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, 0.0.4 and 0.0.22. Notation 0.0... 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 0.0.2. (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 0.0.3. (orrey and amanato saces) Let Ω R n be a bounded domain (oen and connected), and λ 0.
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 0.0.4..,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 0.0.5. [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 0.0.6. 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 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 0.0.7. 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 0.0.8. 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 Definition 0.0.9. For f L,λ (Ω), let f,n = su Ω(x, ) λ n x Ω 0<<d f / Lemma 0.0.0.,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 0.0.. (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 0.0.2.. 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.
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 0.0.3. 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 0.0.4.. 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 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 0.0.5. 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 0.0.6.,λ + defines yet another norm on L,λ (Ω) which is equivalent to,λ.
7 Lemma 0.0.7. 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 0.0.8. 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 0.0.7 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 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 0.0.9. 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 0.0.8 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 0.0.20. 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 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 0.0.8 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 0.0.7 (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
0 Now take K = K 3 + K 2 (n λ). 2 Lemma 0.0.2. 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 0.0.22. 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).
. 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 0.0.6. 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 0.0.20 (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.
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 0.0.9 F (x) = lim(f) and F = f a.e. r 0 in (Ω) Let x, y Ω, r = 2 x y. We have
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 0.0.2 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
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.
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