decompositions oflpand hardy spaces of polyharmonic functions
TRANSCRIPT
Ž .JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 216, 499]509 1997ARTICLE NO. AY975675
Decompositions of L p and Hardy Spaces ofPolyharmonic Functions
Miroslav Pavlovic
Matematicki fakultet, Studentski trg 16, 11001, Belgrade, Serbia, Yugosla iaˇ
Submitted by L. Debnath
Received January 16, 1996
p, q, a Ž .Let H 0 - p, q F `, y` - a - ` denote the space of those polyhar-kmonic functions f of order k on the unit n-ball for which the function r ¬Ž .ay1r q Ž . qŽ .1 y r M f , r belongs to L 0, 1 . Our main result is that, when k G 2 andp
Ž .a ) y1, the operator f ¬ Pf, D f , where Pf is the Poisson integral of f , acts asan isomorphism of H p, q, a onto the direct sum of H p, q, a and H p, q, aq2. Anotherk 1 ky1decomposition theorem, closely related to the Almansi representation theorem, isalso given. Q 1997 Academic Press
1. INTRODUCTION AND RESULTS
Throughout the paper we fix two positive integers, n and k, and letŽ . � `Ž . k 4H s H B s f g C B : D f ' 0 , where B denotes the unit ball in thek k
euclidean n-space and Dk is the k th power of the Laplacian. In particular,H is the class of all functions harmonic in B. In this paper we are1concerned with the spaces
H p , q , a s H l L p , q , a 0 - p , q F `, y` - a - ` ,Ž .k k
where L p, q, a denotes the class of those Borel functions f on B for which1rq
qay12q5 5 5 5 < < < <f s f [ M f , x 1 y x dx - `.Ž . Ž .a p , q , a H p½ 5B
Here as usual1rp
pM f , r s f ry dd y 0 F r - 1 ,Ž . Ž . Ž . Ž .Hp ½ 5
S
where dd is the normalized surface measure on S s B.
499
0022-247Xr97 $25.00Copyright Q 1997 by Academic Press
All rights of reproduction in any form reserved.
MIROSLAV PAVLOVIC500
p, q, a pŽ Ž < < 2 . pay1 .If q s p - `, then L s L B, 1 y x dx , while if q s `,then
a25 5f s sup 1 y r M f , r .Ž . Ž .a pr-1
In particular, if a s 0 and q s `, then H p, q, a is a ‘‘polyharmonic’’ Hardykspace.
Much is known about the structure of the harmonic spaces, especially inw x p, q, athe case n s 2; see, for example, 3 . In this paper we prove that H ,k
a ) y1, can be decomposed into the direct sum of k harmonic spaces. Inorder to state the results we need some simple facts on polyharmonicfunctions.
w xBy the Almansi representation theorem 1 , the class H coincides withkthe class of functions of the form
ky12 j< <f x s x u x , u g H . 1Ž . Ž . Ž .Ý j j 1
js0
It will be convenient to rewrite this as
ky1j2< <f x s 1 y x f x , f g H . 2Ž . Ž . Ž .Ž .Ý j j 1
js0
It is easily verified that f s Pf, where0
< < 21 y xPf x s lim f ry dd y f g H , x g B .Ž . Ž . Ž . Ž .H n ky < <x y yrª1 S
Observe that if f is continuous on the closed unit ball, then Pf coincideswith the Poisson integral of the boundary function.
There is another way to identify a function f g H with a sequence of kkŽ .kharmonic functions. To see this define the operator Q : H ª H [k k 1
Ž .H [ ??? [ H k times by1 1
Q f s Pf , PD1 f , . . . , PDky1 f . 4Ž .Ž .k
w x Ž .From the proof of Proposition 1.3 in 1 it follows that the operator P, DŽ .acts as an isomorphism from H onto H [ H k G 2 . Using this onek 1 ky1
Ž .kproves, by induction on k, that Q is an isomorphism of H onto H .k k 1Our main result here is the following theorem.
Ž . Ž .THEOREM 1. Let a ) y1, k G 2, and 0 - p, q F `. Then i P, Dacts as an isomorphism from H p, q, a onto H p, q, a [ H p, q, aq2, andk 1 ky1Ž . p, q, aii H is isomorphic, ¨ia the operator Q , to the direct sum of the spacesk kH p, q, aq2 j, j s 0, 1, . . . , k y 1.1
POLYHARMONIC FUNCTIONS 501
Our second decomposition theorem is a formal consequence the factthat the operator P acts as a projection of H p, q, a onto H p, q, a, which isk 1part of Theorem 1.
THEOREM 2. Let a ) y1, 0 - p, q F `. Then the operator f ¬Ž . Ž . p, q, af , f , . . . , f , defined by 2 , acts as an isomorphism from H onto0 1 ky1 kthe direct sum of the spaces H p, q, aqj, 0 F j F k y 1.1
In other words, f g H p, q, a if and only if f g H p, q, aqj for e¨ery j.k j 1
We shall deduce these theorems, in Section 2, from the following factsŽ . 1which are of independent interest. For a complex valued C function f
let1r22n f
=f x s , x s x , . . . , x g B.Ž . Ž .Ý 1 nž / xiis1
If s ) 0, we let
1 sy1I f x s f tx t dt , x g B. 5Ž . Ž . Ž .Hs0
p, q, g < <LEMMA 1. Let y` - a - `, 0 - p, q F `. If f g H , then =f gkp, q, aq1 5 < < 5L and there is a constant C independent of f such that =f Faq15 5C f .a
Ž .LEMMA 2. If a ) 0, s ) 0, then I acts as a bounded operator fromsH p, q, aq1 to H p, q, a.k k
The proofs of the lemmas are in Sections 3 and 4. The first results ofw x w xthis kind were proved by Hardy and Littlewood 4 , and we refer to 3 for
information about the case n s 2, k s 1. The case of holomorphic func-w xtions on the ball was considered in 6 . Reduced to the harmonic case our
w xproof is much shorter than that in 6 and is based on a result of Hardy andw x w xLittlewood 4 and Fefferman and Stein 2 on subharmonic behavior of
< < p Ž .f . See Section 3.
2. PROOF OF THE THEOREMS
For a C`-function f on B let
R f s sf q Rf s ) 0 ,Ž .s
wheren f
Rf x s x .Ž . Ý i xiis1
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We have thatR I f s I R f s f ,s s s s
which is easily verified by considering homogeneous polynomials. Thisshows that R and I are isomorphisms of H onto H .s s k k
For the proof of Theorems 1 and 2 we need a simple formula.
Ž .LEMMA 3. If a function f g H is gi en by 2 , thenk
ky1ky1D f s y1 c R R . . . R f ,Ž . k s sq1 sqky2 ky1
ky1Ž .where s s nr2 and c s 4 k y 1 !.k
ky1 Ž .ky1 ky1Ž < < 2 ky2 .Proof. Since D f s y1 D x f it suffices to show thatky1
ky1 < < 2 ky2D x u s c R . . . R u , u g H .Ž . k s sqky2 1
This is deduced from the formula
< < 2 ky2 < < 2 ky4D x u s 4 k y 1 x R uŽ .Ž . sqky2
by induction.
An application of Lemma 1 to the partial derivatives of f yields thefollowing.
LEMMA 4. If y` - a - `, 0 - p, q F `, then the operator D is ap, q, a p, q, aq2 Ž .bounded operator from H into H k G 2 .k ky1
The following result is a reformulation of Theorem 2. For a function fŽ .given by 2 we define
jy1m2< <P f x s 1 y x f x , 1 F j F k y 1 k G 2 .Ž . Ž . Ž .Ž .Ýj m
ms0
Ž Ž ..Note that P f s Pf s f see 3 .1 0
THEOREM 3. If a ) yj, 1 F j F k y 1, then P acts as a boundedjprojection of H p, q, a onto H p, q, a.k j
Ž . 5 5Proof. We omit the indices p, q. We have to prove that P f Faj5 5 a Ž . aC f for f g H C s const. . Let f g H . By Lemma 3 we have thata k k
Ž .ky1 y1P f s f y y1 c g, whereky1 k
ky12 ky1< <g x s 1 y x I I . . . I D f x .Ž . Ž .Ž . s sq1 sqky2
Hence, by successive applications of Lemmas 2 and 4,
5 5 5 ky1 5g s I . . . I D fa aqky1s sqky2
5 ky1 5F C D f aq2Žky1.1
5 5F C f a2
POLYHARMONIC FUNCTIONS 503
Žfor some constants C and C . Lemma 2 is applicable because a q k y1 2. 5 5 5 51 ) 0. This implies that P f F C f , which concludes the proof ina aky1
the case k s 2. If k G 3, we apply the above proof to the functionaP f g H to get the required result for j F k y 2.ky1 ky1
The fact that the operator D maps H onto H aq2 for a ) y1 isk ky1contained in the following.
THEOREM 4. Let 1 F j F k y 1 and a ) yj. Then the operator D j
maps H p, q, a onto H p, q, aq2 j.k kyj
Proof. That D j is into follows from Lemma 4. To prove the rest denoteŽ . aq2 jby Th m, j , 1 F j F m y 1, the following statement: If u g H andmy j
a ) yj, then there is f g H a such that D j f s u.mŽ . aq2Žky1.First we prove Th k, k y 1 , k G 2. Let u g H , a q k y 1 ) 0.1
Define f by
ky1 ky12 y1< <f x s 1 y x y1 c I . . . I u x .Ž . Ž . Ž .Ž . k s sqky2
Then Dky1 f s u, by Lemma 3, and f g H a, by Lemma 2.kŽ . Ž .Thus it remains to deduce Th k, j from Th k, j q 1 for 1 F j F k y 2.
Let u g H aq2 j, a ) yj. Then Du g H aq2 jq2, by Lemma 4. Hence, byky j kyjy1Ž . jq1 jTh k, j q 1 , there exists an f g H such that D f s Du. Then D f gk
aq2 j Ž . j aq2 j Ž j .H Lemma 4 and hence u y D f g H because D u y D f s 0.ky j 1Ž . a j jNow we apply Th j q 1, j to find g g H so that u y D f s D g. Thusjq1
j aŽ .we have u s D f q g and f q g g H , which was to be proved.k
Ž .Proof of Theorem 1. i If D f s 0, then f s Pf and therefore theŽ . a aq2operator P, D is one-to-one. Let u g H and ¨ g H . By Theorem 4,1 ky1
there is g g H a such that ¨ s D g. Taking f s g y Pg q u we get u s Pfka Ž .and D f s ¨ , and f g H by Theorem 3 j s 1 . Combining this withk
Ž .Theorem 3 and Lemma 4 we obtain the assertion i .Ž . Ž . aii That the operator Q , defined by 4 , is an operator from Hk k
into H a [ H aq2 q ??? follows from Theorems 3 and 4. If k s 2, then1 1Ž . Ž . Ž .Q s P, D and therefore assertion ii coincides with i . Let k G 3 andk
assume that ¨ g H aq2 j, 0 F j F k y 1. Then by the induction hypothesisj 1aq2 Ž . Ž .there is g g H such that Q g s ¨ , . . . , ¨ . By i , Pf s ¨ andky1 ky1 1 ky1 0
Ž .D f s g for some f g H . Hence Q f s ¨ , ¨ , . . . , ¨ . The uniquenessk k 0 1 ky1of f is easily verified.
Remark 1. Let p G 1. If either q s ` and a - 0 or q - ` and a F 0,p, q, a � 4then H s 0 . On the other hand, for all p - 1 and q F ` there exist1
a - 0 such that H p, q, a is not trivial, which can be shown by considering1Poisson’s kernel.
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Remark 2. The restriction a ) y1 is in a sense necessary for thevalidity of Theorem 1. For instance, it is easy to see that the conditionŽ p, `, y1. p, `, 1 Ž . Ž p, `, 0.D H s H p G 1 is equivalent to the condition R H2 1 n r2 1
p, `, 1 Ž . Ž .s H . The latter means that M f , r s O 1 if and only if1 pŽ . Ž .y1M R f , r s O 1 y r , which is false.p n r2
3. PROOF OF LEMMA 1
Ž .Let B x denote the ball centered at x and of radius r. We alwaysrŽ .assume that B x ; B. By dn we denote the Lebesgue measure normal-r
Ž .ized so that n B s 1. The heart of our proofs is the following lemma.
LEMMA 5. Let f g H and 0 - p - `. Then there is a constant C skŽ .C p, k, n such that
p pyn < <f x F Cr f dn 6Ž . Ž .HŽ .B xr
and
< < p p yn < < p=f x r F Cr f dn . 7Ž . Ž .H
Ž .B xr
Ž . Ž .Proof. Inequality 7 is implied by 6 and
y1 < <=f x F C r sup f , 8Ž . Ž .1Ž .B xr
C s const. In view of the inequality1
n nr< < < <=f x F sup f q sup D fŽ .
r n q 1Ž . Ž .B x B xr r
Ž w x. Ž .cf. 1, Proposition 3.1 , proving 8 reduces to proving
y2 < <D f x F C r sup f ,Ž . 2Ž .B xr
Ž .where C is a constant. The validity of 6 for p s 1 implies its validity for2Ž w x. Ž . Ž .all p ) 0 cf. 5 . Thus we have to prove 9 and 6 for p s 1.
Ž .Assuming f is given by 1 we define the polynomial h by
ky12 jh r s u 0 rŽ . Ž .Ý j
js0
s f ry dd y .Ž . Ž .HS
POLYHARMONIC FUNCTIONS 505
It is clear that there is a constant C such that
1 ny1u 0 q u 0 F Cn r h r drŽ . Ž . Ž .H0 10
1 ny1F Cn r dr f ry ddŽ .H H0 S
s C f x dn x .Ž . Ž .HB
Ž . Ž . Ž . Ž .Since f 0 s u 0 and D f 0 s 2nu 0 we obtain the result.0 1
Lemma 5 can also be stated as follows.q Ž .LEMMA 5 . Let y` - a - `, 0 - p, q F `. To each « g 0, 1 there
Ž .corresponds a constant C « independent of f g H and such thatk
< < 5 5sup f F C « f . 10Ž . Ž .p , q , aŽ .B 0«
< < < <There holds the analogous inequality in which sup f is replaced by sup Df ,where Df is any partial deri ati e of any order.
This shows that the norm convergence in H p, q, a implies the uniformkconvergence, on compact subsets of B, of all the partial derivatives. Usingthen Fatou’s lemma in the standard way one proves that the space H p, q, a
kis complete. Moreover the closed balls in H p, q, a are compact in thektopology of uniform convergence on compact subsets of B.
Now we shall prove Lemma 1 as well as a simple maximal theorem. Fora continuous function f on B let
qf ry s sup f ty : r - t - 1 q r r2 ,� 4Ž . Ž . Ž .
0 F r - 1, y g S.p, q, a Ž .LEMMA 6. Let f g H a ) y` and letk
2q < <F x s f x or F x s 1 y x =f x .Ž . Ž . Ž . Ž .Ž .p, q, a Ž .Then F g L and there is a constant C s C p, k, n, q, a such that
5 5 5 5F F C f .p , q , a p , q , a
Proof. We shall consider the case where p - ` and q - `. In theremaining cases the proof is simpler. Let
< < < <D x s z g B : z y x - « 1 y x .� 4Ž . Ž .«
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Ž . Ž . Ž . Ž .Since D ty ; D ry for r - t - 1 q r r2, y g S, it follows from 61r4 3r4that, when F s fq,
ynq q< < < <F x F C 1 y x f dn D s D . 11Ž . Ž . Ž .Ž .H 3r4Ž .D x
Ž . Ž . Ž < < 2 . < <Inequality 7 shows that 11 holds when F s 1 y x =f as well.Ž .Replacing x in 11 by Ux, where U is an arbitrary orthogonal transforma-
tion of the n-space, and then applying the change z ¬ Uz, we find that
qynq< <F Ux F C 1 y x f Uz dn z .Ž . Ž . Ž . Ž .H
Ž .D x
pr qŽ .Now, for p G q, we use Minkowski’s inequality for the norm in L dUto deduce that
nq q< < < < < <M F , x F C 1 y x M f , z dn z . 12Ž . Ž . Ž . Ž .Ž .Hp pŽ .D x
Here we also used the identity
pp < <M f , z s f Uz dU,Ž .Ž . Hp
where the integral is taken over the orthogonal group. If q ) p, then weŽ .use the case q s p of 12 together with Jensen’s inequality to conclude
Ž .that 12 holds for all p and q. Since
1 5< < < < < <1 y x F 1 y z F 1 y x , x g D x , 13Ž . Ž . Ž . Ž .4 4
we obtain
b bynq q< < < < < < < <M F , x 1 y x F C M f , z 1 y z dn z ,Ž . Ž . Ž . Ž .Ž .Hp pŽ .D x
b s a q y 1. Integrating this inequality over B and changing the order of5 5 qintegration we find that F is dominated by a constant multiple ofa
bynq < < < <M f , z 1 y z dn z dn x ,Ž . Ž . Ž .Ž .H HpŽ .B E z
where
< < < <E z s x g B : x y z - 3r4 1 y x .� 4Ž . Ž . Ž .
POLYHARMONIC FUNCTIONS 507
Ž Ž .. nŽ < <.n Ž .Finally one has n E z F 3 1 y z , by 13 , and this concludes theproof of Lemma 6 and Lemma 1.
Remark. Through the above proof we used C to denote a positiveconstant which may vary from line to line.
4. PROOF OF LEMMA 2
For the proof we need a technical lemma.
LEMMA 7. Let f be a nonnegati e continuous function on the inter alw . Ž .0, 1 , and let g ) 0, l ) 0. Then there is a constant C s C g , l such that
1 l gy1 lf r 1 y r dr F C sup f rŽ . Ž . Ž .H
0 r-1r2
1 q r1 gy1q C f y f r 1 y r dr .Ž . Ž .H ž /20
14Ž .
Proof. Assume as we may that f is bounded. Denote by I and I the1 2Ž .first and the second integral in 14 , respectively. Then
l1 q r1 gy1ygI s 2 f 1 y r drŽ .H1 ž /2y1
F 2yg I q K ,3
where
l1 q r1 gy1I s f 1 y r dr ,Ž .H3 ž /20
ly1K s g sup f r .Ž .r-1r2
Let l G 1. Then I 1r l F I 1r l q I 1r l, by Minkowski’s inequality, and3 2 1hence
I 1r l F 2yg rlI 1r l q K 1r l1 3
F 2yg rl I 1r lI q I 1r l q K 1r l .Ž .1 2
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Since I - ` and 2yg rl - 1 we get1
y11rl yg rl 1rl 1rlI F 1 y 2 I q K ,Ž . Ž .1 2
which concludes the proof in the case l G 1. If l - 1, the proof is similarl l lŽ . Ž .and is based on the inequality a q b F a q b a, b ) 0 .
ŽProof of Lemma 2. Let q - ` and a ) 0. In the case q s ` the proof.is simpler and is based on the obvious modification of Lemma 7. Let
p, q, aq1 Ž .f g H . By 5 , there are constants C and C such thatk 1 2
r< <I f ry F C sup f q C f ty dt « s 1r2 .Ž . Ž . Ž .Hs 1 2
0Ž .B 0«
Ž . 5 5 5 5This and 10 show that it suffices to prove that g F C f , wherea aq1
rg ry s f ty dt y g S, 0 F r - 1 .Ž . Ž . Ž .H
0
Let p - 1. Then
p1 q r p p pqg y y g ry F 1 y r f ry .Ž . Ž . Ž .ž /2
Hence, by integration over S,
1 q r p p qf y f r F 1 y r M f , r ,Ž . Ž . Ž .pž /2
where
f r s M p g , r .Ž . Ž .p
Ž . 5 5 qCombining this with Lemma 7 l s qrp, g s qa we see that g isa
dominated by a constant multiple of
1 Ž .q aq1 y1q qM f , r 1 y r dr \ A.Ž . Ž .H p0
Now we use the formula
1 ny1< <h x dn x s n h r r drŽ . Ž . Ž .H HB 0
Ž . 5 q5 q Ž q.and 10 to show that A F C f . Finally Lemma 6 F s f com-aq1
pletes the proof in the case p - 1.
POLYHARMONIC FUNCTIONS 509
If p G 1, we use Minkowski’s inequality to get
1 q rqM g , y M g , r F 1 y r M f , rŽ . Ž . Ž .p p pž /2
and then proceed as in the case p - 1.
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