts q sr q rt . , y Ž sr . Ž r 2 y s 2 . q Ž tr . Ž r 2 y t 2 . y Ž ts . Ž s 2 y t 2 . ¥ s yŽ t y s . Ž s y r . Ž r y t . Ž t q s q r . , Ž sr . Ž r y s . y Ž tr . Ž r y t . q Ž ts . Ž s y t . s Ž t y s. Ž s y r . Ž r y t .
Ž 14 .
§
hold. Therefore from formulas Ž12. ] Ž120 ., Ž13., and the triangle inequality, we get inequalities Ž7. ] Ž70 .. This completes the proof of Theorem 1. Proof of Theorem 2. Setting s s m1 t ,
r s m2 t ,
m 2 ) m1 ) 1,
t)0
Ž 15 .
in Ž7. ] Ž70 ., we obtain the following inequalities 5 Ax 5 F a1
1 t
q b1 t 3 ,
5 A2 x 5 F a 2
1 t
2
q b2 t 2 ,
5 A3 x 5 F a 3
1 t3
q b3 t ,
Ž 16 .
LANDAU’S TYPE INEQUALITIES
287
where a1 s
2 m12 q Ž m1 m 2 . q m22 . q m12 Ž m 2 y m1 m 2 y m1 . Ž M
m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . q b1 s M
a2 s 2 M
m1 m 2 24
m1 q m1 m 2 q m 2 m1 m 2
5 x 5,
5 A4 x 5 ,
Ž m22 q m1 m22 q m1 q m 2 . q m1 Ž m 2 y m12 . m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . q b2 s M
a3 s 6 M
m1 q m1 m 2 q m 2 12
1 q m1 q m 2 m1 m 2
5 x 5,
5 A4 x 5 ,
1 Ž m1 q m1 m 2 q m 2 . y m12 5 x 5, q m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . m1 m 2 b3 s M
1 q m1 q m 2 4
5 A4 x 5 .
Minimizing the right-hand side functions of t of Ž16., we get the sharper inequalities 5 Ax 5 4 F
256 27
a13 b1 ,
5 A2 x 5 4 F 16 a22 b 22 ,
5 A3 x 5 4 F
256 27
a3 b 33 . Ž 17 .
But a13 b1 s
M 24
g 1 Ž m1 , m 2 . 5 x 5 3 5 A4 x 5 ,
a22 b 22 s
M2 36
g 2 Ž m1 , m 2 . 5 x 5 2 5 A4 x 5 2 ,
and a3 b 33 s
3M 3 32
g 3 Ž m1 , m 2 . 5 x 5 5 A4 x 5 3 .
Therefore from Ž15. ] Ž17., we obtain inequalities Ž8. ] Ž80 .. This completes the proof of Theorem 2.
288
JOHN M. RASSIAS
Proof of Theorem 3. Taking M s 1, we have g 1 Ž m 1 , m 2 . s 8 gq 1 Ž m1 , m 2 . ,
g 2 Ž m1 , m 2 . s 4 gq 2 Ž m1 , m 2 . ,
g 3 Ž m 1 , m 2 . s 2 gq 3 Ž m1 , m 2 . , where gq 1
m1 Ž 1 q m 2 q m22 y m1 y m1 m 2 .
Ž m1 , m 2 . s Ž m1 m 2 .
3
,
m 2 Ž m1 y 1 . Ž m 2 y m1 .
gq 2 Ž m1 , m 2 . s Ž m1 q m1 m 2 q m 2 . gq 3 Ž m1 , m 2 . s Ž 1 q m1 q m 2 .
2
1 q m 2 q m22 y m12
2
,
m 2 Ž m1 y 1 . Ž m 2 y m1 . 1 q m 2 y m1
3
m 2 Ž m1 y 1 . Ž m 2 y m1 .
.
Hence inequalities Ž8. ] Ž80 . are written, as 5 Ax 5 4 F 5 A2 x 5 4 F 5 A3 x 5 4 F
256 81 16 9 16 9
5 5 3 5 A4 x 5 , gq 1 Ž m1 , m 2 . x
Ž 18 .
5 5 2 5 A4 x 5 2 , gq 2 Ž m1 , m 2 . x
Ž 189.
5 5 5 A4 x 5 3 , gq 3 Ž m1 , m 2 . x
Ž 180 .
for some m1 , m 2 g Rq: m 2 ) m1 ) 1. qŽ . All functions gq i s g i m1 , m 2 , i s 1, 2, 3, attain their minimum at the same m1 , m 2 : m1 s 2 q '2 , m 2 s 3 q 2'2 , so that q min gq 1 Ž m1 , m 2 . s 108 s min g 3 Ž m1 , m 2 . ,
min gq 2 Ž m1 , m 2 . s 625.
Ž 19 . Therefore inequalities Ž18. ] Ž180 . and minima Ž19. yield the even sharper inequalities Ž9. ] Ž90 .. This completes the proof of Theorem 3.
3. GROUPS Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t .5 F M - `, t g R s Žy`, `.. strongly continuous group of linear operators on X with infinitesimal generator A. It is clear that analogous inequalities Žas those in
LANDAU’S TYPE INEQUALITIES
289
the aforementioned Theorems 1]3. hold for every t, s, r g Rys Žy`, 0., t - s - r - 0. s - 0 - t - r. Denote
Case I.
s s m1 t ,
r s m2 t ,
m1 - 0,
m 2 ) 1,
t ) 0,
Ž 159.
and x 1 s 6 tAx,
x 2 s 3t 2A2 x,
x 3 s t 3A3 x,
Ž 20 .
as well as a s 6T Ž t . x y 6 x y b s 6T Ž m1 t . x y 6 x y
ž
3
m1 t
H0
s 6T Ž m1 t . x y 6 x y m14
c s 6T Ž m 2 t . x y 6 x y
t
H0 Ž t y u . T Ž u . A x du, t
4
3 Ž m1 t y u . T Ž u . A4 x du
3
H0 Ž t y u . T Ž m u . A x du 4
1
m2 t
H0
/
,
3 Ž m 2 t y u . T Ž u . A4 x du.
Then system Ž10. takes the form x 1 q x 2 q x 3 s a,
m1 x 1 q m12 x 2 q m13 x 3 s b,
m 2 x 1 q m22 x 2 q m32 x 3 s c.
Ž 109.
Solving system Ž109., we find the unique solution 2
x1 s x2 s x3 s
Ž m1 m 2 . Ž m 2 y m1 . a y m22 Ž m 2 y 1 . b q m12 Ž m1 y 1 . c m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . y Ž m1 m 2 . Ž m22 y m12 . a q m 2 Ž m22 y 1 . b y m1 Ž m12 y 1 . c m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
Ž m1 m 2 . Ž m 2 y m1 . a y m 2 Ž m 2 y 1 . b q m1 Ž m1 y 1 . c . m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
Ž 21 . Ž 219. Ž 210 .
THEOREM 4. Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t .5 F M - `, t g R. strongly continuous group of linear operators on complex Banach space
290
JOHN M. RASSIAS
X with infinitesimal generator A, such that A4 x / 0. Then 5 Ax 5 F M
2 Ž m1 m 2 . q m12 q m22 y m1 q m1 m 2 y m 2 m1 m 2 Ž m1 y 1 . Ž m 2 y 1 .
q qM
ž
M 12
5 A3 x 5 F 6 M
5 x5
m1 m 2
1 t
Ž m1 m 2 . Ž 1 q m1 y m1 m 2 q m 2 . 4 3 5 A x5t , 24 Ž m1 y 1 . Ž m 2 y 1 .
5 A2 x 5 F 2 Ž M q 1 . y q
m1 q m1 m 2 q m 2
1 q m1 q m 2 m1 m 2
/
5 x5
1 t2
Ž y Ž m 1 q m 1 m 2 q m 2 . . 5 A4 x 5 t 2 , m1 q m1 m 2 q m 2 y m22
m1 m 2 Ž m 2 y 1 . Ž m 2 y m1 .
Ž 22 .
y
1 m1 m 2
Ž 229. 5 x5
1 t3
2
qM
ym12 m 2 y m1 m 2 q m1 m22 q Ž m1 m 2 . q m22 q m42 4 m 2 Ž m 2 y 1 . Ž m 2 y m1 .
5 A4 x 5 t ,
Ž 220 . hold for e¨ ery x g DŽ A , for e¨ ery t g R , and for some m1 g Ry, m 2 g Rq, m2 y Ž m 2 q 1 . - m1 - y , m 2 ) 1. m2 q 1 THEOREM 5. Let t ª T Ž t . be a contraction Ž5 T Ž t .5 F 1, t g R. strongly continuous group of linear operators on complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 256 5 Ax 5 4 F f Ž m , m . 5 x 5 3 5 A4 x 5 , Ž 23 . 81 1 1 2 16 5 A2 x 5 4 F f Ž m , m . 5 x 5 2 5 A4 x 5 2 , Ž 239. 9 2 1 2 16 5 A3 x 5 4 F f Ž m , m . 5 x 5 5 A4 x 5 3 , Ž 230 . 9 3 1 2 hold for e¨ ery x g DŽ A4 ., and for some m1 g Ry, m 2 g Rq, m2 y Ž m 2 q 1 . - m1 - y , m 2 ) 1, m2 q 1 4.
q
LANDAU’S TYPE INEQUALITIES
291
where f 1 Ž m1 , m 2 . s
žŽ
4
m1 m 2
/Ž
m1 y 1 . Ž m 2 y 1 .
f 2 Ž m1 , m 2 . s
ž
1 q m1 q m 2 m1 m 2
1 q m1 y m1 m 2 q m 2 . ,
2
/
2 Ž m1 q m1 m 2 q m 2 . ,
f 3 Ž m1 , m 2 . s
2 Ž 1 q m1 y m 2 . Ž ym12 m 2 y m1 m 2 q m1 m22 q Ž m1 m 2 . q m22 q m42 .
Ž m1 m32 . Ž Ž m 2 y 1. Ž m 2 y m1 . .
4
3
.
THEOREM 6. Let t ª T Ž t . be a strongly continuous contraction Ž5 T Ž t .5 F 1, t g R. group of linear operators on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 4 5 Ax 5 4 F 10 5 x 5 3 5 A4 x 5 , Ž 24 . 6
ž /
5 A2 x 5 4 F
16 9
5 A3 x 5 4 F 10
5 x 5 2 5 A4 x 5 2 , 54 13 3 2 5 36
5 x 5 5 A4 x 5 3 ,
Ž 249. Ž 240 .
hold for e¨ ery x g DŽ A4 .. THEOREM 7. Let t ª T Ž t . be a strongly continuous contraction Ž5 T Ž t .5 F 1, t g R. group of linear operators on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 Ax 5 4 F 5 A2 x 5 4 F 5 A x5 F 3
4
m520
32
81 Ž m 20 y 1 . 4 16 9
5 x 5 3 5 A4 x 5 ,
5 x 5 2 5 A4 x 5 2 ,
16 m420 Ž 1 q m220 . 9
Ž
m220
m 20 s
(
y 1.
Ž 259. 3
4
5 x 5 5 A4 x 5 3 ,
hold for e¨ ery x g DŽ A4 ., where 7 q '57 2
Ž 25 .
.
Ž 250 .
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JOHN M. RASSIAS
Proof of Theorem 4. In fact, from Ž20. and Ž21. ] Ž210 ., we get x1 Ax s 6t 2
Ž m1 m 2 . Ž m 2 y m1 . T Ž t . x y m22 Ž m 2 y 1 . T Ž m1 t . x s m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
ž
q
m12 Ž m1 y 1 . T Ž m 2 t . x m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . 2
Ž m1 m 2 . Ž m 2 y m1 . y m22 Ž m 2 y 1 . q m12 Ž m1 y 1 . 1 y x m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . t
/
2
t
y
H0 Ž t y u .
3
Ž m1 m 2 . Ž m 2 y m1 . T Ž u . 6 m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
ž
y q A2 x s
m22 Ž m 2 y 1 . m14 T Ž m1 u . 6 m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
m12 Ž m1 y 1 . m42 T Ž m 2 u . 6 m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
/
1
A4 x du
t
, Ž 26 .
x2 3t 2
s2
ym1 m 2 Ž m22 y m12 . T Ž t . x q m 2 Ž m22 y 1 . T Ž m1 t . x
ž
m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . y y
y
m1 Ž m12 y 1 . T Ž m 2 t . x m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . ym1 m 2 Ž m22 y m12 . q m 2 Ž m22 y 1 . y m1 Ž m12 y 1 . m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . t
H0 Ž t y u .
3
ž
/
1 t2
ym1 m 2 Ž m22 y m12 . T Ž u . 3m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . q
y
x
m 2 Ž m22 y 1 . m14 T Ž m1 u . 3m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
m1 Ž m12 y 1 . m42 T Ž m 2 u . 3m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
/
A4 x du
1 t2
, Ž 269 .
LANDAU’S TYPE INEQUALITIES
A3 x s
293
x3 t3
s6
m1 m 2 Ž m 2 y m1 . T Ž t . x y m 2 Ž m 2 y 1 . T Ž m1 t . x
ž
m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . q y
y
m1 Ž m1 y 1 . T Ž m 2 t . x m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . m 1 m 2 Ž m 2 y m 1 . y m 2 Ž m 2 y 1 . q m1 Ž m1 y 1 . m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . t
H0
Ž t y u. q
3
ž
x
/
1 t3
m1 m 2 Ž m 2 y m1 . T Ž u . y m 2 Ž m 2 y 1 . m14 T Ž m1 u . m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
m1 Ž m1 y 1 . m42 T Ž m 2 u . m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
/
A4 x du
1 t3
.
Ž 260 .
But it is clear that the following identities 2
Ž m1 m 2 . Ž m 2 y m1 . y m22 Ž m 2 y 1 . q m12 Ž m1 y 1 . s Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . Ž m1 q m1 m 2 q m 2 . , ym1 m 2 Ž m22 y m12 . q m 2 Ž m22 y 1 . y m1 Ž m12 y 1 . s y Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . Ž 1 q m1 q m 2 . , ym1 m 2 Ž m22 y m12 . q m 2 Ž m22 y 1 . m14 y m1 Ž m12 y 1 . m42 s ym1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . Ž m1 q m1 m 2 q m 2 . hold. Applying these identities and formulas Ž26. ] Ž260 ., we obtain inequalities Ž22. ] Ž220 .. This completes the proof of Theorem 4. Proof of Theorem 5. In fact, from inequalities Ž22. ] Ž220 . we get 1 1 3 q 2 5 Ax 5 F a1q q bq 5 A2 x 5 F aq 1 t , 2 2 q b2 t , t t 5 A3 x 5 F aq 3 where a1q s 2 bq 1 s
1 t3
q bq 3 t,
m1 m 2
Ž m1 y 1 . Ž m 2 y 1 .
Ž 27 . 5 x 5,
Ž m1 m 2 . Ž 1 q m1 y m1 m 2 q m 2 . 4 5 A x 5, 24 Ž m1 y 1 . Ž m 2 y 1 .
294
JOHN M. RASSIAS
aq 2 s Ž y4 . bq 2 sy
1 12
aq 3 s 12
1 q m1 q m 2 m1 m 2
5 x 5,
Ž m1 q m1 m 2 q m 2 . 5 A4 x 5 , 1 q m1 y m 2
m1 Ž m 2 y 1 . Ž m 2 y m1 .
5 x 5,
2
bq 3 s
ym12 m 2 y m1 m 2 q m1 m22 q Ž m1 m 2 . q m22 q m42 4 m 2 Ž m 2 y 1 . Ž m 2 y m1 .
5 A4 x 5 ,
and new identities m1 m 2 Ž m1 y m 2 . q m 2 Ž m 2 y 1 . q m1 Ž m1 y 1 . s Ž m1 y 1 . Ž m1 q m1 m 2 q m 2 y m22 . , m1 m 2 Ž m1 y m 2 . q m 2 Ž m 2 y 1 . m14 q m1 Ž m1 y 1 . m42 2
s m1 Ž m1 y 1 . Ž ym12 m 2 y m1 m 2 q m1 m22 q Ž m1 m 2 . q m22 q m42 . . Minimizing the right-hand side of Ž27., we find 5 Ax 5 4 F
256 27
2 q 2 5 A2 x 5 4 F 16 Ž aq 2 . Ž b2 . ,
3
Ž a1q . Ž bq 1 ., 5 A3 x 5 4 F
256 27
q 3 Ž aq 3 . Ž b3 . ,
Ž 28 .
where 3
Ž a1q . Ž bq 1 . s
8 24
f 1 Ž m1 , m 2 . ,
1
2 q 2 f 2 Ž m1 , m 2 . , Ž aq 2 . Ž b2 . s
9
q 3 Ž aq 3 . Ž b3 . s
3 16
f 3 Ž m1 , m 2 . .
Therefore inequalities Ž28. yield inequalities Ž23. ] Ž230 .. This completes the proof of Theorem 5. Proof of Theorem 6. Setting m1 s y1, we get min f 1Žy1, m 2 . s 10Ž 58 . 4 at m 2 s 5. Hence f 2 Žy1, 5. s 1, f 3 Žy1, 5. s 5 4 Ž13. 3r2 9 3 4 . Therefore from formulas Ž23. ] Ž230 . and m1 s y1, m 2 s 5, we get inequalities Ž24. ] Ž240 .. This completes the proof of Theorem 6.
LANDAU’S TYPE INEQUALITIES
295
Proof of Theorem 7. In fact, min f 3 Žy1, m 2 . s f 3 Žy1, m 20 . s m420 Ž1 q m 20 s Ž 7 q '57 . r2 Žs root Ž) 1. of equaWe have
'
m220 . 3rŽ m220 y 1. 4 , where tion m42 y 7m22 y 2 s 0..
f 1 Ž y1, m 20 . s
m520
1
8 Ž m 20 y 1 . 4
.
Therefore from formulas Ž24. ] Ž240 ., and m1 s y1, m 2 s m 20 , we obtain inequalities Ž25. ] Ž250 .. This completes the proof of Theorem 7. Case II. r - s - 0 - t. Denote s s m1 t, r s m 2 t, m 2 s m - y1, m1 s y1, t ) 0. From Ž21. ] Ž210 ., we have m2 Ž m q 1 . a y m2 Ž m y 1 . b y 2 c
x1 s
2 m Ž m2 y 1 . x2 s
aqb
, 2 ym Ž m q 1 . a y m Ž m y 1 . b q 2 c
x3 s
,
2 m Ž m2 y 1 .
.
Ž 29 .
Therefore from Ž29. and Ž20., we obtain Ax s
ž
ym2 Ž m q1 . T Ž t . x q m2 Ž m y1 . T Ž y t . x q 2T Ž mt . x 2 m Ž 1 ym2 .
q
t
žH
Ž t y u.
0
3
ž
y A3 x s 6
ž
q
2 t
0
Ž t y u.
3
yx
/
2 m4 T Ž mu . 12 m Ž 1 y m . 2
A4 x du
T Ž u . q T Ž yu . 6
A4 x du
/
1 t2
3
/
1 t
,
, Ž 30 .
,
Ž 309.
2 mŽ 1 y m .
ž
t
t2
2
t
/
1
1
m Ž m q 1 . T Ž t . x q m Ž m y 1 . T Ž yt . x y 2T Ž mt . x
H0 Ž t y u .
m
x
12 m Ž 1 y m2 .
T Ž t . x q T Ž yt . x
žH
1
m2 Ž m q 1 . T Ž u . y m2 Ž m y 1 . T Ž yu .
y A2 x s 2
y
q
1 m
x
/
1 t3
ym Ž m q 1 . T Ž u . y m Ž m y 1 . T Ž yu . 2 m Ž 1 y m2 . q
2 m4 T Ž mu . 2 mŽ 1 y m . 2
A4 x du
/
1 t3
. Ž 300 .
296
JOHN M. RASSIAS
Thus from formulas Ž30. ] Ž300 . we get 5 Ax 5 F
m mq1
5 A2 x 5 F 4 5 x 5 5 A3 x 5 F 12
5 x5
1 t
2
q
m
1 t
ž
q y
1 12
m2 24 Ž m q 1 .
/
5 A4 x 5 t 3 ,
5 A4 x 5 t 2 , 1
Ž 319.
1 m Ž 1 q m2 .
5 A4 x 5 t. Ž 310 . 1 y m2 Minimizing the right-hand side of inequalities Ž31. ] Ž310 ., we find 1ym
5 x5 2
Ž 31 .
5 Ax 5 4 F 5 A2 x 5 4 F 5 A3 x 5 4 F
32 81 16 9 16 9
t3
q
4
h 1 Ž m . 5 x 5 3 5 A4 x 5 ,
Ž 32 .
5 x 5 2 5 A4 x 5 2 ,
Ž 329.
h 3 Ž m . 5 x 5 5 A4 x 5 3 ,
Ž 320 .
where h1 Ž m . s y
m5
Ž m q 1.
4
,
h3 Ž m. s
m4 Ž 1 q m2 .
Ž 1 y m2 .
3
4
,
m - y1.
First, minimizing h1Ž m., m - y1, we get m s y5. Then inequalities Ž32. ] Ž320 . Žwith m s y5. are the same as the inequalities Ž24. ] Ž240 .. Finally, minimizing h 3 Ž m., m - y1, we obtain m s m 0 s y Ž 7 q '57 . r2 . Then inequalities Ž32. ] Ž320 . Žwith m s m 0 . are the same as the inequalities Ž25. ] Ž250 ..
'
4. COSINE FUNCTIONS Let t ª T Ž t . Ž t G 0. be a uniformly bounded Ž5 T Ž t .5 F M - `, t G 0. strongly continuous cosine function with infinitesimal operator A, such that T Ž0. s I Ž[ identity. in B Ž X ., lim t x 0 T Ž t . x s x, ; x, and A is defined as the strong second derivatives of T at zero, Ax s T 0 Ž 0 . x
Ž 33 .
for every x in a linear subspace DŽ A., which is dense in X w5x. For every x g DŽ A., we have the formula T Ž t. x s x q
t
H0 Ž t y u . T Ž u . Ax du.
Ž 34 .
LANDAU’S TYPE INEQUALITIES
297
Using integration by parts, we get from Ž34. the formula t
u
H0 Ž t y u . H0
ž
Ž u y ¨ . f Ž ¨ . d¨ du s
/
1
t
H Žty¨. 6 0
3
f Ž ¨ . d¨ ,
Ž 35 .
where f Ž ¨ . s T¨ A2 x. Note the Leibniz formula: d du
u
žH Ž 0
n
u
u y ¨ . f Ž ¨ . d¨ s n
/ žH Ž 0
uy¨.
ny1
f Ž ¨ . d¨ .
/
Ž 36 .
Employing Ž35. ] Ž36. and iterating Ž34., we find for every x g DŽ A2 . that
T Ž t. x s x q
t2 2!
Ax q
1
t
3
H Ž t y u . T Ž u . A x dx. 3! 0 2
Ž 349.
Similarly iterating Ž349. we obtain for every x g DŽ A4 . that
T Ž t. x s x q
t2 2!
Ax q
t4 4!
A2 x q
t6 6!
A3 x q
1 7!
t
H0 Ž t y u .
7
T Ž u . A4 x du.
Ž 340 . THEOREM 8. Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t . F- M - `, t G 0. strongly continuous cosine function on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 2 2 2 ts q sr q rt Ž ts . q Ž sr . q Ž rt . q s 2 Ž rt y sr y st . 5 Ax 5 F 2 M 5 x5 q tsr Ž t y s . Ž s y r . tsr
qM 5 A2 x 5 F 24 M qM
tsr 20160
5 A4 x 5 ,
Ž 37 .
tqsqr Ž ts 2 q sr 2 q t 2 s q t 2 r . q s Ž rt y s 2 . 5 x5 q tsr Ž t y s . Ž s y r . tsr ts q sr q rt 1680
5 A4 x 5 ,
Ž 379.
298
JOHN M. RASSIAS
5 A3 x 5 F 720 M qM
1 Ž ts q sr q rt . y s 2 5 x5 q tsr Ž t y s . Ž s y r . tsr
tqsqr 56
5 A4 x 5 ,
Ž 370 .
hold for e¨ ery x g DŽ A4 ., and for e¨ ery t, s, r g Rq, 0 - t - s - r. THEOREM 9. Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t .5 F M - `, t G 0. strongly continuous cosine function on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 Ax 5 4 F 5 A2 x 5 4 F 5 A3 x 5 4 F
32 8505 4 1225 40 1029
Mg 1 Ž m1 , m 2 . 5 x 5 3 5 A4 x 5 ,
Ž 38 .
M 2 g 2 Ž m1 , m 2 . 5 x 5 2 5 A4 x 5 2 ,
Ž 389.
M 3 g 3 Ž m1 , m 2 . 5 x 5 5 A4 x 5 3 ,
Ž 380 .
hold for e¨ ery x g DŽ A4 ., and for some m1 , m 2 g Rq, m 2 ) m1 ) 1, where g i s g i Ž m1 , m 2 . are the same as those g i , i s 1, 2, 3, in Theorem 2. THEOREM 10. Let t ª T Ž t . be a strongly continuous contraction Ž5 T Ž t .5 F 1, t G 0. cosine function on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 Ax 5 4 F 5 A2 x 5 4 F 5 A3 x 5 4 F
1024
5 x 5 3 5 A4 x 5 ,
Ž 39 .
5 x 5 2 5 A4 x 5 2 ,
Ž 399.
315 400 49
2880 343
5 x 5 5 A4 x 5 3 ,
Ž 390 .
hold for e¨ ery x g DŽ A4 .. Proof of Theorem 8. In fact, setting t Ž) 0. instead of t 2 in Ž340 ., we get T Ž 't . x s x q q
t 2 1
Ax q
t2 24
A2 x q 7
t3 720
A3 x
H't Ž't y u . T Ž u . A x du. 5040 0 4
Ž 34-.
LANDAU’S TYPE INEQUALITIES
299
Formula Ž34-. yields 2520tAx q 210t 2A2 x q 7t 3A3 x s 5040T Ž 't . x y 5040 x y
7
H0't Ž't y u . T Ž u . A x du 4
2520 sAx q 210 s 2A2 x q 7s 3A3 x s 5040T Ž 's . x y 5040 x y
H0's Ž's y u . T Ž u . A x du 7
4
Ž 109.
2520rAx q 210r 2A2 x q 7r 3A3 x s 5040T Ž 'r . x y 5040 x y
H0'r Ž'r y u . T Ž u . A x du. 7
4
The coefficient determinant Dq of system Ž109. is Dqs 3704400tsr Ž t y s . Ž s y r . Ž r y t . Ž s Dr205800. .
Ž 119.
It is clear that Dq) 0 because 0 - t - s - r. Therefore there is a unique solution of system Ž109. of the form 2
Ax s 2
2
Ž sr . Ž r y s . T Ž 't . x y Ž tr . Ž r y t . T Ž 's . x tsr Ž t y s . Ž s y r . Ž r y t .
ž
q
H0'r K
q 1
y A x s 24 2
ž
2 ts q sr q rt Ž ts . Ž s y t . T Ž 'r . x y x tsr Ž t y s . Ž s y r . Ž r y t . tsr
Ž t , s, r ; u . T Ž u . A4 x du,
ysr Ž r 2 y s 2 . T Ž 't . x q tr Ž r 2 y t 2 . T Ž 's . x tsr Ž t y s . Ž s y r . Ž r y t . ts Ž s 2 y t 2 . T Ž 'r . x
y
tsr Ž t y s . Ž s y r . Ž r y t .
H0'r K
q 2
q
A3 x s 720
Ž 40 .
ž
q
tqsqr tsr
x
/
Ž t , s, r ; u . T Ž u . A4 x du,
Ž 409.
sr Ž r y s . T Ž 't . x y tr Ž r y t . T Ž 's . x q ts Ž s y t . T Ž 'r . x tsr Ž t y s . Ž s y r . Ž r y t . y
1 tsr
/
x y
H0'r K
q 3
Ž t , s, r ; u . T Ž u . A4 x du, Ž 400 .
/
300
JOHN M. RASSIAS
where
¡ Ž sr . Ž r y s . Ž't y u . 2
7
2
y Ž tr . Ž r y t . Ž 's y u .
7
2520tsr Ž t y s . Ž s y r . Ž r y t . 7
2
Kq 1
s
Ž ts . Ž s y t . Ž 'r y u . q , ~ 2520tsr Ž t y s . Ž s y r . Ž r y t . 7
2
0 F u F 't ,
2
y Ž tr . Ž r y t . Ž 's y u . q Ž ts . Ž s y t . Ž 'r y u . 2520tsr Ž t y s . Ž s y r . Ž r y t .
¢
2
,
't F u F 's ,
7
2
Ž ts . Ž s y t . Ž 'r y u . , 2520tsr Ž t y s . Ž s y r . Ž r y t .
¡sr Ž r
7
's F u F 'r ,
7
y s 2 . Ž 't y u . y Ž tr . Ž r 2 y t 2 . Ž 's y u .
7
210tsr Ž t y s . Ž s y r . Ž r y t .
Kq 2
s
~
q
ts Ž s 2 y t 2 . Ž 'r y u .
7
210tsr Ž t y s . Ž s y r . Ž r y t .
0 F u F 't ,
,
7
ytr Ž r 2 y t 2 . Ž 's y u . q ts Ž s 2 y t 2 . Ž 'r y u . 210tsr Ž t y s . Ž s y r . Ž r y t . ts Ž s 2 y t 2 . Ž 'r y u .
7
,
't F u F 's ,
7
¢210tsr Ž t y s . Ž s y r . Ž r y t . , ¡sr Ž r y s . Ž't y u .
7
's F u F 'r ,
y tr Ž r y t . Ž 's y u .
7
7tsr Ž t y s . Ž s y r . Ž r y t .
Kq 3 s
~
q
ts Ž s y t . Ž 'r y u .
7
7tsr Ž t y s . Ž s y r . Ž r y t . 7
ytr Ž r y t . Ž 's y u . q ts Ž s y t . Ž 'r y u . 7tsr Ž t y s . Ž s y r . Ž r y t . ts Ž s y t . Ž 'r y u .
0 F u F 't ,
, 7
,
't F u F 's ,
7
¢7tsr Ž t y s . Ž s y r . Ž r y t . ,
's F u F 'r .
LANDAU’S TYPE INEQUALITIES
301
w ' x Ž0 - t - s - r . and Note that Kq i G 0, i s 1, 2, 3, for every u g 0, r
H0'r K
q 1
du s
tsr 20160
H0'r K
,
q 3
H0'r K
q 2
du s
du s
tqsqr 56
.
ts q sr q rt 1680
,
Ž 139.
From Ž40. ] Ž400 ., triangle inequality, Ž139., Ž15. and similarly as in the previous Section 2 on semigroups, we get inequalities Ž37. ] Ž370 .. This completes the proof of Theorem 8. Proof of Theorem 9. From Ž15. and Ž37. ] Ž370 . and similar calculations as in Section 2 on semigroups, we find inequalities Ž38. ] Ž380 .. This completes the proof of Theorem 9. Proof of Theorem 10. Setting M s 1, using Ž38. ] Ž380 ., and minimizing Ž g i m1 , m 2 ., i s 1, 2, 3, as in Section 2 on semigroups, we obtain inequalities Ž39. ] Ž390 .. This completes the proof of Theorem 10.
REFERENCES 1. Z. Ditzian, Some remarks on inequalities of Landau and Kolmogorov, Aequationes Math. 12 Ž1975., 145]151. 2. E. Hille and R. S. Phillips, ‘‘Functional Analysis and Semigroups,’’ Amer. Math. Soc. Colloq. Publ., Vol. 31, AMS, Providence, Rhode Island, 1957. 3. R. R. Kallman and G.-C. Rota, On the inequality 5 f 9 5 2 F 4 5 f 5 5 f 0 5, in ‘‘Inequalities, II’ ŽO. Shisha, Ed.., pp. 187]192, Academic Press, New YorkrLondon, 1970. 4. H. Kraljevic´ and S. Kurepa, Semigroups on Banach spaces, Glas. Mat. 5 Ž1970., 109]117. 5. H. Kraljevic´ and J. Pecaric, ˇ ´ Some Landau’s type inequalities for infinitesimal generators, Aequationes Math. 40 Ž1990., 147]153. 6. E. Landau, Einige Ungleichungen fur ¨ zweimal differentzierbare Funktionen, Proc. London Math. Soc. (2) 13 Ž1913., 43]49.
202, 280]301 Ž1996.
0317
Landau’s Type Inequalities John M. Rassias Pedagogical Department, E. E. National Uni¨ ersity of Athens, 4, Agamemnonos Str., Aghia Paraske¨ i, Attikis 15342, Greece Submitted by A. M. Fink Received May 8, 1995
Let X be a complex Banach space, and let t ª T Ž t . Ž5 T Ž t .5 F 1, t G 0. be a strongly continuous contraction semigroup Žon X . with infinitesimal generator A. This paper proves that 5 Ax 5 4 F
1024 3
5 x 5 3 5 A4 x 5 ,
5 A2 x 5 4 F
10 4 9
5 x 5 2 5 A4 x 5 2 ,
5 A3 x 5 4 F 192 5 x 5 5 A4 x 5 3 hold for every x g DŽ A4 .. Inequalities are established also for uniformly bounded strongly continuous semigroups, groups, and cosine functions. Q 1996 Academic Press, Inc.
1. INTRODUCTION Edmund Landau w6x initiated the following extremum problem: The sharp inequality between the supremum-norms of derivatives of twice differentiable functions f such that 5 f 95 2 F 4 5 f 5 5 f 0 5
Ž q.
holds with norm referring to the space C w0, `x. Then R. R. Kallman and G.-C. Rota w3x found the more general result that inequality 5 Ax 5 2 F 4 5 x 5 5 A2 x 5
Ž 1.
holds for every x g DŽ A , and A the infinitesimal generator Ži.e., the strong right derivative of T at zero. of t ª T Ž t . Ž t G 0.: a semigroup of linear contractions on a complex Banach space X 2.
280 0022-247Xr96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
LANDAU’S TYPE INEQUALITIES
281
Z. Ditzian w1x achieved the better inequality 5 Ax 5 2 F 2 5 x 5 5 A2 x 5
Ž 2.
for every x g DŽ A2 ., where A is the infinitesimal generator of a group t ª T Ž t . Ž5 T Ž t .5 s 1, t g R. of linear isometries on X. Moreover H. Kraljevic´ and S. Kurepa w4x established the even shaper inequality 5 Ax 5 2 F
4 3
5 x 5 5 A2 x 5
Ž 3.
for every x g DŽ A2 ., and A the infinitesimal generator Ži.e., the strong right second derivative of T at zero. of t ª T Ž t . Ž t G 0.: a strongly continuous cosine function of linear contractions on X. Therefore the best Landau’s type constant is 43 Žfor cosine functions.. The above-mentioned inequalities Ž1. ] Ž3. were extended by H. Kraljevic´ and J. Pecaric ˇ ´ w5x so that new Landau’s type inequalities hold. In particular, they proved that 5 Ax 5 3 F
243 8
5 x 5 2 5 A3 x 5 ,
5 A2 x 5 3 F 24 5 x 5 5 A3 x 5 2 .
Ž 19 .
hold for every x g DŽ A3 ., where A is the infinitesimal generator of a strongly continuous contraction semigroup on X, Besides they obtained the analogous but better inequalities 5 Ax 5 3 F
9 8
5 x 5 2 5 A3 x 5 ,
5 A2 x 5 3 F 3 5 x 5 5 A3 x 5 2
Ž 29 .
which hold for every x g DŽ A3 ., where A is the infinitesimal generator of a strongly continuous contraction group on X. Moreover they got the set of analogous inequalities 5 Ax 5 3 F
81 40
5 x 5 2 5 A3 x 5 ,
5 A2 x 5 3 F
72 25
5 x 5 5 A3 x 5 2
Ž 39 .
for every x g DŽ A3 ., where A is the infinitesimal generator of a strongly continuous cosine function on X. In this paper, we extend the above inequalities Ž19. ] Ž39. so that other Landau’s inequalities hold for every x g DŽ A4 ., where A is infinitesimal generator of a uniformly bounded continuous semigroup Žresp. group, or cosine function..
282
JOHN M. RASSIAS
2. SEMIGROUPS Let t ª T Ž t . be uniformly bounded Ž5 T Ž t .5 F M - `, t G 0. strongly continuous semigroup of linear operators on X with infinitesimal generator A, such that T Ž0. s I Ž[ Identity. in B Ž X . [ the Banach algebra of bounded linear operators on X, lim t x 0 T Ž t . x s x, for every x, and Ax s lim
T Ž t. y I t
t x0
Ž s T 9 Ž 0. x .
x
Ž 4.
for every x in a linear subspace DŽ A. Ž[ Domain of A., dense in X w2x. For every x g DŽ A., we have the formula T Ž t. x s x q
t
H0 T Ž u . Ax du.
Ž 5.
Using integration by parts, we get the formula t
ž
u
H0 H0
T¨ A2 x d¨ du s
/
t
H0 Ž t y u . TuA x du. 2
Ž 6.
Employing Ž6. and iterating Ž5., we find for every x g DŽ A2 . that T Ž t . x s x q tAx q
t
H0 Ž t y u . TuA x du. 2
Ž 59 .
Similarly iterating Ž59., we obtain for every x g DŽ A4 . that T Ž t . x s x q tAx q
t2 2
A2 x q
t3 6
1
A3 x q
t
3
H Ž t y u . TuA x du. 6 0 4
Ž 50 .
THEOREM 1. Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t .5 F M - `, t G 0. strongly continuous semigroup of linear operators on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 Ax 5 F M
Ž Ž ts . 2 q Ž sr . 2 q Ž rt . 2 . q s 2 Ž rt y sr y st . tsr Ž t y s . Ž s y r . q
ts q sr q rt tsr
5 x5 q M
tsr 24
5 A4 x 5 ,
Ž 7.
LANDAU’S TYPE INEQUALITIES
5 A2 x 5 F 2 M
tqsqr Ž tr 2 q sr 2 q t 2 s q t 2 r . q s Ž rt y s 2 . 5 x5 q tsr Ž t y s . Ž s y r . tsr
qM 5 A3 x 5 F 6 M
283
ts q sr q rt 12
5 A4 x 5 ,
Ž 79 .
1 tqsqr Ž ts q sr q rt . y s 2 5 x5 q M 5 A4 x 5 , Ž 70 . q tsr Ž t y s . Ž s y r . tsr 4
hold for e¨ ery x g DŽ A4 . and for e¨ ery t, s, r g Rqs Ž0, `., 0 - t - s - r. THEOREM 2. Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t .5 F M - `, t G 0. strongly continuous semigroup of linear operators on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 32 5 Ax 5 4 F Mg 1 Ž m1 , m 2 . 5 x 5 3 5 A4 x 5 , Ž 8. 81 5 A2 x 5 4 F 5 A3 x 5 4 F
4 9
M 2 g 2 Ž m 1 , m 2 . 5 x 5 2 5 A4 x 5 2 ,
Ž 89 .
8
M 3 g 3 Ž m1 , m 2 . 5 x 5 5 A4 x 5 3 , Ž 80 . 9 hold for e¨ ery x g DŽ A4 ., and for some m1 , m 2 g Rq, m 2 ) m1 ) 1, where g 1 Ž m1 , m 2 . s Ž m1 m 2 . =
2 m12 q Ž m1 m 2 . q m22 . q m12 Ž m 2 y m1 m 2 y m1 . Ž M
m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . q
g 2 Ž m1 , m 2 . s Ž m1 q m1 m 2 q m 2 . = M
m1 q m1 m 2 q m 2 m1 m 2
3
,
2
Ž m22 q m1 m22 q m1 q m 2 . q m1 Ž m 2 y m12 . m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . q
g 3 Ž m1 , m 2 . s Ž 1 q m1 q m 2 .
3
M
1 q m1 q m 2 m1 m 2
2
,
1 Ž m1 q m1 m 2 q m 2 . y m12 q . m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . m1 m 2
284
JOHN M. RASSIAS
THEOREM 3. Let t ª T Ž t . be a strongly continuous contraction Ž5 T Ž t .5 F 1, t G 0. semigroup of linear operators on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 Ax 5 4 F 5 A2 x 5 4 F
1024
5 x 5 3 5 A4 x 5 ,
Ž 9.
5 x 5 2 5 A4 x 5 2 ,
Ž 99 .
3 10 4 9
5 A3 x 5 4 F 192 5 x 5 5 A4 x 5 3 ,
Ž 90 .
hold for e¨ ery x g DŽ A4 .. Proof of Theorem 1. In fact, formula Ž50 . yields the system
H0 Ž t y u . T Ž u . A x du¦ s 6 sAx q 3s A x q s A x s 6T Ž s . x y 6 x y H Ž s y u . T Ž u . A x du¥ t
6 tAx q 3t 2A2 x q t 3A3 x s 6T Ž t . x y 6 x y 2
2
3
3
3
4
3
4
0
6 rAx q 3r 2A2 x q r 3A3 x s 6T Ž r . x y 6 x y
H0 Ž r y u . T Ž u . A x du.§ r
3
4
Ž 10 . The coefficient determinant D of system Ž10. is D s 18tsr Ž t y s . Ž s y r . Ž r y t . .
Ž 11 .
It is clear that D is positive because of the hypothesis 0 - t - s - r. Therefore there is a unique solution of system Ž10. of the form 2
2
2
Ž sr . Ž r y s . T Ž t . x y Ž tr . Ž r y t . T Ž s . x q Ž ts . Ž s y t . T Ž r . x Ax s tsr Ž t y s . Ž s y r . Ž r y t . y
ts q sr q rt tsr
x y
r
4
H0 K Ž t , s, r ; u . T Ž u . A x du, 1
Ž 12 .
LANDAU’S TYPE INEQUALITIES
A2 x s 2
y Ž sr . Ž r 2 y s 2 . T Ž t . x q Ž tr . Ž r 2 y t 2 . T Ž s . x tsr Ž t y s . Ž s y r . Ž r y t . y q
A3 x s 6
285
Ž ts . Ž s 2 y t 2 . T Ž r . x tsr Ž t y s . Ž s y r . Ž r y t . tqsqr
x q
tsr
r
H0 K
2
Ž t , s, r ; u . T Ž u . A4 x du,
Ž 129.
Ž sr . Ž r y s . T Ž t . x y Ž tr . Ž r y t . T Ž s . x q Ž ts . Ž s y t . T Ž r . x y tsr Ž t y s . Ž s y r . Ž r y t . 1 tsr
x y
r
4
H0 K Ž t , s, r ; u . T Ž u . A x du, 3
Ž 120 .
where
¡ Ž sr . Ž r y s . Ž t y u. 2
3
2
y Ž tr . Ž r y t . Ž s y u .
3
6 tsr Ž t y s . Ž s y r . Ž r y t . 2
K1
s~
3
Ž ts . Ž s y t . Ž r y u . q , 6 tsr Ž t y s . Ž s y r . Ž r y t . 2
3
2
0FuFt
y Ž tr . Ž r y t . Ž s y u . q Ž ts . Ž s y t . Ž r y u .
3
6 tsr Ž t y s . Ž s y r . Ž r y t . 2
t F u F s,
,
3
Ž ts . Ž s y t . Ž r y u . , 6 tsr Ž t y s . Ž s y r . Ž r y t .
¢ ¡ Ž sr . Ž r
2
sFuFr
3
y s 2 . Ž t y u . y Ž tr . Ž r 2 y t 2 . Ž s y u .
3
3tsr Ž t y s . Ž s y r . Ž r y t . 3
K2
s~
Ž ts . Ž s 2 y t 2 . Ž r y u . q , 3tsr Ž t y s . Ž s y r . Ž r y t . 3
y Ž tr . Ž r 2 y t 2 . Ž s y u . q Ž ts . Ž s 2 y t 2 . Ž r y u . 3tsr Ž t y s . Ž s y r . Ž r y t .
¢
0FuFt 3
,
t F u F s,
3
Ž ts . Ž s 2 y t 2 . Ž r y u . , 3tsr Ž t y s . Ž s y r . Ž r y t .
sFuFr
286
JOHN M. RASSIAS
¡ Ž sr . Ž r y s . Ž t y u.
3
y Ž tr . Ž r y t . Ž s y u .
3
tsr Ž t y s . Ž s y r . Ž r y t . 3
K3 s
~
q
Ž ts . Ž s y t . Ž r y u . , tsr Ž t y s . Ž s y r . Ž r y t .
0FuFt
3
y Ž tr . Ž r y t . Ž s y u . q Ž ts . Ž s y t . Ž r y u .
3
,
tsr Ž t y s . Ž s y r . Ž r y t .
¢
t F u F s,
3
Ž ts . Ž s y t . Ž r y u . , tsr Ž t y s . Ž s y r . Ž r y t .
s F u F r.
It is obvious that K i s K i Ž t, s, r; u. G 0, i s 1, 2, 3, for every u g w0, r x Ž0 - t - s - r ., and that the following equalities r
H0 K
1
du s
tsr 24
r
,
H0 K
2
du s
ts q sr q rt 12
r
,
H0 K
3
du s
tqsqr 4
,
Ž 13 . hold. Note that Ž12. ] Ž120 . hold because the identities 2
2
¦
2
Ž sr . Ž r y s . y Ž tr . Ž r y t . q Ž ts . Ž s y t . s Ž t y s . Ž s y r . Ž r y t . Ž ts q sr q rt . , y Ž sr . Ž r 2 y s 2 . q Ž tr . Ž r 2 y t 2 . y Ž ts . Ž s 2 y t 2 . ¥ s yŽ t y s . Ž s y r . Ž r y t . Ž t q s q r . , Ž sr . Ž r y s . y Ž tr . Ž r y t . q Ž ts . Ž s y t . s Ž t y s. Ž s y r . Ž r y t .
Ž 14 .
§
hold. Therefore from formulas Ž12. ] Ž120 ., Ž13., and the triangle inequality, we get inequalities Ž7. ] Ž70 .. This completes the proof of Theorem 1. Proof of Theorem 2. Setting s s m1 t ,
r s m2 t ,
m 2 ) m1 ) 1,
t)0
Ž 15 .
in Ž7. ] Ž70 ., we obtain the following inequalities 5 Ax 5 F a1
1 t
q b1 t 3 ,
5 A2 x 5 F a 2
1 t
2
q b2 t 2 ,
5 A3 x 5 F a 3
1 t3
q b3 t ,
Ž 16 .
LANDAU’S TYPE INEQUALITIES
287
where a1 s
2 m12 q Ž m1 m 2 . q m22 . q m12 Ž m 2 y m1 m 2 y m1 . Ž M
m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . q b1 s M
a2 s 2 M
m1 m 2 24
m1 q m1 m 2 q m 2 m1 m 2
5 x 5,
5 A4 x 5 ,
Ž m22 q m1 m22 q m1 q m 2 . q m1 Ž m 2 y m12 . m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . q b2 s M
a3 s 6 M
m1 q m1 m 2 q m 2 12
1 q m1 q m 2 m1 m 2
5 x 5,
5 A4 x 5 ,
1 Ž m1 q m1 m 2 q m 2 . y m12 5 x 5, q m1 m 2 Ž m1 y 1 . Ž m 2 y m1 . m1 m 2 b3 s M
1 q m1 q m 2 4
5 A4 x 5 .
Minimizing the right-hand side functions of t of Ž16., we get the sharper inequalities 5 Ax 5 4 F
256 27
a13 b1 ,
5 A2 x 5 4 F 16 a22 b 22 ,
5 A3 x 5 4 F
256 27
a3 b 33 . Ž 17 .
But a13 b1 s
M 24
g 1 Ž m1 , m 2 . 5 x 5 3 5 A4 x 5 ,
a22 b 22 s
M2 36
g 2 Ž m1 , m 2 . 5 x 5 2 5 A4 x 5 2 ,
and a3 b 33 s
3M 3 32
g 3 Ž m1 , m 2 . 5 x 5 5 A4 x 5 3 .
Therefore from Ž15. ] Ž17., we obtain inequalities Ž8. ] Ž80 .. This completes the proof of Theorem 2.
288
JOHN M. RASSIAS
Proof of Theorem 3. Taking M s 1, we have g 1 Ž m 1 , m 2 . s 8 gq 1 Ž m1 , m 2 . ,
g 2 Ž m1 , m 2 . s 4 gq 2 Ž m1 , m 2 . ,
g 3 Ž m 1 , m 2 . s 2 gq 3 Ž m1 , m 2 . , where gq 1
m1 Ž 1 q m 2 q m22 y m1 y m1 m 2 .
Ž m1 , m 2 . s Ž m1 m 2 .
3
,
m 2 Ž m1 y 1 . Ž m 2 y m1 .
gq 2 Ž m1 , m 2 . s Ž m1 q m1 m 2 q m 2 . gq 3 Ž m1 , m 2 . s Ž 1 q m1 q m 2 .
2
1 q m 2 q m22 y m12
2
,
m 2 Ž m1 y 1 . Ž m 2 y m1 . 1 q m 2 y m1
3
m 2 Ž m1 y 1 . Ž m 2 y m1 .
.
Hence inequalities Ž8. ] Ž80 . are written, as 5 Ax 5 4 F 5 A2 x 5 4 F 5 A3 x 5 4 F
256 81 16 9 16 9
5 5 3 5 A4 x 5 , gq 1 Ž m1 , m 2 . x
Ž 18 .
5 5 2 5 A4 x 5 2 , gq 2 Ž m1 , m 2 . x
Ž 189.
5 5 5 A4 x 5 3 , gq 3 Ž m1 , m 2 . x
Ž 180 .
for some m1 , m 2 g Rq: m 2 ) m1 ) 1. qŽ . All functions gq i s g i m1 , m 2 , i s 1, 2, 3, attain their minimum at the same m1 , m 2 : m1 s 2 q '2 , m 2 s 3 q 2'2 , so that q min gq 1 Ž m1 , m 2 . s 108 s min g 3 Ž m1 , m 2 . ,
min gq 2 Ž m1 , m 2 . s 625.
Ž 19 . Therefore inequalities Ž18. ] Ž180 . and minima Ž19. yield the even sharper inequalities Ž9. ] Ž90 .. This completes the proof of Theorem 3.
3. GROUPS Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t .5 F M - `, t g R s Žy`, `.. strongly continuous group of linear operators on X with infinitesimal generator A. It is clear that analogous inequalities Žas those in
LANDAU’S TYPE INEQUALITIES
289
the aforementioned Theorems 1]3. hold for every t, s, r g Rys Žy`, 0., t - s - r - 0. s - 0 - t - r. Denote
Case I.
s s m1 t ,
r s m2 t ,
m1 - 0,
m 2 ) 1,
t ) 0,
Ž 159.
and x 1 s 6 tAx,
x 2 s 3t 2A2 x,
x 3 s t 3A3 x,
Ž 20 .
as well as a s 6T Ž t . x y 6 x y b s 6T Ž m1 t . x y 6 x y
ž
3
m1 t
H0
s 6T Ž m1 t . x y 6 x y m14
c s 6T Ž m 2 t . x y 6 x y
t
H0 Ž t y u . T Ž u . A x du, t
4
3 Ž m1 t y u . T Ž u . A4 x du
3
H0 Ž t y u . T Ž m u . A x du 4
1
m2 t
H0
/
,
3 Ž m 2 t y u . T Ž u . A4 x du.
Then system Ž10. takes the form x 1 q x 2 q x 3 s a,
m1 x 1 q m12 x 2 q m13 x 3 s b,
m 2 x 1 q m22 x 2 q m32 x 3 s c.
Ž 109.
Solving system Ž109., we find the unique solution 2
x1 s x2 s x3 s
Ž m1 m 2 . Ž m 2 y m1 . a y m22 Ž m 2 y 1 . b q m12 Ž m1 y 1 . c m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . y Ž m1 m 2 . Ž m22 y m12 . a q m 2 Ž m22 y 1 . b y m1 Ž m12 y 1 . c m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
Ž m1 m 2 . Ž m 2 y m1 . a y m 2 Ž m 2 y 1 . b q m1 Ž m1 y 1 . c . m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
Ž 21 . Ž 219. Ž 210 .
THEOREM 4. Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t .5 F M - `, t g R. strongly continuous group of linear operators on complex Banach space
290
JOHN M. RASSIAS
X with infinitesimal generator A, such that A4 x / 0. Then 5 Ax 5 F M
2 Ž m1 m 2 . q m12 q m22 y m1 q m1 m 2 y m 2 m1 m 2 Ž m1 y 1 . Ž m 2 y 1 .
q qM
ž
M 12
5 A3 x 5 F 6 M
5 x5
m1 m 2
1 t
Ž m1 m 2 . Ž 1 q m1 y m1 m 2 q m 2 . 4 3 5 A x5t , 24 Ž m1 y 1 . Ž m 2 y 1 .
5 A2 x 5 F 2 Ž M q 1 . y q
m1 q m1 m 2 q m 2
1 q m1 q m 2 m1 m 2
/
5 x5
1 t2
Ž y Ž m 1 q m 1 m 2 q m 2 . . 5 A4 x 5 t 2 , m1 q m1 m 2 q m 2 y m22
m1 m 2 Ž m 2 y 1 . Ž m 2 y m1 .
Ž 22 .
y
1 m1 m 2
Ž 229. 5 x5
1 t3
2
qM
ym12 m 2 y m1 m 2 q m1 m22 q Ž m1 m 2 . q m22 q m42 4 m 2 Ž m 2 y 1 . Ž m 2 y m1 .
5 A4 x 5 t ,
Ž 220 . hold for e¨ ery x g DŽ A , for e¨ ery t g R , and for some m1 g Ry, m 2 g Rq, m2 y Ž m 2 q 1 . - m1 - y , m 2 ) 1. m2 q 1 THEOREM 5. Let t ª T Ž t . be a contraction Ž5 T Ž t .5 F 1, t g R. strongly continuous group of linear operators on complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 256 5 Ax 5 4 F f Ž m , m . 5 x 5 3 5 A4 x 5 , Ž 23 . 81 1 1 2 16 5 A2 x 5 4 F f Ž m , m . 5 x 5 2 5 A4 x 5 2 , Ž 239. 9 2 1 2 16 5 A3 x 5 4 F f Ž m , m . 5 x 5 5 A4 x 5 3 , Ž 230 . 9 3 1 2 hold for e¨ ery x g DŽ A4 ., and for some m1 g Ry, m 2 g Rq, m2 y Ž m 2 q 1 . - m1 - y , m 2 ) 1, m2 q 1 4.
q
LANDAU’S TYPE INEQUALITIES
291
where f 1 Ž m1 , m 2 . s
žŽ
4
m1 m 2
/Ž
m1 y 1 . Ž m 2 y 1 .
f 2 Ž m1 , m 2 . s
ž
1 q m1 q m 2 m1 m 2
1 q m1 y m1 m 2 q m 2 . ,
2
/
2 Ž m1 q m1 m 2 q m 2 . ,
f 3 Ž m1 , m 2 . s
2 Ž 1 q m1 y m 2 . Ž ym12 m 2 y m1 m 2 q m1 m22 q Ž m1 m 2 . q m22 q m42 .
Ž m1 m32 . Ž Ž m 2 y 1. Ž m 2 y m1 . .
4
3
.
THEOREM 6. Let t ª T Ž t . be a strongly continuous contraction Ž5 T Ž t .5 F 1, t g R. group of linear operators on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 4 5 Ax 5 4 F 10 5 x 5 3 5 A4 x 5 , Ž 24 . 6
ž /
5 A2 x 5 4 F
16 9
5 A3 x 5 4 F 10
5 x 5 2 5 A4 x 5 2 , 54 13 3 2 5 36
5 x 5 5 A4 x 5 3 ,
Ž 249. Ž 240 .
hold for e¨ ery x g DŽ A4 .. THEOREM 7. Let t ª T Ž t . be a strongly continuous contraction Ž5 T Ž t .5 F 1, t g R. group of linear operators on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 Ax 5 4 F 5 A2 x 5 4 F 5 A x5 F 3
4
m520
32
81 Ž m 20 y 1 . 4 16 9
5 x 5 3 5 A4 x 5 ,
5 x 5 2 5 A4 x 5 2 ,
16 m420 Ž 1 q m220 . 9
Ž
m220
m 20 s
(
y 1.
Ž 259. 3
4
5 x 5 5 A4 x 5 3 ,
hold for e¨ ery x g DŽ A4 ., where 7 q '57 2
Ž 25 .
.
Ž 250 .
292
JOHN M. RASSIAS
Proof of Theorem 4. In fact, from Ž20. and Ž21. ] Ž210 ., we get x1 Ax s 6t 2
Ž m1 m 2 . Ž m 2 y m1 . T Ž t . x y m22 Ž m 2 y 1 . T Ž m1 t . x s m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
ž
q
m12 Ž m1 y 1 . T Ž m 2 t . x m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . 2
Ž m1 m 2 . Ž m 2 y m1 . y m22 Ž m 2 y 1 . q m12 Ž m1 y 1 . 1 y x m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . t
/
2
t
y
H0 Ž t y u .
3
Ž m1 m 2 . Ž m 2 y m1 . T Ž u . 6 m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
ž
y q A2 x s
m22 Ž m 2 y 1 . m14 T Ž m1 u . 6 m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
m12 Ž m1 y 1 . m42 T Ž m 2 u . 6 m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
/
1
A4 x du
t
, Ž 26 .
x2 3t 2
s2
ym1 m 2 Ž m22 y m12 . T Ž t . x q m 2 Ž m22 y 1 . T Ž m1 t . x
ž
m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . y y
y
m1 Ž m12 y 1 . T Ž m 2 t . x m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . ym1 m 2 Ž m22 y m12 . q m 2 Ž m22 y 1 . y m1 Ž m12 y 1 . m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . t
H0 Ž t y u .
3
ž
/
1 t2
ym1 m 2 Ž m22 y m12 . T Ž u . 3m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . q
y
x
m 2 Ž m22 y 1 . m14 T Ž m1 u . 3m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
m1 Ž m12 y 1 . m42 T Ž m 2 u . 3m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
/
A4 x du
1 t2
, Ž 269 .
LANDAU’S TYPE INEQUALITIES
A3 x s
293
x3 t3
s6
m1 m 2 Ž m 2 y m1 . T Ž t . x y m 2 Ž m 2 y 1 . T Ž m1 t . x
ž
m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . q y
y
m1 Ž m1 y 1 . T Ž m 2 t . x m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . m 1 m 2 Ž m 2 y m 1 . y m 2 Ž m 2 y 1 . q m1 Ž m1 y 1 . m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . t
H0
Ž t y u. q
3
ž
x
/
1 t3
m1 m 2 Ž m 2 y m1 . T Ž u . y m 2 Ž m 2 y 1 . m14 T Ž m1 u . m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
m1 Ž m1 y 1 . m42 T Ž m 2 u . m1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 .
/
A4 x du
1 t3
.
Ž 260 .
But it is clear that the following identities 2
Ž m1 m 2 . Ž m 2 y m1 . y m22 Ž m 2 y 1 . q m12 Ž m1 y 1 . s Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . Ž m1 q m1 m 2 q m 2 . , ym1 m 2 Ž m22 y m12 . q m 2 Ž m22 y 1 . y m1 Ž m12 y 1 . s y Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . Ž 1 q m1 q m 2 . , ym1 m 2 Ž m22 y m12 . q m 2 Ž m22 y 1 . m14 y m1 Ž m12 y 1 . m42 s ym1 m 2 Ž m1 y 1 . Ž m 2 y 1 . Ž m 2 y m1 . Ž m1 q m1 m 2 q m 2 . hold. Applying these identities and formulas Ž26. ] Ž260 ., we obtain inequalities Ž22. ] Ž220 .. This completes the proof of Theorem 4. Proof of Theorem 5. In fact, from inequalities Ž22. ] Ž220 . we get 1 1 3 q 2 5 Ax 5 F a1q q bq 5 A2 x 5 F aq 1 t , 2 2 q b2 t , t t 5 A3 x 5 F aq 3 where a1q s 2 bq 1 s
1 t3
q bq 3 t,
m1 m 2
Ž m1 y 1 . Ž m 2 y 1 .
Ž 27 . 5 x 5,
Ž m1 m 2 . Ž 1 q m1 y m1 m 2 q m 2 . 4 5 A x 5, 24 Ž m1 y 1 . Ž m 2 y 1 .
294
JOHN M. RASSIAS
aq 2 s Ž y4 . bq 2 sy
1 12
aq 3 s 12
1 q m1 q m 2 m1 m 2
5 x 5,
Ž m1 q m1 m 2 q m 2 . 5 A4 x 5 , 1 q m1 y m 2
m1 Ž m 2 y 1 . Ž m 2 y m1 .
5 x 5,
2
bq 3 s
ym12 m 2 y m1 m 2 q m1 m22 q Ž m1 m 2 . q m22 q m42 4 m 2 Ž m 2 y 1 . Ž m 2 y m1 .
5 A4 x 5 ,
and new identities m1 m 2 Ž m1 y m 2 . q m 2 Ž m 2 y 1 . q m1 Ž m1 y 1 . s Ž m1 y 1 . Ž m1 q m1 m 2 q m 2 y m22 . , m1 m 2 Ž m1 y m 2 . q m 2 Ž m 2 y 1 . m14 q m1 Ž m1 y 1 . m42 2
s m1 Ž m1 y 1 . Ž ym12 m 2 y m1 m 2 q m1 m22 q Ž m1 m 2 . q m22 q m42 . . Minimizing the right-hand side of Ž27., we find 5 Ax 5 4 F
256 27
2 q 2 5 A2 x 5 4 F 16 Ž aq 2 . Ž b2 . ,
3
Ž a1q . Ž bq 1 ., 5 A3 x 5 4 F
256 27
q 3 Ž aq 3 . Ž b3 . ,
Ž 28 .
where 3
Ž a1q . Ž bq 1 . s
8 24
f 1 Ž m1 , m 2 . ,
1
2 q 2 f 2 Ž m1 , m 2 . , Ž aq 2 . Ž b2 . s
9
q 3 Ž aq 3 . Ž b3 . s
3 16
f 3 Ž m1 , m 2 . .
Therefore inequalities Ž28. yield inequalities Ž23. ] Ž230 .. This completes the proof of Theorem 5. Proof of Theorem 6. Setting m1 s y1, we get min f 1Žy1, m 2 . s 10Ž 58 . 4 at m 2 s 5. Hence f 2 Žy1, 5. s 1, f 3 Žy1, 5. s 5 4 Ž13. 3r2 9 3 4 . Therefore from formulas Ž23. ] Ž230 . and m1 s y1, m 2 s 5, we get inequalities Ž24. ] Ž240 .. This completes the proof of Theorem 6.
LANDAU’S TYPE INEQUALITIES
295
Proof of Theorem 7. In fact, min f 3 Žy1, m 2 . s f 3 Žy1, m 20 . s m420 Ž1 q m 20 s Ž 7 q '57 . r2 Žs root Ž) 1. of equaWe have
'
m220 . 3rŽ m220 y 1. 4 , where tion m42 y 7m22 y 2 s 0..
f 1 Ž y1, m 20 . s
m520
1
8 Ž m 20 y 1 . 4
.
Therefore from formulas Ž24. ] Ž240 ., and m1 s y1, m 2 s m 20 , we obtain inequalities Ž25. ] Ž250 .. This completes the proof of Theorem 7. Case II. r - s - 0 - t. Denote s s m1 t, r s m 2 t, m 2 s m - y1, m1 s y1, t ) 0. From Ž21. ] Ž210 ., we have m2 Ž m q 1 . a y m2 Ž m y 1 . b y 2 c
x1 s
2 m Ž m2 y 1 . x2 s
aqb
, 2 ym Ž m q 1 . a y m Ž m y 1 . b q 2 c
x3 s
,
2 m Ž m2 y 1 .
.
Ž 29 .
Therefore from Ž29. and Ž20., we obtain Ax s
ž
ym2 Ž m q1 . T Ž t . x q m2 Ž m y1 . T Ž y t . x q 2T Ž mt . x 2 m Ž 1 ym2 .
q
t
žH
Ž t y u.
0
3
ž
y A3 x s 6
ž
q
2 t
0
Ž t y u.
3
yx
/
2 m4 T Ž mu . 12 m Ž 1 y m . 2
A4 x du
T Ž u . q T Ž yu . 6
A4 x du
/
1 t2
3
/
1 t
,
, Ž 30 .
,
Ž 309.
2 mŽ 1 y m .
ž
t
t2
2
t
/
1
1
m Ž m q 1 . T Ž t . x q m Ž m y 1 . T Ž yt . x y 2T Ž mt . x
H0 Ž t y u .
m
x
12 m Ž 1 y m2 .
T Ž t . x q T Ž yt . x
žH
1
m2 Ž m q 1 . T Ž u . y m2 Ž m y 1 . T Ž yu .
y A2 x s 2
y
q
1 m
x
/
1 t3
ym Ž m q 1 . T Ž u . y m Ž m y 1 . T Ž yu . 2 m Ž 1 y m2 . q
2 m4 T Ž mu . 2 mŽ 1 y m . 2
A4 x du
/
1 t3
. Ž 300 .
296
JOHN M. RASSIAS
Thus from formulas Ž30. ] Ž300 . we get 5 Ax 5 F
m mq1
5 A2 x 5 F 4 5 x 5 5 A3 x 5 F 12
5 x5
1 t
2
q
m
1 t
ž
q y
1 12
m2 24 Ž m q 1 .
/
5 A4 x 5 t 3 ,
5 A4 x 5 t 2 , 1
Ž 319.
1 m Ž 1 q m2 .
5 A4 x 5 t. Ž 310 . 1 y m2 Minimizing the right-hand side of inequalities Ž31. ] Ž310 ., we find 1ym
5 x5 2
Ž 31 .
5 Ax 5 4 F 5 A2 x 5 4 F 5 A3 x 5 4 F
32 81 16 9 16 9
t3
q
4
h 1 Ž m . 5 x 5 3 5 A4 x 5 ,
Ž 32 .
5 x 5 2 5 A4 x 5 2 ,
Ž 329.
h 3 Ž m . 5 x 5 5 A4 x 5 3 ,
Ž 320 .
where h1 Ž m . s y
m5
Ž m q 1.
4
,
h3 Ž m. s
m4 Ž 1 q m2 .
Ž 1 y m2 .
3
4
,
m - y1.
First, minimizing h1Ž m., m - y1, we get m s y5. Then inequalities Ž32. ] Ž320 . Žwith m s y5. are the same as the inequalities Ž24. ] Ž240 .. Finally, minimizing h 3 Ž m., m - y1, we obtain m s m 0 s y Ž 7 q '57 . r2 . Then inequalities Ž32. ] Ž320 . Žwith m s m 0 . are the same as the inequalities Ž25. ] Ž250 ..
'
4. COSINE FUNCTIONS Let t ª T Ž t . Ž t G 0. be a uniformly bounded Ž5 T Ž t .5 F M - `, t G 0. strongly continuous cosine function with infinitesimal operator A, such that T Ž0. s I Ž[ identity. in B Ž X ., lim t x 0 T Ž t . x s x, ; x, and A is defined as the strong second derivatives of T at zero, Ax s T 0 Ž 0 . x
Ž 33 .
for every x in a linear subspace DŽ A., which is dense in X w5x. For every x g DŽ A., we have the formula T Ž t. x s x q
t
H0 Ž t y u . T Ž u . Ax du.
Ž 34 .
LANDAU’S TYPE INEQUALITIES
297
Using integration by parts, we get from Ž34. the formula t
u
H0 Ž t y u . H0
ž
Ž u y ¨ . f Ž ¨ . d¨ du s
/
1
t
H Žty¨. 6 0
3
f Ž ¨ . d¨ ,
Ž 35 .
where f Ž ¨ . s T¨ A2 x. Note the Leibniz formula: d du
u
žH Ž 0
n
u
u y ¨ . f Ž ¨ . d¨ s n
/ žH Ž 0
uy¨.
ny1
f Ž ¨ . d¨ .
/
Ž 36 .
Employing Ž35. ] Ž36. and iterating Ž34., we find for every x g DŽ A2 . that
T Ž t. x s x q
t2 2!
Ax q
1
t
3
H Ž t y u . T Ž u . A x dx. 3! 0 2
Ž 349.
Similarly iterating Ž349. we obtain for every x g DŽ A4 . that
T Ž t. x s x q
t2 2!
Ax q
t4 4!
A2 x q
t6 6!
A3 x q
1 7!
t
H0 Ž t y u .
7
T Ž u . A4 x du.
Ž 340 . THEOREM 8. Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t . F- M - `, t G 0. strongly continuous cosine function on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 2 2 2 ts q sr q rt Ž ts . q Ž sr . q Ž rt . q s 2 Ž rt y sr y st . 5 Ax 5 F 2 M 5 x5 q tsr Ž t y s . Ž s y r . tsr
qM 5 A2 x 5 F 24 M qM
tsr 20160
5 A4 x 5 ,
Ž 37 .
tqsqr Ž ts 2 q sr 2 q t 2 s q t 2 r . q s Ž rt y s 2 . 5 x5 q tsr Ž t y s . Ž s y r . tsr ts q sr q rt 1680
5 A4 x 5 ,
Ž 379.
298
JOHN M. RASSIAS
5 A3 x 5 F 720 M qM
1 Ž ts q sr q rt . y s 2 5 x5 q tsr Ž t y s . Ž s y r . tsr
tqsqr 56
5 A4 x 5 ,
Ž 370 .
hold for e¨ ery x g DŽ A4 ., and for e¨ ery t, s, r g Rq, 0 - t - s - r. THEOREM 9. Let t ª T Ž t . be a uniformly bounded Ž5 T Ž t .5 F M - `, t G 0. strongly continuous cosine function on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 Ax 5 4 F 5 A2 x 5 4 F 5 A3 x 5 4 F
32 8505 4 1225 40 1029
Mg 1 Ž m1 , m 2 . 5 x 5 3 5 A4 x 5 ,
Ž 38 .
M 2 g 2 Ž m1 , m 2 . 5 x 5 2 5 A4 x 5 2 ,
Ž 389.
M 3 g 3 Ž m1 , m 2 . 5 x 5 5 A4 x 5 3 ,
Ž 380 .
hold for e¨ ery x g DŽ A4 ., and for some m1 , m 2 g Rq, m 2 ) m1 ) 1, where g i s g i Ž m1 , m 2 . are the same as those g i , i s 1, 2, 3, in Theorem 2. THEOREM 10. Let t ª T Ž t . be a strongly continuous contraction Ž5 T Ž t .5 F 1, t G 0. cosine function on a complex Banach space X with infinitesimal generator A, such that A4 x / 0. Then the following inequalities 5 Ax 5 4 F 5 A2 x 5 4 F 5 A3 x 5 4 F
1024
5 x 5 3 5 A4 x 5 ,
Ž 39 .
5 x 5 2 5 A4 x 5 2 ,
Ž 399.
315 400 49
2880 343
5 x 5 5 A4 x 5 3 ,
Ž 390 .
hold for e¨ ery x g DŽ A4 .. Proof of Theorem 8. In fact, setting t Ž) 0. instead of t 2 in Ž340 ., we get T Ž 't . x s x q q
t 2 1
Ax q
t2 24
A2 x q 7
t3 720
A3 x
H't Ž't y u . T Ž u . A x du. 5040 0 4
Ž 34-.
LANDAU’S TYPE INEQUALITIES
299
Formula Ž34-. yields 2520tAx q 210t 2A2 x q 7t 3A3 x s 5040T Ž 't . x y 5040 x y
7
H0't Ž't y u . T Ž u . A x du 4
2520 sAx q 210 s 2A2 x q 7s 3A3 x s 5040T Ž 's . x y 5040 x y
H0's Ž's y u . T Ž u . A x du 7
4
Ž 109.
2520rAx q 210r 2A2 x q 7r 3A3 x s 5040T Ž 'r . x y 5040 x y
H0'r Ž'r y u . T Ž u . A x du. 7
4
The coefficient determinant Dq of system Ž109. is Dqs 3704400tsr Ž t y s . Ž s y r . Ž r y t . Ž s Dr205800. .
Ž 119.
It is clear that Dq) 0 because 0 - t - s - r. Therefore there is a unique solution of system Ž109. of the form 2
Ax s 2
2
Ž sr . Ž r y s . T Ž 't . x y Ž tr . Ž r y t . T Ž 's . x tsr Ž t y s . Ž s y r . Ž r y t .
ž
q
H0'r K
q 1
y A x s 24 2
ž
2 ts q sr q rt Ž ts . Ž s y t . T Ž 'r . x y x tsr Ž t y s . Ž s y r . Ž r y t . tsr
Ž t , s, r ; u . T Ž u . A4 x du,
ysr Ž r 2 y s 2 . T Ž 't . x q tr Ž r 2 y t 2 . T Ž 's . x tsr Ž t y s . Ž s y r . Ž r y t . ts Ž s 2 y t 2 . T Ž 'r . x
y
tsr Ž t y s . Ž s y r . Ž r y t .
H0'r K
q 2
q
A3 x s 720
Ž 40 .
ž
q
tqsqr tsr
x
/
Ž t , s, r ; u . T Ž u . A4 x du,
Ž 409.
sr Ž r y s . T Ž 't . x y tr Ž r y t . T Ž 's . x q ts Ž s y t . T Ž 'r . x tsr Ž t y s . Ž s y r . Ž r y t . y
1 tsr
/
x y
H0'r K
q 3
Ž t , s, r ; u . T Ž u . A4 x du, Ž 400 .
/
300
JOHN M. RASSIAS
where
¡ Ž sr . Ž r y s . Ž't y u . 2
7
2
y Ž tr . Ž r y t . Ž 's y u .
7
2520tsr Ž t y s . Ž s y r . Ž r y t . 7
2
Kq 1
s
Ž ts . Ž s y t . Ž 'r y u . q , ~ 2520tsr Ž t y s . Ž s y r . Ž r y t . 7
2
0 F u F 't ,
2
y Ž tr . Ž r y t . Ž 's y u . q Ž ts . Ž s y t . Ž 'r y u . 2520tsr Ž t y s . Ž s y r . Ž r y t .
¢
2
,
't F u F 's ,
7
2
Ž ts . Ž s y t . Ž 'r y u . , 2520tsr Ž t y s . Ž s y r . Ž r y t .
¡sr Ž r
7
's F u F 'r ,
7
y s 2 . Ž 't y u . y Ž tr . Ž r 2 y t 2 . Ž 's y u .
7
210tsr Ž t y s . Ž s y r . Ž r y t .
Kq 2
s
~
q
ts Ž s 2 y t 2 . Ž 'r y u .
7
210tsr Ž t y s . Ž s y r . Ž r y t .
0 F u F 't ,
,
7
ytr Ž r 2 y t 2 . Ž 's y u . q ts Ž s 2 y t 2 . Ž 'r y u . 210tsr Ž t y s . Ž s y r . Ž r y t . ts Ž s 2 y t 2 . Ž 'r y u .
7
,
't F u F 's ,
7
¢210tsr Ž t y s . Ž s y r . Ž r y t . , ¡sr Ž r y s . Ž't y u .
7
's F u F 'r ,
y tr Ž r y t . Ž 's y u .
7
7tsr Ž t y s . Ž s y r . Ž r y t .
Kq 3 s
~
q
ts Ž s y t . Ž 'r y u .
7
7tsr Ž t y s . Ž s y r . Ž r y t . 7
ytr Ž r y t . Ž 's y u . q ts Ž s y t . Ž 'r y u . 7tsr Ž t y s . Ž s y r . Ž r y t . ts Ž s y t . Ž 'r y u .
0 F u F 't ,
, 7
,
't F u F 's ,
7
¢7tsr Ž t y s . Ž s y r . Ž r y t . ,
's F u F 'r .
LANDAU’S TYPE INEQUALITIES
301
w ' x Ž0 - t - s - r . and Note that Kq i G 0, i s 1, 2, 3, for every u g 0, r
H0'r K
q 1
du s
tsr 20160
H0'r K
,
q 3
H0'r K
q 2
du s
du s
tqsqr 56
.
ts q sr q rt 1680
,
Ž 139.
From Ž40. ] Ž400 ., triangle inequality, Ž139., Ž15. and similarly as in the previous Section 2 on semigroups, we get inequalities Ž37. ] Ž370 .. This completes the proof of Theorem 8. Proof of Theorem 9. From Ž15. and Ž37. ] Ž370 . and similar calculations as in Section 2 on semigroups, we find inequalities Ž38. ] Ž380 .. This completes the proof of Theorem 9. Proof of Theorem 10. Setting M s 1, using Ž38. ] Ž380 ., and minimizing Ž g i m1 , m 2 ., i s 1, 2, 3, as in Section 2 on semigroups, we obtain inequalities Ž39. ] Ž390 .. This completes the proof of Theorem 10.
REFERENCES 1. Z. Ditzian, Some remarks on inequalities of Landau and Kolmogorov, Aequationes Math. 12 Ž1975., 145]151. 2. E. Hille and R. S. Phillips, ‘‘Functional Analysis and Semigroups,’’ Amer. Math. Soc. Colloq. Publ., Vol. 31, AMS, Providence, Rhode Island, 1957. 3. R. R. Kallman and G.-C. Rota, On the inequality 5 f 9 5 2 F 4 5 f 5 5 f 0 5, in ‘‘Inequalities, II’ ŽO. Shisha, Ed.., pp. 187]192, Academic Press, New YorkrLondon, 1970. 4. H. Kraljevic´ and S. Kurepa, Semigroups on Banach spaces, Glas. Mat. 5 Ž1970., 109]117. 5. H. Kraljevic´ and J. Pecaric, ˇ ´ Some Landau’s type inequalities for infinitesimal generators, Aequationes Math. 40 Ž1990., 147]153. 6. E. Landau, Einige Ungleichungen fur ¨ zweimal differentzierbare Funktionen, Proc. London Math. Soc. (2) 13 Ž1913., 43]49.