Trace Theorems for Sobolev spaces

Transcription

1 ational and Kapodistrian University of Athens Department Of Mathematics Master Thesis Trace Theorems for Sobolev spaces Author: Maria Kampanou Supervisors: Prof. Pier Domenico Lamberti Prof. Gerasimos Barbatis A thesis submitted in fulfillment of the requirements for the degree of Master of Science in Applied Mathematics Athens, July 218

2 Abstract For R an open set with boundary satisfying certain smoothness assumptions, we consider the Besov spaces Bp l 1/p ( ) as the trace spaces of the Sobolev spaces Wp(). l More specifically, first we consider the traces in R 1 of functions defined on R and we prove that the trace operator T : Wp 1 (R ) L p (R 1 ) satisfies T Wp(R 1 + ) = Bp 1 1/p (R 1 ). Then we prove that T Wp() l = Bp l 1/p ( ), where R an open set with C l -boundary. For the case p = 2, we approach the definition of the trace spaces with two other methods, namely using the Fourier transformation and using the spectral definition given by Auchmuty [3]. Finally, we use the previous results to prove the existence of solution for the Dirichlet problem. 1

3 Acknowledgements The Erasmus Program gave me the great opportunity to live in Padova, Italy and work on this thesis. I would like to thank my supervisor, Prof. Pier Domenico Lamberti for his guidance during the whole process of writing this thesis. Even though his time was very limited, he was always available to discuss and to give me his insight on any problems I might have encountered. My sincere thanks to Prof. Gerasimos Barbatis for bringing me in touch with Prof. Lamberti and mostly for the help he has provided from the first day I contacted him regarding my thesis till the very last. Many thanks to to Prof. I. Stratis and all the Professors of the master Program of Applied Mathematics who do their best for the students, despite all the difficulties they encounter due to the longstanding effects of the economical crisis in Greece. My sincere thanks to Ms. Alkistis tai and George Papadakis for their invaluable help. one of these would have been possible without the love and support by my parents and sister throughout all these years. A heartfelt thanks to my best friend Thomas Pazios who has always been there for me and Jozefien De Praetere for all her encouragement. The greatest thanks to my boyfriend ikos Psallidakis for his support in every level imaginable and above all for believing in me even when I didn t. 2

4 Contents 1 Introduction 4 2 Preliminaries Some useful results Sobolev Spaces Trace of a function Besov spaces Trace Spaces of Sobolev Spaces 19 4 The case p = 2: H s spaces and their traces Characterization via Fourier transform Auchmuty s method The harmonic Steklov eigenproblem Spectral representation of the trace and the extension operator The HA s ( ) spaces Existence of solution for the Poisson Problem with Dirichlet boundary conditions Existence via the classical definition of H s spaces With use of Auchmuty s Definition References 52 3

5 Chapter 1 Introduction In many problems of mathematical physics and the calculus of variations it is not sufficient to deal with classical solutions of differential equations, especially when questions are raised about the regularity of the solutions. In many cases it is necessary to introduce the notion of weak derivatives and work in the so called Sobolev spaces. In , S.L.Sobolev introduced spaces of integrable functions having weak derivatives in L p. These function spaces have turned out to be very useful when studying partial differential equations on smooth and nonsmooth domains and their boundary value problems. A key component when using Sobolev spaces as a framework to deal with boundary value problems in PDEs is Trace theory. When we work with Sobolev spaces we have a certain knowledge regarding the regularity and smoothness of the solutions. But what about the regularity and smoothness of the functions defining the boundary conditions? In order to better understand the importance of this question for a domain R let us consider the following problem: { u = F, x, (1.1) u = g, x, which is also know as Poisson problem with non-homogeneous Dirichlet boundary conditions and let us suppose that we wish to find a solution u H s (), where H s () = W s 2 (). 4

6 Introduction There are several approaches to the solution of such problem. One approach is to look for a minimizer of the corresponding energy functional. Then the minimizer satisfies the associated Euler- Lagrange equation which coincides with the given problem. For this approach it is essential to know which functions g are admissible. If we think of the trace of a function u H s () as an element of L 2 ( ) and in a more broad sense as the restriction of u to the boundary of (in fact, it is the restriction if u is continuous) equivalently to problem (1.1) we can write: { u = F, x, (1.2) T u = g, x, where T denotes the trace operator. Thus, it becomes more clear why describing the trace operator s image from a Sobolev space is a fundamental problem. In fact this problem has challenged mathematicians for a long time. It was first solved in the case p = 2 via Fourier Transform methods. The solution for p 2 and l = 1 was given by Gagliardo in the 5 s. The general case was sorted out by Besov and ikolskii. The solution involves the so-called Besov-ikolskii spaces Bp( ) l (in the following chapters we will be referring to them simply as Besov spaces). A very interesting and recent approach, for the case p = 2, involves a spectral representation of the trace spaces and was given by G. Auchmuty in 26. For this approach, Auchmuty generates an orthonormal basis for L 2 ( ) with use of the Steklov eigenfunctions and uses this basis to describe the trace spaces. In the last chapter, after we present the trace theorems we will go back to the Poisson problem with Dirichlet boundary conditions that motivated us in the first place and we will prove the existence of solutions: first with use of the classical definition of trace spaces (i.e. via Besov spaces) and then with use of the spectral definition as given by Auchmuty. Lastly, another aspect regarding the importance of describing the trace image of Sobolev spaces lies in the study of Sobolev spaces themselves and not just in the frame of PDEs, since they constitute an individual field of study. 5

7 Chapter 2 Preliminaries We present here a number of results widely used in the following chapters. For further information we refer to [7], [8], [9], [1]. 2.1 Some useful results Lemma 2.1 (Minkowski s inequality). Let (X, M, µ) be a measure space. Let 1 p and let u, υ : X [, ] be measurable functions. Then u + υ L p u L p + υ L p. Lemma 2.2 (Minkowski s inequality for integrals). Let (X, M, µ) and (Y,, ν) be two measure spaces. Assume that µ, ν are complete and σ-finite. Let u : X Y [, ] be a (M )-measurable function and let 1 p. Then u(x, ) dµ(x) u(x, ) L p (Y,,ν)dµ(x). L p (Y,,ν) x Lemma 2.3 (Minkowski s inequality for sums). Let (X, M, µ) and (Y,, ν) be two measure spaces. Assume that µ, ν are complete and σ-finite. Let u, υ : X Y [, ] be a (M )-measurable function and let 1 p. Then ( ) 1/p ( ) 1/p ( ) 1/p u(x) + υ(x) p u(x) p + υ(x) p. x X x X 6 x x X

8 Preliminaries Theorem 2.1 (Tonelli s Theorem). Let (X, M, µ) and (Y,, ν) be two measure spaces. Assume that µ and ν are complete and σ-finite and let u : X Y [, ] be a (M )-measurable function. Then for µ-a.e., x X the function u(x, ) is measurable and the function u(, y)dν(y) is measurable. Similarly, for ν-a.e. and y Y the function u(, y) is measurable and the function u(x, )dµ(y) is measur- Y Y able. Moreover, ( ) u(x, y)dν(y) dµ(x) = X Y Y ( X ) u(x, y)dµ(x) dν(y) Theorem 2.2 (Fubini s Theorem). Let (X, M, µ) and (Y,, ν) be two measure spaces. Assume that µ and ν are complete and let u : X Y [, ] be (µ ν)-integrable. Then for µ-a.e. x Xthe function u(x, ) is ν-integrable and the function u(, y)dν(y) is µ-integrable. Y Theorem 2.3 (Hölder s inequality). Let (X, M, µ) be a measure space, 1 p and and let q be its Hölder conjugate exponent, i.e. 1 p + 1 q = 1. If u, υ : X [, ] are measurable functions, then uυ L 1 (X) u L p (X) υ L q (X). Lemma 2.4 (Fatou s lemma). Let (X, M, µ) be a measure space. Let u n : X [, ] be a sequence of measurable functions, then lim inf u ndµ lim inf u n dµ. n n X Theorem 2.4. Let (X, M, µ) be a measure space. Let Y be a metric space, and let u : X Y R be a function. Assume that for each fixed y Y the function x X u(x, y) is measurable and that there exists y Y such that X lim u(x, y) = u(x, y ), y y for every x X. Assume also that there exists an integrable function g : X [, ] such that u(x, y) g(x), 7

9 Preliminaries for µ a.e. x X and for all y Y. Then the function F : Y R, defined by F (y) = u(x, y)dµ(x), y Y, is well-defined and continuous at y. X Theorem 2.5. Let Y be an interval of R and assume that for each fixed x X the function y Y u(x, y) is differentiable and that for each fixed y Y the functions x X u(x, y) and x X u (x, y) y are measurable. Assume also that for some y Y the function x X u(x, y ) is integrable and that there exists an integrable function h : X [, ] such that u (x, y) y h(x) for µ a.e. x X and for all y Y.Then the function F : Y R, defined by F (y) = u(x, y)dµ(x), y Y, is well-defined and differentiable, with F u (y) = (x, y)dµ(x). y X ow we need to define the functions called mollifiers. φ(x) := X Definition 2.1. Let φ L 1 (R ) be a non-negative, bounded function with suppφ B(, 1), φ(x)dx = 1. R (2.1) For every ε > we define φ ε (x) = 1 ε φ( x), x R. ε The functions φ ε are called mollifiers. In the case that φ is of class Cc defined by { { } α exp, if x < 1 1 x 2 1 1, if x 1, (2.2) where α > is such that (2.1) is satisfied. We call φ ε standard mollifiers. 8

10 Preliminaries Remark. We note that suppφ ε B(, ε). Definition 2.2. Let R be an open set and let u L 1 (). We consider the open set ε := {x : dist(x, ) > ε}. For x ε we define the function u ε : ε R, which we will call a mollification of u: u ε (x) := (u φ ε )(x) = u(x y)φ ε (y)dy, where φ ε the standard mollifiers. Theorem 2.6. Let R be an open set, let φ L 1 (R ) be a nonnegative bounded function satisfying (2.1), and let u L 1 (). 1. If u C(), then u ε u as ε + uniformly on compact subsets of. 2. For every Lebesque point x of u we have u ε (x) u(x) as ε If 1 p, then for every ε > and u ε L p ( ε) u L p (), 4. If u L p (), 1 p <, then u ε L p ( ε) u L p () as ε +. (2.3) ( ) 1/p lim u ε u p dx =. ε + ε In particular, for any open set with dist(, ) >, u ε u in L p ( ). Theorem 2.7. Let R be an open set, let φ L 1 (R ) be defined as in (2.2) and let u L 1 (). Then u ε C ( ε ) for all < ε < 1. Moreover, for every multi-index α we have for all x ε. a u ε x a (x) = ( u a φ ε x a ) (x) = 9 R a φ ε (x y)u(y)dy, xa

11 Preliminaries Definition 2.3. Let (H,, ) be a Hilbert space and let T : H H be a symmetric operator, i.e. T x, y = y, T x, where H a Hilbert space. We say that T is non-negative if for all u H. T u, u, (2.4) Proposition 2.1. Let (H,, ) be a Hilbert space and let T : H H be a self-adjoint operator. Then T is non-negative if and only if σ(t ) [, ], where σ(t ) denotes the spectrum of T. Theorem 2.8 (Courant-Rayleigh minmax principle). Let ( H,, ) be a Hilbert space and let T : H H be a compact, self-adjoint operator, whose positive eigenvalues are listed in decreasing order λ n λ 1. Then λ n = min V H dimv=n 1 max x V x T x, x x, where V H is an (n 1)-dimensional subspace. Theorem 2.9 (Riesz s Representation Theorem). Let M be a bounded linear functional on a Hilbert space H equipped with the inner product,. Then there exists some g H such that for every u H Mu = u, g. Moreover, M = g, where the inducted norm. Theorem 2.1. (Spectral Theorem for compact and self-adjoint operators) Let H be a Hilbert space and T : H H a compact and selfadjoint operator on H. Then there exists a finite or infinite sequence of real eigenvalues {λ n } n=1 with {λ n } and a corresponding orthonormal sequence of eigenfunctions {e n } n=1 in H such that 1. T e n = λ n e n for all 1 n, 2. ImT = span {e n } n=1, 3. if =, that is {λ n } n=1 is infinite, then λ n as n. 1

12 Preliminaries Theorem Let R a bounded set. The space L 2 () with the standard norm is a Banach space. Moreover, let {φ n } n=1 be an orthonormal sequence in L 2 (), then given a sequence of numbers {c n } n=1 such that c 2 n < there exists a function u L 2 () such that n=1 u(x) 2 dx = c 2 n, c n = n=1 2.2 Sobolev Spaces φ n (x)u(x)dx. Let R be an open set, s, 1 p. Let Cc () be the space of functions in C () with compact support in, i.e. the space of test functions. Definition 2.4. Let u, υ L 1 Loc () and α = (α 1,..., α ) a multiindex. We say that υ is the α th -weak partial derivative of u, and we write D α u = υ if ud α φdx = ( 1) α υφdx, for all φ C c (). Definition 2.5. We define the Sobolev Space Wp s () to be the space of all the functions u L p () such that for all α with α s, the weak derivative D α u exists and belongs to L p (). In the special case p = 2 we write H s () = W2 s (). We also introduce the following norm in Wp s (): ( 1/p u W s p () := D α u L ()) p, p α s u W s () := D α u L (). α s Definition 2.6. Let {u m } m=1, R and u W s p (). We say that u m converges to u in W s p () and write u m u in W s p () provided that lim u m u W s m p () =. 11

13 Preliminaries Definition 2.7. We denote by W s,p () the closure of C c () in W s p () and respectively H s () for the case p = 2. Thus we have that u W s,p () for R if and only if there exist functions u m Cc () such that u m u in Wp s (). We can interpret W s,p () as comprising all those functions u Wp s () such that D α u = on the boundary for all α s 1. Proposition. For all s, 1 p, the Sobolev Space W s p () is a Banach space. Proof. Let {u j } j=1 a Cauchy sequence in W s p (). Then for all α s {D α u j } j=1 is a Cauchy sequence in L p () and since L p () is complete, there exist functions u, u a L p () such that u j u, as j in L p () D α u j u a, as j in L p (), for all α s. Let φ Cc (), then by Hölder s inequality it follows (u u j )D α φ p dx u u j p dx D α φ q dx, where q the conjugate of p. Hence, u W s p () and D α u = u α for all α s. This means that D α u j D α u for all α s and as a result we have u j u in W s p (). Definition 2.8. For an arbitrary nonempty set R we denote by C b () the Banach space of functions u continuous and bounded on with the norm u C() = sup u(x). x Definition 2.9. Let l. We denote by Cb l () the Banach space of functions u C b () such that for all α where a = l and for all x the derivatives (D a u)(x) exist and D a u C b (), with the norm u C l () = u C() + α =l D α u C(). 12

14 Preliminaries Lemma 2.5. Let 1 q < p and let R be a bounded open set with boundary of class C,1. Then the embedding is compact. W 1 p () L q () Lemma 2.6. Let l, 1 p and let R be an open set with Lipschitz boundary, i.e. of class C,1. Let also φ Cb l (). Then for all u Wp() l uφ W l p () c u W l p (suppφ ), where c > is independent of u. Lemma 2.7. Let l, 1 p and let R be an open set having a quasi-resolved boundary 1. Moreover, let g = (g 1,..., g ) : R, g K C l (), K = 1,...,. We suppose that for all α, satisfying 1 α l the derivatives D α g K are bounded on and the Jacobian J g (x) = Dg is such that inf Dx x Dg (x) >. Furthermore, let Dx g() be also an open set with a quasi-resolved boundary. Then for all u Wp() l c 1 u W l p (g()) ug W l p () c 2 u W l p (g()), where c 1, c 2 > are independent of u and p. ext we present two approximation by smooth functions theorems. For the proofs we refer to [5]. Theorem 2.12 (Global approximation by smooth functions). Let R be an open set and let u W l p() for some 1 p <. Then there exist functions u m C () W l p() such that u m u in W l p(). Theorem 2.13 (Global approximation by functions smooth up to the boundary). Let R be an open bounded set with C 1 boundary. We suppose that u W l p() for some 1 p <. Then there exist functions u m C () W l p() such that u m u in W l p(). Definition 2.1. Let R be an open set and let 1 < p <. The Sobolev space L 1,p () is the space of all functions u L 1 Loc () whose gradient u (in the sense of distributions) belongs to L p (). 1 The precise definition of this notion is given in the next chapter. 13

15 Preliminaries 2.3 Trace of a function In this section we present the definition, as well as several results, regarding the Trace operator. Let u L 1 Loc (R ), > 1. We want to define the trace g of the function u on R M, where 1 M <. In order to do so we will represent each x R as x = (y, z),where y = (x 1,..., x M ) R M, z = (x M+1,..., x ) R M and we will consider R M as the M-dimensional space of all points (y, z), where z = (,..., ) and y runs through any possible values. If u is continuous we can define the trace to be the restriction of the function on R M. However, this definition obviously does not make sense for every u L 1 Loc (R ). In order to define the trace g for a function u L 1 Loc (R ) we have to make certain requirements: 1. g L 1 Loc (RM ), 2. if g L 1 Loc (RM ) is a trace of u, then ψ L 1 Loc (RM ) is also a trace of u iff ψ is equivalent to g on R M, 3. if g L 1 Loc (RM ) is a trace of u and h is equivalent to u on R, then g is also a trace of h, 4. if u is continuous then u(y, ) is a trace of u. The following definition, as given by Burenkov [1], satisfies all of the requirements above. Definition Let u L 1 Loc (R ) and g L 1 Loc (RM ). We call the function g a trace of the function u if there exists a function h L 1 Loc (R ) such that h is equivalent to u on R and h(, z) g( ) in L 1 Loc(R M ) as z. Theorem (Existence of Trace) Let s, m,, m < n and 1 p. Then traces on R M exist for all u W s p (R ) iff equivalently, iff s > n m p for 1 < p, s M for p = 1, W s p (R M ) C(R M ). 14

16 Preliminaries Proof. We refer to [1]. Definition We say that a domain R is a bounded domain with a resolved boundary with the parameters d, D satisfying < d D < if = {x R : α < x < φ(x), x A}, (2.5) where diam D, x = (x 1,..., x 1 ) R 1, A = {x R 1 : a i < x i < b i, i = 1,..., 1}, < a i < b i < and a + d φ(x), x A. Let R be an open set with C 1 -boundary. ow we wish to extend Definition 2.11 to the case of R, R M being replaced with, respectively. In order to do so, we will use the same method as in [1]. We start with R being a bounded domain having the form (2.5), with φ of class C 1. Let u L 1 (). In the spirit of Definition 2.11 we say that the function g L 1 (Γ), where Γ = {x R : x = φ(x), x A}, is a trace of the function u on Γ if there exists a function h equivalent to u on such that h( + te ) g( ) in L 1 (Γ) as t, where e = (,...,, 1). Since there is no guarantee that the boundary will be flat near a chosen point x, we can use a proper transformation Φ of (with inverse Φ 1 ), which straightens out near x. Then with use of the transformation Φ of (as chosen in [1]) we have that g(φ 1 ) is a trace of u(φ 1 ) on Φ(Γ). ext we suppose that R is an open set such that for a certain map λ, which is a composition of rotations, reflections and translations, the set λ() is a bounded domain with C 1 -boundary and Γ is such that λ(γ) = {x R : x = φ(x), x A} (for further details we refer to [1]). In this case we say that g is a trace of u on Γ if g(λ 1 ) is a trace of u(λ 1 ) on λ(γ). Finally, let R be an open set with C 1 -boundary and let V j be open parallelepipeds defined as in [1]. Then there exists a partition of unity ψ j C (R ) such that ψ j 1, suppψ j (V j ) d, 2 15

17 Preliminaries j = 1,..., s, s ψ j = 1 on and j=1 D α ψ j (x) cd α, x R, α, j = 1,..., s, where c > is independent of x, j and d (for the proof we refer to [1]). Having said so, we have the following definition. Definition Let R be an open set with C 1 -boundary and u : R with u L 1 (B ), for each ball B R. We suppose that u = s j=1 u j, where suppu j V j and u j L 1 (V j ). If the functions g j are traces of the functions u j on V j, j = 1,..., s, then the function g = s g j is said to be a trace of the function u on. j=1 Theorem 2.15 (Trace Theorem). Let R bounded with C 1 boundary. Then there exists a bounded linear operator such that T : W 1 p () L p ( ) 1. T u = u 2. T u L p ( ) C u W 1 p (), for all u W 1 p () and C = C(p, ) independent of u. Proof. For the proof we refer to [5]. Theorem Let 2 and R be an open set whose boundary is uniformly Lipschitz, let 1 p <, and let u W 1 p (). Then T u = if and only if u W 1,p (). Proof. For the proof we refer to [2]. Lastly we present the following Theorem regarding the compactness of the Trace operator. For the proof we refer to [6]. Theorem (Compact Trace Theorem) Let R be a bounded open set with C 1 boundary and let 1 p <. Then the trace operator T : W 1 p () L p ( ) is compact. 16

18 Preliminaries 2.4 Besov spaces In order to determine the Trace Spaces of the Sobolev Spaces W s p it is necessary to introduce the Besov Spaces B s p. Definition Let u : R R. We have the following definitions. For all h R, i = 1,..., and x R we define h i u(x) := u(x + he i ) u(x) = u(x i, x i + h) u(x i, x i ), where e i is the i-th vector of the canonical basis in R. For h R and σ we define σ ( ) σ σ hu(x) = ( 1) σ k u(x + kh). k k= Thereinafter, the use of each definition given above will be clear from the context. Definition Let s >, σ, σ > s and 1 p, θ. The function u L 1 Loc () belongs to the Besov space Bs p,θ () if u is measurable on R and u B s p,θ (R ) := u L p (R ) + u B s p,θ (R ) <, where if 1 θ < and u B s p,θ (R ) := ( u B s (R ) := R ( σ ) h u θ L p (R ) dh h l h σ h sup u L p (R ). h R,h h l ) 1 θ, This definition is independent of σ > l as the following lemma shows. (For the proofs of the following Lemma and Propositions we refer to [1].) Lemma 2.8. Let l >, 1 p, θ. Then the norms B s p,θ (R ) corresponding to different σ satisfying σ > l are equivalent. Proposition 2.2. Let 1 p, θ, s >. () is a Banach space. B s p,θ Then the Besov space 17

19 Preliminaries Remark. For θ = s we denote B s p,p = B s p. Proposition 2.3. Let < s < 1, 1 p. For any u L 1 Loc (R ) let u ε := φ ε u, where φ ε is a a standard mollifier. Then u ε B s p (R ) u B s p (R ), for all ε > and lim u ε B ε + s p (R ) = u B s p (R ). Moreover if p < and u Bp(R s ), then lim u ε u B ε + s p (R ) =. In particular, if p < then C (R ) B s p(r ) is dense in B s p(r ). ow we need to define the spaces B s p( ), where s >, 1 p. In order to do so, we use similar arguments as in the case of defining the Trace in R instead of R. If R is a bounded domain with C 1 -boundary and u is defined on Γ, we say that u B s p(γ) if u(x, φ(x) B s p(a) and we set u B s p (Γ) = u(φ 1 ) B s p (Φ(Γ)) = u(x, φ(x) B s p (A), where Φ, A defined as above. If R is such that for a certain map λ, which is a composition of rotations,reflections and translations, the set λ() is bounded with C 1 -boundary, then u B s p(γ) if u(λ 1 ) B s p(λ(γ)) and u B s p (Γ) = u(λ 1 ) B s p (λ(γ)) = u(λ) B s p (Λ), where Λ = Φ(λ). Hence we have the following definition. Definition Let s >, 1 p and let R be an open set with with C 1 -boundary. We say that u Bp( ) s if uψ j Bp(V s j ), j = 1,..., l and ( l ) 1/p u B s p ( ) = j=1 uψ j p B s p (V j ) ( l ) 1/p = uψ j (Λ 1 ) p Bp s(φ(γ)), j=1 where Λ j = Φ j (λ j ) and ψ j is the partition of unity defined above. 18

20 Chapter 3 Trace Spaces of Sobolev Spaces In this chapter we will define the Trace spaces of the Sobolev spaces with use of the Besov spaces. More specifically we have the following Theorems. Theorem 3.1. Let 1 p, l, M, with M <, l > M p the trace operator T : Wp(R l ) L p (R M ). Then and l M T Wp(R l p ) = Bp (R M ). Proof. For the proof we refer to [1]. We will prove the theorem for the case l = 1 and M = 1. For the inclusion T Wp 1 (R +) Bp 1 1/p (R 1 ) it suffices to show that for all u Wp 1 (R +) the trace g = T u Bp 1 1/p (R 1 ). More specifically we will prove that there exists a constant C = C(, p) such that for all u L 1 (R +). g B 1 1/p p (R 1 ) C u L p (R ), Theorem 3.2. Let 1 < p <, 2, R + = {(x, x ) R 1 R : x > )}. Then, T Wp 1 (R +) Bp 1 1/p (R 1 ). 19

21 Trace Spaces of Sobolev Spaces Proof. Let u L 1 (R +) C (R ). For h i u(x) := u(x + he i ) u(x) = u(x i, x i + h) u(x i, x i ), we write u(x, ) = u(x, x ) (u(x, x ) u(x, )) = u(x, x ) x u(x, ), for all x R 1, x >. By integrating in x over the interval (, h), h > we get h Therefore, u(x, ) dx u(x, ) 1 h h We replace u with h i u h u(x, x ) dx + u(x, x ) dx + 1 h h h x u(x, ) dx. x u(x, ) dx. h i u(x, ) 1 h 1 h + 1 h h h h h i u(x, x ) dx + 1 h h i u(x, x ) dx h x h i u(x, ) dx x u(x + he i, ) + x u(x, ) dx, (3.1) where e i = (e i, ), i = 1,..., 1. By the Fundamental Theorem of Calculus we have: h i u(x, x ) = u(x + he i, x ) u(x, x ) = h x u(x +e i, ) = u(x +he i, x ) u(x +e i, ) = x x u(x, ) = u(x, x ) u(x, ) = x Then by 3.1 it follows h i u(x, ) 1 h + 1 h h h h x u x (x, z)dz u x i (x + ξe i, x )dξ u x (x +he i, z)dz u (x + ξe x i, x ) i dξdx u (x + he x i, z) + u (x, z) x dzdx. 2

22 Trace Spaces of Sobolev Spaces For J = h x u x (x + he i, z) + u x (x, z) dzdx we have h h J u (x + he x i, z) + u (x, z) x dzdx }{{} I and the Tonelli-Fubini Theorem yields h h I = u (x + he x i, z) + u (x, z) x dx dz h = h u (x + he x i, z) + u (x, z) x dz. Therefore, h i u(x, ) 1 h + h h h u (x + ξe x i, x ) i dξdx u (x + he x i, z) + u (x, z) x dz. By Minkowski s inequality for integrals we obtain h i u(x, ) L p (R 1 ) c h Hence, + c h h h = c h h h + c h i u(, ) L p (R 1 ) c h h For fixed small ε > we write h u x i (, x ) L p (R 1 )dx = u x i (x + ξe i, x ) L p (R 1 )dξdx u x (x + he i, z) L p (R 1 ) + u x (x, z) L p (R 1 )dz u x i (x + ξe i, x ) L p (R 1 )dx u x (x + he i, x ) L p (R 1 ) + u x (x, x ) L p (R 1 )dx u x i (, x ) L p (R 1 ) + u x (, x ) L p (R 1 )dx. h 21 x ε x ε u x i (, x ) L p (R 1 )dx. (3.2)

23 Trace Spaces of Sobolev Spaces With use of Hölder s inequality we get h u ( h (, x ) x L p (R 1 )dx x εq dx i where 1 p + 1 q h = 1. Similarly, = ch 1 p ε ( h u ( h (, x ) x L p (R 1 )dx ch 1/q ε Then by (3.2) it follows h i u(, ) p L p (R 1 ) chp/q εp h Hence, h i u(, ) p L p (R 1 ) dh C h p By Tonelli s Theorem we get ) 1/q ( h x εp u (, x ) p x dx L p (R 1 ) i x εp u x i (, x ) p L p (R 1 ) dx ) 1/p, x εp u ) 1/p (, x ) p. L x p (R 1 ) ) 1/p ( x εp u (, x ) p x + u ) (, x L p (R 1 ) ) p dx L i x p (R 1 ). h p/q εp h h p ( x εp u u ) (, x ) p dx L x p (R 1 ) dh. h p/q εp ( x εp u x i (, x ) p L p (R 1 ) + h i u(, ) p L p (R 1 ) dh C (, x h p x h p ) p x + L p (R 1 ) i u ) (, x ) p dhdx L x p (R 1 ) ( = C x εp u (, x ) p x + L p (R 1 ) i u ) (, x ) p h p/q εp dhdx L x p (R 1 ) h p. Since x h p/q εp dh = 1 h p x dh = 1 h εp+1 h i u(, ) p L p (R 1 ) dh C h p εp = C εp εp x εp x, we have ( u x i (, x ) p L p (R 1 ) + u ( u x i p L p (R + ) + u x p L p (R + ) 22 ) (, x ) p dx L x p (R 1 ) ).

24 Trace Spaces of Sobolev Spaces ow we want to remove the assumption that u C (R ). First we can extend via reflection the function u L 1,p (R +) to the whole L 1,p (R ). For ε > we consider the sequence {u ε } with u ε := u φ ε, where φ ε are the standard mollifiers. As ε + we have: u ε u in L 1 Loc (R ), u ε u in L p (R ) and u ε (, ) T u in L 1 Loc (R 1 ). We can select a subsequence such that u ε (x, ) T u(x ). Then by Fatou s Lemma for it follows I = R 1 T u(x + he i) T u(x ) p dx dh R 1 u ε (x + he I lim inf i, ) u ε (x, ) p ε + dx dh h p C(, p) lim u ε (x) p dx = C(, p) u(x) p dx. ε + R + h p R + Hence, T W 1 p (R +) B 1 1/p p (R 1 ). (3.3) For the inverse inclusion Bp 1 1/p (R 1 ) T Wp 1 (R +) we will prove that for all g Bp 1 1/p (R 1 ) there exists a function u Wp 1 (R +) such that T u = g and where C 2 = C 2 (, p). u W 1 p (R + ) C 2 g B 1 1/p p (R 1 ), Theorem 3.3. Let 1 < p <, 2, R + = {(x, x ) R 1 R : x > )}. Then, Bp 1 1/p (R 1 ) T Wp 1 (R +). Proof. Let g B 1 1/p p (R 1 ). Let φ C c (R 1 ) be such that suppφ B 1 (, 1) and R 1 φ(x )dx = 1. 23

25 Trace Spaces of Sobolev Spaces For x > we define υ(x, x ) := 1 x 1 R 1 φ ( x y x ) g(y )dy, for x R 1.Then for all x > we have ( ) υ(x, x ) p dx 1 x y p φ g(y )dy dx R 1 R 1 x 1 R x 1 ( ( ) 1 x y q ) p/q φ dy g(y ) p dy dx, R 1 R x 1 R 1 x 1 where in the last inequality we have used Hölder s inequality. By Theorem 2.6, where x plays the role of ε, we get ( ) p/q υ(x, x ) p dx φ(x y ) q dy g(y ) p dy dx R 1 R 1 R 1 R 1 g(y ) p dy. R 1 (3.4) We will use the notation x = (x i, x i ) R 2 R, for i = 1,..., 1. Then by Theorem 2.7 we have ( x y Since R υ x i (x) = 1 x R 1 φ x i x ) g(y )dy. ( ) φ x y ( ) g(y i, x i )dy i = g(y φ x y i, x i ) dy i = x i x R x i x we can write υ (x) = 1 x i x = 1 x Therefore, R 2 R 1 υ x i (x) = 1 x ( ( φ x y R x i x ( φ x y x i x R 1 ( φ y x i ) g(y )dy i R ) [g(y ) g(y i, x i )]dy. x φ ( x y x i x ) [g(x y ) g(x i y i, x i )]dy. 24 ) g(y i, x i )dy i ) dy

26 Trace Spaces of Sobolev Spaces Since suppφ B 1 (, 1) it follows υ (x) x i C g(x y ) g(x i y i, x i ) dy. x B 1 (,x ) We raise both sides to the power p and integrate in x over R + p ( υ (x) 1 R x + i dx C R x p + 1 C B 1 (, x ) p/q R + x p ) p g(x y ) g(x i y i, x i ) dy dx B 1 (,x ) B 1 (,x ) g(x y ) g(x i y i, x i ) p dy dx, where in the second inequality we have used Hölder s inequality. Then, p υ (x) (x 1 R x + i dx C )p 1 g(x y ) g(x R x p i y i, x i ) p dy dx, + B 1 (,x ) and by Tonelli s theorem we get p υ x (x) g(x y ) g(x i y i, x i ) p R x + i dx C dy R + B 2 (,x ) x x p+ 1 i dy i dx. }{{} I (3.5) For the calculation of I we consider the change of variables z i = x i y i and z i = x i y i. Then, x g(z I = i, z i ) g(z i, x i ) p dz R + B 2 (,x ) x x p+ 1 i dz i dx x y i i = g(z I, x i y i ) p dx Idx i dz i dz i dx, R R 2 B 2 (,x ) x x p+1 x x p+ 1 where in the second equality we have used Tonelli s theorem. By Hölder s inequality we get x 2 x I C y i x p+ 1 i g(z ) p dz dy i dx x R 1 1 x = C y i i g(z ) p dz dy i dx R 1 } {{ } J 25

27 Trace Spaces of Sobolev Spaces and once more with use of Tonelli s theorem we have J = y i i g(z ) p 1 R 1 y i x p+1 dz dy i dx = C y i i g(z ) p dy i dx. R 1 Therefore by 3.5 it follows υ (x) p dx C x i R + for all i = 1,..., 1. In order to estimate υ x g(y ) g(x ) = 1 i=1 y p i R 1 g(x + he i) g(x ) p we write: Since R 1 φ(x )dx = 1 we have υ(x, x ) = 1 i=1 h p dx dh, (3.6) [ g(y1,..., y i, x i+1,..., x 1 ) g(y 1,..., y i 1, x I,..., x 1 ) ]. 1 R 1 x 1 ( ) x y [g(y1 φ,..., y i, x i+1,..., x 1) x g(y 1,..., y i 1, x I,..., x 1 ) ] dy + g(x ). Then by Theorem 2.5 it follows 1 υ ( ( )) (x 1 x y [g(y1, x ) = φ,..., y x R x 1 x 1 i, x i+1,..., x 1) x i=1 g(y 1,..., y i 1, x I,..., x 1 ) ] dy. Hence, 1 υ (x, x ) x C i=1 1 x B 1 (,x ) g(y 1,..., y i 1, x I,..., x 1 ) dy. g(y1,..., y i, x i+1,..., x 1 ) By raising to the power p, integrating over R + and by following the exact same procedure as for υ x i (x ) we obtain: p υ R 1 (x) g(z + he x dx C i) g(z ) p dz dh. (3.7) h p R + 26

28 Trace Spaces of Sobolev Spaces For x = (x, x ) R + we define u(x) := e x p υ(x). Then by (3.4) and Tonelli s theorem we get u(x) p dx = e x υ(x, x ) p dx dx R + R 1 e x dx g(x ) p dx R 1 = g(x ) p dx. R 1 since for i = 1,..., 1 we have u (x) x i = e x p υ (x) x i υ (x) x i, by (3.6) it follows u (x) x i R + Moreover, p dx C u (x) = e x p x so by (3.4) and (3.7) we have ( R + u (x) x p R 1 g(x + he i) g(x ) p dx dh. h p υ (x) 1 x p x p e υ(x), 1/p ( p 1/p ( dx) υ (x) R x + dx) + 1 R + { ( R 1 g(z + he C i) g(z ) p In other words u W 1 p (R +) and h p ( ) 1/p } + g(x ) p dx <. R 1 x p p e ) p 1/p υ(x) dx ) 1/p dz dh u W 1 p (R + ) C g B 1 1/p p (R 1 ). (3.8) It remains to show that T u = g. By Proposition (2.3) for p we know that there exists a sequence 27

29 Trace Spaces of Sobolev Spaces {g n } in C (R 1 ) Bp 1 1/p (R 1 ) such that g g n. 1 1/p B p (R 1 ) Let u n := e x p υ n (x) be the corresponding sequence. We have that υ n C (R +) with υ n (x, ) = g n (x ), which follows from the fact that υ n = 1 φ x R ( ) x y 1 1 gn x (y )dy is defined via convolusion with parameter ε = 1 x (as ε ). By (3.8) we have that u n u in Wp 1 (R +) and since T : Wp 1 (R +) L p (R 1 ) is a continuous operator we have that T u = g. Therefore, B 1 1/p p (R 1 ) T W 1 p (R +). (3.9) Therefore by (3.3), (3.9) it follows T W 1 p(r + ) = B 1 1/p p (R 1 ). ow we wish to define the trace spaces for the case in which R is replaced with R an open set. Once more we will divide the proof in two parts. For the inclusion T Wp() l Bp l 1/p ( ) we will prove that for all u Wp() l there exists a trace g Bp l 1/p (, ), l > 1/p. Theorem 3.4. Let l, 1 < p < and R an open set with C l boundary. Then, for l > 1/p we have T W l p() B l 1/p p ( ). (3.1) Proof. Let V j be parallelepipeds defined as by Burenkov [1] (pg.149) and partition of unity ψ j C (V j ), j = 1,..., s. First we will prove that the trace g j of uψ j exists on V j and g j Bp l 1/p (V j ), j = 1,..., s. By Lemma 2.6 we get: uψ j W l p () c 1 u W l p (suppψ j ) c 1 u W l p (V j ), (3.11) where c 1 > independent of u. In other words we have that uψ j W l p(v j ), for all j = 1,..., s. By Lemma 2.7 it follows (uψ j )(Λ 1 j ) W l p (Λ j (V j )) c 2 (uψ j )(Λ 1 j )Λ j W l p (V j ) = c 2 uψ j W l p (V j ), 28 (3.12)

30 Trace Spaces of Sobolev Spaces where Λ j = Φ j (λ j ) (see Definition 2.16). Therefore we have that (uψ j )(Λ 1 j ) Wp(Λ l j (V j )). We extend by to R the function (uψ j )(Λ 1 j ) and for this extension we have that (uψ j )(Λ 1 j ) Wp(R l ), since supp(ψ j )(Λ 1 j ) Λ j (V j ). Then by Theorem 2.14 we know that there exists a trace h j (of the extended function) on R 1 and therefore on Λ j (V j ). That means that g j := h j (Λ j ) is a trace of uψ j on V j and as a result by Definition 2.13 it follows that g := s uψ j = u s ψ j = u). j=1 j=1 For l > 1/p we have that g j B l 1/p p s g j j=1 is a trace of u on (since = h (V j ) j h l 1/p B (Λ j (V j )) j c l 1/p B (R 1 ) 3 uψ j l 1/p B p where in the last inequality we have used (3.12). Then by Definition 2.16 it follows g B l 1/p p ( ) = = ( s j=1 ( s j=1 c 3 ( s j=1 p gψ j p Bp l 1/p (V j ) g j p Bp l 1/p (V j ) uψ j p W l p(v j ) ) 1 p ) 1 p ) 1 p. p (V j ), Therefore, Lemma 2.6 gives us g B l 1/p p ( ) c 4 ( s j=1 u p W l p(v j ) ) 1 p s c 5 j=1{ u p L p (V j ) + a u p L p (V j ) } a =l where in the last inequality we have used an equivalent norm to the standard one in W l p(v j ). 1 p, 29

31 Trace Spaces of Sobolev Spaces With use of Minkowski s inequality for sums we get g B l 1/p p ( ) c 5 { ( s j=1 u p L p (V j ) ) 1 p + a =l c 5 ( s 1 p u p L p () + a =l s 1 p a u p L p () c 6 s 1 p u W l p (). Hence, it follows that g Bp l 1/p ( ). s a u p L p (V j ) j=1 ) 1 p } For the inverse inclusion it suffices to show that for all g Bp l 1/p ( ) there exists a function u Wp() l such that g is a trace of u on. Hence we have the following theorem. Theorem 3.5. Let l, 1 < p < and R an open set with C l boundary. Then, for l > 1/p we have B l 1/p p ( ) T W l p(). (3.13) Proof. Let g Bp l 1/p ( ). We consider the functions g j = (gψ j )(Λ 1 on Λ j (V j ) and we extend them by to R 1. We define E : B l 1/p p ( ) W l p() with j ) Eg = s j=1 ( E ( (gψj )(Λ 1 j ) )) (Λ j ), where E is a modification of the extension operator such that supp(e g j (Λ j )) V j and supp(e g j ) Λ j (V j ). Then by Lemma 2.7 it follows E g j (Λ) W l p (R ) = (E g j )(Λ j ) W l p (V j ) M 1 E g j W l p (Λ j (V j )) = M 1 E g j W l p (R ), 3

32 Trace Spaces of Sobolev Spaces where M 3 is independent of g and j. We have that g j is a trace of E g j on R 1 (see details in [1]), therefore g j (Λ 1 j ) = (gψ j )(Λ 1 j ) on Λ j (V j ). Then ( (gψ j )(Λ 1 j ) ) (Λ j ) = gψ j is a trace of ( E (gψ j )(Λ 1 j ) ) (Λ j ) on V j and by definition 3 g = s gψ j j=1 s ( is a trace of E (gψ j )(Λ 1 ) ) (Λ j ) = Eg on s V j =. We have j=1 j=1 E g j W l p (R ) M 2 g j l 1/p B p (R 1 ) M 3 g j l 1/p B p (V j ) M 4 gψ j (Λ 1 j ) B l 1/p p (Λ j (V j )), where the last two inequalities follow from Lemma 2.5 and 2.7 respectively and M 2, M 3, M 4 are constants independent of g, j. Then Hence, Eg W l p (R ) = M 3 s (E g j (Λ j )) j=1 W l p (R ) s E g j W l p (R ) j=1 M 3 M 4 s j=1 gψ j l 1/p B p (V j ) Eg W l p (R ) M 5 g B l 1/p p ( ). This means that the extension operator E is bounded (and linear) and since T Eg = g we have that Bp l 1/p ( ) T Wp(). l Therefore it follows T Wp() l = Bp l 1/p ( ). 31

33 Trace Spaces of Sobolev Spaces Remark. For l = 1 and p we have that T W1 1 () = L 1 ( ) (for the proof we refer to [1]). For the case p = let us consider for simplicity that l = 1, i.e. the space W (). 1 Then since is open, bounded and with of class C 1 it follows that W () 1 is the space of all Lipschitz functions (this follows from the characterization of W () 1 in [5]). ow let u : R be such that u W (). 1 Since u is bounded and Lipschitz, it admits a unique Lipschitz extension to. Hence, the trace of u can be defined as the restriction of the extension of u to. Thus, the trace of u is a Lipschitz function on. Therefore, it follows that T W () 1 Lip( ), where Lip( ) the space of Lipschitz continuous functions defined on. As a matter of fact T W () 1 = Lip( ). Indeed let u Lip( ), then as proven in [11], u can be extended to a Lipschitz function on the whole R therefore on and we have that W () 1 is exactly the space of all Lipschitz functions defined on. 32

34 Chapter 4 The case p = 2: H s spaces and their traces As mentioned above, when p = 2 we denote H s = W s 2. We will see in the following sections that for this case we can skip the classical definition of the Trace Spaces, that is with use of Besov spaces, and use alternative ways to define the trace spaces. 4.1 Characterization via Fourier transform A way to characterize the trace spaces is with use of Fourier transform. In order to do so we will need several useful definitions and results which are presented bellow. Definition 4.1. We call Schwartz class and write S(R ) the class of functions φ C (R ) such that for any multi-index α and any k sup (1 + x ) k α φ(x) <. x R Definition 4.2. (Fourier Transform) Let u S(R ). We define the Fourier transformation F : u û by F {u(x)} = û(ξ) = (2π) 2 u(x)e ix ξ dx. R 33

35 The case p = 2: H s spaces and their traces It is known that if u S(R ) then û S(R ). Therefore F : S(R ) S(R ) is a linear operator. For the inverse transformation F 1 : S(R ) S(R ) we have the following formula: u(x) = (2π) 2 û(ξ)e ix ξ dξ. R The Fourier transform can be extended by continuity to L 2 (R ) and in fact F : L 2 (R ) L 2 (R ) is a unitary operator : u(x) 2 dx = û(ξ) 2 dξ, R R that is u L 2 (R ) = F u L 2 (R ), which is also known as Parseval s equality. ow we will use the Fourier transform in order to characterize the Sobolev spaces H s (). For the derivatives α u(x) we have F { α u(x)} = α u(ξ) = (iξ) α û(ξ) = i α ξ α û(ξ). Hence, u 2 H l (R ) = α u 2 dx = α l R R ( ξ α ) û(ξ) 2 2 dξ. (4.1) Here we notice that if we expand A(ξ) = (1 + ξ 2 l by the binomial expansion, that is A(ξ) = (1 + ξ 2 ) l = )( ) ξ 2 k, we can choose appropriate constants c 1, c 2 such that c 1 (1 + ξ 2 ) l α l α l k l ( l k ξ α 2 c 2 (1 + ξ 2 ) l, which when applied in (4.1) gives us c 1 R (1 + ξ 2 ) l û(ξ) 2 dξ u 2 H l (R ) c 2 R (1 + ξ 2 ) l û(ξ) 2 dξ. Thus, the norm ( (1+ ξ 2 ) l û(ξ) 2 dξ ) 1 2 is equivalent to the standard R one in Wp(R l ). With that being said we have the following definition of H s (R ) spaces. 34

36 The case p = 2: H s spaces and their traces Definition 4.3. Let u L 2 (R ) and s R. We say that u H s (R ) if R (1 + ξ 2 ) s û(ξ) 2 dξ <. As we have proven in the previous chapter, Besov spaces allow us to define the traces of Sobolev spaces and naturally this also applies in the case of H s = W s 2 spaces. However, the previous representation of H s spaces via Fourier transform allows us to have an additional approach on how to define the trace spaces. More specifically, we have the following theorems. Theorem 4.1. Let u H s (R ) and s > 1/2. Then the trace operator T : C (R ) C (R 1 ) can be extended uniquely by continuity to a linear continuous operator T : H s (R ) H s 1/2 (R 1 ). In particular, T u H s 1/2 (R 1 ) C u H s (R ). Proof. Let u C (R ). Let x = (x 1, x 2,..., x 1 ) R 1 then we have the inverse Fourier transform formula: u(x) = u(x, x ) = (2π) /2 û(ξ, ξ )e ix ξ e ix ξ dξ dξ. R By definition of the Trace operator we get T u(x ) = u(x, ) = (2π) /2 û(ξ, ξ )e ix ξ dξ dξ. R That is u(x, ) = (2π) /2 û(ξ, ξ )e ix ξ dξ dξ R + = (2π) /2 û(ξ, ξ )e ix ξ dξ dξ R 1 and by Fubini-Tonelli Theorem we get + u(x, ) = (2π) /2 û(ξ, ξ )dξ e ix ξ dξ. We write T u(x ) = (2π) ( 1)/2 R 1 R 1 ( 1 + û(ξ, ξ )dξ )e ix ξ i dξ, 2π 35

37 The case p = 2: H s spaces and their traces which by the inverse Fourier formula gives us We have T u(ξ ) 2 = 1 2π T u(ξ ) = 1 + û(ξ, ξ )dξ. 2π + + (1 + ξ 2 ) s/2 (1 + ξ 2 ) s/2 û(ξ, ξ )dξ 2 û(ξ) 2 (1 + ξ 2 ) s dξ } {{ } I 1 + (1 + ξ 2 + ξ) 2 s dξ, } {{ } I 2 (4.2) where in the last inequality we have used Hölder s inequality. For the calculation of I 2 we set 1 + ξ 2 = α 2 and get I 2 = y ξ = α + α α 2s 1 (α 2 + ξ dξ 2 )s = (1 + y 2 ) s dy = c sα 1 2s, where c s = + 1 < since s > 1. (1+y 2 ) s 2 Therefore, I 2 = c s (1 + ξ 2 ) 1 2s. Then by (4.2) we have 1 α 2s (1 + ( ξ α )2 ) dξ s + (1 + ξ 2 ) s 1/2 T u(ξ ) 2 c s û(ξ) 2 (1 + ξ 2 ) s dξ. We integrate over R 1 R 1 (1 + ξ 2 ) s 1/2 T u(ξ ) 2 dξ c s Thus for all u C (R ) R û(ξ) 2 (1 + ξ 2 ) s dξ. T u 2 H s 1/2 (R 1 ) c s u 2 H s (R ). (4.3) We note that the norm u(ξ) = ( R (1+ ξ 2 ) s û(ξ) 2 dξ) 1/2 is equivalent to the standard norm in H s (R ). 36

38 The case p = 2: H s spaces and their traces ow we will extend the previous result for functions u in C (R ) to functions u in H s (R ). In order to do so, we will use the fact that C (R ) is dense in H s (R ), as follows. Let u H s (R ). Then there exists {u j } j=1 in C (R ) such that Then (4.3) gives us that as m, l. u j u H s (R ), j. T u m T u l 2 H s 1/2 (R 1 ) c s u m u l 2 H s (R ), In other words {T u j } j is a Cauchy sequence in H s 1/2 (R 1 ) and since H s 1/2 (R 1 ) is complete in R 1 there exists a limit υ H s 1/2 (R 1 ), T u j υ, j in H s 1/2 (R 1 ) and by definition we set υ = T u. We note that υ does not depend on the choice of the sequence {u j }. Indeed, let {w j } j=1 in C (R ) be such that w j u H s (R ) as j. Then, as above, {w j } j=1 is a Cauchy sequence and for w H s 1/2 (R 1 ) such that T w j w, j, we have: w j u j H s (R ) w j u H s (R ) + u u j H s (R ), as j. Hence, T w j T u j 2 H s 1/2 (R 1 ) = T (w j u j ) 2 H s 1/2 (R 1 ) c s w j u j H s (R ). Since u j C (R ) for all j, by 4.3 it follows We take the limit T u j 2 H s 1/2 (R 1 ) c s u j 2 H s (R ). lim T u j 2 j H s 1/2 (R 1 ) c s lim, u j 2 j H s (R ) which gives us with u H s (R ). T u 2 H s 1/2 (R 1 ) c s u 2 H s (R ), 37

39 The case p = 2: H s spaces and their traces As a result of the previous Theorem we have H s 1/2 (R 1 ) T H s (R ). For the inverse inclusion we have the following Theorem. Theorem 4.2 (Extension Theorem). For k Z +, s > k + 1/2 we set H s 1/2 (R 1 ) = H s 1/2 (R 1 ) H s 3/2 (R 1 ) H s k 1/2 (R 1 ). Then there exists a linear continuous operator E : H s 1/2 (R 1 ) H s (R ) such that if u = Eg with g = (g, g 1,, g k ) H s 1/2 (R 1 ) then g j = T j u, for all j =, 1,, k, where T j := T j : H s (R ) x j H s j 1/2 (R 1 ). Moreover, k Eg 2 H s (R ) c g 2 H s 1/2 (R 1 ) := c g j 2 H s j 1/2 (R 1 ) Proof. Let h C (R) with h(t) 1 and h(t) = 1 for t 1. For ξ R 1, x R we consider: j= V (ξ, x ) = k j= 1 ) j! xj ĝj(ξ )h (x 1 + ξ 2, where ĝ j (ξ ) is the Fourier transform of g j (x ) over g = (g,..., g k ) H s 1/2 (R 1 ). We will prove that V (ξ, x ) is the Fourier transform of a function u(x, x ) H s (R ) with respect to x. For x = we have V (ξ, ) = ĝ (ξ ) and j x V (ξ, ) = ĝ j (ξ ). We use the following identities regarding the Fourier transform: F {x j g(x )} = i j ĝ (j) (ξ ) = i j j ĝ(ξ ξ j ) F {g(αx )} = 1 αĝ( ξ α ), α R F {x j g(αx )} = i j 1 α j+1 ĝ (j) ( ξ α ), α R. 38

40 The case p = 2: H s spaces and their traces It follows that the Fourier transform of V (ξ, x ) with respect to x is F {V (ξ, x )} = = k j= k j= 1 j!ĝj(ξ )F {x (x j h )} 1 + ξ 2 i j ( j!ĝj(ξ )(1 + ξ 2 ) j+1 2 ĥ (j) ξ ) 1 + ξ 2 For û(ξ) = V (ξ, ξ )(= F {V (ξ, x )}) we have that u H s (R ). Indeed, u 2 H s (R ) = R û(ξ) 2 (1 + ξ 2 ) s dξ We write I = R n 1 c R k ( ĝ j (ξ ) 2 (1 + ξ 2 ) j 1 ξ ) 2 ĥ(j) (1 + ξ 2 ) s dξ. R 1 + ξ 2 }{{} I j= ( ĝ j (ξ ) 2 (1 + ξ 2 ) j 1 ĥ (j) ξ ) 2 (1 + ξ 2 ) s dξ n dξ 1 + ξ 2 = ĝ j (ξ ) 2 (1 + ξ 2 ) j 1 R 1 For the calculation of I 1 we set τ = ( ĥ(j) R ξ ) 2 (1 + ξ 2 ) s dξ dξ. 1 + ξ 2 } {{ } I 1 ξ 1+ ξ 2. We have dξ = 1 + ξ 2 dτ and 1 + ξ 2 = 1 + ξ 2 + ξ 2 = 1 + ξ 2 + τ 2 (1 + ξ 2 ) = (1 + ξ 2 )(1 + τ 2 ), then I 1 = ĥ(j) (τ) 2 (1 + ξ 2 ) s (1 + τ 2 ) s (1 + ξ 2 ) 1/2 dτ R = (1 + ξ 2 ) s+1/2 (τ) 2 (1 + τ 2 ) s dτ, R } {{ } C(j, s) 39

41 The case p = 2: H s spaces and their traces where C(j, s) < since ĥ S(R) for h C (R). Hence, u 2 H s (R ) C = C k j= R 1 ĝ j (ξ ) 2 (1 + ξ 2 ) s j 1/2 dξ k g j 2 H s j 1/2 (R 1 ). j= Therefore the operator E : H s 1/2 (R 1 ) H s (R ) with Eg = u is linear and continuous and we now prove that T u j = g j, for all j =, 1,..., k. This is equivalent to proving that the Fourier Transform of j u x j with respect to x is ĝ j (ξ ). Denote by F the Fourier Transform on the variable x. We have F ( j u ) = j F u x j x j resumed. As a result we have T H s (R ) H s 1/2 (R 1 ). Therefore, T H s (R ) = H s 1/2 (R 1 ). = j V (ξ,x ) x j = ĝ j (ξ ), as 4.2 Auchmuty s method Let be a bounded region in R and let its boundary be a finite union of disjoint closed Lipschitz surfaces, each surface having finite surface area and unique outward normal ν( ) defined almost everywhere. We consider the inner product: u, υ θ = u υdx + uυdσ and the corresponding norm θ which is equivalent to the standard one in H 1 (). A function u H 1 () is called harmonic in provided that it is a solution of Laplace s equation in the weak sense. amely: u φdx =, (4.4) 4

42 The case p = 2: H s spaces and their traces for all φ C 1 c (), or equivalently for all φ H 1 (). We define H() to be the space of all harmonic functions u in. When is defined as above we have by definition C 1 c () = H 1 () and by (1) we have (C 1 c ()) = H(). Then by the Projection Theorem on the Hilbert space H 1 () endowed with, it follows Therefore, H 1 () = C 1 c () θ (C 1 c ()) H 1 () = H 1 () θ H(), (4.5) where θ is a θ-orthogonal direct sum. Auchmuty s method gives us a spectral definition of the trace spaces HA s ( ), s R, where the subscript A is here used to emphasize the fact here the definition is given for this method. The idea is based on the fact that the harmonic Steklov eigenfunctions provide an orthogonal basis in H() as well as an orthogonal basis in L 2 (, dσ). More specifically we can define an orthonormal basis in L 2 (, dσ) by means of the Steklov eigenfunctions and then use this basis to represent the trace spaces. In the end the definition of HA s ( ) spaces is reduced to certain summability conditions regarding the harmonic Steklov coefficients which, as described above, help us represent the functions g HA s ( ) The harmonic Steklov eigenproblem Let be a region in R defined as above. We consider the boundary value problem: { s =, x (4.6) D ν s = δs, x for δ R. Recall that D ν s denotes the normal derivative of s on. In order to find the weak formulation of (4.6) we argue as follows. We assume that s H 1 () is a classical solution to (4.6). Then by integrating over we get sυdx =, 41

43 The case p = 2: H s spaces and their traces for all υ H 1 (). By Green s Formula it follows s υdx = υd ν sdσ, Finally, by using the boundary condition we get the weak form of (4.6): s υdx = δ υsdσ, (4.7) for all υ H 1 (). For υ = s we get that δ. We rewrite (4) in the following form s υdx + υsdσ = (δ + 1) υsdσ. (4.8) We consider the Laplace operator as an operator from H 1 () to its dual H 1 () defined by [s][υ] = s υdx, for all υ H1 () and the operator I : H 1 () H 1 () defined ny I[s][υ] = sυdσ, for all υ H 1 (). The operator I : H 1 () H 1 () with (I )[s][υ] = s υdx + sυdσ = s, υ θ is an isometry and there exists (I ) 1 (by Riesz s Representation Theorem). We also define the operator J : L 2 ( ) H 1 () with J[s][υ] = sυdσ, for all υ L 2 ( ), and lastly we consider the trace operator T : H 1 () L 2 ( ). Then (4.8) can be written as Therefore, ote that (I )s = (δ + 1)(J T )s. (I ) 1 J T s = 1 δ + 1 s. H 1 () T L 2 ( ) J H 1 () (I ) 1 H 1 (). Hence we can define the operator M : H 1 () H 1 () with M = (I ) 1 J T. Then the weak form (4.7) of the Steklov eigenproblem is equivalent to Ms = µs, (4.9) 42

44 The case p = 2: H s spaces and their traces where µ = 1 δ+1 >. We notice that M is compact by definition, since the trace operator T : H 1 () L 2 ( ) is compact, and also M is non-negative. Let s KerM, then (I ) 1 J T s =, which, since (I ) is an isometry, is equivalent to J T s = and by definition of J T s =. By Theorem (2.16) we have that T s = iff s H(). 1 KerM = H 1 (). Hence, by (4.5) we obtain Therefore, H 1 () = KerM θ H(). (4.1) Moreover, we have that the codimension of KerM, codimkerm = dim(h 1 ()/H()), is infinite and since M is a non-negative, compact, self-adjoint operator in a Hilbert space we have that the spectrum σ(m) is discrete and σ(m) \ {} consists of eigenvalues {µ j } of finite multiplicity with µ j, as j. More specifically by the Courant-Raylegh minmax principle we have µ j = max min V H() s V \{} dimv=j = max min V H() s V \{} dimv=j Ms, s θ s θ (I 1 ) J T s, s θ s θ (I ) (I ) 1 J[T s][s] = max min V H() s s θ V H 1()={} dimv=j Therefore for the Steklov eigenvalues δ j = 1 µ j 1 we have δ j = min max s 2 dx V H() s s2 dσ. V H 1()={} dimv=j 43

45 The case p = 2: H s spaces and their traces By the Spectral Theorem for compact and self-adjoint operators there exists an orthonormal set of eigenfunctions {s j } j= corresponding to the eigenvalues {µ j } j= such that (KerM) = span s j : j. Then by (4.1) it follows that S = {s j basis for H(). : j } is an orthonormal Definition 4.4. We call a Steklov expansion an expression of the form u(x) = c j s j, where c j := u, s j θ, for u H 1 (). j= Since S = {s j : j } is an orthonormal basis in H() by Theorem (2.11) (which implies the existence of limiting function iff the coefficients are square summable) we have that a Steklov expression represents a H 1 -harmonic function on iff c j 2 < Spectral representation of the trace and the extension operator The Steklov eigenfunctions s j (as described above) have L 2 traces on the boundary whenever is defined as in the beggining of this section. We now define j= ŝ j (x) := 1 + δ j T s j (x), for x and j. Then, as proven in [4], Ŝ = {ŝ j : j } is an orthonormal basis in L 2 (, dσ). Since T is continuous for all u H 1 () we have T u = T u, s j θ s j = j= u, s j θ T s j, j= that is T u = (1 + δ j ) 1/2 u, s j θ ŝ j, j= 44