Sobolev Spaces. Chapter Hölder spaces

Transcription

1 Chapter 2 Sobolev Spaces Sobolev spaces turn out often to be the proper setting in which to apply ideas of functional analysis to get information concerning partial differential equations. Here, we collect a few basic results about Sobolev spaces. A general reference to this topic is Adams [1], Gilbarg- Trudinger [29], or Evans [26]. 2.1 Hölder spaces Assume R n is open and 0 < γ 1. Definition write (i) If u : R is bounded and continuous, we u C( ) := sup u(x). x (ii) The γ th -Hölder semi-norm of u : R is [u] C 0,γ ( ) := and the γ th -Hölder norm is { u(x) u(y) } sup x,y, x =y x y γ, u C 0,γ ( ) := u C( ) + [u] C 0,γ ( ). (iii) The Hölder space C k,γ ( ) 11

2 12 CHAPTER 2. SOBOLEV SPACES consists of all functions u C k ( ) for which the norm u C k,γ ( ) := D α u C( ) + [D α u] C 0,γ ( ) (2.1.1) is finite. α k α =k Theorem The space of functions C k,γ ( ) is a Banach space. Proof. It is easy to verify that C k,γ ( ) is a norm (Exercise). It suffices to check the completeness of C k,γ ( ). 1. Construction of limit functions. Let {u m } m=1 be a Cauchy sequence in Ck,γ ( ), i.e., u m u n C k,γ ( ) 0, as m,n +. Since any Cauchy sequence is a bounded sequence, so u m C k,γ ( ) is uniformly bounded. Particularly, u m C( ) is uniformly bounded, that is, {u m (x)} is a uniformly bounded sequence. Furthermore, the boundedness of Du m or C( ) [Dα u m ] C 0,γ ( ) implies the equicontinuity of the sequence {u m (x)}. The Arzela-Ascoli theorem implies there exists a subsequence, still denoted by {u m } m=1, which converges to a bounded continuous function u(x) uniformly in. The same argument can be repeated for any derivative of D α u m of order up to and including k. That is, there exist functions u α C( ), such that 2. D α u = u α. D α u m u α, uniformly in, α k. Deduce by induction, first let α = (1, 0,, 0), since the series u 1 (x) + + j=1 uniformly in and the series D α u 1 (x) + + j=1 (u j+1 u j ) = lim u m+1(x) = u(x), m (D α u j+1 D α u j ) = lim m Dα u m+1 (x) = u α (x),

3 2.1. HÖLDER SPACES 13 uniformly in. The theory of series implies that u(x) is differentiable in and D α u = u α. The induction method shows that D α u = u α is valid for all multi-index α such that α k. 3. u C k,γ ( ). It suffices to show that for any multi-index α such that α = k, there holds [D α u] C 0,γ ( ) < +. For any x,y, x y, since D α u m D α u uniformly in, there exists a N large enough, such that: Calculate: D α u N (z) D α u(z) 1 2 x y γ, z. D α u(x) D α u(y) D α u(x) D α u N (x) + D α u N (x) D α u N (y) + D α u N (y) D α u(y) x y γ + x y γ D α u N C 0,γ ( ) (1 + sup D α u m C 0,γ ( ) ) x y γ, m Thus the boundedness of D α u m C 0,γ ( ) implies [Dα u] C 0,γ ( ) < u m u in C k,γ ( ). It also suffices to show that for any multi-index α such that α = k, there holds [D α u m D α u] C 0,γ ( ) 0, as m. In fact, for any x,y, x y, there holds D α u m (x) D α u(x) (D α u m (y) D α u(y)) lim sup D α u m (x) D α u n (x) (D α u m (y) D α u n (y)) n x y γ lim n [Dα u m D α u n ] C 0,γ ( ). Since {u m } m=1 is a Cauchy sequence in Ck,γ ( ), [D α u m D α u] C 0,γ ( ) lim n [Dα u m D α u n ] C 0,γ ( ) 0 as m 0.

4 14 CHAPTER 2. SOBOLEV SPACES 2.2 Sobolev spaces L p spaces The L p () space consists of all the function u : R such that u p dx <, where 1 p ; with norm ( u p dx ) 1/p, if 1 p <, u L p () = esssup u, if p =. In order to deduce some properties of L p space, let us start from some basic inequalities, which will be used latter. 1 Lemma (Young inequality) Let 1 < p, q <, p + 1 q = 1, then ab ap p + bq, a,b > 0. q Proof. Method 1. The mapping x e x is convex, and consequently, ab = e log a+log b = e 1 p log ap + 1 q log bq 1 p elog ap + 1 q elog bq = ap p + bq q. Method 2. Let y = x p 1,x 0, then x = y 1 p 1 = y q 1, Consider the area of the region restricted by y = 0, x = a, and curve y = x p 1 is I = a 0 x p 1 dx = ap p ; and the area of the region restricted by x = 0, y = b, and curve x = y q 1 is II = b 0 y q 1 dy = bq q ; The area of the rectangle [0,a] [0,b] is ab, Thus ab I+II ap p + bq q, and the equality holds if and only if b = a p 1. It is easy to obtain the following Young inequality with ε (Exercise):

5 2.2. SOBOLEV SPACES 15 Lemma (Young inequality with ε) Let 1 < p, q <, 1 1 q = 1, then ab ε ap p + 1 b q ε q/p, a,b > 0. q p + Lemma (Hölder inequality) Let 1 p, q, 1 p + 1 q = 1, then if u L p (), v L q (), we have uv dx u L p () v L q (). Proof. 1. In the cases where p = or q =, it is easy, because there exists a subset, with =, such that sup u = u L () or sup v = v L (). 2. In the cases where 1 < p,q <. By the homogeneity of the inequality, we may assume that u L p () = v L q () = 1. Then the Young inequality implies that uv dx 1 p u p dx + 1 q u q dx = 1 = u L p () v L q (). By induction argument, it is easy to show the following Generalized Hölder inequality (Exercise): Lemma (Generalized Hölder inequality) Let 1 p 1,,p m 1, p p m = 1, then if u k L p k() for k = 1,,m, we have u 1 u m dx m Π k=1 u k L p k(). Lemma (Minkowski inequality) Assume 1 p, u, v L p (), then u + v L p () u L p () + v L p ().

6 16 CHAPTER 2. SOBOLEV SPACES Proof. From the Hölder inequality, we have u + v p L p () = u + v p dx u + v p 1 ( u + v )dx ( ) u + v (p 1) p p 1 ( p p 1 dx ( = u + v p 1 L p () ( u L p () + v L p ()). u p dx) 1/p + ( v p dx) 1/p) Lemma (Interpolation inequality for L p - norms) Assume 1 s r t, and 1 r = θ s + (1 θ) t. Suppose also u L s () L t (). Then u L r (), and u L r () u θ L s () u (1 θ) L t (). Proof. From the Hölder inequality, we have u r dx = u θr u (1 θ)r dx ( u θr ) s θr ( θr dx s u (1 θ)r t (1 θ)r dx ) (1 θ)r t. Theorem (Completeness of L p ()) L p () is a Banach space if 1 p. Proof. It is easy to check that L p () is a norm (Exercise). It suffices to verify the completeness of L p () 1. Construction of the limit function. Assume that 1 p < and let {u m } be a Cauchy sequence in L p (). There is a subsequence {u mj } of {u m } such that u mj+1 u mj p 1/2 j, j = 1,2.

7 2.2. SOBOLEV SPACES 17 Let v m (x) = m u mj+1 (x) u mj (x), then j=1 v m p m u mj+1 u mj p j=1 m 1/2 j < 1. Putting v(x) = lim v m(x), which may be infinite for some x, we obtain m by the Fatou Lemma that v(x) p dx lim inf v m (x) p dx 1. m Hence v(x) < a.e. in and its absolutely convergence implies the series u m1 + (u mj+1 (x) u mj (x)) (2.2.1) j=1 converges to a limit u(x) a.e. in. Let u(x) = 0 whenever it is undefined as the limit of series (2.2.1). Thus j=1 lim u m j (x) = u(x) a.e. in. j 2. u L p () and u m u in L p (). For any ε > 0, there exists N such that if m,n N, then u m u n p < ε. Hence by the Fatou Lemma again, we have u m (x) u(x) p dx = lim u m(x) u mj (x) p dx j lim inf u m (x) u mj (x) p dx j ε p. If m N, thus u = (u u m ) + u m L p (), and u m u p 0 as m. 3. In the case where p =, let {u m } be a Cauchy sequence in L (), then there exists a subset A, with A = 0, such that if x \ A, then for n,m = 1,, there holds u m (x) u m, u m (x) u n (x) u m u n. That is, u m (x) is uniformly bounded in \ A, and u m (x) is a Cauchy sequence for any x \A, thus u m (x) converges uniformly on \A to

8 18 CHAPTER 2. SOBOLEV SPACES a bounded function u(x). Set u(x) = 0 if x A. We have u L () and u m u 0 as m. Theorem (Riesz Representation Theorem for L p ()) Let 1 < p < and L [L p ()] 1. Then there exists v L p (), p + 1 p = 1 such that for all u L p (), there holds L(u) = u(x)v(x) dx. Moreover, v L p () = L [L p ()], that is, [L p ()] L p (). Theorem (Riesz Representation Theorem for L 1 ()) Let L [L 1 ()]. Then there exists v L () such that for all u L 1 (), there holds L(u) = u(x)v(x) dx. Moreover, v L () = L [L 1 ()], that is, [L 1 ()] L (). Theorem (Reflexivity of L p ) L p () is reflexive if and only if 1 < p <. Theorem C 0 () is dense in Lp () if 1 p <. Theorem (Separability of L p ()) L p () is separable 1 p < Weak derivatives Let C0 () denotes the space of infinitely differentiable functions φ : R, with compact support in. We will call a function φ belonging to C0 () a test function.

9 2.2. SOBOLEV SPACES 19 Motivation for definition of weak derivatives. Assume we are given a function u C 1 (). Then if ϕ C0 (), we see from the integration by parts that uφ xi dx = u xi φdx (i = 1,,n). (2.2.2) More general now, if k is a positive integer, u C k (), and α = (α 1,,α n ) is a multi-index of order α = α α n = k, then u(d α φ)dx = ( 1) α (D α u)φdx. (2.2.3) We next examine that formula (2.2.3) is valid for u C k (), and ask whether some variant of it might still be true even if u is not k times continuously differentiable. Note that the left-hand side of (2.2.3) make sense if u is only locally summable: the problem is rather that if u is not C k, then the expression D α u on the right-hand side of (2.2.3) has no obvious meaning. We overcome this difficulty by asking if there exists a locally summable function v for which formula (2.2.3) is valid, with v replacing D α u: Definition Suppose u,v L 1 loc (), and α is a multi-index. We say that v is the α th -weak partial derivative of u, written provided that D α u = v, ud α φdx = ( 1) α for all test functions ϕ C 0 (). vφ dx (2.2.4) Lemma (Uniqueness of weak derivative) The weak partial derivative of u, if it exists, is uniquely defined up to a set of measure zero. Example Let n = 1, = (0,2), and { x, if 0 < x 1, u(x) = 1, if 1 x < 2.

10 20 CHAPTER 2. SOBOLEV SPACES Define { 1, if 0 < x 1, v(x) = 0, if 1 x < 2. From the definition , u = v. (Exercise) Example Let n = 1, = (0,2), and { x, if 0 < x 1, u(x) = 2, if 1 x < 2. We assert u does not exist in the weak sense. (Exercise) Definition of Sobolev spaces Fix 1 p and let k be a nonnegative integer. We define now certain function spaces, whose members have weak derivatives of various orders lying in various L p spaces. Definition i) The Sobolev space W k,p () consists of all locally summable functions u : R such that for each multi-index α with α k, D α u exists in the weak sense and belongs to L p (). ii) If u W k,p (), we define its norm to be ( D α u p dx ) 1/p, (1 p < ), u W k,p () := α k ess sup D α u, (p = ). α k Definition (Convergence in W k,p ()) i) Let {u m } m=1, u W k,p (). We say u m converges (strongly) to u in W k,p (), written u m u in W k,p (), provided that lim u m u m W k,p () = 0.

11 2.2. SOBOLEV SPACES 21 ii) We write to mean for each U. u m u in W k,p loc () u m u in W k,p (U), Definition We denote by the closure of C 0 () in W k,p (). W k,p 0 () Remark (i) If p = 2, we usually write H k () = W k,2 (), H k 0() = W k,2 0 (), (k = 0,1, ). The letter H is used, since - as we will see - H k () is a Hilbert space. Note that H 0 () = L 2 (). (ii) We henceforth identity functions in W k,p () which agree almost everywhere. Example Let = B(0,1), the open unit ball in R n, and u(x) = x α (x,x 0), α > 0. Then u W 1,p () if and only if 0 < (α + 1)p < n. In particular u W 1,p () for each p n. (Exercise) Example Let {r k } k=1 B(0,1) R n. Set be a countable, dense subset of = u(x) = k=1 1 2 k x r k α (x ). Then u W 1,p () if and only if (α + 1)p < n. If 0 < (α + 1)p < n, we see that u belongs to W 1,p () and yet is unbounded on each open subset of. (Exercise) Theorem (Properties of weak derivatives) Assume u,v W k,p (), α k. Then

12 22 CHAPTER 2. SOBOLEV SPACES (i) D α u W k α,p () and D β (D α u) = D α (D β u) = D α+β u for all multi-indices α, β with α + β k; (ii) For each λ, µ R, λu + µv W k,p () and D α (λu + µv) = λd α u + µd α v, α k; (iii) If is an open subset of, then u W k,p ( ); (iv) If ξ C 0 (), then ξu W k,p () and D α (ξu) = β α ( ) α D β ξd α β u (Leibniz formula), β where ( ) α = β α! β!(α β)!. Theorem (Sobolev spaces as function spaces) For each k = 0,1, and 1 p, Sobolev space W k,p () is a Banach space; W k,p () is reflexive if and only if 1 < p < ; For 1 p <, the subspace W k,p C () is dense in W k,p (), thus W k,p () is separable; Moreover, W k,2 () is a Hilbert space with scalar product (u, v) W k,2 = α k D α ud α v dx. Proof. 1. It is easy to check that W k,p () is a norm (Exercise). Next to show that W k,p () is complete. 2. Construction of limit functions. Assume {u m } m=1 is a Cauchy sequence in W k,p (). Then for each α k, {D α u m } m=1 is a Cauchy sequence in Lp (). Since L p () is complete, there exist functions u α L p (), such that for each α k. In particular, D α u m u α, in L p (), u m u (0,,0) =: u in L p (). 3. u W k,p () and D α u = u α, α k.

13 2.2. SOBOLEV SPACES 23 In fact, let ψ C0 (), then ud α ψ dx = lim m 4. u m u in W k,p (). u m D α ψ dx = lim m ( 1) α (D α u m )ψ dx = u α ψ dx. Since D α u m u α = D α u in L p () for all α k, it follows that u m u in W k,p (). 0 α k 5. Regard W k,p () as a closed subspace of a Cartesian product space of spaces L p (): Let N = N(n,k) = 1 be the number of multiindexes α satisfying 0 α k. For 1 p, let L p N = Π N L p (), j=1 the product norm of u = (u 1,,u N ) in L p N being given by ( N u j p 1/p, p) if 1 p <, u L p = j=1 N max u j, if p =. 1 j N Since a closed subspace of a reflexive (Resp. separable, uniformly convex) Banach space is also reflexive (Resp. separable, uniformly convex), thus W k,p () is reflexive if and only if 1 < p < ; and W k,p () is separable if 1 p <. Furthermore, ( C () ) N is dense in L p N and thus in W k,p () if 1 p <. 6. If p = 2, define the scalar product as (u,v) W k,2 = D α ud α v dx. α k It is easy to verify W k,2 = H k is a Hilbert space Inequalities Theorem (Gagliardo-Nirenberg-Sobolev inequality) Let 1 p < n, p = np. There exists a constant C, depending only on n p

14 24 CHAPTER 2. SOBOLEV SPACES p and n, such that for all u C 1 0 (Rn ). Proof. 1. First assume p = 1. u L p (R n ) C Du L p (R n ), (2.2.5) Since u has compact support, for each i = 1,2,n and x R n we have and so u(x) u(x) = + Consequently n n u(x) n 1 Π xi u xi (x 1,,x i 1,y i,x i+1,,x n )dy i, Du(x 1,,x i 1,y i,x i+1,,x n ) dy i, i = 1,2,n. i=1 ( + Integrate this inequality with respect to x 1 : + u(x) + n Π i=1 ( + = + n n 1 dx1 n Π i=2 ( + ( + Du(x 1,,x i 1,y i,x i+1,,x n ) dy i ) 1 n 1. Du(x 1,,x i 1,y i,x i+1,,x n ) dy i ) 1 n 1 dx 1 Du(y 1,x 2,,x n ) dy 1 ) 1 n 1 ( + Du dy 1 ) 1 n 1 ( n Π i=2 Du(x 1,,x i 1,y i,x i+1,,x n ) dy i ) 1 n Du dx 1 dy i ) 1 n 1, (2.2.6) the last inequality follows from the Generalized Hölder inequality. Now integrate (2.2.6) with respect to x 2 : + + ( + n u(x) n 1 dx1 dx 2 + ) 1 + n 1 Du dx 1 dy 2 n Π i=1,i 2 1 n 1 Ii dx 2, dx 1

15 2.2. SOBOLEV SPACES 25 where I 1 = + Du dy 1, I i = + + Du dx 1 dy i,i = 3,,n. Applying once more the Generalized Hölder inequality, we find + + ( + n Π i=3 u(x) n n 1 dx1 dx 2 + ( + Du dx 1 dy 2 ) ( + + n 1 Du dx 1 dx 2 dy i ) 1 n 1. Du dy 1 dx 2 ) 1 n 1 We continue by integrating with respect to x 3,,x n, eventually to find n n ( + + ) 1 n 1 u(x) n 1 dx Π Du dx 1 dx n R n i=1 ( ) n n 1 = Du dx. R n This is conclusion for p = 1. (2.2.7) 2. Consider now the case that 1 < p < n. We apply (2.2.7) to v := u γ, where γ > 1 is to be selected. Then ( Rn ) u(x) γn n 1 n n 1 dx We choose γ so that D u γ dx = γ u γ 1 Du dx R n R n γ ( R n u (γ 1) p p 1 dx ) p 1 p ( R n Du p) 1 p. γn n 1 = (γ 1) p. That is, we get p 1 γ = p(n 1) n p γn in which case n 1 = (γ 1) p p 1 = becomes (2.2.5). > 1, (2.2.8) np n p = p. Thus, (2.2.8)

16 26 CHAPTER 2. SOBOLEV SPACES Remark The exponent p = np is called critical Sobolev n p exponent, because it is the unique exponent q = p to make the following inequality: u L q (R n ) C Du L p (R n ) be true for all u C 1 0 (Rn ). Proof. For any u C 1 0 (Rn ) and λ > 0, define the scaled function u λ (x) = u(λx). Direct calculations show that and u λ L q (R n ) = λ n q u L q (R n ) Du λ L p (R n ) = λ 1 n p Du L p (R n ). If the inequality in remark is true for all u C 1 0 (Rn ), replacing u by u λ, there holds u L q (R n ) Cλ 1 n p + n q Du L p (R n ). If 1 n p + n q 0, upon sending λ to either 0 or, we will get C =. Theorem (Poincare inequality for W 1,p 0 ()) Assume is a bounded, open subset of R n, Suppose u W 1,p 0 () for some 1 p < n. Then there exists a constant C, depending only on p, n and, such that u L q () C Du L p (), (2.2.9) for each q [1,p ], the constant C depending only on p,q,n and. Proof. Since u W 1,p 0 (), then exists a sequence of functions {u m } m=1 C0 1,p () converging to u in W0 (). We extend each function u m (x) to be 0 on R n \ and apply Theorem to discover u L p () C Du L p (). Since <, the Hölder inequality implies u L q () C u L p () if 1 q p. Remark In view of Theorem , on W 1,p 0 (), the norm Du L p () is equivalent to u W 1,p 0 () if is bounded.

17 2.2. SOBOLEV SPACES 27 Theorem (Morrey inequality) Assume n p <, γ := 1 n. Then there exists a constant C, depending only on p and n, p such that for all u C 1 (R n ). u C 0, γ (R n ) C u W 1,p (R n ), (2.2.10) Proof. 1. First choose any ball B(x,r) R n. We claim there exists a constant C, depending only on n, such that Du(y) u(x) u(y) dy C dy. (2.2.11) y x n 1 B(x,r) B(x,r) To prove this, fix any point w B(0,1). Then if 0 < s < r, there holds u(x + sw) u(x) = s d dt u(x + tw)dt Hence B(0,1) = u(x + sw) u(x) ds = 0 s 0 s 0 s 0 s 0 Du(x + tw) w dt Du(x + tw) dt. B(0,1) B(0,1) Du(x + tw) dsdt Du(x + tw) tn 1 dsdt. tn 1 Let y = x + tw, so that t = x y. Then converting from polar coordinates, we have Du(y) u(x + sw) u(x) ds dy B(0,1) B(x,s) x y n 1 Du(y) dy. x y n 1 B(x,r) Multiply by s n 1 and integrate from 0 to r with respect to s: u(x) u(y) dy rn Du(y) dy. n y x n 1 B(x,r) B(x,r)

18 28 CHAPTER 2. SOBOLEV SPACES It follows (2.2.11). 2. sup u C u W 1,p (R n ). R n Now fix x R n. We apply (2.2.11) as follows: u(x) u(x) u(y) dy + u(y) dy B(x,1) B(x,1) Du(y) C B(x,1) y x n 1 dy + C u L p (B(x,1)) ( dy C Du L p (R n ) B(x,1) x y (n 1) p p 1 C u W 1,p (R n ). ) p 1 p The last estimate holds since p > n implies (n 1) p p 1 B(x,1) dy x y (n 1) p p 1 As x R n is arbitrary, (2.2.12) implies 3. [u] C 0, 1 n/p (R n ) C Du L p (R n ). <. + C u L p (R n ) (2.2.12) < n; so that sup R n u C u W 1,p (R n ). (2.2.13) Choose any two points x,y R n and write r := x y. Let W := B(x,r) B(y,r). Then u(x) u(y) u(x) u(z) dz + u(y) u(z) dz. (2.2.14) W On the other hand, (2.2.11) allows us to estimate u(x) u(z) dz C u(x) u(z) dz W B(x,r) C Du L p (R n ) ( W dz B(x,1) x z (n 1) p p 1 C ( r n (n 1) p ) p 1 p 1 p Du L p (R n ) ) p 1 p = Cr 1 n p Du L p (R n ). (2.2.15)

19 2.2. SOBOLEV SPACES 29 Similarly, there holds u(y) u(z) dz Cr 1 n p Du L p (R n ). W Substituting this estimate and (2.2.15) into (2.2.14) yields u(x) u(y) Cr 1 n p Du L p (R n ) = C x y 1 n p Du L p (R n ). Thus [u] C 0, 1 n/p (R n ) = sup { u(x) u(y) } C Du L x =y x y 1 n/p p (R n ). This inequality and (2.2.13) complete the proof Embedding theorems and Trace theorems Definition Let (X, X ), (Y, Y ) be Banach spaces. X is (continuously) embedded into Y, denoted X Y, if there exists an injective linear map i : X Y and a constant C such that i(x) Y C x X, x X. X is compactly embedded into Y, denoted X Y, if i maps bounded subsets of X into precompact subsets of Y. Theorem (Exercise) (i) For R n with Lebesgue measure L n () <, 1 p < q, we have L q () L p (). This ceases to be true if L n () = ; (ii) Suppose is a precompact domain in R n, let m = 0,1,, and 0 α < β 1. Then C m,β ( ) C m,α ( ) compactly. Theorem (Sobolev embedding theorem) Let R n be a bounded domain with Lipschitz boundary, k N, 1 p. Then the following hold: (i) If kp < n, then W k,p () L q () for 1 q p = embedding is compact, if q < np n kp ; np n kp ; the

20 30 CHAPTER 2. SOBOLEV SPACES (ii) If 0 m < k n p < m + 1, then W k,p () C m,α ( ), for 0 α k m n p ; the embedding is compact, if α < k m n p. Proof. 1. The embedding of the case where kp < n. Assume that kp < n and u W k,p (). Then D α u L p () for all α = k. The Gagliardo-Nirenberg-Sobolev inequality implies D β u L p 1() C u W k,p (), for β = k 1, p 1 = np n p, so u W k 1,p 1 (). Similarly, we find u W k 2,p 2 (), where p 2 = np 1 = np. Continuing, we eventually discover after k steps n p 1 n 2p that W 0,p k() = L p k(), for p k = np n 1 = np n p n 1 n kp, and u L p k() C u W k,p (). Since is bounded, thus for any 1 q p = np n kp, W k,p () L q (). 2. The compactness of embedding. That is to show that if {u m } m=1 is a bounded sequence in W k,p (), there exists a subsequence {u mj } j=1 which converges in Lq (). For simplicity, we just prove the case k = 1. By induction argument, one get the proof of general case. By extension argument, without loss of generality assume that = R n and the functions {u m } m=1 all have compact support in some bounded open set U R n, moreover, Let us first study the smooth functions: sup u m W k,p (U) <. (2.2.16) m u ε m := η ε u m (ε > 0,m = 1,2, ), where η ε := 1 ε nη(x ) denotes the usual mollifier and η satisfies ε η C0 (Rn ), η 0, η(x)dx = 1. R n A concrete example of η is 1 Cexp( η(x) := x 2 ), if x < 1, 1 0, if x 1,

21 2.2. SOBOLEV SPACES 31 and constant C > 0 is selected so that η(x)dx = 1. We may R n suppose that the functions {u ε m} m=1 all have support in U as well. 3. Claim u ε m u m in L q (U) as ε 0, uniformly in m. (2.2.17) To prove this claim, we first note that if u m is smooth, then u ε m (x) u m(x) = η(y)(u m (x εy) u m (x))dy = B(0,1) B(0,1) = ε Thus u ε m(x) u m (x) dx ε U ε 1 η(y) B(0,1) B(0,1) U 0 η(y) η(y) d dt u m(x εty)dtdy Du m (z) dz. Du m (x εty) y dtdy. U Du m (x εty) dxdtdy By approximation this estimate holds if u m W 1,p (U). Hence u ε m u m L 1 (U) ε Du m L 1 (U) ε Du m L p (U), the latter inequality holds since U is bounded. Owing to (2.2.16) we thereby discover u ε m u m in L 1 (U) as ε 0, uniformly in m. (2.2.18) But then since 1 q < p, we using the Interpolation inequality for L p -norm that u ε m u m L q (U) u ε m u m θ L 1 (U) uε m u m 1 θ L p (U), where 1 (1 θ) = θ + q p, 0 < θ 1. Consequently, (2.2.16) and the Gagliardo-Nirenberg-Sobolev inequality imply u ε m u m L q (U) C u ε m u m θ L 1 (U) ;

22 32 CHAPTER 2. SOBOLEV SPACES whence assertion (2.2.17) follows from (2.2.18). 4. Next we claim for each fixed ε > 0, the sequence {u ε m } m=1 is uniformly bounded and equi-continuous. (2.2.19) Indeed, if x R n, then u ε m (x) for m = 1,2,. Similarly Du ε m(x) B(x,ε) η ε (x y) u m (y) dy η ε L (R n ) u m L 1 (U) C ε n <, B(x,ε) Dη ε (x y) u m (y) dy Dη ε L (R n ) u m L 1 (U) C <, εn+1 for m = 1, 2,. Thus the claim (2.2.19) follows these two estimates. 5. Now fix δ > 0. We will show that there exists a subsequence {u mj } j=1 of {u m} m=1 such that lim sup u mj u mk L q (U) δ. (2.2.20) j,k To see this, let us first employ assertion (2.2.17) to select ε > 0 so small that u ε m u m L q (U) δ/2, (2.2.21) for m = 1,2,. We now observe that since functions {u m } m=1, and thus functions {u ε m } m=1, have support in some fixed bounded set U Rn, we may use (2.2.19) and the Arzela-Ascoli theorem, to obtain a subsequence {u ε m j } j=1 {uε m} m=1 which converges uniformly on U. In particular therefore lim sup u ε m j u ε m k L q (U) = 0. (2.2.22) j,k Then (2.2.21) and (2.2.22) imply (2.2.20).

23 2.2. SOBOLEV SPACES We next employ assertion (2.2.20) with δ = 1, 1 2, 1, and use 3 the standard diagonal argument to extract a subsequence {u ml } l=1 of {u m } m=1 such that lim sup u ml u mk L q (U) = 0. l,k Conclusion (i) of theorem is proved by steps 1-6. Next to prove conclusion (ii) of theorem. 7. Assume that 0 m < k n p < m + 1. Then as above we see u W k l,r (), (2.2.23) for r = np n lp provided that lp < n. We choose the integer l 0 = [ n p ]. Consequently, (2.2.23) holds for r = np > n. Hence (2.2.23) and n pl 0 Morrey inequality imply that D α u C 0,1 n r () for all α k l 0 1. Observe also that 1 n r = 1 n p +l 0 = [ n p ]+1 n p. Thus u Cm, γ ( ) with m = k l 0 1 = k [ n p ] 1, γ = 1 n r = [n p ] + 1 n, and the p stated estimate follows easily. 8. The Arzela-Ascoli theorem implies the compactness of embedding W k,p () C m,α ( ) provided that α < γ = k m n p. Denote (u) = udx = average of u over. Theorem (Poincare inequality for W 1,p ()) Let R n be bounded and connected, with C 1 boundary. Assume 1 p. Then there exists a constant C, depending only on p, n and, such that u (u) L p () C Du L p (), (2.2.24) for all u W 1,p (R n ). Proof. Argue by contradiction. If the stated estimate is false, that is, for each integer k = 1,2,, there would exist a function u k W 1,p () such that u k (u k ) L p () > k Du k L p (). (2.2.25)

24 34 CHAPTER 2. SOBOLEV SPACES Normalize by defining v k := u k (u k ), k = 1,2,. u k (u k ) L p () Then, there holds and from (2.2.25), we have (v k ) = 0, v k L p () = 1, (2.2.26) Dv k L p () < 1, k = 1,2,. (2.2.27) k Particularly, {v k } k=1 is a bounded in W 1,p (). From the compact embedding, there exists a subsequence {v kj } j=1 {v k} k=1, and a function v L p () such that v kj v in L p (). From (2.2.26), it follows that (v) = 0, v L p () = 1. On the other hand, (2.2.27) implies that for every ϕ C0 (), there holds vϕ xi dx = lim v kj ϕ xi dx = lim (v kj ) xi ϕdx = 0, j j consequently, v W 1,p () and Dv = 0 a.e. in. Thus v is constant, since is connected, furthermore, (v) = 0, hence v c = 0, which contradicts the fact that v L p () = 1. Theorem (Poincare inequality for a ball) Suppose 1 p < n. Then there exists a constant C, depending only on p and n, such that u (u) x,r L p (B(x,r)) Cr Du L p (B(x,r)), (2.2.28) holds for each ball B(x,r) R n and each function u W 1,p 0 (B(x,r)). Proof. The case = B(0,1) follows from Theorem In general, if u W 1,p 0 (B(x,r)), set v(y) := u(x + ry), (y B(0,1)).

25 2.2. SOBOLEV SPACES 35 Then v W 1,p 0 (B(0,1)), and there holds v (v) 0,1 L p (B(0,1)) C Dv L p (B(0,1)), Changing variables, we recover estimate (2.2.28). Lemma (Poincare-Sobolev inequality) For any ε > 0 there exists a C = C(ε,n) such that for u H 1 (B 1 ) with {x B1 ;u = 0} ε B1, there holds u 2 C Du 2. B 1 B 1 Proof. Suppose not. Then there exists a sequence {u m } H 1 (B 1 ) such that {x B1 ;u m = 0} ε B1, and B 1 u 2 m = 1, B 1 Du m 2 0 as m. Hence we may assume u m u 0 H 1 (B 1 ) strongly in L 2 (B 1 ) and weakly in H 1 ( (B 1 ). Then lim Dum 2 D(u m u 0 ) 2) = Du 0 2 and m Du 0 2 lim Du m 2 = 0. Thus u 0 is a nonzero constant. So m 0 = lim u m u 0 2 lim u m u 0 2 m B m 1 {u m=0} u 0 2 inf {um = 0} > 0. m This contradiction completes the proof. Definition For a domain with C k -boundary = Γ, k N,1 < p <, denote W k 1 p,p (Γ) as the set of equivalent classes { {u}+ W k,p 0 (); u W k,p () }, endowed with norm u Γ W k 1 p,p (Γ) = inf { v W k,p () : u v W k,p 0 () }. The elements of W k 1 p,p (Γ) is call the traces u Γ of functions u W k,p ().

26 36 CHAPTER 2. SOBOLEV SPACES Theorem (Trace Theorem) Assume is a domain with C k - boundary = Γ, k N,1 < p <. Then W k 1 p,p (Γ) is a Banach space. Theorem (Extension Theorem) For any domain with C k - boundary = Γ, k N,1 < p <, there exists a continuous linear extension operator ext : W k 1 p,p (Γ) W k,p () such that ( ext(u) ) Γ = u, for all u W k 1 p,p (Γ). Theorem Suppose is a domain with C k -boundary = Γ, k N,1 < p <. Then W k,p (Γ) W k 1 p,p (Γ) W k 1,p (Γ) and both embeddings are compact. EXERCISES 1. Prove The Young inequality with ε - Lemma Prove the generalized Hölder inequality - Lemma Complete Example Complete Example Complete Example Complete Example Verify C k,γ ( ), L p () and W k,p () are norms. 8. Prove Theorem