Mizuta, Y., Shimomura, T. and Sobukawa, T. Osaka J. Math. 46 (2009), 255 27 SOOLEV S INEQUALITY FOR RIESZ POTENTIALS OF FUNCTIONS IN NON-DOULING MORREY SPACES YOSHIHIRO MIZUTA, TETSU SHIMOMURA and TAKUYA SOUKAWA (Received July 6, 2007, revised December 20, 2007) Abstract Our aim in this paper is to give Sobolev s inequality and Trudinger exponential integrability for Riesz potentials of functions in non-doubling Morrey spaces.. Introduction The space introduced by Morrey [3] in 938 has become a useful tool of the study for the existence and regularity of solutions of partial differential equations. In the present paper, we aim to establish Sobolev s inequality for the Riesz potentials of functions in generalized Morrey spaces in the non-doubling setting, as extensions of Gogatishvili-Koskela [4], Orobitg-Pérez [4] and Sawano-Sobukawa-Tanaka [9]. Let X be a separable metric space with a nonnegative Radon measure. For simplicity, write x y for the distance of x and y. We assume that (x) = 0 and 0 ((x, r)) ½ for x ¾ X and r 0, where (x, r) denotes the open ball centered at x of radius r 0. In this paper, may or may not be doubling. Let G be an open set in X. We define the Riesz potential of order «for a nonnegative measurable function f on G by U «f (x) = G x y «f (y) ((x, 4x y)) d(y). Here we introduce the family L p,;k (G) of all measurable functions f on G such that f p p,,g;k = sup x¾g,0rd G r ((x, kr)) G f (y) p d(y) ½, where p ½, 0, k and d G denotes the diameter of G. In case X = R n with a nonnegative Radon measure, we know that L p,;k (G) = L p,;k 2 (G) 2000 Mathematics Subject Classification. 35, 46E35.
256 Y. MIZUTA, T. SHIMOMURA AND T. SOUKAWA when k 2 k, but we have an example (see Remark 2.) in which L p,; (G) = L p,;2 (G); see also Sawano-Tanaka [20]. The space L p,;2 (G) is referred to as a generalized Morrey space. To obtain Sobolev type inequalities for Riesz potentials of functions belonging to generalized Morrey spaces, we consider a generalized maximal function defined by M k f (x) = sup r 0 ((x, kr)) G f (y) d(y) for k and a locally integrable function f on G. In view of Sawano [8, Corollary 2.]), M 2 is bounded in L p (X). Further it is useful to remark that in a certain metric measure space X, M k fails to be bounded in L p (X) if and only if k 2 (see Sawano [8, Proposition.]). y applying the fact that M 2 is a bounded mapping from L p,;2 (X) to L p,;4 (X) (see Sawano-Tanaka [20, Theorem 2.3]), we first show that U «f ¾ L p,;4 (X) for f ¾ L p,;2 (X), where p = p «0; in the borderline case = «p, we consider the exponential integrability. For this, we also refer the reader to Sawano-Sobukawa- Tanaka [9, Theorem 3.]. Finally, in our Morrey space setting, we establish an exponential integrability for functions satisfying a Poincaré inequality, as an extension of Gogatishvili-Koskela [4] and Orobitg-Pérez [4]. For related results, see Adams [], Chiarenza-Frasca [3, Theorem 2], Nakai [5, Theorem 2.2] and the authors [, 2] in the doubling case. 2. Sobolev s inequality Throughout this paper, let C denote various constants independent of the variables in question. For a nonnegative measurable function f on G and k, define the maximal function M k f (x) = sup r 0 ((x, kr)) = sup 0r d G ((x, kr)) G G f (y) d(y) for x ¾ G, where d G denotes the diameter of G. Recall that f p p,,g;k = sup x¾g,0rd G r ((x, kr)) G f (y) d(y) f (y) p d(y).
SOOLEV S INEQUALITY FOR RIESZ POTENTIALS 257 When X = R n with a nonnegative Radon measure, we see that if G is an open set of R n and k 2, then f p,,g;k C f p,,g;2 for all f ¾ L p,;2 (G), where C is a constant depending only on k and n; for this fact, see e.g. [20, Proposition.]. REMARK 2.. L p,;k (G) = We set f sup x¾g,0rd G r f (y) p d(y) ½. ((x, kr)) G When G R n, L p,;k (G) = L p,;2 (G) for all k. We show by an example that L p,; (G) = L p,;k (G) when k. For this, consider a measure given by d(y) = e y dy on R. For 0, letting f (y) = y p for y 0 and f (y) = 0 for y 0, we note the following: (i) if 0 x r, then r ((x, 2r)) (ii) if x r 0, then r ((x, 2r)) (iii) if x 0 and r 0, then f (y) p d(y) f (y) p d(y) r ((x, r)) r 2r e x (e 2r ) ex+r C r + e r ; r x+r y e y dy e x (e 2r ) x r r e x (e 2r ) C r + e r ; 0 y dy (x + r) (x r) f (y) p d(y) r (x + r). e x+r
258 Y. MIZUTA, T. SHIMOMURA AND T. SOUKAWA If 0 and, then (i) and (ii) imply that f ¾ L p,;2 (R ), and if 0 and, then (iii) implies that lim sup r½ r ((x, r)) for every fixed x 0, so that f ¾ L p,; (R ). f (y) p d(y) = ½ In what follows, if f is a function on G, then we assume that f = 0 outside G. First we present the boundedness of maximal functions in the Morrey space L p,;2 (G) due to Sawano-Tanaka [20, Theorem 2.3]. Lemma 2.2. If 0, then M 2 f p,,g;4 C f p,,g;2 for all f ¾ L p,;2 (G). Proof. Let f p,,g;2, and fix x ¾ G and 0 r d G. Write A 0 = (x, 2r) and A j = (x, 2 j+ r) Ò (x, 2 j r) for each positive integer j. We set f j = f A j, where E denotes the characteristic function of E. Note that M 2 f (z) p d 2 p 2 p (I + I 2 ), M 2 f 0 (z) p d + M 2 g 0 (z) p d where g 0 = È ½ j= f j. We have by Sawano [8, Theorem.2 and Proposition.] I = C M 2 f 0 (z) p d C (x,2r) Next we see that for z ¾ (x, r) M 2 f j (z) C sup t:(2 j )r t(2 j+ +)r f 0 (z) p d f (z) p d Cr ((x, 4r)). f (y) d ((z, 2t)) (z,t)
SOOLEV S INEQUALITY FOR RIESZ POTENTIALS 259 p sup f (y) p d t:(2 j )r t(2 j+ +)r ((z, 2t)) (z,t) C C(2 j r) p, so that M 2 g 0 (z) ½ j= M 2 f j (z) C ½ j= (2 j r) p Cr p. Hence it follows that Thus we obtain I 2 Cr d Cr ((x, r)). r M 2 f (z) p d C, ((x, 4r)) which proves the lemma. Lemma 2.3., then If f is a nonnegative measurable function on G such that f p,,g;2 (x,æ) x y «f (y) ((x, 4x y)) d(y) CÆ«M 2 f (x) for x ¾ G and Æ 0. Proof. (x,æ) as required. We have x y «f (y) ((x, 4x y)) d(y) = ½ j= ½ j= (x,2 j+ Æ)Ò(x,2 j Æ) (2 j+ Æ) «((x, 2 j+2 Æ)) Æ «M 2 f (x) ½ j= = CÆ «M 2 f (x), 2 ( j+)«x y «f (y) ((x, 4x y)) d(y) (x,2 j+ Æ) f (y) d(y)
260 Y. MIZUTA, T. SHIMOMURA AND T. SOUKAWA Lemma 2.4. Let p «. Let f be a nonnegative measurable function on G such that f p,,g;2. In case p «, GÒ(x,Æ) and in case p = «and G is bounded, GÒ(x,Æ) for x ¾ G and small Æ 0. x y «f (y) p d(y) CÆ«((x, 4x y)) x y «f (y) ((x, 4x y)) d(y) C log Æ Proof. Let j 0 be the smallest integer such that 2 j 0 Æ d G, where d G is the diameter of G as before. y using Hölder s inequality, we have = GÒ(x,Æ) j 0 x y «f (y) ((x, 4x y)) d(y) j=0 (x,2 j+ Æ)Ò(x,2 j Æ) j 0 j=0 CÆ «CÆ «(2 j+ Æ) «((x, 2 j+2 Æ)) j 0 2 «j j=0 j 0 j=0 x y «f (y) ((x, 4x y)) d(y) ((x, 2 j+2 Æ)) 2 «j (2 j+ Æ) p j 0 = CÆ «p 2 («p) j, j=0 which proves the required inequality. (x,2 j+ Æ) (x,2 j+ Æ) f (y) d(y) p f (y) p d(y) With the aid of Lemmas 2.2, 2.3 and 2.4, we can apply Hedberg s trick (see [6]) to obtain a Sobolev type inequality for Riesz potentials due to Adams [, Theorem 3.], Chiarenza and Frasca [3, Theorem 2], Nakai [5, Theorem 2.2] and Sawano-Tanaka [20, Theorem 3.3].
SOOLEV S INEQUALITY FOR RIESZ POTENTIALS 26 Theorem 2.5. c such that Let p = p «0. Then there exists a positive constant r U «f (x) p d(x) c ((z, 4r)) (z,r) for all z ¾ X and r 0, whenever f is a nonnegative measurable function on X satisfying f p,, X;2. Proof. U «f (x) = We see from Lemmas 2.3 and 2.4 that (x,æ) x y «f (y) ((x, 4x y)) XÒ(x,Æ) d(y) + x y «f (y) ((x, 4x y)) d(y) CÆ «M 2 f (x) + CÆ «p for all Æ 0. Here, letting we have Æ = M 2 f (x) p, U «f (x) CM 2 f (x) «p = CM 2 f (x) pp, which yields U «f (x) p d(x) C M 2 f (x) p d(x) (z,r) (z,r) for z ¾ X and r 0. Hence Lemma 2.2 gives r ((z, 4r)) for such z and r, as required. U «f (x) p d(x) C (z,r) REMARK 2.6. Theorem 2.5 implies that the mapping f U «f is bounded from L p,;2 (X) to L p,;4 (X). REMARK 2.7. When X = R n, consider the potential U «,k f (x) = x y «f (y) ((x, kx y)) d(y) for k. Then we can show that the mapping f U «,k f is bounded from L p,;2 (R n ) to L p,;2 (R n ), when p = p «0.
262 Y. MIZUTA, T. SHIMOMURA AND T. SOUKAWA REMARK 2.8. We show by an example that the mapping f U «, f fails to be bounded in L p,;2 (R n ). For this purpose, consider d(y) = e y dy and y p when y 0; f (y) = 0 when y 0. In view of Remark 2., we see that f ¾ L p,;2 (R ) when 0. Further we see that U «, f (x) ½ x t «(x + t) p e x (e t e t ) ex+t dt = ½ for all x 0 when «p + 0. L p,;2 (R ) when 0. This implies that U «, f does not belong to 3. Exponential integrability Our aim in this section is to discuss the exponential integrability. Theorem 3.. Let G be bounded and = «p. Then there exists a positive constant c such that r exp(cu «f (x)) d(x) ((z, 4r)) G(z,r) for all z ¾ G and 0 r d G, whenever f is a nonnegative measurable function on G satisfying f p,,g;2. Proof. U «f (x) = Let f p,,g;2. We see from Lemmas 2.3 and 2.4 that G(x,Æ) x y «f (y) ((x, 4x y)) GÒ(x,Æ) d(y) + x y «f (y) ((x, 4x y)) d(y) CÆ «M 2 f (x) + C log Æ for x ¾ G and small Æ 0. Here, letting Æ = M 2 f (x) «log(m 2 f (x)) «when M 2 f (x) is large enough, we have exp(u «f (x)) C + CM 2 f (x) p,
SOOLEV S INEQUALITY FOR RIESZ POTENTIALS 263 so that Lemma 2.2 yields G(z,r) exp(u «f (x)) d(x) C(G (z, r)) + C C(G (z, r)) + C for z ¾ G and r 0. Hence we find c 0 such that r ((z, 4r)) G(z,r) G(z,r) ((z, 4r)) r exp(cu «f (x)) d(x) M 2 f (x) p d(x) for z ¾ G and 0 r d G, whenever f p,,g;2. Thus the required result is obtained. REMARK 3.2. In Theorem 3., we can not add an exponent q such that r ((z, 4r)) G(z,r) For this, consider the potential U(x) = exp(cu «f (x) q ) d(x). x y «n y «dy, where = (0, ) R n. If = «p n and f (y) = y «(y), then r n f (y) p dy r n x y «p dy Cr n r n «p = C for all x ¾ and r 0, so that f ¾ L p,; (). On the other hand, we see that U(x) Ò(x,x2) 3 «C log 2 x Ò(x,x2) x y «n f (y) dy x y n dy for x ¾, and hence for c 0 and q. exp(cu(x) q ) dx = ½
264 Y. MIZUTA, T. SHIMOMURA AND T. SOUKAWA Consider the function e N (t) = e t t t 2 2! t N (N )!. Theorem 3.3. Let = «p. For, take a positive integer N such that N p «p = p. Then there exists a positive constant c such that r ((z, 4r)) (z,r) e N (cu «f ) d(x) for all z ¾ X and r 0, whenever f is a nonnegative measurable function on X satisfying f p,, X;2 + f p,, X;2. Proof. U «f (x) = We see from Lemmas 2.3 and 2.4 that (x,æ) x y «f (y) ((x, 4x y)) XÒ(x,Æ) d(y) + x y «f (y) ((x, 4x y)) d(y) CÆ «M 2 f (x) + C log Æ for small Æ 0. Here, letting when M 2 f (x) is large enough, we have Æ = M 2 f (x) «log(m 2 f (x)) «U «f (x) C log(2 + M 2 f (x)). We write G = x ¾ X : M 2 f (x) 2 and G 2 = x ¾ X : M 2 f (x) 2. Then we find c 0 such that e N (c U «f (x)) d(x) M 2 f (x) p d(x) G (z,r) (z,r) and G 2 (z,r) e N (c U «f (x)) d(x) G 2 (z,r) U «f (x) p d(x) M 2 f (x) p d(x) (z,r)
SOOLEV S INEQUALITY FOR RIESZ POTENTIALS 265 for z ¾ X and r 0. Hence Lemma 2.2 gives r ((z, 4r)) (z,r) e N (c 2 U «f (x)) d(x) for such z and r, whenever f p,, X;2 + f p,, X;2. This gives the the required result. REMARK 3.4. Let «p and f be a nonnegative measurable function on X belonging to L p,;2 (X). Then U «f (x) is seen to be continuous at x 0 ¾ X where ( (x 0, r)) = 0 for r 0 and (3.) In fact, for Æ 0, we write U «f (x) = (x 0,3Æ) = U (x) + U 2 (x). The proof of Lemma 2.3 implies that x 0 y «f (y) d(y) ½. ((x 0, 2x 0 y)) x y «f (y) ((x, 4x y)) XÒ(x d(y) + x y «f (y) 0,3Æ) ((x, 4x y)) d(y) U (x) CÆ «p for x ¾ (x 0, Æ), when «p 0. Note that ((x, 4x y)) ((x 0, 4x 0 y)) as x x 0 for fixed y ¾ X Ò (x 0, 3Æ) by the assumption that ( (x 0, r)) = 0 for r 0. Since x y «((x, 4x y)) Cx 0 y «((x 0, 2x 0 y)) for x ¾ (x 0, Æ) and y ¾ X Ò (x 0, 3Æ), we have by Lebesgue s dominated convergence theorem lim xx 0 U 2 (x) = U 2 (x 0 ), which shows that U «f (x) is continuous at x 0. 4. Poincaré inequality Let be a nonnegative measure on an open set G. For a measurable function u on G, we define the integral mean over a measurable set E G of positive measure by u E = E u(x) d = (E) E u(x) d(x). In this section, we assume that satisfies the lower Ahlfors s-regularity condition (4.) c r s () ½
266 Y. MIZUTA, T. SHIMOMURA AND T. SOUKAWA for all balls = (x, r) G, where s 0 and c is a positive constant. We say that a couple (u, g) satisfies a strong (, p 0 ) Poincaré inequality (in G) if (4.2) u(x) u d(x) c P (diam ) +s (2) g(y) p 0 d(y) p0 for each ball = (x, r) with 2 = (x, 2r) G, where p 0 p and c P positive constant. Set p = p 0. is a Theorem 4.. Let be a nonnegative measure on G satisfying (4.), and assume that a couple (u, g) satisfies a strong (, p 0 ) Poincaré inequality (4.2). If g p,,g;2, then r u(x) u p d(x) C (2) for every ball = (x, r) with 2 G. Orobitg-Pérez [4] gave a version of Sobolev s inequalities in the L p space setting. Proof of Theorem 4.. Let 2 = (x 0, 2r) G. For x ¾, set i (x) = (x, 2 i r). y the Lebesgue differentiation theorem we have lim u i (x) = u(x) i½ for -a.e. x and hence we may assume that our fixed point x has this property. Let N = N(x) be a positive integer whose value will be determined later. Letting i = i (x) for i =, 2,, we have by (4.) u(x) u u u + ( ) + ½ i= ½ i= ( i+ ) u i u i+ u u d + i u u i d u u d ( )
SOOLEV S INEQUALITY FOR RIESZ POTENTIALS 267 Cr s + C ½ i=n+ = I 0 + I + I 2. u u d + C N i= (2 i r) s i u u i d (2 i r) s i u u i d y using Hölder s inequality and a strong (, p 0 ) Poincaré inequality, we find p0 I 0 C(diam ) g p 0 d (2) p C(diam ) g p d (2) Cr p and I C C C N i=0 N i=0 N i=0 p0 2 i r g p 0 d (2 i ) i 2 i r (2 i ) (2 i r) p C(2 N r) p. i g p d p According to the estimation of I, we obtain ½ I 2 C 2 i r (2 i ) C i=n+ ½ i=n+ 2 i rm 2 g 0 (x) p 0 C2 N rm 2 g 0 (x) p 0, i g p 0 d p0 where g 0 (y) = g(y) p 0 (y) with denoting the characteristic function of. Now, considering N to be the integer part of (2 N r) p = M 2 g 0 (x) p 0, we establish (4.3) u(x) u C[r p + M 2 g 0 (x) p(p p 0 ) ].
268 Y. MIZUTA, T. SHIMOMURA AND T. SOUKAWA Therefore, it follows from Lemma 2.2 that as required. u(x) u p d(x) C [r + M 2 g 0 (x) pp 0 ]d(x) C r () + = C r () + Cr (2), g 0 (y) pp 0 d(y) g(y) p d(y) Corollary 4.2. Let be a nonnegative measure on G satisfying (4.), and assume that a couple (u, g) satisfies a strong (, p 0 ) Poincaré inequality. Then u(x) u p d(x) p Cr sp () p for every = (x, r) with 2 G, where p = p s 0. g(y) p d(y) To show this, first suppose Ê g(y)p d(y). Then the decay condition (4.) implies that g p,s,g; is bounded. Now we see from the inequalities after (4.3) that Hence we obtain u(x) u p d(x) u(x) u p d(x) Cr s (). p for a general g, which gives the required result. C(r s ()) p p p g(y) p d(y) REMARK 4.3. Let G be an open set in R n. We assume that a couple (u, g) satisfies a (, p 0 ) Poincaré inequality in G, that is, p0 (4.4) u(x) u d(x) c ¼ P ()s g(y) p 0 d(y) for all balls G, where p 0 p and c ¼ P (u, g). We further assume that ( ) = 0 and () s g(y) p d(y) is a positive constant independent of
SOOLEV S INEQUALITY FOR RIESZ POTENTIALS 269 for each ball G, where p. Then sup G () s u(x) u p d(x) C. For this, we also refer to Hajłasz-Koskela [5]. For a proof of this fact, let x ¾ G be a Lebesgue point of u. As in the proof of Theorem. by Gogatishvili-Koskela [4], we take a sequence of balls j such that x ¾ j+ j and ( j ) = 2 j (). Then, as in (4.3), we can prove u(x) u p C[() s + M g 0 (x) pp 0 ], which gives the required inequality by the boundedness of the maximal operator M. Finally we discuss the exponential integrability in the same manner as in Theorem 4.. Theorem 4.4. Let G be bounded and = p. Let be a nonnegative measure on G satisfying (4.), and assume that a couple (u, g) satisfies a strong (, p 0 ) Poincaré inequality. Then there exists a positive constant c such that r (2) exp(cu(x) u ) d(x) for every ball = (z, r), whenever 2 G and g p,,g;2. REMARK 4.5. Let be a nonnegative measure on R n satisfying (4.), and assume that a couple (u, g) satisfies a strong (, p 0 ) Poincaré inequality. If g ¾ L p,;2 (R n ) and p, then u can be corrected almost everywhere to be continuous on R n ; for this, see [0]. In fact, the first part of the proof of Theorem 4. implies that for almost every x ¾, which proves for almost every x, y ¾. (p )p u(x) u Cr (p )p u(x) u(y) Cx y ACKNOWLEDGMENT. We would like to thank Dr. Yoshihiro Sawano and the referee for their kind comments and suggestions.
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SOOLEV S INEQUALITY FOR RIESZ POTENTIALS 27 Yoshihiro Mizuta Department of Mathematics Graduate School of Science Hiroshima University Higashi-Hiroshima 739 852 Japan e-mail: yomizuta@hiroshima-u.ac.jp Tetsu Shimomura Department of Mathematics Graduate School of Education Hiroshima University Higashi-Hiroshima 739 8524 Japan e-mail: tshimo@hiroshima-u.ac.jp Takuya Sobukawa Department of Mathematics Education Faculty of Education Okayama University Tsushima-naka 700 8530 Japan e-mail: sobu@cc.okayama-u.ac.jp