Sobolev Trace Theorem and the Dirichlet Problem in a Ball

Transcription

1 Inernaional Journal of Mahemaical Analysis Vol. 1, 216, no. 24, HIKARI Ld, hp://dx.doi.org/ /ijma Sobolev Trace Theorem and he Dirichle Problem in a Ball Yoichi Miyazaki School of Denisry, Nihon Universiy Kanda-Surugadai, Chiyoda-ku Tokyo , Japan Copyrigh c 216 Yoichi Miyazaki. This aricle is disribued under he Creaive Commons Aribuion License, which permis unresriced use, disribuion, and reproducion in any medium, provided he original work is properly cied. Absrac In an elemenary and direc way we prove he race heorem as well as solve he Dirichle boundary value problem for he uni ball in he framework of L p -based Sobolev spaces. Mahemaics Subjec Classificaion: 46E35, 35J5 Keywords: Sobolev space, Besov space, race heorem, Dirichle boundary value problem, Laplacian 1 Inroducion In our previous paper [9], we gave an elemenary proof of he race heorem for he uni disk in he framework of L p -based Sobolev spaces. We also solved he Dirichle boundary value problem for he Laplacian in he uni disk in an elemenary way. The mehod used in [9] is direc in he sense ha i does no use he echnique of reducing he problem o he half space by localizing and flaering he boundary. Boh he race heorem and he Dirichle boundary problem have been fully culivaed in he more general seings. For example, see [5, 11] for he race heorem wih p = 2, [12] for he race heorem wih general p, [11] for he Dirichle problem of he Laplacian wih p = 2, and [3, 7] for he Dirichle problem of higher-order ellipic equaions wih general p. See

2 1174 Y. Miyazaki also he references cied in [9]. However, i would be meaningful o invesigae in he framework of L p spaces wheher here is a mehod ha makes use of he specialiy of he uni ball, and corresponds o he classical heory for C 2 funcions, as described in [1, 4, 5, 6]. In his paper, we do he same hings as in [9] for he uni ball in R d, allowing d o be greaer han 2. Whereas for d = 2 we can consider a group acion of R o he uni circle or define he ranslaion of a poin x = (cos θ, sin θ) by h R o be (cos(θ + h), sin(θ + h)), we canno do so for he uni sphere if d 3. Since we mus rea wo poins x and y on he uni sphere as equals, he problem for d 3 is a lile more difficul han ha for d = 2. For he proof of he race heorem we carry over one of he proofs given in [8] for he whole space, which uses Muramau s inegral formula [1], o he uni ball. Le us formulae he wo heorems ha we will prove in an elemenary way. Throughou his paper we denoe by B he uni ball in R d wih d 2. The closure of B is denoed by B. For 1 < p < le W 1 p (B) be he L p -based Sobolev space of order 1: W 1 p (B) = {f L p (B) : j f L p (B) (j = 1,..., d)}. As a definiion of Besov spaces we adop he classical one. Le < s < 1 and 1 < p <. For a measurable funcion f on, he boundary of B or (d 1)-sphere of radius 1, we se [ f B s pp () = ] f(x) f(y) p 1/p ds sp+d 1 x ds y, (1.1) f B s pp () = f Lp() + f B s pp (), (1.2) where ds is he surface elemen on. The Besov space B s pp() is defined o be he space of all funcions f L p () saisfying f B s pp () <. Equipped wih he norm B s pp (), he space B s pp() is a Banach space. Theorem 1.1. Le 1 < p <. The race operaor Tr : W 1 p (B) C( B) C() defined by he resricion o he uni sphere, namely Tr f(x) = f(x) for x exends o a bounded linear operaor from Wp 1 (B) ono Bpp 1 1/p (), and saisfies Tr f Lp() C(d, p) f W 1 p (B), Tr f C(d, p) f 1 1/p B pp () L p(b), (1.3) where C(d, p) is a consan depending only on d and p.

3 Sobolev race heorem in a ball 1175 Theorem 1.2. Le 1 < p <. For a given g Bpp 1 1/p () he Dirichle boundary problem for he Laplace equaion { u(x) = in B, (1.4) Tr u = g has a unique soluion u in Wp 1 (B), where he differenial operaor should be considered in he disribuional sense. This soluion u also belongs o C (B). The map g u is a bounded linear operaor from Bpp 1 1/p () o Wp 1 (B), and saisfies u Lp(B) C(d, p) g Lp(), u Lp(B) C(d, p) g B 1 1/p pp (). In [9] we deal wih he case d = 2, and adoped as he Besov seminorm f B s pp() = [ π dh π f( + h) f() p d h sp+1 insead of (1.1), where f is he periodic funcion on R defined by f() = f(cos, sin ). We can verify ha he norm f B s pp ( D) defined in (1.2) is equivalen o f Lp( D) + f Bpp s ( D) by using he inequaliy h /2 eih 1 h for h 1, and hence ha he Besov space defined in his paper coincides wih ha defined in [9]. This paper is organized as follows. In Secion 2 we review Muramau s inegral formula, which plays a cenral role in he proof of Theorem 1.1. Secion 3 is devoed o he evaluaion of some inegrals ha will be used in he proof of Theorems 1.1 and 1.2. We prove Theorem 1.1 in Secion 4, and Theorem 1.2 in Secion 5. 2 Muramau s inegral formula In his secion, we inroduce Muramau s inegral formula [1] in he form ha mees our purpose. We wish o express a funcion f belonging o L p (B) or Wp 1 (B) by inegral operaors wih he help of modified convoluions. To his end, we choose a funcion ϕ C (R d ) wih supp ϕ B and ϕ(x) dx = 1. B For f L p (B) we define he funcion M f of (, x) wih > and x R d by ( ) x z M f (, x) = d ϕ x f(z) dz. (2.1) B Clearly M f C ((, ) R d ). Unlike he convoluion d ϕ( /) f(x), we use (x z)/ x insead of (x z)/ in he above definiion. The advanage ] 1/p

4 1176 Y. Miyazaki of he exra erm x is ha he suppor of he funcion ϕ((x )/ x) is conained in (1 )x (supp ϕ) B if < 1 and x 1, and hence ha we can handle derivaives for disribuions on B. Here and in he following we use he convenion x + V = {x + v : v V }, V = {v : v V } for a subse V of R d, x R d and >. We se K j (x) = x j ϕ(x) for j = 1,..., d, and have { d ϕ((x y)/ x)} = d 1 n ( j K j )((x y)/ x). Inegraing / {M f (, x)} over he inerval [ɛ, 1] wih < ɛ < 1 and noing M f (1, x) = R d ϕ( z)f(z) dz, we ge M f (ɛ, x) = n j=1 1 ɛ j=1 ( ) x z d d 1 ( j K j ) x f(z) dz + ϕ f() (2.2) B for f L p (B). If we assume f W 1 p (B), hen we can express he inegral in (2.2) in erms of he derivaives of f. Indeed, in (2.2) we wrie ( j K j )((x z)/ x) = zj {K j ((x z)/ x)}, which is suppored in B as a funcion of z, and move zj o f(z) by he definiion of disribuional derivaive. Then we ge n 1 ( ) x z M f (ɛ, x) = d d K j x j f(z) dz + ϕ f() (2.3) for f W 1 p (B). j=1 ɛ B Proposiion 2.1. Le < ɛ < 1. (i) Le f belong o C( B), he space of coninuous funcions on he closure of B. Then M f (ɛ, ) C( B), and M f (ɛ, ) converges o f uniformly on B as ɛ. (ii) Le f L p (B). Then M f (ɛ, ) L p (B), and M f (ɛ, ) f in L p (B) as ɛ. (iii) Le f W 1 p (B). Then M f (ɛ, ) W 1 p (B) C (R d ), and M f (ɛ, ) f in W 1 p (B) as ɛ. Remark 2.2. By Proposiion 2.1 we know ha f Wp 1 (B) (resp. f L p (B)) is wrien as he limi of he righ-hand side of (2.3) (resp. (2.2)) in Wp 1 (B) (resp. L p (B)). We call hese expressions of f Muramau s inegral formula.

5 Sobolev race heorem in a ball 1177 Proof. (i) Le f C( B). M f (ɛ, ) C( B) follows from ϕ C (B). The change of variables y = (x z)/ɛ x gives M f (ɛ, x) f(x) = ϕ(y){f((1 ɛ)x ɛy) f(x)} dy B for x 1 and < ɛ < 1. We noe ha (1 ɛ)x ɛy B in he above inegral. The uniform coninuiy of f yields he uniform convergence of M f (ɛ, ). (ii) Le f L p (B). Since ϕ((x z)/ɛ x) implies (x z)/ɛ x+supp ϕ 2B, we have M f (ɛ, x) ϕ L (B) B ɛ d χ 2B ((x z)/ɛ) f(z) dz, where χ 2B is he characerisic funcion of 2B. By Young s inequaliy M f (ɛ, ) Lp(B) ϕ L (B) χ 2B L1 (R d ) f Lp(B). (2.4) Thus M f (ɛ, ) L p (B). The riangle inequaliy and (2.4) give M f (ɛ, ) f Lp(B) M g (ɛ, ) g Lp(B) + C f g Lp(B) (2.5) wih C = 1 + ϕ L (B) χ 2B L1 (R d ) for any g C (B), he space of coninuous funcions suppored in B. Since C (B) is dense in L p (B), we see from (i) and (2.5) ha M f (ɛ, ) f in L p (B) as ɛ. (iii) Le f W 1 p (B). M f (ɛ, ) C (R d ) follows from ϕ C (B). We differeniae (2.1) in x j and noe ha he suppor of xj {ϕ((x z)/ x)} = ( 1)(1 ) zj {ϕ((x z)/ x)} as a funcion of z is conained in (1 )x + B B for x B and < 1. Then he definiion of disribuional derivaive yields xj M f (, x) = (1 )M j f(, x). Applying (ii) o f and j f, we see ha M f (ɛ, ) W 1 p (B) and ha M f (ɛ, ) f and xj M f (ɛ, ) j f in L p (B) as ɛ. Hence M f (ɛ, ) f in W 1 p (B). 3 Some inequaliies In his secion, we collec some lemmas for he proof of Theorems 1.1 and 1.2. Lemma 3.1. Le 1 < p <. Le (X, M, µ) and (Y, N, ν) be σ-finie measure spaces, and le K be an (M N )-measurable funcion on X Y. If wo quaniies M := sup K(x, y) dν(y), M 1 := sup K(x, y) dµ(x) x X Y y X X

6 1178 Y. Miyazaki are finie, hen he inegral operaor T f(x) = K(x, y)f(y) dν(y) is a bounded Y operaor from L p (Y ) o L p (X), and saisfies T f Lp(X) M 1 1/p M 1/p 1 f Lp(Y ) for f L p (Y ). Proof. This is he well-known boundedness heorem for inegral operaors. For he proof see [2, Theorem 6.18]. In he res of his paper, we denoe by c d he surface area of he uni sphere, i.e. c d = ds, (3.1) and by χ he characerisic funcion of he inerval [, 1], i.e. χ() = 1 for 1, χ() = for / [, 1]. (3.2) Lemma 3.2. Le x = y = 1 and r 1. Then rx y 2 = x ry 2 = (1 r) 2 + r 2. (3.3) Proof. Every expression in (3.3) equals r 2 2rx y + 1. Lemma 3.3. Le x = 1, r 1 and >. Then ( ) ( ) ( ) x ry x ry 1 r χ ds x = χ ds y C(d) d 1 χ. Proof. By roaion we see ha he wo inegrals depend only on and r. So he equaliy follows from x ry = rx y by (3.3). I remains o show he inequaliy. Le r 1/2. Since x ry 1 r, we have χ( x ry /) χ((1 r)/). Then he inegral is bounded by c d χ((1 r)/). If 1/2, hen 2 1 d d 1. If < < 1/2, hen χ((1 r)/) = since 1 r 1/2 >. Hence we ge he desired inequaliy for r 1/2. Le 1/2 r 1. Since x ry implies 1 r and 2 by (3.3), we have χ( x ry /) χ((1 r)/)χ(/2). So he problem reduces o showing he inequaliy ( ) χ ds y C(d) d 1 for >. (3.4) If 1/2, hen (3.4) is obvious, since he inegral is bounded by c d. Le < 1/2. By roaion we may assume x = (,...,, 1). We noe ha χ(/) implies y and y d > for y = (y, y d ). Using ds y = dy / 1 y 2 and he subsiuion y = z, we have ( ) 1 χ ds y dy = 2d 1 dz, 3 which gives (3.4). y z 1

7 Sobolev race heorem in a ball 1179 Lemma 3.4. Le x = 1 and < 1. Then { } 1 min, 1 ds d 1/p y C(d, p) 1+1/p. Proof. By roaion we may assume x = (,...,, 1). We divide he sphere ino wo pars, and wrie he inegral of he lemma as I + I 1, where I and I 1 are he inegrals over /2 and over /2, respecively. Since /2 implies y /2 and y d > for y = (y, y d ), we have, wih he change of variables y = z, I = x y /2 1+1/p d+1+1/p ds y z 1/2 z d+1+1/p dz. y /2 1 y d+1+1/p dy 1 y 2 As for I 1 we noe ha he condiions /2 1/2 and y = 1 imply y /4 and y d >, since x d y d = 1 1 y 2 = y 2 /(1 + 1 y 2 ) y and hence 2 y. Therefore 1 I 1 x y /2 ds d 1/p y = /2 x y 1/2 y d+1/p dy d 1/p d 1 c d y /4 = 2 1+1/p 3 z 1/4 z d+1/p dz + 2 d 1/p d 1 c d. The lemma follows from he above inequaliies for I and I 1. + x y 1/2 Lemma 3.5. Le x = y = 1, r < 1 and a < 1. Then a rx y ds a d x = rx y ds d y C(d, a)(1 r) a 1. x y 1/2 Proof. By roaion and (3.3) we see ha he wo inegrals depend only on r and a, and equal each oher. To show he inequaliy we consider wo cases. Le r 1/2. The inegral is bounded by 2 a+d c d, and he desired inequaliy follows from 1 r 1. Le 1/2 r < 1. We wrie he inegral as a I + I 1 := rx y ds a d y + rx y ds y. d x y 1/2

8 118 Y. Miyazaki Since rx y 2 1 x y by (3.3), we have I 2 a+d x y 1/2 x y d ds y 2 a+2d c d. This combined wih 1 r 1/2 gives I C(1 r) a 1. For I 1 we may assume ha x = (,,, 1) by roaion. By (3.3) I 1 x y 1/2 1 2 d {(1 r) } (d a)/2 ds y. Using ds y = dy / 1 y 2, and changing he variables y = (1 r)z, we have 2 d 1 I y 1/2 {(1 r) 2 + y 2 dy }(d a)/2 2 d+1 (1 r) a 1 1 R (1 + z d 1 2 ) (d a)/2 dz. The desired inequaliy follows from he above inequaliies for I and I 1. Lemma 3.6. Le x = y = 1. Then 1 Proof. The inegral is evaluaed by 1/2 (1 r) 1+1/p rx y d dr C(d, p) d+1/p. (3.5) (2 1 ) 1+1/p 2 d dr + 1 1/2 2 d 1/p + 2 d d+1/p (1 r) 1+1/p dr 2 d {(1 r) } d/2 s 1+1/p ds, (s 2 + 1) d/2 where we have made he change of variables 1 r = s. The desired inequaliy follows from 2. 4 Proof of he race heorem In he proof of he race heorem he main ask is o evaluae he inegrals ha appear in Muramau s inegral formula. Proposiion 4.1. Le K be a non-negaive valued funcion in C (R d ) saisfying supp K B and K(x) dx = 1. For < ɛ < 1 and g L R d p (B) we define he funcion G ɛ (x) by 1 ( ) x z G ɛ (x) = d d K x g(z) dz (4.1) ɛ B for x = 1. Then G ɛ Bpp 1 1/p (), and here exiss G Bpp 1 1/p () such ha G ɛ G in L p () as ɛ, and ha G Lp() C(d, p) g Lp(B), G B 1 1/p pp () C(d, p) g L p(b). (4.2)

9 Sobolev race heorem in a ball 1181 Proof. Sep 1. By K C (R d ), he righ-hand side of (4.1) belongs o C (R d ) and hence G ɛ L p (). We will show ha G ɛ converges in L p () as ɛ, and evaluae he L p norm of he limi funcion. Le x = 1. Since K((x z)/ x) implies (x z)/ x+supp K 2B, we have K((x z)/ x) K L (B)χ( x z /2) and hence G ɛ (x) C 1 ɛ d 1 r d 1 dr d χ ( x rω 2 ) g(rω) ds ω by he change of variables z = rω. Here we recall ha χ is he characerisic funcion of [, 1], as defined in (3.2). Minkowski s inequaliy, Young s inequaliy and Lemma 3.3 yield 1 1 ( ) 1 r G ɛ Lp() C d 1 χ g(r)r d 1 dr 2 ɛ wih g(r) = [ g(rω) p ds ω ] 1/p. Le q be he conjugae exponen of p, i.e. p 1 + q 1 = 1. Using Hölder s inequaliy wih g p L = 1 p(b) g(r)p r d 1 dr and 1 χ((1 r)/2)q r d 1 dr 1 χ(s/2)q ds 2, we have G ɛ Lp() C 1 ɛ 1/p g Lp(B) d C(1 ɛ 1 1/p ) g Lp(B). (4.3) If we replace 1 ɛ d by δ ɛ d in he above calculaion, hen we ge G δ G ɛ Lp() C(δ 1 1/p ɛ 1 1/p ) g Lp(B) for < ɛ < δ < 1. Since {G ɛ } saisfies he Cauchy crierion, G ɛ converges in L p () as ɛ. Le G be he limi funcion. Leing ɛ in (4.3), we ge he firs inequaliy in (4.2). wih Sep 2. We will evaluae G 1 1/p B pp () For x = y = 1 we wrie G ɛ G ɛ (x) G ɛ (y) = 1 ( x z K(x, y, z, ) = K ɛ o show G B1 1/p pp (). d d K(x, y, z, )g(z) dz B ) ( ) y z x K y. In view of K(X) K(Y ) = (X Y ) 1 K(θX + (1 θ)y ) dθ and he fac ha K((x z)/ x) implies (x z)/ x + B 2B, we have { } { ( ) ( )} K(x, x z y z y, z, ) C min, 1 χ + χ 2 2

10 1182 Y. Miyazaki wih C = max{2 K L (B), K L (B)}. Thus we ge wih g (z) = (1 z ) 1/p g(z) and G ɛ (x) G ɛ (y) H(x, y, z)g (z) C 1 1/p dz (4.4) B 1 z ( ) 1 1/p 1 z H(x, y, z) = 1 { } { ( ) ( )} x z y z min, 1 d χ + χ d. 2 2 Le us suppose for he momen ha he inegrals H(x, y, z) M := dz, M 1 := 1 z B H(x, y, z) d 1 ds x ds y are esimaed by a consan independen of x, y and z. Applying Lemma 3.1 o he inegral operaor in (4.4), we find ha he map g (z) G ɛ (x) G ɛ (y) / 1 1/p is a bounded operaor from L p (B, dz/(1 z )) o L p (, ds x ds y / d 1 ), and ge G ɛ C g 1 1/p B pp () Lp(B,dz/(1 z )) = C g Lp(B). Since G ɛ (x) G(x) a.e. x by Sep 1, Faou s lemma yields G 1 1/p B lim inf pp () ɛ G ɛ C g 1 1/p B pp () L p(b). Therefore we obain G B 1 1/p pp () wih he second inequaliy in (4.2). Thus i remains o evaluae M and M 1. Using polar coordinaes z = rω wih r < 1 and ω = 1, and Lemma 3.3, we have M = 1 C 1 r d 1 dr 1 r dr 1 r 1 H(x, y, rω) ds ω ( ) 1 1/p { } 1 r min, 1 1 χ ( ) 1 r d. 2 Changing he variables (1 r)/2 = s, and replacing by, we have M C C 1 s 1/p χ(s) ds 1 1 1/p min{ 1, 1} 1 d <. ( ) 1 1/p { } min, 1 1 d

11 Sobolev race heorem in a ball 1183 Paying aenion o he symmery of he inegral defining M 1 wih respec o x and y, we find ha M 1 is equal o ds x ds 1 ( ) 1 1/p { } ( ) y 1 z x z 2 min, 1 d χ d. d 1 2 Using Lemmas 3.4 and 3.3, and changing he variables (1 z )/2 = s, we have 1 ( ) 1 1/p ( ) 1 z x z M 1 C d d χ ds x 2 1 ( ) 1 1/p ( ) 1 z 1 z C 1 χ d 2 C 1 s 1/p χ(s) ds <. This complees he proof of he proposiion. We are now ready o prove Theorem 1.1. Le f Wp 1 (B) and invoke (2.3). We noe ha he consan ϕ f(), he las erm in (2.3), belongs o () and ha B 1 1/p pp ϕ f() Lp() c d ϕ Lq(B) f Lp(B), ϕ f() B 1 1/p pp () = (4.5) wih p 1 + q 1 = 1. Combining his fac wih Proposiion 4.1, we see ha Tr M f (ɛ, ) converges in L p (B) o a funcion belonging o Bpp 1 1/p (). So we define Tr f o be he limi funcion: Tr f := lim ɛ Tr M f (ɛ, ) in L p (B). This is well defined, since if f Wp 1 (B) C( B) hen Tr M f (ɛ, ) converges o Tr f in C( B) by Proposiion 2.1(i), and since Wp 1 (B) C( B) is dense in Wp 1 (B) by Proposiion 2.1(iii). By definiion Tr f Bpp 1 1/p (). We also ge (1.3) by Proposiion 4.1 and (4.5). Thus we have proved he asserions in Theorem 1.1 excep he surjeciviy of he race operaor, which follows from Theorem The Dirichle problem The aim of his secion is o prove Theorem 1.2. inegral formula wih kernel We rely on he Poisson P (X, y) = 1 X 2 c d X y d for X < 1, y = 1, (5.1) where c d is he surface area of he uni sphere given by (3.1).

12 1184 Y. Miyazaki Proposiion 5.1. If a funcion u belongs o C 2 (B) C 1 ( B) and saisfies u = in B, hen u(x) = P (X, y) Tr u(y) ds y. (5.2) Proof. This proposiion is well known (see [1, Chaper 2], [4, Chaper 2], [5, Chaper 1], [6, Chaper 2]). For he sake of convenience we skech he proof. For X B and y R d we consider he fundamenal soluion { (2π) 1 log y X (d = 2), G (y, X) = (2 d) 1 c 1 d y X 2 d (d 3) for he Laplacian in R d, and define he Green funcion G(y, X) by G(y, X) = G (y, X) y 2 d G ( y 2 y, X). Taking ino accoun ha y G(y, X) = for y X and G(y, X) = for y = 1, we apply Green s formula o u(y) and G ( y, X) in he region B \B ɛ (X), where B ɛ (X) is he ball of radius ɛ cenered a X, and ake he limi as ɛ. Then we ge (5.2) wih P (X, y) = νy G(y, X), (5.3) where νy denoes he ouer normal derivaive wih respec o y. A calculaion shows ha P (X, y) is given by (5.1). We prove Theorem 1.2 in five seps. Sep 1. We show ha he Poisson operaor P defined by Pg(X) = P (X, y) g(y) ds y is a bounded operaor from L p () o L p (B). Le u = Pg and wrie X = rx wih r < 1 and x = 1. By Lemma 3.5 wih a = and Young s inequaliy Then u(r ) Lp() C(1 r 2 )(1 r) 1 g Lp() 2C g Lp(). 1 u p L = p(b) u(r ) p L p() rd 1 dr d 1 (2C) p g p L, p() which is he desired resul. Sep 2. We show ha in Proposiion 5.1 we may replace he assumpion u C 2 (B) C 1 ( B) by u W 1 p (B), ha is, (5.2) holds for u W 1 p (B) saisfying u = in B in he disribuional sense.

13 Sobolev race heorem in a ball 1185 We invoke (2.2) wih f replaced by u. By he equaliy xi {K((x z)/ x)} = ( 1) zi {K((x z)/ x)} and he definiion of disribuional derivaive we have n 1 ( )} x z x M u (ɛ, x) = d d 1 ( 1) 2 z {( j K j ) x u(z) dz ɛ B j=1 =. Since M u (ɛ, ) C (R d ), we can apply Proposiion 5.1 o M u (ɛ, ), and ge M u (ɛ, X) = P (X, y) Tr M u (ɛ, y) ds y. Le us ake he limi as ɛ in he above equaion. By Proposiion 2.1 M u (ɛ, X), he lef-hand side, ends o u in Wp 1 (B). On he oher hand, by he boundedness of he race operaor and he asserion in Sep 1, we see ha he righ-hand side ends o P (X, y) Tr u ds y in L p (B). Hence we obain (5.2) for u Wp 1 (B) saisfying u =. Sep 3. From he formula for u Wp 1 (B) in Sep 2 we conclude ha for a given g Bpp 1 1/p () he soluion u o he Dirichle boundary problem (1.4) mus be wrien as u(x) = P (X, y)g(y) ds y, (5.4) if he soluion exiss. This implies he uniqueness of he soluion. In he res of he proof of Theorem 1.2 we define u(x) by (5.4) and show ha u(x) is he soluion o (1.4) ha saisfies he properies saed in Theorem 1.2. Since X α P (X, y) is wrien in he form of a bounded funcion muliplied by X y d α, we can inerchange he order of differeniaion and inegraion o ge α u(x) = α X P (X, y)g(y) ds y. Hence u C (B). By (5.3) and X G(y, X) = we find ha X P (X, y) = and hence u =. Sep 4. We show ha u(x) defined by (5.4) belongs o Wp 1 (B). We have already obained u L p (B) wih u Lp(B) C g Lp() in Sep 1. In order o show j u L p (B) for j = 1,..., d we need some properies of he Poisson kernel. The subsiuion u 1 in (5.2) gives P (X, y) ds y = 1. (5.5) Se Q j (X, y) = Xj P (X, y). Differeniaing (5.5), we have Q j (X, y) ds y =. Therefore j u(x) = Q j (X, y){g(y) g(x)} ds y.

14 1186 Y. Miyazaki From c d Q j (X, y) = and X y 1 X i follows ha 2X j X y d(1 X 2 ) d X y Xj y j d+1 X y c d Q j (X, y) (2d + 2) X y d. (5.6) We wrie X = rx wih r < 1 and x = 1, apply Hölder s inequaliy o he produc 1 1/p g(x) g(y) Q j (X, y), 1 1/p and use (5.6) and Lemma 3.5 wih a = 1 1/p. Then we have j u(x) p ( C C(1 r) 1+1/p ) 1 1/p p 1 ds X y d y 1 1/p rx y d ( ) 1 1/p p g(x) g(y) ds X y d 1 1/p y ( ) p g(x) g(y) ds y. 1 1/p This combined wih Lemma 3.6 gives 1 j u(x) p dx = ds x j u(rx) p r d 1 dr B ( ) p g(x) g(y) ds x ds y C 1 1/p, d 1 which implies j u Lp(B) C g. Thus we conclude u W 1 1 1/p B pp () p (B). Sep 5. I remains o show ha u defined by (5.4) saisfies he boundary condiion, ha is, Tr u = g. Since u(rx) u(x) in Wp 1 (B) as r 1 (see [9, Lemma 6] for he proof), we know Tr u(r ) Tr u in Bpp 1 1/p (). So i suffices o show ha By (5.5) we have F (x) := u(rx) g(x) = Tr u(r ) g in L p () as r 1. (5.7) P (rx, y){g(y) g(x)} ds y for x = 1. Hölder s inequaliy gives ( ) p 1 F (x) p P (rx, y) 1 1/p ds y ( ) p g(x) g(y) P (rx, y) d 1/p ds y 1 1/p. d 1

15 Sobolev race heorem in a ball 1187 I follows from Lemma 3.5 wih a = 1 1/p and he inequaliy P (rx, y) 2c 1 d (1 r)/ rx y d ha he firs inegral is bounded by (1 r) 1 1/p. To evaluae he second inegral we may assume 1/2 r < 1, since we ake he limi as r 1. By (3.3) we have rx y max{(1 r), 2 1 }, and hence rx y d 2 d+1/p (1 r) 1/p d 1/p. Therefore P (rx, y) d 1/p 2c 1 d (1 r)d 1/p rx y d 2 d+1 1/p c 1 d (1 r)1 1/p. Thus we ge F (x) p ds x C(1 r) p 1 g p, Bpp 1 1/p which implies (5.7). This complees he proof of Theorem 1.2. Acknowledgemens. This research was suppored by JSPS Gran-in-Aid for Scienific Research (C) References [1] L.C. Evans, Parial Differenial Equaions, Second ediion, AMS, Providence, RI, 21. hp://dx.doi.org/1.19/gsm/19 [2] G.B. Folland, Real Analysis. Modern Techniques and Their Applicaions, Second ediion, Wiley, New York, [3] J. Franke, T. Runs, Regular ellipic boundary value problems in Besov- Triebel-Lizorkin spaces, Mah. Nachr., 174 (1995), hp://dx.doi.org/1.12/mana [4] D. Gilbarg, N.S. Trudinger, Ellipic Parial Differenial Equaions of Second Order, Springer, Berlin, 21. hp://dx.doi.org/1.17/ [5] D.D. Haroske, H. Triebel, Disribuions, Sobolev Spaces, Ellipic Equaion, European Mah. Soc., Zürich, 28. hp://dx.doi.org/1.4171/42 [6] L.L. Helms, Poenial Theory, Second ediion, Springer, London, 214. hp://dx.doi.org/1.17/ [7] V. Maz ya, M. Mirea, T. Shaposhnikova, The Dirichel problem in Lipschiz domains for higher order ellipic sysems wih rough coefficiens, J. Anal. Mah., 11 (21), hp://dx.doi.org/1.17/s

16 1188 Y. Miyazaki [8] Y. Miyazaki, New proofs of he race heorem of Sobolev spaces, Proc. Japan Acad. Ser. A, 84 (28), hp://dx.doi.org/1.3792/pjaa [9] Y. Miyazaki, Sobolev race heorem and he Dirichle problem in he uni disk, Milan J. Mah., 82 (214), hp://dx.doi.org/1.17/s x [1] T. Muramau, On Besov spaces and Sobolev spaces of generalized funcions defined on a general region, Publ. Res. Ins. Mah. Sci., 9 (1973), hp://dx.doi.org/1.2977/prims/ [11] M.E. Taylor, Parial Differenial Equaions I, Basic Theory, Second ediion, Springer, New York, 211. hp://dx.doi.org/1.17/ [12] H. Triebel, Inerpolaion Theory, Funcion Spaces, Differenial Operaors, Second ediion, Johann Ambrosius Barh, Heidelberg, Received: Augus 2, 216; Published: Ocober 16, 216