0 <β 1 let u(x) u(y) kuk u := sup u(x) and [u] β := sup

Transcription

1 456 BRUCE K. DRIVER 24. Hölder Spaces Notation Let Ω be an open subset of R d,bc(ω) and BC( Ω) be the bounded continuous functions on Ω and Ω respectively. By identifying f BC( Ω) with f Ω BC(Ω), we will consider BC( Ω) as a subset of BC(Ω). For u BC(Ω) and 0 <β 1 let u(x) u(y) kuk u := sup u(x) and [u] β := sup x Ω x,y Ω x y β. x6=y If [u] β <, then u is Hölder continuous with holder exponent 43 β. The collection of β Hölder continuous function on Ω will be denoted by C 0,β (Ω) :={u BC(Ω) :[u] β < } and for u C 0,β (Ω) let (24.1) kuk C 0,β (Ω) := kuk u +[u] β. Remark If u : Ω C and [u] β < for some β>1, then u is constant on each connected component of Ω. Indeed, if x Ω and h R d then u(x + th) u(x) t [u] βt β /t 0 as t 0 which shows h u(x) =0for all x Ω. If y Ω is in the same connected component as x, then by Exercise 17.5 there exists a smooth curve σ :[0, 1] Ω such that σ(0) = x and σ(1) = y. So by the fundamental theorem of calculus and the chain rule, Z 1 Z d 1 u(y) u(x) = 0 dt u(σ(t))dt = 0 dt =0. 0 This is why we do not talk about Hölder spaces with Hölder exponents larger than 1. Lemma Suppose u C 1 (Ω) BC(Ω) and i u BC(Ω) for i =1, 2,...,d, then u C 0,1 (Ω), i.e. [u] 1 <. The proof of this lemma is left to the reader as Exercise Theorem Let Ω be an open subset of R d. Then (1) Under the identification of u BC Ω with u Ω BC (Ω),BC( Ω) is a closed subspace of BC(Ω). (2) Every element u C 0,β (Ω) has a unique extension to a continuous function (still denoted by u) on Ω. Therefore we may identify C 0,β (Ω) with C 0,β ( Ω) BC( Ω). (In particular we may consider C 0,β (Ω) and C 0,β ( Ω) to bethesamewhenβ>0.) (3) The function u C 0,β (Ω) kuk C 0,β (Ω) [0, ) is a norm on C 0,β (Ω) which make C 0,β (Ω) into a Banach space. Proof. 1. The first item is trivial since for u BC( Ω), the sup-norm of u on Ω agrees with the sup-norm on Ω and BC( Ω) is complete in this norm. 43 If β =1,uis is said to be Lipschitz continuous.

2 ANALYSIS TOOLS WITH APPLICATIONS Suppose that [u] β < and x 0 Ω. Let {x n } Ω be a sequence such that x 0 = lim x n. Then u(x n ) u(x m ) [u] β x n x m β 0 as m, n showing {u(x n )} is Cauchy so that ū(x 0) := lim u(x n ) exists. If {y n } Ω is another sequence converging to x 0, then u(x n ) u(y n ) [u] β x n y n β 0 as n, showing ū(x 0 ) is well defined. In this way we define ū(x) for all x Ω and let ū(x) =u(x) for x Ω. Since a similar limiting argument shows ū(x) ū(y) [u] β x y β for all x, y Ω it follows that ū is still continuous and [ū] β =[u] β. In the sequel we will abuse notation and simply denote ū by u. 3. For u, v C 0,β (Ω), v(y)+u(y) v(x) u(x) [v + u] β = sup x,y Ω x y β x6=y v(y) v(x) + u(y) u(x) sup x,y Ω x y β [v] β +[u] β x6=y and for λ C it is easily seen that [λu] β = λ [u] β. This shows [ ] β is a semi-norm on C 0,β (Ω) and therefore k k C 0,β (Ω) defined in Eq. (24.1) is a norm. To see that C 0,β (Ω) is complete, let {u n } be a C0,β (Ω) Cauchy sequence. Since BC( Ω) is complete, there exists u BC( Ω) such that ku u n k u 0 as n. For x, y Ω with x 6= y, u(x) u(y) u n (x) u n (y) x y β = lim x y β lim sup[u n ] β lim ku nk C 0,β (Ω) <, and so we see that u C 0,β (Ω). Similarly, u(x) u n (x) (u(y) u n (y)) (u m u n )(x) (u m u n )(y) x y β = lim m x y β lim sup[u m u n ] β 0 as n, m showing [u u n ] β 0 as n and therefore lim ku u n k C 0,β (Ω) =0. Notation Since Ω and Ω are locally compact Hausdorff spaces, we may define C 0 (Ω) and C 0 ( Ω) as in Definition We will also let C 0,β 0 (Ω) :=C 0,β (Ω) C 0 (Ω) and C 0,β 0 ( Ω) :=C 0,β (Ω) C 0 ( Ω). It has already been shown in Proposition that C 0 (Ω) and C 0 ( Ω) are closed subspaces of BC(Ω) and BC( Ω) respectively. The next proposition describes the relation between C 0 (Ω) and C 0 ( Ω). Proposition Each u C 0 (Ω) has a unique extension to a continuous function on Ω given by ū = u on Ω and ū =0on Ω and the extension ū is in C 0 ( Ω). Conversely if u C 0 ( Ω) and u Ω =0, then u Ω C 0 (Ω). In this way we may identify C 0 (Ω) with those u C 0 ( Ω) such that u Ω =0.

3 458 BRUCE K. DRIVER Proof. Any extension u C 0 (Ω) to an element ū C( Ω) is necessarily unique, since Ω is dense inside Ω. So define ū = u on Ω and ū =0on Ω. We must show ū is continuous on Ω and ū C 0 ( Ω). For the continuity assertion it is enough to show ū is continuous at all points in Ω. For any > 0, by assumption, the set K := {x Ω : u(x) } is a compact subset of Ω. Since Ω = Ω \ Ω, Ω K = and therefore the distance, δ := d(k, Ω), between K and Ω is positive. So if x Ω and y Ω and y x <δ,then ū(x) ū(y) = u(y) < which shows ū : Ω C is continuous. This also shows { ū } = { u } = K is compact in Ω and hence also in Ω. Since >0 was arbitrary, this shows ū C 0 ( Ω). Conversely if uª C 0 ( Ω) such that u Ω = 0 and > 0, then K := x Ω : u(x) is a compact subset of Ω which is contained in Ω since Ω K =. Therefore K is a compact subset of Ω showing u Ω C 0 ( Ω). Definition Let Ω be an open subset of R d,k N {0} and β (0, 1]. Let BC k (Ω) (BC k ( Ω)) denote the set of k times continuously differentiable functions u on Ω such that α u BC(Ω) ( α u BC( Ω)) 44 for all α k. Similarly, let BC k,β (Ω) denote those u BC k (Ω) such that [ α u] β < for all α = k. For u BC k (Ω) let kuk C k (Ω) = X k α uk u and α k kuk C k,β (Ω) = X k α uk u + X [ α u] β. α k α =k Theorem The spaces BC k (Ω) and BC k,β (Ω) equipped with k k C k (Ω) and k k C k,β (Ω) respectively are Banach spaces and BCk ( Ω) isaclosedsubspaceof BC k (Ω) and BC k,β (Ω) BC k ( Ω). Also C k,β 0 (Ω) =C k,β 0 ( Ω) ={u BC k,β (Ω) : α u C 0 (Ω) α k} is a closed subspace of BC k,β (Ω). Proof. Suppose that {u n } BCk (Ω) is a Cauchy sequence, then { α u n } is a Cauchy sequence in BC(Ω) for α k. Since BC(Ω) is complete, there exists g α BC(Ω) such that lim k α u n g α k u =0for all α k. Letting u := g 0, we must show u C k (Ω) and α u = g α for all α k. This will be done by induction on α. If α = 0 there is nothing to prove. Suppose that we have verified u C l (Ω) and α u = g α for all α l for some l<k.then for x Ω, i {1, 2,...,d} and t R sufficiently small, a u n (x + te i )= a u n (x)+ Letting n in this equation gives a u(x + te i )= a u(x)+ Z t 0 Z t 0 i a u n (x + τe i )dτ. g α+ei (x + τe i )dτ from which it follows that i α u(x) exists for all x Ω and i α u = g α+ei. This completes the induction argument and also the proof that BC k (Ω) is complete. 44 To say α u BC( Ω) means that α u BC(Ω) and α u extends to a continuous function on Ω.

4 ANALYSIS TOOLS WITH APPLICATIONS 459 It is easy to check that BC k ( Ω) is a closed subspace of BC k (Ω) and by using Exercise 24.1 and Theorem 24.4 that that BC k,β (Ω) is a subspace of BC k ( Ω). The fact that C k,β 0 (Ω) is a closed subspace of BC k,β (Ω) is a consequence of Proposition To prove BC k,β (Ω) is complete, let {u n } BCk,β (Ω) be a k k C k,β (Ω) Cauchy sequence. By the completeness of BC k (Ω) just proved, there exists u BC k (Ω) such that lim ku u n k C k (Ω) =0. An application of Theorem 24.4 then shows lim k α u n α uk C 0,β (Ω) =0for α = k and therefore lim ku u n k C k,β (Ω) =0. The reader is asked to supply the proof of the following lemma. Lemma The following inclusions hold. For any β [0, 1] BC k+1,0 (Ω) BC k,1 (Ω) BC k,β (Ω) BC k+1,0 ( Ω) BC k,1 ( Ω) BC k,β (Ω). Definition Let A : X Y be a bounded operator between two (separable) Banach spaces. Then A is compact if A [B X (0, 1)] is precompact in Y or equivalently for any {x n } X such that kx n k 1 for all n the sequence y n := Ax n Y has a convergent subsequence. Example Let X = 2 = Y and λ n C such that lim λ n =0, then A : X Y defined by (Ax)(n) =λ n x(n) is compact. Proof. Suppose {x j } j=1 2 such that kx j k 2 = P x j (n) 2 1 for all j. By Cantor s Diagonalization argument, there exists {j k } {j} such that, for each n, x k (n) =x jk (n) converges to some x(n) C as k. Since for any M <, x(n) 2 = lim x k (n) 2 1 k P we may conclude that x(n) 2 1, i.e. x 2. Let y k := A x k and y := A x. We will finish the verification of this example by showing y k y in 2 as k. Indeed if λ M =max λ n, then n M X ka x k A xk 2 = λ n 2 x k (n) x(n) 2 = X λ n 2 x k (n) x(n) 2 + λ M 2 x k (n) x(n) 2 M+1 λ n 2 x k (n) x(n) 2 + λ M 2 k x k xk 2 λ n 2 x k (n) x(n) 2 +4 λ M 2. Passing to the limit in this inequality then implies lim sup ka x k A xk 2 4 λ M 2 0 as M. k

5 460 BRUCE K. DRIVER Lemma If X A Y B Z are continuous operators such the either A or B is compact then the composition BA : X Z is also compact. Proof. If A is compact and B is bounded, then BA(B X (0, 1)) B(AB X (0, 1)) which is compact since the image of compact sets under continuous maps are compact. Hence we conclude that BA(B X (0, 1)) is compact, being the closed subset of the compact set B(AB X (0, 1)). If A is continuos and B is compact, then A(B X (0, 1)) is a bounded set and so by the compactness of B, BA(B X (0, 1)) is a precompact subset of Z, i.e. BA is compact. Proposition Let Ω o R d such that Ω is compact and 0 α<β 1. Then the inclusion map i : C β (Ω) C α (Ω) is compact. Let {u n } C β (Ω) such that ku n k C β 1, i.e. ku n k 1 and u n (x) u n (y) x y β for all x, y Ω. By Arzela-Ascoli, there exists a subsequence of {ũ n } of {u n } and u C o ( Ω) such that ũ n u in C 0. Since u(x) u(y) = lim ũ n(x) ũ n (y) x y β, u C β as well. Define g n := u ũ n C β, then [g n ] β + kg n k C 0 = kg n k C β 2 and g n 0 in C 0. To finish the proof we must show that g n 0 in C α. Given δ>0, g n (x) g n (y) [g n ] α =sup x6=y x y α A n + B n where gn (x) g n (y) A n =sup x y α : x 6= y and x y δ gn (x) g n (y) =sup x y β x y β α : x 6= y and x y δ δ β α [g n ] β 2δ β α and gn (x) g n (y) B n =sup x y α : x y >δ 2δ α kg n k C 0 0 as n. Therefore, lim sup [g n ] α lim sup A n + lim sup B n 2δ β α +0 0 as δ 0. This proposition generalizes to the following theorem which the reader is asked to proveinexercise24.2below. Theorem Let Ω be a precompact open subset of R d,α,β [0, 1] and k, j N 0. If j + β>k+ α, then C j,β Ω is compactly contained in C k,α Ω.

6 ANALYSIS TOOLS WITH APPLICATIONS Exercises. Exercise Prove Lemma Exercise Prove Theorem Hint: First prove C j,β C j,α Ω is compact if 0 α<β 1. Then use Lemma repeatedly to handle all of the other cases.