J.evol.equ. 1 (2001) 115–142 1424–3199/01/010115–28 $ 1.50 + 0.20/0 © Birkh¨auser Verlag, Basel, 2001
Lp -Theory of the Stokes equation in a half space Wolfgang Desch, Matthias Hieber and Jan Pruss ¨
Abstract. In this paper, we investigate Lp -estimates for the solution of the Stokes equation in a half space H where 1 ≤ p ≤ ∞. It is shown that the solution of the Stokes equation is governed by an analytic semigroup on BU Cσ (H ), C0,σ (H ) or L∞ σ (H ). From the operatortheoretical point of view it is a surprising fact that the corresponding result for L1σ (H ) does not hold true. In fact, there exists an L1 -function f satisfying divf = 0 such that the solution of the corresponding resolvent equation with right hand side f does not belong to L1 . Taking into account however a recent result of Kozono on the nonlinear Navier-Stokes equation, the L1 -result is not surprising and even natural. We also show that the Stokes operator admits a R-bounded H ∞ -calculus on Lp for 1 < p < ∞ and obtain as a consequence maximal Lp –Lq -regularity for the solution of the Stokes equation.
1. Introduction In this paper, we consider the Stokes equation in a half space H := Rn+1 + , i.e. we consider the set of equations ut − 1u + 5p = f
in H × (0, ∞)
div u = 0
in H × (0, ∞)
u = 0
on ∂H × (0, ∞)
(1.1)
u(0) = u0 where u = (u1 , . . . , un , un+1 )T is interpreted as the velocity field and p as the pressure. We are interested in the Lp -theory of the solution of (1.1) where 1 ≤ p ≤ ∞. If 1 < p < ∞, p one usually defines the subspace Lσ (H ) of Lp (H ) consisting of all solenoidal functions f p in Lp and associates to (1.1) the so-called Stokes operator A in (Lσ (H ))n+1 defined as Au := P 1u D(A) := (W 2,p (H ) ∩ W0 (H ) ∩ Lpσ (H ))n+1 . 1,p
Here P denotes the Helmholtz projection in Lp (H ). Then it is well known that the Stokes p operator A generates an analytic semigroup on Lσ (H ) for 1 < p < ∞ (see for instance Received August 24, 2000; accepted September 30, 2000. 2000 Mathematics Subject Classification: 35Q30, 35K55. Key words and phrases: Stokes system, resolvent estimates, analytic semigroups, H ∞ -calculus, maximal Lp -regularity, R-boundedness.
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[Sol77], [McC81], [Uka87]). It seems to be unknown whether system (1.1) is also solvable in L1 (H ) or L∞ (H ). Observe that the Helmholtz projection is not bounded in L1 (H ) or p L∞ (H ). Thus the usual decomposition of Lp (H ) in Lσ (H ) and its orthogonal complement which is true for 1 < p < ∞ is no longer possible if p = 1 or p = ∞. The strategy we use to solve the Stokes system and to get sharp regularity results on the resolvent is the following: Taking Fourier transforms we obtain a representation of the solution of the corresponding resolvent equation as a sum of two terms, the first being the resolvent of the Dirichlet Laplacian on Lp (Rn+1 + ), the second one being a remainder term. In Section 3 we derive pointwise upper bounds on the remainder term which allow to prove L∞ -estimates for the solution of the corresponding resolvent problem, i.e. we obtain estimates of the form kuk∞ ≤
M kf k∞ |λ|
for the equation defined below in (2.1), where λ belongs to a suitable sector in the complex plane. The above estimate allows us to prove that the Stokes operator generates an analytic semigroup on the spaces BU Cσ (H ), C0,σ (H ), L∞ σ (H ), which is, of course, not strongly continuous in the latter case. The case p = 1 is of special interest, see [GMS99], [Koz98] and [Miy97]. In Section 5 n+1 satisfying div f = 0 such that u 6 ∈ we give an example of a function f ∈ L1 (Rn+1 + ) n+1 n+1 1 L (R+ ) , where uˆ is the solution of the resolvent equation (2.1) defined below. This implies in particular that there is no estimate of the form ku(t, ·)k1 ≤ Cku0 k1
(1.2)
for the solution of the Stokes Problem (1.1) with f = 0. This is remarkable because the solution of the Stokes equation on all of Rn satisfies an estimate of form (1.2). Taking into account however a recent result of Kozono [Koz98] on the equations of Navier-Stokes for exterior domains one does not expect an estimate of form (1.2) to hold true. Indeed, he showed that the existence of a local strong solution of the Navier-Stokes equations in L1 implies that no force could act on the boundary of the domain which would mean that the Navier-Stokes equations are physically meaningless. Thus, from this point of view one does not expect that (1.2) holds true. However, the pointwise upper bound on the remainder term allows us to prove an estimate of the form k∇u(t, ·)k1 ≤ C
1 t 1/2
ku0 k1
for the solution of (1.1) with f = 0 which was proved first in [GMS99]. The pointwise upper bound on the remainder term allows us also to show that for 1 < p < ∞ the Stokes operator admits a bounded H ∞ -calculus on Lp (H )n+1 . In fact, even a
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stronger result is true: In Section 7 we prove that the Laplacian on Rn and the remainder term even admit an R-bounded H ∞ -calculus on Lp (H )n+1 for 1 < p < ∞. Thus the Stokes operator admits an R-bounded H ∞ -calculus on Lp (H )n+1 . As a consequence, we obtain maximal Lp –Lq -regularity for the solution of the Stokes equation (1.1). For more information on the role of the Stokes semigroup in the theory of the NavierStokes equations, we refer to the recent article of Amann [Ama00] and the survey article by Wiegner [Wie99]. NOTATIONS AND CONVENTIONS. Throughout this paper, we define for 0 < θ ≤ π the sector 6θ in the complex plane by 6θ := {z ∈ C\{0}; |argz| < θ }. If X and Y are Banach spaces, L(X, Y ) denotes the space of all bounded, linear operators from X to Y ; moreover, L(X) := L(X, X). The spectrum of a linear operator A in X is denoted by σ (A). Given p ∈ [1, ∞), we denote by Lpσ (H ) := {f ∈ Cc∞ (H ); div f = 0}−k·kLp the Banach space of all solenoidal functions in Lp (H ). If 1 < p < ∞, then Lp (H ) = Lpσ (H ) ⊕ Gp (H ), p
where Gp (H ) consists of all functions f ∈ Lp (H ) for which there exists g ∈ Lloc (H ) such p that f = ∇g. The projection P : Lp (H ) onto Lσ (H ) is called the Helmholtz Projection. By C, M and c we denote various constants which may differ from line to line, but which are always independent of the free variables. 2. The resolvent problem for the Stokes problem In this section we consider the resolvent equation for the Stokes problem in the half P := {(x, y) ∈ Rn+1 ; x ∈ Rn , y > 0}. Given p ∈ [1, ∞], λ ∈ π and space H := Rn+1 + n+1 with div f = 0, find a velocity field u = (u , . . . , u , u T f ∈ Lp (Rn+1 1 n n+1 ) and a + ) pressure field p such that λu − 1u + 5p = f
in H,
div u = 0
in H,
(2.1)
u|∂H = 0. For x ∈ Rn and y > 0 we write u = (v, w)T with v = (v1 , . . . , vn )T and f = (fv , fw )T with fv = ((fv )1 , . . . , (fv )n )T . Assume that fw (ξ, 0) = 0 for all ξ ∈ Rn . Applying the Fourier transform with respect to x we obtain ˆ y) − ∂y2 v(ξ, ˆ y) = fˆv (ξ, y) − iξ · p(ξ, ˆ y), (λ + |ξ |2 )v(ξ, (λ + |ξ
2
|)w(ξ, ˆ y) − ∂y2 w(ξ, ˆ y)
= fˆw (ξ, y) − ∂y p(ξ, ˆ y),
ξ ∈ Rn , y > 0 n
ξ ∈ R ,y > 0
(2.2) (2.3)
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iξ · v(ξ, ˆ y) + ∂y w(ξ, ˆ y) = 0, v(ξ, ˆ 0) = 0, w(ξ, ˆ 0) = 0,
ξ ∈ Rn , y > 0
J.evol.equ.
(2.4)
n
(2.5)
n
(2.6)
ξ ∈R
ξ ∈R .
Multiplying equation (2.2) by iξ , applying ∂y to (2.3) and adding them yields ˆ y) − ∂y2 p(ξ, ˆ y) = −∂y fˆw (ξ, y) − iξ fˆv (ξ, y) |ξ |2 p(ξ, = −( div f )ˆ(ξ, y) = 0 for ξ ∈ Rn and y > 0, where we already took into account equation (2.4). Hence p(ξ, ˆ y) = e−|ξ |y pb0 (ξ ),
ξ ∈ Rn ,
y>0
for some function pˆ 0 . We thus obtain for vˆ and wˆ the following representations 1 v(ξ, ˆ y) = 2ω(ξ )
Z∞ [e−ω(ξ )|y−s| − e−ω(ξ )(y+s) ] 0
(2.7)
[fˆv (ξ, s) − iξ e−|ξ |s pˆ 0 (ξ )] ds, Z∞ 1 [e−ω(ξ )|y−s| − e−ω(ξ )(y+s) ] w(ξ, ˆ y) = 2ω(ξ ) 0
(2.8)
[fˆw (ξ, s) + |ξ |e−|ξ |s pˆ 0 (ξ )] ds,
for ξ ∈ Rn , y > 0 and where ω(ξ ) := (|λ| + |ξ |2 ) 2 for ξ ∈ Rn . In order to determine pˆ 0 consider ∂y w(ξ, ˆ y) at y = 0, i.e. 1
Z∞ ˆ 0) = ∂y w(ξ,
e−ω(ξ )s [fˆw (ξ, s) + |ξ |e−|ξ |s pˆ 0 (ξ )] ds = 0,
ξ ∈ Rn .
0
This implies that (ω(ξ ) + |ξ |) pˆ 0 (ξ ) = − |ξ |
Z∞
e−ω(ξ )s fˆw (ξ, s) ds,
ξ 6 = 0.
(2.9)
0
By assumption, iξ · fˆv (ξ, y) + ∂y fˆw (ξ, y) = 0 for ξ ∈ Rn and y > 0. Integrating by parts yields Z∞ e
−ω(ξ )
−ω(ξ )s
fˆw (ξ, s) ds =
0
Z∞ 0
Z∞ = 0
d −ω(ξ )s ˆ (e )fw (ξ, s) ds ds e−ω(ξ )s (iξ ) · fbv (ξ, s) ds.
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Thus fˆwL (ξ )
Z∞ :=
e−ω(ξ )s fˆw (ξ, s) ds =
0
iξ − ω(ξ )
Z∞ 0
(2.10) −iξ ˆL f (ξ ) e−ω(ξ )s fˆv (ξ, s) ds =: ω(ξ ) v
n
for all ξ ∈ R . Inserting (2.9) and (2.10) into (2.8) and (2.7) we obtain vˆ = vˆ1 + vˆ2 (2.11)
wˆ = wˆ 1 + wˆ 2 with vˆ1 , vˆ2 and wˆ 1 , wˆ 2 given for ξ ∈ Rn and y > 0 by 1 vˆ1 (ξ, y) = 2ω(ξ ) vˆ2 (ξ, y) =
1 2ω(ξ )
Z∞ [e−ω(ξ )|y−s| − e−ω(ξ )(y+s) ]fˆv (ξ, s) ds 0
Z∞ [e−ω(ξ )|y−s| − e−ω(ξ )(y+s) ] 0
(iξ ) −|ξ |s e (ω(ξ ) + |ξ |) ds fˆwL (ξ ) |ξ | =
1 iξ [e−y|ξ | − e−ω(ξ )y ]fˆwL (ξ ) |ξ | ω(ξ ) − |ξ |
1 [e−y|ξ | − e−w(ξ )y ](ξ · fˆvL (ξ ))ξ (w(ξ ) − |ξ |)|ξ |w(ξ ) Z∞ 1 [e−ω(ξ )|y−s| − e−ω(ξ )(y+s) ]fˆw (ξ, s) ds wˆ 1 (ξ, y) = 2ω(ξ ) =
0
wˆ 2 (ξ, y) = =
−1 [e−y|ξ | − e−ω(ξ )y ]fˆwL (ξ ) ω(ξ ) − |ξ | 1 iξ · [e−y|ξ | − e−ω(ξ )y ]fˆvL (ξ ). ω(ξ ) ω(ξ ) − |ξ |
Observe that v1 = (λ − 1D )−1 fv ,
w1 = (λ − 1D )−1 fw ,
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where 1D denotes the Laplacian in Rn+1 + subject to zero Dirichlet boundary conditions. It is well known that for 1 ≤ p ≤ ∞ and λ ∈ 6θ with θ < π there exists a constant M > 0 such that M kfv kLp (Rn+1 )n + |λ| M ≤ kfw kLp (Rn+1 ) . + |λ|
kv1 kLp (Rn+1 )n ≤ +
kw1 kLp (Rn+1 ) +
Hence, in order to obtain Lp -estimates for v and w we may restrict ourselves in the following to the cases v2 and w2 . The Lp -estimates for v2 and w2 will be derived from pointwise upper bounds for the inverse Fourier transform of vˆ2 and wˆ 2 . 3. Pointwise upper bounds for the remainder term We proved in the previous Section 2 that v = (λ − 1D )−1 fv + v2 w = (λ − 1D )−1 fw + w2 where v2 and w2 are defined as in (2.11). In this section we derive pointwise upper estimates for v2 and w2 . Let θ < π and define rˆv : Rn × R+ × R+ × 6θ → C by rˆv (ξ, y, y 0 , λ) := p
1 e−|ξ |y − e−ω(ξ )y 0 ξ ξ T e−ω(ξ )y , ω(ξ ) − |ξ | |ξ |ω(ξ )
|λ| + |ξ |2 . Define Z 1 eix·ξ rˆv (ξ, y, y 0 , λ)dξ. rv (x, y, y 0 , λ) = (2π)n
where ω(ξ ) =
(3.1)
(3.2)
Rn
Note that rv is well defined since the above integral is absolutely convergent for (y, y 0 ) 6 = (0, 0). Observe first that by the following scaling argument it suffices to consider the case |λ| = 1 and |arg λ| ≤ θ < π. In fact, the change of coordinates 1
ξ → |λ| 2 ξ,
y→
y |λ|
1 2
,
y0 →
y0 |λ|
1 2
,
x→
x 1
|λ| 2
yield rˆv (ξ, y, y 0 , λ) =
1 1
|λ| 2
rˆv
λ , |λ| y, |λ| y 0 , 1 |λ| |λ| 2 ξ
1 2
1 2
!
(3.3)
Lp -Theory of the Stokes equation in a half space
Vol. 1, 2001
and hence 0
rv (x, y, y , λ) = |λ|
n−1 2
rv
121
λ . |λ| x, |λ| y, |λ| y , |λ| 1 2
1 2
1 2
0
For this reason, we may suppose now that λ ∈ 6θ with |λ| = 1. For z ∈ C consider next the function φ given by φ(z) =
1 − e−z , z
z ∈ C\{0}
and note that C , 1 + |z|
|φ(z)| ≤
Re z ≥ 0
(3.4)
for a suitable constant C > 0. Thus rˆv (ξ, y, y 0 , λ) = ye−|ξ |y e−ω(ξ )y
0
1 ξ ξ T φ((ω(ξ ) − |ξ |)y). |ξ |ω(ξ )
Choose now a rotation Q in Rn such that Qx = (|x|, 0, . . . , 0) and write Qξ = (a, rb),
a ∈ R,
r > 0,
b ∈ Rn−1 ,
|b| = 1.
By this coordinate transformation we obtain rv (x, y, y 0 , λ) = cn
Z∞
r n−2
·e−y
0
λ+r 2 +a 2
Z∞
Sn−1 −∞
0
√
Z
√ y 2 2 ei|x|a √ e−y r +a √ 2 2 2 2 λ+r +a r +a
p p a φ(y( λ + r 2 + a 2 − r 2 + a 2 )) ( a rb ) dadbdr. rbT
(3.5)
where Sn−1 is the unit sphere in Rn−1 . Next, for ε > 0 small enough, we shift the path of integration for a to the contour a → s + iε(r + |s|), s ∈ R, without changing the value of the integral thanks to Cauchy’s theorem. Then r 2 + a 2 = r 2 + s 2 − ε2 (r + |s|)2 + 2iε(r + |s|)s. Hence, given ε0 > 0, there exists a constant c > 0 such that for ε ∈ (0, ε0 ) we have c|r 2 + a 2 | ≤ (r + |s|)2 ≤
1 2 |r + a 2 |, c
and |arg (r 2 + a 2 )| ≤ cε,
r > 0,
s ∈ R.
r > 0,
s∈R
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We thus obtain for x ∈ Rn , y > 0, y 0 > 0 and λ ∈ 6θ with |λ| = 1 the bounds |ei|x|a | √ 2 2 |e−y r +a | √ 0 2 2 |e−y |λ|+r +a | √ r 2 + a 2 p |λ| + r 2 + a 2 q p | |λ| + r 2 + a 2 − r 2 + a 2 |
= e−ε|x|(r+|s|) , ≤ e−cy(r+|s|) , 0
≤ e−cy (1+r+|s|) , ≤ c
|λ| = 1,
|arg λ| ≤ θ < π,
r + |s| 1 + r + |s|
1 1 . = p ≥c √ 2 2 2 2 1 + r + |s| |λ| + r + a + r + a
Inserting these bounds into (3.5) yields for a multiindex α |(∂x )α rv (x, y, y 0 , λ)| Z∞ Z∞ 1+|α| 0 0 (r + s) n−2 dsdr e−c(r+s)(|x|+y+y ) ye−cy ≤M r 1+r +s+y 0
= Mye
0
−cy 0
Z∞ r
0
Z∞ r
0
= Mye−cy
n−2
Z∞
s n+|α|
0
0
e−cs(|x|+y+y ) 1+|α| dsdr s 1+s+y 0
e−cs(|x|+y+y ) ds 1+y+s
for some constant M > 0 independent of x ∈ Rn , y, y 0 > 0 and λ ∈ C with |λ| = 1 and |arg λ| ≤ θ < π. We thus have proved the following result. PROPOSITION 3.1. Let θ ∈ (0, π) and α be a multiindex. Then there exist constants M, c > 0 such that α
0
|(∂x ) rv (x, y, y , λ)| ≤ Mye
−cy 0
Z∞ 0
s n+|α| −cs(|x|+y+y 0 ) ds, e 1+y+s
where x ∈ Rn , y, y 0 > 0 and λ ∈ C with |λ| = 1 and |arg λ| ≤ θ < π. REMARK 3.2. For θ < π we define rˆw : Rn × R+ × R+ × 6θ → Cn by rˆw (ξ, y, y 0 , λ) :=
e−|ξ |y − e−ω(ξ )y iξ −ω(ξ )y 0 e ω(ξ ) − |ξ | ω(ξ )
(3.6)
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Copying the above proof we see that the assertion of Proposition 3.1 remains true if rv is replaced by rw , where Z 1 0 eix·ξ rˆw (ξ, y, y 0 , λ)dξ, x ∈ Rn , y, y 0 > 0. (3.7) rw (x, y, y , λ) = (2π)n Rn
The kernel estimates given in Proposition 3.1 and Remark 3.2 allow us to derive Lp -estimates for v2 and w2 via the following lemma on Lp -continuity of integral operators acting in half spaces.
LEMMA 3.3. Suppose that 1 ≤ p ≤ ∞ and let operator in
Lp (Rn+1 + ) Z∞ Z
(Tf )(x, y) =
1 p
+
k(x − x 0 , y, y 0 )f (x 0 , y 0 )dx 0 dy 0 ,
x ∈ Rn , y > 0,
where k : Rn × R+ × R+ → C is a measurable function. a) Let 1 < p < ∞. If p0 p1 p Z∞ Z∞ 0 0 p 0 kk(·, y, y )k1 dy dy =: M1 < ∞,
0
≤ M1 . then T ∈ L(Lp (Rn+1 + )) and kT kL(Lp (Rn+1 + )) b) Let p = 1. If Z∞ sup
y 0 >0
kk(·, y, y 0 )k1 dy =: M2 < ∞,
0
≤ M2 . then T ∈ L(L1 (Rn+1 + )) and kT kL(L1 (Rn+1 + )) c) Let p = ∞. If Z∞ sup
y>0
= 1. Let T be an integral
of the form
0 Rn
0
1 p0
kk(·, y, y 0 )k1 dy 0 =: M3 < ∞,
0
then T ∈ L(L∞ (Rn+1 + )) and kT kL(L∞ (Rn+1 )) ≤ M3 . +
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Proof. a) By Young’s inequality and H¨older’s inequality we have Z∞ Z
|Tf (x, y)|p dxdy
0 Rn
p
Z∞ Z∞
0 0 0
= (k(·, y, y ) ∗ f (·, y ))(·)dy dy
0 0 p p Z∞ Z∞ ≤ kk(·, y, y 0 )k1 kf (·, y 0 )kp dy 0 dy 0
≤
0
p kf k p n+1 L (R+ )
p0 p Z∞ Z∞ 0 p kk(·, y, y 0 )k dy. 1 0
0
¨
The assertions b), c) are proved in a simliar way.
Combining the estimates obtained in Proposition 3.1 and Remark 3.2 with Lemma 3.3 we obtain the following estimates for v2 and w2 . PROPOSITION 3.4. Let 1 < p ≤ ∞ and θ ∈ (0, π). Let v2 and w2 be defined as in (2.11). Then there exists a constant M > 0 such that M kfv kLp (Rn+1 )n + |λ| M kfv kLp (Rn+1 )n ≤ + |λ|
kv2 kLp (Rn+1 )n ≤ +
kw2 kLp (Rn+1 ) +
for all λ ∈ 6θ . Proof. Observe that for λ ∈ 6θ Z∞ Z v2 (x, y) =
rv (x − x 0 , y, y 0 , λ)fv (x 0 , y 0 )dx 0 dy 0 ,
x ∈ Rn , y > 0,
0 Rn
where rv (x, y, y 0 , λ) = |λ|
n−1 2
1 1 1 λ , rv |λ| 2 x, |λ| 2 y, |λ| 2 y 0 , |λ|
x ∈ Rn , y, y 0 > 0
and rv is satisfying the estimate given in Proposition 3.1. In order to prove the assertion, we verify the conditions of Lemma 3.3 a) and c).
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To this end, note that Z Z n−1 1 1 1 α 0 α 0 λ 2 2 2 2 |(∂x ) rv (x, y, y , λ)|dx = |λ| (∂x ) rv |λ| x, |λ| y, |λ| y , |λ| dx Rn
Rn
≤ Cn |λ|
n−1 2
|λ| 2 ye−c|λ| 1
Z∞ Z∞
1 2 y0
1
0 n
≤ Cn |λ| 2 ye−c|λ|
1 2 y0
Z∞
≤ Cn ye
1
1 + |λ| 2 y + s
0
s |α|
1 + |λ| y + s |λ|
Z∞
1
0
e−cs|λ| 2 (y+y ) 1 2
1 + |λ| y
0
≤ Cn e−c|λ|
1
1 2
0 −c|λ| 2 y 0
s n+|α| ρ n−1
y
1 2 y0
1 2
1 + |λ| y
·
n 2
e−cs|λ|
e−cs|λ|
1 2 (ρ+y+y 0 )
1 2 (y+y 0 )
ds
dρds
(3.8)
s |α| ds 1
1 2
(|λ| (y + y 0 ))1+|α|
.
Hence, if 1 < p < ∞ and |α| = 0, then Z∞
1
p0
krv (·, y, y 0 , λ)k1 dy 0 ≤ Cn,p
0
|λ|
1 2
·
1 |λ|
p0 2
1 1
(1 + |λ| 2 y)p
0
and
p0 p Z∞ Z∞ 0 krv (·, y, y 0 , λ)kp dy 0 dy 1 0
0
≤ Cnp
1 p
|λ| 2p0
·
1
Z∞
1 p
1
|λ| 2 |λ| 2
0
1 dσ = Cnp (1 + σ )p
1 |λ|
p
.
If p = ∞, then Z∞ Z
|rv (·, y, y 0 , λ)|dxdy 0
0 Rn
Z∞ e
≤ Cn 0
1
−c|λ| 2 y 0
y
(3.9)
1 1 · ≤ Cn dy · 1 1 |λ| 1 + |λ| 21 y 1 + |λ| 2 y |λ| 2 y 0
1
and hence Z∞ Cn sup krv (·, y, y 0 , λ)k1 dy 0 ≤ |λ| y>0 0
The estimate for rw follows in exactly the same way.
¨
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Note that the kernel estimate given in Proposition 3.1 does not allow to verify the assumptions of Lemma 3.3 b) for the case p = 1. We investigate this point in detail in Section 5. Summing up we proved the following result: n+1 such THEOREM 3.5. Let 1 < p ≤ ∞, 0 < θ < π and λ ∈ 6θ . Let f ∈ Lp (Rn+1 + ) T that div f = 0 and fn+1 |∂H = 0. Let u = (v, w) be defined as in (2.11). Then there exists a constant M > 0 such that M kf kLp (Rn+1 )n+1 . kukLp (Rn+1 )n+1 ≤ + + |λ|
For 1 < p < ∞ and λ > 0 consider the mapping R(λ) : Lpσ (H ) → Lpσ (H ),
f 7 → uλ , p
where uλ is defined as in (2.11). Let A be the Stokes operator in Lσ (H ) defined as in Section 1. Then R(λ)(λ − A)f = f for all f ∈ D(A) and (λ − A)R(λ)f = f for all p f ∈ Lσ (H ). Thus R(λ) = (λ − A)−1 ,
λ > 0.
Theorem 3.5 implies now the following well known result (see e.g. [Sol77], [McC81], [Gig81], [FS96], [BS87], [Uka87], [GS89]). COROLLARY 3.6. Let 1 < p < ∞. Then the Stokes operator A defined as in Section 1 p generates an analytic strongly continuous semigroup on Lσ (H )n+1 . 4. The Stokes operator in BU Cσ (H ), C0,σ (H ), L∞ σ (H ) In this section we define the Stokes operator in BU Cσ (H ), C0,σ (H ), L∞ σ (H ) and show that it is the generator of an analytic semigroup on these spaces (which is not strongly continuous in the case of L∞ σ (H )). To this end, define BU Cσ (H ) := {f ∈ BU C(H ); divf = 0, f (x1 , . . . , xn , 0) = 0 for all
x1 , . . . , xn ∈ R}
and C0,σ (H ) := {f ∈ Cc∞ (H ); divf = 0}−k·kL∞ . Let Xσ (H ) be one of the spaces BU Cσ (H ) or C0,σ (H ). For θ ∈ (0, π ) and λ ∈ 6θ consider the mapping R(λ) : Xσ (H )n+1 → L∞ (H )n+1 ,
f 7 → uλ ,
where uλ is the solution of the Stokes equation given in (2.11). Theorem 3.5 and a direct calculation show that {R(λ); λ > 0} is a pseudo-resolvent in Xσ (H )n+1 .
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LEMMA 4.1. Let f ∈ Xσ (H )n+1 . Then lim λR(λ)f = f.
λ→∞
Proof. Notice first that R(λ)fv = (λ − 1D )−1 fv + v2 (λ) R(λ)fw = (λ − 1D )−1 fw + w2 (λ), where 1D denotes the Dirichlet Laplacian and v2 , w2 are defined as in (2.11). Since 1D generates a C0 -semigroup on BU C(H ) or C0 (H ), respectively, it follows that lim λ(λ − 1D )−1 f = f
λ→∞
for all f ∈ Xσ (H ). It thus remains to prove that limλ→∞ λv2 (λ) = 0 in Xσ (H )n and limλ→∞ λw2 (λ) = 0 in Xσ (H ). In order to do so, note first that limλ→∞ λv2 (λ) = 0 in BU Cσ (H )n if and only if Z ∞ Z λ(n+1)/2 rv (λ1/2 x 0 , λ1/2 y, λ1/2 y 0 , 1) lim sup λ→0 y>0
Rn
0
fv (x − x , y )dx dy = 0. 0
Since Z Rn
0
0
(4.1)
0
λ(n+1)/2 rv (λ1/2 x 0 , λ1/2 y, λ1/2 y 0 , 1) dx 0 = λrˆv (0, y, y 0 , λ) = 0
it follows that (4.1) is satisfied provided Z ∞Z |λ(n+1)/2 rv (λ1/2 x 0 , λ1/2 y, λ1/2 y 0 , 1)| lim sup λ→0 y>0 0
Rn
|fv (x − x 0 , y 0 ) − fv (x, 0)|dx 0 dy 0 = 0. But Z
∞Z 0
|λ(n+1)/2 rv (λ1/2 x 0 , y, λ1/2 y 0 , 1)||fv (x − x 0 , y 0 ) − fv (x, 0)|dx 0 dy 0 Z ∞Z y0 x0 0 0 = |rv (x , y, y , 1)| fv x − 1/2 , 1/2 − fv (x, 0) dx 0 dy 0 n λ λ R 0 Z ∞Z ≤ |rv (x 0 , y, y 0 , 1)|dx 0 dy 0 · sup Rn
0
Rn
|x 0 |≤R,|y 0 |≤S
0 0 fv x − x , y (x, 0) − f v 1/2 1/2 λ λ
wolfgang desch, matthias hieber and jan pruss ¨
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Z + Z +
∞Z
Rn S ∞Z
≤M
|rv (x 0 , y, y 0 , 1)|dx 0 dy 0
|x 0 |≥R
0
0
0
0
|rv (x , y, y , 1)|dx dy
sup |x 0 |≤
R S ,|y 0 |≤ 1/2 λ λ1/2
0
0
· 2|fv |∞
0
|fv (x − x , y ) − fv (x, 0)| + 2|fv |∞ M
for all S, R > 0 by (3.9) and since Z Z ∞Z 0 0 0 0 |rv (x , y, y , 1)|dx dy ≤ M S
J.evol.equ.
Rn
S
∞
1 1 + S R
0
M e−cy dy 0 ≤ 0 y S
by (3.8) and Z ∞ Z ∞ y sn 0 ds |rv (x 0 , y, y 0 , 1)|dy 0 ≤ M e−cs(|x |+y) 1 + y + s c + s 0 0 Z ∞ 0 e−cs|x | s n ds ≤M 0
by the estimate given in Proposition 3.1 which implies that Z ∞Z M |rv (x 0 , y, y 0 , 1)|dx 0 dy 0 ≤ . 0 R |x |≥R 0 This implies the assertion if Xσ (H ) = BU Cσ (H ). The case where of Xσ (H ) = C0,σ (H ) is proved in a similar way. ¨ The above lemma shows that kerR(λ) = 0 for all λ > 0. Hence, there exists a closed, densely defined operator AXσ in Xσ (H )n+1 such that R(λ) = (λ − AXσ )−1 , λ > 0. DEFINITION 4.2. The operator AXσ is called the Stokes operator in Xσ (H )n+1 . Theorem 3.5 implies now the following result. THEOREM 4.3. The Stokes operator AXσ generates a strongly continuous analytic semigroup on Xσ (H )n+1 . Finally, we consider the solution of the Stokes equation in L∞ σ (H ). This space is defined as ˆ follows: Observe that ∇ acts as a bounded operator from W 1,1 (H ) into L1 (H )n+1 , where Wˆ 1,1 (H ) = {f ∈ L1loc (H ); ∇f ∈ L1 (H )}. Hence Div := −∇ ∗ is a bounded operator from L∞ (H )n+1 into Wˆ 1,1 (H )∗ . We define n+1 := kerDiv. L∞ σ (H )
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R 1,1 (H ). Consider the mapping Thus f ∈ L∞ σ (H ) if and only if H ∇ϕf = 0 for all ϕ ∈ W ∞ n+1 ∞ n+1 → L (H ) defined as before. Theorem 3.5 and a direct calculation R(λ) : Lσ (H ) implies that {R(λ); λ > 0} is a pseudo-resolvent. In contrast to the situation of Xσ (H ) we do not have that limλ→∞ λv2 (λ) = 0 in L∞ σ (H ). However, the representation of the remainder term given in Section 3 allows us to show that kerR(λ) = 0 in L∞ σ (H ) for all ∞ (H )n+1 such that λ > 0. Thus there exists a closed operator AL∞ in L σ σ )−1 , λ > 0. R(λ) = (λ − AL∞ σ n+1 . Note that A ∞ is not densely We call the operator AL∞ the Stokes operator in L∞ Lσ σ (H ) σ defined; however Theorem 3.5 implies the following result.
generates an analytic semigroup on THEOREM 4.4. The Stokes operator AL∞ σ (which is not strongly continuous in 0).
n+1 L∞ σ (H )
5. The case p = 1 n+1 satisfying div f = 0 In this section we give an example of a function f ∈ L1 (Rn+1 + ) n+1 1 such that w2 6 ∈ L (R+ ) where w2 is defined as in Section 2. This is rather surprising n+1 whenever f ∈ Lp (Rn+1 )n+1 with div f = 0 for all p ∈ (1, ∞] since u ∈ Lp (Rn+1 + ) + as we have seen in Section 2. More precisely, we have the following result.
THEOREM 5.1. Let 0 < θ < π and λ ∈ 6θ . Then there exists f = (f1 , . . . , fn+1 ) ∈ n n+1 satisfying div f = 0 and f L1 (Rn+1 n+1 (x, 0) = 0 for all x ∈ R such that u 6 ∈ + ) n+1 ˆ w) ˆ defined as in (2.11) is the solution of the resolvent equation L1 (R+ )n+1 , where uˆ = (v, (2.2)–(2.6). We base the construction of our counterexample on well known properties of the Hardy space H 1 (Rn ). We remind the reader that H 1 (Rn ) := {f ∈ L1 (Rn ) : f ∗ ∈ L1 (Rn )} where f ∗ is given by f ∗ (x) := sup |(kt ∗ f )(x)|, t>0
x ∈ Rn ,
and kt denotes the Gaussian kernel given by kt (x) =
2
1 − |x| 4t (x ∈ Rn , t > 0). n e (4π t) 2 space H 1 (Rn ) becomes a Banach
Equipped with the norm kf kH 1 (Rn ) := kf ∗ kL1 (Rn ) , the space. It is well known that an L1 -function f belongs to H 1 (Rn ) if and only if its Riesz transforms Rj f belong to L1 (Rn ) for all j ∈ {1, . . . , n}. This property of H 1 (Rn ) will be of crucial importance for the following.
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J.evol.equ.
Proof of Theorem 5.1. Consider the Gaussian kernel kr for some r > 0. Then kr 6 ∈ H 1 (Rn ) because kr∗ 6 ∈ L1 (Rn ). Thus, since kr ∈ L1 (Rn ), there exists j ∈ {1, . . . , n} such that Rj kr 6 ∈ L1 (Rn ). Fix j ∈ {1, . . . , n} with this property and define for x ∈ Rn , y > 0 and r > 0 the function fw : Rn+1 + → R by fw (x, y) := 4yTω1 (y)
∂ (1 − 1)3/2 kr (x), ∂xj
x ∈ Rn ,
y > 0.
by the pseudoHere Tw1 denotes the C0 -semigroup on Lp (Rn ), 1 ≤ p < ∞ generated p differential operator A given by Af = −F −1 (ω1 fˆ), where ω1 (ξ ) = 1 + |ξ |2 for ξ ∈ Rn . Note that kTω1 (y)k ≤ Me−y for suitable M > 0 and all y ≥ 0. Hence Z∞
p kfw (·, y)kLp (Rn ) dy
Z∞ ≤C
0
y p e−yp dy < ∞,
0
for suitable C > 0, which means that fw ∈ Lp (Rn+1 + ) for all p ∈ [1, ∞). Observe moreover that 2 fˆw (ξ, y) = 4ye−ω1 (ξ )y (iξj )ω13 (ξ )e−r|ξ | ,
ξ ∈ Rn , y > 0
and 2 ∂y fˆw (ξ, y) = (iξj )4ω13 (ξ )e−r|ξ | (e−ω1 (ξ )y − ω1 (ξ )ye−ω1 (ξ )y ) =: iξj gˆ r (ξ, y)
T with gr ∈ Lp (Rn+1 + ) for 1 ≤ p < ∞. Set now f := (fv , fw ) with fv := (0, . . . , −gr , n+1 0, . . . , 0) so that the j -th component of fv is −gr . Thus div f = 0, f ∈ Lp (Rn+1 + ) and (f · ν)|∂H = 0, where ν denotes the outer unit normal. Observe next that the functions
ξ 7 → fˆwL (ξ ) = 4
Z∞ 2 2 se−2ω1 (ξ )s (iξj )ω13 (ξ )e−r|ξ | ds = (iξj )e−r|ξ | ω1 (ξ ) 0
ξ 7 → wˆ 2 (ξ, s) = −(iξj )ω1 (ξ )e−r|ξ | (ω1 (ξ ) + |ξ |)[e−s|ξ | − e−ω1 (ξ )s ], Z∞ 1 1 iξj −r|ξ |2 2 (iξj )ω1 (ξ )e−r|ξ | = − e ξ 7 → wˆ 2 (ξ, s)ds = − |ξ |ω1 (ξ ) 2 |ξ | 2
0
all belong to L2 (Rn ). It thus follows from Plancherel’s theorem that Z∞ 0
w2 (x, y)dy = −F −1 (rj (·)e−r|·| )(x), 2
s>0
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ξ
where rj (ξ ) = i |ξj| for ξ 6 = 0. Hence Z Z∞ kw2 kL1 (Rn+1 ) = +
|w2 (x, y)| dydx Rn
0
Z Z∞ w2 (x, y)dy dx ≥ Rn 0 Z 2 |F −1 (rj (·)e−r|·| )(x)| dx = Rn
= kRj kr kL1 (Rn ) = ∞, by the choice of j at the beginning of this section. This implies the assertion. REMARK 5.2. It is an open problem to decide whether the assertion of Theorem 5.1 remains true if the half space H is replaced by a bounded domain with smooth boundary. We now turn our attention to gradient estimates in the L1 -norm of the solution of the Stokes equation as they were proved in [GMS99] in 1999. To this end let |λ| = 1 and note that by (3.8) we have for |α| = 1 Z ∞Z |(∂x )α rv (x, y, y 0 , λ)|dxdy sup y 0 >0 0
Z
Rn
∞Z ∞
sy e−csy dsdy 1+y+s 0 0 Z ∞Z y sy e−csy dsdy < ∞. = 2C 1 + y+s 0 0 ≤C
Moreover, note that ξξT ω(ξ ) ξξT 0 + e−ω(ξ )(y+y ) . ω(ξ )|ξ | 0
∂y rˆv (ξ, y, y 0 , λ) = −e−ω(ξ )y e−|ξ |y 1 − e−(ω(ξ )−|ξ |)y ω(ξ ) − |ξ |
The first term on the right hand side above is estimated exactly in the same way as ∂x rv . In order to treat the second term denote its inverse Fourier transform by rv1 . For |λ| = 1, we obtain Z ∞ Z ∞ r +s 0 r n−2 e−c|x|(r+s) e−c(y+y )(1+r+s) dsdr |rv1 (x, y, y 0 , λ)| ≤ M 1 + r +s 0 0
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Thus sup
Z
y 0 >0 0
∞Z Rn
|rv1 (x, y, y 0 , λ)|dxdy ≤ M
Z
∞Z ∞
0
0
J.evol.equ.
1 −cy(1+s) e dsdy < ∞. 1+s
Lemma 3.3 b), a scaling argument and a simliar argument for rw implies now the following result which was proved first in [GMS99]. PROPOSITION 5.3. There exists a constant M > 0 such that the solution of the Stokes equation (1.1) with right hand side f = 0 satisfies k∇u(t, ·)k1 ≤
M ku0 k1 , t 1/2
t > 0.
6. Bounded H ∞ -calculus Given θ ∈ (0, π), we denote by H ∞ (6θ ) the Banach algebra of all bounded holomorphic functions f : 6θ → C equipped with the supremum norm. We also denote by H0∞ (6θ ) the set of all g ∈ H ∞ (6θ ) such that there exist constants C ≥ 0, s > 0 with |g(z)| ≤ C
|z|s , 1 + |z|2s
z ∈ 6θ .
Let now ω ∈ (0, π) and let A be a closed, densely defined operator in a Banach space X which is one to one and has dense range. Assume that σ (A) ⊂ 6ω and that for every ω0 ∈ (ω, π ) there exists M > 0 such that k(λ − A)−1 k ≤
M , |λ|
λ ∈ C\6ω0 .
Let ω < θ < π. Then, given g ∈ H0∞ (6θ ), the operator Z 1 g(λ)(λ − A)−1 dλ g(A) := 2πi 0
is a well defined element of L(X), where 0 denotes the positively oriented contour {λ = 0 te±iω ; t ≥ 0} for some ω0 ∈ (ω, θ). Moreover, for z ∈ 6ϑ set h(z) := z(1 + z)−2 . Then f (A) := [h(A)]−1 (f h)(A),
f ∈ H ∞ (6ϑ )
is a well defined operator in X. Let 0 < ω < θ < π . We say that A admits a bounded H ∞ -calculus on the sector 6θ if f (A) ∈ L(X) and there exists a constant M > 0 such that kf (A)kL(X) ≤ Mkf k∞ ,
f ∈ H ∞ (6ϑ ).
(6.1)
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It is well known, that A satisfies (6.1) if there exists M > 0 such that kg(A)kL(X) ≤ Mkgk∞ ,
g ∈ H0∞ (6ϑ ).
For details see [CDMY96]. Thus in deriving estimates for f (A), it suffices to establish estimates for g(A), where g ∈ H0∞ (6ϑ ). For 1 < p < ∞ let A be the Stokes operator in Lp (H )n+1 defined as in Section 1. We show in the following that −A admits a bounded H ∞ -calculus on Lp (H )n+1 for every sector 6θ with 0 < θ < π. We start with the following lemma. LEMMA 6.1. Let T be an integral operator of the form Z∞ (Tf )(y) =
k(y, y 0 )f (y 0 )dy 0 ,
y > 0,
(6.2)
0
where k : R+ × R+ → C is a measurable function such that the above integral is well defined. Suppose that for some p ∈ (1, ∞) there exists a constant M > 0 such that |(Tf )(y)| ≤
M 1
yp
kf kLp (R+ ) ,
y > 0.
If T ∈ L(Lq0 (R+ )) for some q0 ∈ (p, ∞], then T ∈ L(Lq (R+ )) for all q ∈ (p, q0 ]. Proof. By assumption, Tf is dominated pointwise by a function belonging to the weak p p Lp -space Lw (R+ ). Thus T : Lp (R+ ) → Lw (R+ ) is a bounded operator. The assumption and the Marcinkiewicz interpolation theorem imply that T ∈ L(Lq (R+ )) for all q ∈ (p, q0 ]. ¨ COROLLARY 6.2. Let k : R+ × R+ → C be a measurable function. Suppose that there exists M > 0 such that M y 0 log 1 + 0 , y, y 0 > 0. |k(y, y )| ≤ y + y0 y Let T be defined as in (6.2) and let 1 < p ≤ ∞. Then T ∈ L(Lp (R+ )). Proof. Note first that Z∞
0
0
|k(y, y )|dy ≤ M 0
which implies that T ∈
Z∞ log (1 + 0
1+
y y0
L(L∞ (R+ )).
y y0 )
dy 0 =M y0
Z∞ 0
log (1 + s) ds < ∞ (1 + s)s
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If p > 1 let
1 p
+
1 p0
|Tf (y)| ≤ M
= 1. For f ∈ Lp (R+ ) we obtain by H¨older’s inequality
(y
0
M
=
1
yp M
≤
1
yp
1
Z∞ log p0 (1 +
Z∞ 0
J.evol.equ.
y 0 p0 y 0 )dy 0 + y 0 )p
kf kLp (R+ )
0
log p (1 + s) 1 0 0 dskf kLp (R+ ) (1 + s)p s 2−p
kf kLp (R+ ) ,
y > 0. ¨
Thus the assertion follows from Lemma 6.1. Let now h ∈ H0∞ (6θ ) where 0 < θ < π is fixed. Consider the function Z 1 h(λ)rv (x, y, y 0 , −λ)dλ, x ∈ Rn , y, y 0 > 0, kh,v (x, y, y 0 ) = 2πi 0
where rv is defined as in (3.2) and 0 := {ρe±iϕ , ρ ≥ 0} with 0 < ϕ < θ. The estimate for rv given in Proposition 3.1 yields Z∞
0
|kh,v (x, y, y )| ≤ CkhkH ∞
|rv (x, y, y 0 , ρe±i(π−ϕ) )|dρ
0
Z∞ ≤ Ckhk
n 2
ρ ye
H∞
1 −cρ 2 y 0
1
1+ρ2y +s
0
ye
0
s n e−cρ 2 s(|x|+y+y ) 1
0
Z∞ ≤ CkhkH ∞
Z∞
Z∞
1 −cρ 2 y 0
−cσ (|x|+y+y ) ne
0
1
ρ 2 + ρy + σ
=: CkhkH ∞ k1 (x, y, y 0 ). Now Z
0
Z∞
|k1 (x, y, y )|dx ≤ Cn Rn
ye
1
−cρ 2 y 0
y = C y + y0
0
e−cσ (y+y ) 1
0
y ≤ C y + y0
Z∞
ρ 2 + ρy + σ
0
Z∞
1
e−cρ 2 y 1
0
Z∞ 0
0
dσ dρ
0
ρ 2 + ρy
(6.3)
0
σ
0
dsdρ
dρ
e−csy ds, 1 + sy
y, y 0 > 0.
dσ dρ
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Splitting the latter integral at s = Z∞ 0
1 y0 ,
we obtain
1
0 ye−csy
1 + sy
Zy 0 ds ≤ 0
Z∞
y y ds + 1 + sy 1+
≤ log This implies that Z |k1 (x, y, y 0 )|dx ≤ Rn
1+
y y0
y y0
+C
C log y + y0
135
1 y0
y y0
1+
1+
0
e−csy ds
y y0
≤ C log
y y0
,
1+
y y0
.
y, y 0 > 0.
(6.4)
Define now the operator Th,v in Lp (Rn+1 + ) by Z∞ Z (Th,v f )(x, y) := 0
kh,v (x − x 0 , y, y 0 )f (x 0 , y 0 )dx 0 dy 0 ,
x ∈ Rn , y > 0.
(6.5)
Rn
Then, by Young’s inequality, (6.3), (6.4) and Corollary 6.2 we have p Z∞ Z Z∞ Z∞ |(Th,v f )(x, y)|p dxdy ≤ kkh,v (·, y, y 0 )k1 kf (·, y 0 )kp dy 0 dy 0 Rn
Z∞ Z∞ p ≤ Ckhk ∞ H p
0
0
1 log y + y0
0 p . Lp (Rn+1 + )
0
y 1+ 0 y
p kf (·, y 0 )kp dy 0 dy
(6.6)
≤ CkhkH ∞ kf k
Moreover, by Remark 3.2 the function rw defined as in (3.7) satisfies also an estimate of the form given in Proposition 3.1. Thus, the function kh,w defined by Z 1 h(λ)rw (x, y, y 0 , −λ)dλ, x ∈ Rn , y, y 0 > 0 kh,w (x, y, y 0 ) := 2πi 0
also satisfies |kh,w (x, y, y 0 )| ≤ CkhkH ∞ k1 (x, y, y 0 ),
x ∈ Rn , y, y 0 > 0.
Define the operator Th,w as in (6.5) with kh,v replaced by kh,w . We conclude that Th,w satisfies estimate (6.6). Finally note that the operator −1D admits a bounded H ∞ (6θ )-calculus on Lp (Rn+1 + ) for every θ ∈ (0, π). This follows from Corollary 7.3 below or by the results proved in [PS93]. This fact and (6.6) imply the following result.
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THEOREM 6.3. Let 1 < p < ∞ and let A be the Stokes operator in defined as in n+1 . Then −A admits a bounded H ∞ (6 )-calculus on Lp (Rn+1 )n+1 for each Lp (Rn+1 θ + ) + θ ∈ (0, π ). As a consequence we obtain by the Dore-Venni theorem maximal Lp –Lq -estimates for the solution of the Stokes equation (see also [GS91], [Gig85]). 7. R-bounded H ∞ -calculus for −1 on Rn+1 and Rn+1 + R-bounded families of operators play an important role in the variant of the DoreVenni theorem which has been recently proved by Kalton and Weis [KW00]. They are defined as follows: let X and Y be Banach spaces. We call a family of operators T ⊂ L(X, Y ) R-bounded, if there is a constant C > 0 and p ∈ [1, ∞) such that for each N ∈ N, Tj ∈ T , xj ∈ X and all independent, symmetric {−1, 1}-valued random variables εj on a probability space (, M, µ) we have
X
X
N
N
ε T x ≤ C · ε x . j j j j j
j =1
p
p
j =1 L (;Y )
L (;X)
The smallest such C is called the R-bound of T and is denoted by R(T ). For detailed information on this subject and its relation to maximal Lp -regularity and to Lp -Fourier multipliers, we refer to [CdPSW00], [Wei99] and [CP00]. REMARK 7.1. a) Let X and Y be Hilbert spaces. Then T ⊂ L(X; Y ) is R-bounded if and only if T is uniformly bounded. b) Let 1 ≤ p < ∞ and let X = Y = Lp (G) for some G ⊂ Rn open. Then T ⊂ L(X, Y ) is R-bounded if and only if there is a constant M > 0 such that the following square function estimate
1 1
N
N 2 2
X
X
|Tj fj |2 ≤ M |fj |2 (7.1)
j =1
p
p
j =1 L (G)
L (G)
holds for each N ∈ N, f ∈ Lp (G) and Tj ∈ T . This assertion is a coinsequence of Khintchine’s inequality. For p ∈ [1, ∞) there exists a constant Kp > 0 such that
X
N
ε a Kp−1 j j
j =1
Lp ()
1 2
X
N N X
≤ |aj |2 ≤ Kp ε a j j
j =1 j =1
Lp ()
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for all N ∈ N, aj ∈ C and all independent, symmetric {−1, 1}-valued random variables εj on (, M, µ). Let ⊂ Rn open and let 1 < p < ∞. Assume that a given operator A in Lp () admits a bounded H ∞ -calculus on Lp () for some sector 6θ with 0 < θ < π. By the recent results given in [KW00] it is an interesting question to ask whether or not the set {h(A) : h ∈ H0∞ (6θ ), khkH ∞ (6θ ) ≤ R} ⊂ L(Lp ())
(7.2)
is R-bounded. If this holds true, we say that A admits an R-bounded H ∞ -calculus on Lp () for the sector 6θ . Our first result in this section asserts that −1 on Lp (Rn ), where 1 denotes the Laplacian, admits an R-bounded H ∞ -calculus on Lp (Rn ) for each sector 6θ where 0 < θ < π. More precisely, the following holds true. THEOREM 7.2. Let 1 < p < ∞ and denote by 1 the Laplacian in Rn . Then −1 admits an R-bounded H ∞ -calculus on Lp (Rn ) on the sector 6θ for 0 < θ < π. Proof. Let h ∈ H0∞ (6θ ), where 0 < θ < π . Then the Fourier transform of h(−1) is given by h(|ξ |2 ) for ξ ∈ Rn . The kernel kh (·) corresponding to h(| · |2 ) is given by Z 1 eixξ h(|ξ |2 )dξ, x ∈ Rn . kh (x) = (2π)n Rn
Choosing a rotation Q such that Qx = (|x|, 0, . . . , 0) and writing Qξ = (a, rb), with a ∈ R, r > 0, b ∈ Rn−1 , |b| = 1 we obtain Z∞ kh (x) = cn
r 0
n−2
Z∞
h(r 2 + a 2 )ei|x|a dadr,
x ∈ Rn .
−∞
Next we deform the contour of integration via Cauchy’s theorem to a = s + i(r + |s|),
r > 0, s ∈ R.
We then obtain Im(r 2 + a 2 ) 2ε|s|(r + |s|) 4ε(r 2 + s 2 ) Re(r 2 + a 2 ) = r 2 + s 2 − ε2 (r + |s|)2 ≤ (1 − 2ε 2 )(r 2 + s 2 ) ≤ Cε,
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which implies that for ε small enough our new contour stays inside 6θ . Thus Z∞ |kh (x)| ≤ C
r
n−2
0
Z∞
khkH ∞ e−c|x|(r+s) dsdr
0
Z∞ ≤ CkhkH ∞
|x|1−n e−c|x|s ds = C
0
khkH ∞ , |x|n
x ∈ Rn \{0}.
Similarly, |D α kh (x)| ≤ Cα khkH ∞
1 , |x|n+|α|
x ∈ Rn \{0}
for each multiindex α. If |x| ≥ 2|y| we obtain 1 Z d kh (x − ty)dt |kh (x − y) − kh (x)| = dt 0
Z1 ≤ |y| 0
|y| dt khkH ∞ ≤ C n+1 khkH ∞ . n+1 |x − ty| |x|
This implies that for R > 0 the uniform H¨ormander condition is satisfied, i.e.: Z sup |kh (x − y) − kh (x)|dx |x|>2|y|
khkH ∞ ≤R
Z ≤ C|y|R |x|>2|y|
dx = C|y|R |x|n+1
Z∞
2|y|
Since A = −1 is selfadjoint in
L2 (Rn )
(7.3) dr = CR. r2
we see that the set
{h(A) : h ∈ H0∞ (6θ ), khkH ∞ (6θ ) ≤ R} ⊂ L(L2 (Rn )) is uniformly bounded. By Remark 7.1 a) and 7.1 b) it follows that there is a constant C > 0 such that
1 1
N
N 2 2
X
X
|Hj fj |2 ≤ CR |fj |2
j =1
2 n
2 n
j =1 L (R )
L (R )
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for N ∈ N, Hj := hj (A), hj ∈ H0∞ (6θ ), khj kH ∞ (6θ ) ≤ R. Set now X := l2N (N) and define K : Lp (Rn , X) → Lp (Rn , X) by (Kf )i := Hi fi ,
i = 1, . . . , N.
The uniform H¨ormander condition (7.3) implies that Z kK(x − y) − K(x)k dx ≤ CR. |x|>2|y|
Since K acts as a bounded operator on L2 (Rn ; X), the Benedek-Calderon-Panzone theorem (see e.g. [Hie99]) implies that K is Lp -bounded for 1 < p < ∞. This means that there is a constant C > 0 such that
1
N 2
X
2 |Hj fj |
j =1
Lp (Rn )
1
N 2
X
2 ≤C |fj |
j =1
.
Lp (Rn )
Remark 7.1 b) implies that the set {h(A) : h ∈ H0∞ (6θ ), khkH ∞ (6θ ) ≤ R} ⊂ L(Lp (Rn )) is R-bounded for all R > 0 and all 0 < θ < π.
¨
COROLLARY 7.3. Let 1 < p < ∞. Denote by 1D the Dirichlet Laplacian in Rn+1 + . Then −1D admits an R-bounded H ∞ (6θ )-calculus on Lp (Rn+1 ) for each θ ∈ (0, π ). + Proof. The resolvent of the Dirichlet Laplacian 1D in Rn+1 + may be written as (λ + 1D )−1 = P0 (λ + 1)−1 E0 − P0 R(λ + 1)−1 E0 ,
λ ∈ C\6θ ,
where 1 denotes the Laplacian in Rn+1 , E0 denotes extension by 0 to Rn+1 , P0 the projection from Rn+1 to Rn+1 + and R the reflection in the normal coordinate y. Therefore we ∞ have for h ∈ H (6θ ) h(−1D ) = P0 h(−1)E0 − P0 Rh(−1)E0 . The assertion thus follows from Theorem 7.2.
¨
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We finally turn our attention to the remainder term of the Stokes operator as defined in Section 5. We start with the following proposition. PROPOSITION 7.4. Let 1 ≤ p < ∞, G ⊂ Rn be open and let T = {Tµ : µ ∈ M} ⊂ L(Lp ()) be a family of integral operators of the form Z (Tµ f )(x) = kµ (x, y)f (y)dy, x ∈ G, f ∈ Lp (G), G
such that there exists a function k0 with |kµ (x, y)| ≤ k0 (x, y), For x ∈ G set (T0 f )(x) = R-bounded.
f.a.a. x, y ∈ G ∈ M. R
G k0 (x, y)f (y)dy.
If T0 ∈ L(Lp (G)), then T ⊂ L(Lp (G)) is
Proof. Due to Remark 7.1 b) it suffices to verify the square function estimate (7.1). But ¨ this follows easily from the Lp -boundedness of the dominating operator T0 . Combining (6.5), (6.6) and Corollary 6.2 with Proposition 7.4 we obtain the following result. COROLLARY 7.5. Let 1 < p < ∞, 0 < θ < π . Let kh,v and kh,w be defined as in Section 5. For h ∈ H0∞ (6θ ) define operators Th,v and Th,w as in (6.5). Let R > 0. Then the sets
{Th,v : h ∈ H0∞ (6θ ), khkH ∞ (6θ ) ≤ R} ⊂ L(Lp (Rn+1 + ))
{Th,w : h ∈ H0∞ (6θ ), khkH ∞ (6θ ) ≤ R} ⊂ L(Lp (Rn+1 + )) are R-bounded. Summarizing, we proved the following result. p
THEOREM 7.6. Let 1 < p < ∞. Denote by A the Stokes operator in Lσ (Rn+1 + ). Then p ) for each θ ∈ (0, π ). −A admits an R-bounded H ∞ (6θ )-calculus on Lσ (Rn+1 +
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Acknowledgements We would like to thank J¨urgen Saal for thoughtful comments and discussions concerning in particular the content of Section 5. We also grateful to the referee for pointing out reference [Koz98].
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Weis, L., Operator-valued multiplier theorems and maximal Lp -regularity. Preprint, 1999. Wiegner, M., The Navier-Stokes equation - a never ending challenge? Jahresberichte DMV 181 (1999), 1–25. Wolfgang Desch Institut f¨ur Mathematik Universit¨at Graz Heinrichstr. 36 A-8010 Graz Austria e-mail:
[email protected] Matthias Hieber Fachbereich Mathematik Angewandte Analysis Technische Universit¨at Darmstadt Schlossgartenstr. 7 D-64289 Darmstadt Germany e-mail:
[email protected] Jan Pr¨uss Fachbereich f¨ur Mathematik und Informatik Institut f¨ur Analysis Martin-Luther-Universit¨at Halle-Wittenberg Theodor-Lieser-Str. 5 D-06099 Halle Germany e-mail:
[email protected]
