An Introduction to the Morrey and Campanato Spaces

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1 An Introduction to the orrey and amanato Saces Abstract Regularity theorems which allow one to conclude higher regularity of a function from a lower regularity and a differential equation lay a central role in the theory of PDEs. One version of the Sobolev embedding theorem which states that W,k (R n ) r,β (R n ) for k r a = n/ is such an examle. The amanato saces, an enhanced version of the orrey saces, extend the notion of functions of bounded mean oscillation and allow a full characterization 0,β (R n ). The theory of amanato saces may come in useful when the Sobolev embedding theorem is not. The main results of this roject are summarized in Theorems 0.0.2, and Notation Given a set Ω R n and an oen ball B (x), we denote Ω(x, ) := Ω B (x). 2. We will write the Lebesgue measure of a set E by E. 3. V n denotes the volume of the unit ball B n (0) in R n : V n := B n (0). 4. Given a function f : Ω R, we denote the average of f over a set Ω by (f) Ω := f Ω Ω 5. d denotes the diameter of a bounded domain Ω R n : d := diamω. Definition (Hölder continuous functions) Let Ω be a domain in R n and k 0, β > 0. We denote by k,β (Ω) the subset of k (Ω) of functions f satisfying for every multiindex i k. D i f(x) D i f(y) H i,β (f) := su < x,y Ω x y β x y The functions in the sace 0,β (Ω) are called Hölder continuous. Definition (orrey and amanato saces) Let Ω R n be a bounded domain (oen and connected), and λ 0.

2 2. The orrey saces, denoted L,λ (Ω), are the collection of all functions f L (Ω) such that f,λ := su λ x Ω 0<<d f / < 2. The amanato saces, denoted L,λ (Ω), are the collection of all functions f L (Ω) such that [f],λ := su λ x Ω 0<<d f (f) / < Proosition ,λ defines a norm on the orrey saces, making them into a normed vector sace. 2. [ ],λ defines a seminorm on the amanato saces. These can be made into normed vector saces by setting for every f L,λ (Ω), f,λ = f + [f],λ Remark [f],λ = 0 if and only if f = (f) a.e. for every (x, ) Ω (0, d), which imlies that f is a.e. constant on Ω. Theorem The orrey and amanato saces are Banach. We rove comleteness in the orrey sace, the other being very similar. Let {u n } be a auchy sequence in L,λ (Ω). Recall d := diamω. hoose x Ω, then su y Ω dist(x, y) < d because if we had su dist(x, y) = d, then y Ω dist(x, y) > diamω, because x is an interior oint we could find a small ball B δ (x) Ω and t B δ (x) such that su y Ω a contradiction. This means that there is < d such that B (x) Ω. This gives us the relation ( ( ) λ ) / d u u d λ u,λ. Ω(x, ) So in articular the sequence {u n } is auchy sequence in L (Ω), which is comlete, so there exists u L (Ω) such that lim n u j u = 0.

3 3 We show that u L,λ (Ω). By inkowski inequality, for all (x, ) Ω (0, d), λ ( ) / ( ) / ( ) / u λ u u n + λ u n. Since the sequence is auchy in L,λ (Ω), the 2nd term on the RHS is uniformly bounded by, say, K. Then taking the limit when n the st term on the RHS 0 and we get λ ( ) / u K. Taking the suremum over all (x, ) Ω (0, d) gives u,λ K. Remains to show that lim n u n u,λ = 0. Let ɛ > 0 be given. We have λ ( ) / ( ) / ( ) / u u n λ u u m + λ u m u n ( λ ) / u u m + u m u n,λ hoose n o so that n, m n o u m u n,λ < ɛ. Then taking the limit when m the st term on the RHS 0. Finally taking the suremum over all (x, ) Ω (0, d) gives u u n,λ ɛ. Definition A family {E r } r>0 of Borel subsets of R n is said to shrink nicely to x R n if : E r B r (x) for all r > 0. There is a constant α > 0, indeendent of r, such that E r > α B r (x) = αv n r n. Remark Given a bounded domain Ω R n, consider the family of sets {Ω(x, ) : x Ω, 0 < < d} = {Ω(x, ) : 0 < < d}. Suose that each family {Ω(x, ) : 0 < < d} shrinks nicely, then we have strictly x Ω ositive constants {α x } x Ω such that Ω(x, ) > α x B (x) for all x Ω, (0, d). If the α x can be chosen so that α := inf x Ω α x > 0 then we shall say that the domain Ω is tye-α, that is to say, there exists α > 0 such that Ω(x, ) > αv n n for all (x, ) Ω (0, d).

4 4 Definition For f L,λ (Ω), let f,n = su Ω(x, ) λ n x Ω 0<<d f / Lemma ,n also defines a norm on the orrey saces.,n is finer than,n and if Ω is tye-α, then the two norms are equivalent. Showing,n is a norm is straightforward. That,n is finer than,n comes from the fact that λ = V n λ n (V n n ) n λ = B (x) λ n αv n n Ω(x, ), i.e. (αv n ) λ n Ω(x, ) λ n λ.. If Ω is tye-α, then there exists α > 0 such that for all (x, ) Ω (0, d), n λ We recall a fundamental result that will be useful in the roof the subsequent theorem. Theorem (Lebesgue Differentiation Theorem) Suose f L loc (Rn ). Then for Lebesgue-almost all x R n E r E r f(y) f(x) dy = 0 for every family {E r } r>0 that shrinks nicely to x. lim r 0 Theorem For <, L,0 (Ω) = L (Ω), i.e. L,0 (Ω) and L (Ω) are continuously imbedded in each other. 2. For <, L,n (Ω) L (Ω). If further Ω is tye-α (see Remark 0.0.8), then L,n (Ω) L (Ω). 3. For <, λ > n, L,λ (Ω) = {0}. 4. For q < and λ, µ 0 so that λ n µ n q then L q,µ (Ω) L,λ (Ω).. Straight from the definition of the orrey norm we have f,0 = f. So the identity ma from L,0 (Ω) into L (Ω) and vice-versa is continuous. 2. Let f L (Ω). Then for all (x, ) Ω (0, d) we have f n Ω(x, ) n f B (x) n f V n f.

5 where V n is the volume of the unit ball in R n. From this we conclude that f,n Vn / f and the identity ma from L (Ω) into L,n (Ω) is continuous. This ma is also surjective when Ω is tye-α. To see this, suose that there exists f L,n (Ω) \ L (Ω). Then f L (Ω). In articular f L (Ω) and so by the Lebesgue differentiation theorem, lim r 0 f = f(x) for almost all x Ω. We also have f =. So given > 0 arbitrarily large, the set {x Ω : f(x) > / } has strictly ositive measure and almost all oints in that set are Lebesgue oints. Thus we can find (x, ) Ω (0, d) such that f >. We conclude f,n = and using the assumtion that Ω is tye-α, this is equivalent to f,n =, but this is a contradiction. Finally, since the identity ma from L (Ω) to L,0 (Ω) is a continuous linear bijection, we conclude by the bounded inverse theorem that the identity ma from L,n (Ω) to L (Ω) is continous. 3. For λ > n, say λ = n + ɛ, then by the Lebesgue differentiation theorem lim 0 n f V n f(x) for almost all x Ω. From this we see that unless f(x) = 0 (almost everywhere) lim 0 ɛ n f =. 4. By Hölder s inequality with conjugate exonents /q and /q, we have f ( ) /q ( f q ) /q (V n n ) /q ( f q ) /q Vn /q n( /q)+µ/q ( µ f q ) /q. Now λ n 5 µ n q λ n( /q) + µ/q ( d) n( /q)+µ/q ( d) λ n( /q)+µ/q λ d n( /q)+µ/q λ. Putting everything together we have ( λ f ) / ( µ f q ) /q and thus f,λ f q,µ, where = Vn /q d n( /q)+µ/q λ is a constant. It follows that the identity ma from L q,µ (Ω) into L,λ (Ω) is continuous. Remark Theorem.. suggests that for fixed [, ) the orrey saces {L,λ (Ω)} λ [0,n] rovide a certain scaling of the saces between L (Ω) and L (Ω). Also, taking = q in oint 4, we have L,µ (Ω) L,λ (Ω) whenever µ λ, just like for finite L saces. Some, but not all, roerties of the orrey saces also hold for the amanato saces: Theorem For <, L,0 (Ω) = L (Ω). 2. For q < and λ, µ 0 so that λ n µ n q then L q,µ (Ω) L,λ (Ω).

6 6 In order to state the next major results concerning the orrey and amanato saces we must first develo useful tools. In what follows we will be assuming throughout that Ω is tye-α. Lemma f L,λ (Ω) if and only if f L (Ω) and. f,λ := su (inf λ x Ω c R 0<<d ) f c / < learly f,λ [f],λ. This takes care of the only if art of the statement. Now suose that f L (Ω) and f,λ <. By convexity of t t for, we have = 2 ( 2 ( = 2 ( f (f) f c + c (f) ) f c + Ω(x, ) c f ) Ω(x, ) f c + Ω(x, ) (c f) ) 2 ( 2 ( 2 ( f c + f c + f c ) (c f) ) c f ) Where c R is arbitrary and in the 2nd to last ste we used the fact that since we are in a finite measure sace. Since c R is arbitrary we can take the infimum on the RHS and then take the suremum over all (x, ) Ω (0, d) on both side. So we have [f],λ 2 f,λ. orollary ,λ + defines yet another norm on L,λ (Ω) which is equivalent to,λ.

7 7 Lemma Then there exists a constant K = K(, α, n), where α is the constant concerning the regularity of Ω such that ( ) 0 < r < s < d (f) r λ (f) Ω(x,s) K + s λ / [f],λ r n for all f L,λ (Ω) and x Ω. ( (f) (f) Ω(x,s) 2 (f) f ) + f (f) Ω(x,s) (f) (f) Ω(x,s) 2 ( ( Ω(x, r) (f) (f) Ω(x,s) 2 (f) f + (f) f + Ω(x,s) ) f (f) Ω(x,s) ) f (f) Ω(x,s) αv n r n (f) (f) Ω(x,s) 2 ( r λ + s λ) [f],λ where we used the regularity condition in the last ste. Now regrou terms. Lemma There exists a constant K = K(, λ, α) such that (f) (f) Ω(x, 2 k ) K [f],λ (λ n) k m=0 2 m(n λ) whenever f L,λ (Ω) and (x, ) Ω (0, d) and k N. r = Let f L,λ (Ω) and (x, ) Ω (0, d) be given. By Lemma we have for every m N (taking 2 m+ and s = 2 m ): (f) Ω(x, 2 m+ ) (f) Ω(x, 2 m ) K ( ( ) λ + ( ) λ 2 m+ 2 m ( 2 m+ ) n ) / [f],λ = K λ n m(n λ) 2 2 n ( + 2 λ ) / [f],λ (f) Ω(x, 2 m+ ) (f) Ω(x, 2 m ) K λ n m(n λ) 2 [f],λ

8 8 Because K = K2 n ( + 2 λ ) / is indeendent of m, we can sum over m = 0,, 2,...k and use the triangle inequality on the LHS to get the result. Lemma Let λ > n. Then for all f L,λ (Ω) there exists a function F defined on Ω that equals f a.e. in Ω and such that F (x) = lim 0 (f) on Ω, the convergence being uniform. The existence of F is just Lebesgue s differentiation theorem : lim 0 (f) = f(x) a.e. in Ω. We have to show that the convergence is uniform. By Lemma for any n, q N we have (f) Ω(x, 2 n ) (f) Ω(x, 2 n+q ) ( ) (λ n) K [f],λ 2 n, where K is a constant indeendent of x and q. Here we see that the sequence {(f) Ω(x, 2 n )} n= is auchy uniformly with resect to x. So for each x Ω, let F (x) := lim n (f) Ω(x, 2 n ) By Lemma 0.0.8, we have Taking n we get (f) Ω(x,σ) (f) Ω(x, σ 2 k ) K [f],λ σ (λ n) k m=0 (f) Ω(x,σ) F (x) K [f],λ σ (λ n) 2 m(n λ) for some constant K. This says that (f) Ω(x,σ) F (x) uniformly as σ 0. Lemma Let 0 λ < n. Then there exists a constant K = K(α,, λ, n) > 0 such that for all f L,λ (Ω) and (x, ) Ω (0, d) we have : (f) (f) Ω + K [f],λ (λ n)

9 9 Let f L,λ (Ω) and (0, d) be given. hoose k N so that d 2 k+ < d 2 k. We have (f) (f) Ω + (f) Ω (f) Ω(x, d 2 k ) + (f) Ω(x, d 2 k ) (f) To show the result, we must bound aroriately the 2nd and 3rd terms on the RHS. Since in fact (f) Ω = (f) Ω(x,d), we can bound the 2nd term on the RHS using Lemma We have (f) Ω (f) Ω(x, d ) K 2 [f],λ d (λ n) 2 k k m=0 2 m(n λ) = K 2 [f],λ d (λ n) 2 k(n λ) 2 (n λ) K 2 [f],λ (λ n) (λ n) (k+) 2 k(n λ) 2 2 (n λ) K 2 [f],λ (λ n) 2 k(λ n) 2 k(n λ) 2 (n λ) = K 2 [f],λ (λ n) 2 (n λ) ( ) 2 k(n λ) Although the last term in brackets deends on k, it is bounded above by. We bound the 3rd term on the RHS using Lemma (f) Ω(x, d 2 k ) (f) K 3 ( λ + ( d 2 n ) λ n ) / [f],λ ( ) λ + (2) λ / K 3 [f] n,λ K 3 ( λ n (2 λ + ) / ) K 3 ) ( λ n [f],λ [f],λ

10 0 Now take K = K 3 + K 2 (n λ). 2 Lemma There exists a constant K = K(α, n, λ) such that for all f L,λ (Ω) and all x, y Ω, = 2 x y (f) (f) Ω(y,) K [f],λ x y λ n Let f L,λ (Ω) and x, y Ω be given. = 2 x y Ω(x, ) Ω(y, ) contains both Ω(x, ) and Ω(y, ). (f) (f) Ω(y,) (f) u + u (f) Ω(y,) (f) (f) Ω(y,) (f) u + u (f) Ω(y,). Ω(y,) Ω(y,) (f) (f) Ω(y,) Ω(x, ) Ω(y, ) Ω(y,) (f) u + u (f) Ω(y,) Ω(y,) (f) (f) Ω(y,) Ω(x, ) Ω(y, ) 2 λ [f],λ Since Ω(x, 2 ) Ω(x, ) Ω(y, ) and Ω(x, 2 ) αv n ( 2) n by regularity of Ω, we get (f) (f) Ω(y,) K λ n [f],λ for some constant K. Theorem Let <. For λ [0, n), L,λ (Ω) = L,λ (Ω) 2. For λ (n, n + ], L,λ (Ω) = 0,β (Ω), where β = λ n. 3. For λ > n + or β >, the saces L,λ (Ω) and 0,β (Ω) contain only constant functions. 4. For λ = n, L (Ω) L,n (Ω), but L (Ω) L,n (Ω).

11 . Let λ [0, n) and f L,λ (Ω). By orollary 0.0.6, ) f,λ 2 ( f + f,λ = 2 f + su (inf λ x Ω c R 0<<d ) / f c 2 ( f + f,λ ) K f,λ where K is some constant; the last inequality is derived in the roof of Theorem Hence f L,λ (Ω) and the identity oerator from L,λ onversely, suose f L,λ (Ω). We have : (Ω) into L,λ(Ω) is continuous. f 2 ( f (f) + (f) ) The st term on the RHS is bounded above by λ [f],λ and to bound the 2nd term on the RHS we have by Lemma (f) ) ( (f) Ω + K [f],λ (λ n) ( 2 (f) Ω + K [f],λ λ n) Therefore, f 2 ( λ [f],λ + Ω(x, ) 2 ( (f) Ω + K [f],λ λ n)) = 2 ( λ [f],λ + Ω(x, ) 2 (f) Ω + Ω(x, ) 2 K [f],λ λ n) 2 ( λ [f],λ ) Ω(x, ) + 2 f + V n n 2 K [f],λ Ω λ n K ( λ [f],λ + ) n f for some constant K. ultilying both sides by λ, taking the suremum and using the fact that n λ > 0 gives us f,λ K f,λ.

12 2 2. Let λ > n and α = (λ n), f 0,α (Ω).Then f (f) = Ω(x, r) Ω(x, r) (f(y) f(t)dt dy f(y) f(t) dt dy where we have used recisely the following basic relation for finite L saces: f f Ω(x, r) /. Ω(x, r) f(y) f(t) y t α dt y t α dy Ω(x, r) H0,αr α dt dy = Ω(x, r) H 0,αr α r λ f (f) V n r n H 0,αr α λ [f],λ V n H 0,α Also f f Ω / = f (Ω) Ω /. So taking K = max{v n, Ω / } (for examle) we get f,λ K f 0,β (Ω). For the converse, by art 4 of Theorem 0.0.2, L,λ (Ω) L,β+n(Ω) whenever β λ n, so it is enough to show that L,β+n (Ω) 0,β for β (0, ]. Let f L,β+n (Ω). By Lemma F (x) = lim(f) and F = f a.e. r 0 in (Ω) Let x, y Ω, r = 2 x y. We have

13 3 F (x) F (y) F (x) (f) + (f) (f) Ω(y,r) + (f) Ω(y,r) F (y) By the last relation in the roof of Lemma 0.0.9, we have (f) F (x) K [f],β+n r β and (f) Ω(y,r) F (y) K [f],β+n r β and by Lemma we get (f) (f) Ω(y,r) K 2 [f],β+n (r/2) β Putting everything together, we have F (x) F (y) x y β K [f],β+n H 0,β (f) K [f],β+n 3. The result is well known for 0,β (Ω). 4. Let f L (Ω) and (x, ) Ω (0, d). We have: f (f) 2 f + Ω(x, ) (f) 2 Ω(x, ) ( f + f ) V n n 2 f

14 4 [f],n K f So this gives us L (Ω) L,n (Ω). For n =, say Ω = (0, ), then log x is a tyical examle of a function that doesn t belong to L (0, ), but that does belong to L,n (0, ). This can be generalized to higher dimensions and more arbitrary domains.

15 References [] KUFNER, JOHN, FUČIK, Function Saces, Noordhoff International Publishing, Leyden, (977), [2] PEETRE, On the Theory of L,λ Saces, Journal of Functional Analysis 4, 7-87 (969) [3] RUPFLIN, What is a... orrey Sace?, What is a... amanato Sace?, (2008) [4] ADAS, XIAO, orrey saces in Harmonic Analysis, Ark. at., 50 (202),