************************************* Partial Differential Equations II (Math 849, Spring 2019) Baisheng Yan

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1 ************************************* Partial Differential Equations II (Math 849, Spring 2019) by Baisheng Yan Department of Mathematics Michigan State University

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3 Contents Chapter 1. Preliminaries Banach Spaces Bounded Linear Operators Weak Convergence and Compact Operators Spectral Theory for Compact Linear Operators 9 Chapter 2. Sobolev Spaces Weak Derivatives and Sobolev Spaces Approximations and Extensions Sobolev embedding Theorems Compactness Additional Topics Spaces of Functions Involving Time 42 Chapter 3. Second-Order Linear Elliptic Equations Differential Equations in Divergence Form The Lax-Milgram Theorem Energy Estimates and Existence Theory Fredholm Alternatives Regularity Regularity for Linear Systems* 68 iii

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5 Chapter 1 Preliminaries 1.1. Banach Spaces Vector Spaces. A (real) vector space is a set X, whose elements are called vectors, and in which two operations, addition and scalar multiplication, are defined as follows: (a) To every pair of vectors x and y corresponds a vector x + y in such a way that x + y = y + x and x + (y + z) = (x + y) + z. X contains a unique vector 0 (the zero vector or origin of X) such that x+0 = x for every x X, and to each x X corresponds a unique vector x such that x + ( x) = 0. (b) To every pair (α, x), with α R and x X, corresponds a vector αx in such a way that 1x = x, and such that the two distributive laws α(βx) = (αβ)x α(x + y) = αx + αy, (α + β)x = αx + βx hold. A nonempty subset M of a vector space X is called a subspace of X if αx + βy M for all x, y M and all α, β R. A subset M of a vector space X is said to be convex if tx + (1 t)y M whenever t (0, 1), x, y M. (Clearly, every subspace of X is convex.) Let x 1,..., x n be elements of a vector space X. The set of all α 1 x α n x n, with α i R, is called the span of x 1,..., x n and is denoted by span{x 1,..., x n }. The elements x 1,..., x n are said to be linearly independent if α 1 x α n x n = 0 implies that α i = 0 for each i; otherwise, the elements x 1,..., x n are said to be linearly dependent. An arbitrary collection of vectors is said to be linearly independent if every finite subset of distinct elements is linearly independent. The dimension of a vector space X, denoted by dim X, is either 0, a positive integer or. If X = {0} then dim X = 0; if there exist linearly independent {u 1,..., u n } such 1

6 2 1. Preliminaries that span{x 1,..., x n } = X, then dim X = n and {u 1,..., u n } is called a basis for X; in all other cases dim X = Normed Spaces. A (real) vector space X is said to be a normed space if to every x X there is associated a nonnegative real number x, called the norm of x, in such a way that (a) x + y x + y for all x and y in X (Triangle inequality) (b) αx = α x for all x X and all α R (c) x > 0 if x 0. Note that (b) and (c) imply that x = 0 iff x = 0. Moreover, it easily follows from (a) that x y x y for all x, y X Completeness and Banach Spaces. A sequence {x n } in a normed space X is called a Cauchy sequence if, for each ɛ > 0, there exists an integer N such that x m x n < ɛ for all m, n N. We say a sequence {x n } converges to x in X and write x n x if lim n x n x = 0 and, in this case, x is called the limit of {x n }. A normed space X is called complete if every Cauchy sequence in X converges to a limit in X. A complete (real) normed space is called a (real) Banach space. A Banach space is separable if it contains a countable dense set. It can be shown that a subspace of a separable Banach space is itself separable. Example 1.1. Let be an open subset of R n, n 1. The set C() of (real-valued) continuous functions defined on is an infinite dimensional vector space with the usual definitions of addition and scalar multiplication: (f + g)(x) = f(x) + g(x) for f, g C(), x (αf)(x) = αf(x) for α R, f C(), x. C( ) consists of those functions which are uniformly continuous on. Each such function has a continuous extension to. C 0 () consists of those functions which are continuous in and have compact support in. (The support of a function f defined on is the closure of the set {x : f(x) 0} and is denoted by supp(f).) The latter two spaces are clearly subspaces of C(). For each n-tuple α = (α 1,..., α n ) of nonnegative integers, we denote by D α the partial derivative D α 1 1 Dαn n, D i = / x i of order α = α α n. If α = 0, then D 0 = I(identity). For integers m 0, let C m () be the collection of all f C() such that D α f C() for all α with α m. We write f C () iff f C m () for all m 0. For m 0, define C0 m() = C 0() C m () and let C0 () = C 0() C (). The spaces C m (), C (), C0 m(), C 0 () are all subspaces of the vector space C(). Similar definitions can be given for C m ( ) etc. For m 0, define X to be the set of all f C m () for which f m, sup D α f(x) <. α m

7 1.1. Banach Spaces 3 Then X is a Banach space with norm m,. To prove, for example, the completeness when m = 0, we let {f n } be a Cauchy sequence in X, i.e., assume for any ε > 0 there is a number N(ε) such that for all x sup f n (x) f m (x) < ε if m, n > N(ε). x But this means that {f n (x)} is a uniformly Cauchy sequence of bounded continuous functions, and thus converges uniformly to a bounded continuous function f(x). Letting m in the above inequality shows that f n f m, 0. Note that the same proof is valid for the set of bounded continuous scalar-valued functions defined on a nonempty subset of a normed space X. Example 1.2. Let be a nonempty Lebesgue measurable set in R n. For p [1, ), we denote by L p () the set of equivalence classes of Lebesgue measurable functions on for which ( ) 1 f p f(x) p p dx <. (Two functions belong to the same equivalence class, i.e., are equivalent, if they differ only on a set of measure 0.) Let L () denote the set of equivalence classes of Lebesgue measurable functions on for which f ess-sup x f(x) <. Then L p (), 1 p, are Banach spaces with norms p. For p [1, ] we write f L p loc () iff f Lp (K) for each compact set K. For the sake of convenience, we will also consider L p () as a set of functions. With this convention in mind, we can assert that C 0 () L p (). In fact, if p [1, ), then as we shall show later, C 0 () is dense in L p (). The space L p () is also separable if p [1, ). This follows easily, when is compact, from the last remark and the Weierstrass approximation theorem. Finally we recall that if p, q, r [1, ] with p 1 + q 1 = r 1, then Hölder s inequality implies that if f L p () and g L q (), then fg L r () and fg r f p g q. Example 1.3. The Cartesian product X Y, of two vector spaces X and Y, is itself a vector space under the following operations of addition and scalar multiplication: [x 1, y 1 ] + [x 2, y 2 ] = [x 1 + x 2, y 1 + y 2 ] α[x, y] = [αx, αy]. If in addition, X and Y are normed spaces with norms X, Y respectively, then X Y becomes a normed space under the norm [x, y] = x X + y Y. Moreover, under this norm, X Y becomes a Banach space provided X and Y are Banach spaces.

8 4 1. Preliminaries Hilbert Spaces. Let H be a real vector space. H is said to be an inner product space if to every pair of vectors x and y in H there corresponds a real-valued function (x, y), called the inner product of x and y, such that (a) (x, y) = (y, x) for all x, y H (b) (x + y, z) = (x, z) + (y, z) for all x, y, z H (c) (λx, y) = λ(x, y) for all x, y H, λ R (d) (x, x) 0 for all x H, and (x, x) = 0 if and only if x = 0. For x H we set (1.1) x = (x, x) 1/2. Theorem 1.4. If H is an inner product space, then for all x and y in H, it follows that (a) (x, y) x y (b) x + y x + y (Cauchy-Schwarz inequality); (Triangle inequality); (c) x + y 2 + x y 2 = 2( x 2 + y 2 ) (Parallelogram law). Proof. (a) is obvious if x = 0, and otherwise it follows by taking δ = (x, y)/ x 2 in 0 δx + y 2 = δ 2 x 2 + 2δ(x, y) + y 2. This identity, with δ = 1, and (a) imply (b). (c) follows easily by using (1.1). Furthermore, by (d), equation (1.1) defines a norm on an inner product space H. If H is complete under this norm, then H is said to be a Hilbert space. Example 1.5. The space L 2 () is a Hilbert space with inner product (f, g) = f(x)g(x) dx for all f, g L 2 (). Theorem 1.6. Every nonempty closed convex subset S of a Hilbert space H contains a unique element of minimal norm. Proof. Choose x n S so that x n d inf{ x : x S}. Since (1/2)(x n + x m ) S, we have x n + x m 2 4d 2. Using the parallelogram law, we see that (1.2) x n x m 2 2( x n 2 d 2 ) + 2( x m 2 d 2 ) and therefore {x n } is a Cauchy sequence in H. Since S is closed, {x n } converges to some x S and x = d. If y S and y = d, then the parallelogram law implies, as in (1.2), that x = y. If (x, y) = 0, then x and y are said to be orthogonal, written sometimes as x y. For M H, the orthogonal complement of M, denoted by M, is defined to be the set of all x H such that (x, y) = 0 for all y M. It is easily seen that M is a closed subspace of H. Moreover, if M is a dense subset of H then M = {0}. Theorem 1.7. (Projection) Suppose M is a closed subspace of a Hilbert space H. Then for each x H there exist unique y M, z M such that x = y + z. The element y is called the projection of x onto M.

9 1.2. Bounded Linear Operators 5 Proof. Fix x H and let S = {x y : y M}. It is easy to see that S is convex and closed. Theorem 1.6 implies that there exists a y M such that x y x w for all w M. Let z = x y. For an arbitrary w M, w 0, let α = (z, w)/ w 2 and note that z 2 z αw 2 = z 2 2α(z, w) + α 2 w 2 = z 2 (z, w)/ w 2, which implies (z, w) = 0. Therefore z M ; clearly, x = y + z. If x = y + z for some y M, z M, then y y = z z M M = {0}, which implies the uniqueness of y, z. Remark. In particular, if M is a proper closed subspace of H, then there is a nonzero element in M. Indeed, select x H\M, let y be the projection of x on M. Then z = x y is a nonzero element of M Bounded Linear Operators Linear Operators. Let X, Y be real vector spaces. A map T : D(T ) X Y is said to be a linear operator if T (αx + βy) = αt x + βt y x, y D(T ), α, β R. Let X, Y be normed spaces. A linear operator T : X Y is said to be bounded if there exists a constant m > 0 such that (1.3) T x m x for all x X. In this case, we define the operator norm T of T by (1.4) T = sup T x = sup T x. x X, x =1 x X, x 1 The collection of all bounded linear operators T : X Y will be denoted by B(X, Y ). We also set B(X) = B(X, X). Observe that T S T S if S B(X, Y ), T B(Y, Z). Theorem 1.8. If X and Y are normed spaces, then B(X, Y ) is a normed space with norm defined by equation (1.4). If Y is a Banach space, then B(X, Y ) is also a Banach space. Proof. It is easy to see that B(X, Y ) is a normed space. To prove its completeness when Y is complete, assume that {T n } is a Cauchy sequence in B(X, Y ). Since (1.5) T n x T m x T n T m x we see that, for each fixed x X, {T n x} is a Cauchy sequence in Y and therefore we can define a map T by T x = lim n T nx for all x X. Clearly T is a linear operator from X to Y. If ε > 0, then the right side of (1.5) is smaller than ε x provided that m and n are large enough. Thus, (letting n ) T x T m x ε x for all large enough m. Hence, T x ( T m + ε) x, which shows that T B(X, Y ). Moreover, T T m < ε for all large enough m. Hence, lim n T n = T. The following important theorems can be found in any textbook of functional analysis.

10 6 1. Preliminaries Theorem 1.9. (Banach-Steinhaus) Let X be a Banach space and Y a normed space. If A B(X, Y ) is such that sup T A T x < for each fixed x X, then sup T A T <. Theorem (Bounded Inverse) If X and Y are Banach spaces and if T B(X, Y ) is one-to-one and onto, then T 1 B(Y, X) Dual Spaces and Reflexivity. Let X be a (real) normed space. Then the Banach space B(X, R) is called the (normed) dual space of X and will be denoted by X*. Elements of X* are called bounded linear functionals or continuous linear functionals on X. Frequently, we shall use the notation f, x to denote the value f(x) of f X* at x X. Thus f, x f x for all f X*, x X. Example Suppose 1 < p, q < satisfy 1/p + 1/q = 1 and let be a nonempty Lebesgue measurable set in R n. Then L p () = L q (). The case of p = is different. The dual of L is much larger then L 1. Theorem (Hahn-Banach) Let X be a normed space and Y a subspace of X. Assume f Y. Then there exists f X such that f, y = f, y y Y, f X = f Y. Corollary Let X be a normed space and x 0 0 in X. Then there exists f X such that f = 1 and f, x 0 = x 0. The dual space (X ) := X of X is called the second dual space of X and is again a Banach space. Note that to each x X we associate a unique F x X defined by F x (f) = f, x for all f X. Then F x (f) x f and hence F x x. If x 0, then by Corollary 1.13 there exists f X such that f = 1 and F x (f) = f, x = x, which implies F x x. Therefore, F x = x. Thus, the (canonical) mapping J : X X**, given by Jx = F x, is a linear isometry of X onto the subspace J(X) of X. Since J is one-to-one, we can identify X with J(X). A Banach space X is called reflexive if J(X) = X. For example, all L p spaces with 1 < p < are reflexive Bounded Linear Functionals on a Hilbert Space. Theorem (Riesz Representation) If H is a Hilbert space and f H*, then there exists a unique y H such that Moreover, f = y. f(x) = f, x = (x, y) for all x H. Proof. If f(x) = 0 for all x, take y = 0. Otherwise, there is an element z N (f) such that z = 1. (Note that the linearity and continuity of f implies that N (f) is a closed subspace of H.) Put u = f(x)z f(z)x. Since f(u) = 0, we have u N (f). Thus (u, z) = 0, which implies f(x) = f(x)(z, z) = f(z)(x, z) = (x, f(z)z) = (x, y), where y = f(z)z. To prove uniqueness, suppose (x, y) = (x, y ) for all x H. Then in particular, (y y, y y ) = 0, which implies y = y. From the Cauchy-Schwarz inequality we get f(x) x y, which yields f y. The reverse inequality follows by choosing x = y in the representation.

11 1.3. Weak Convergence and Compact Operators 7 Corollary H is reflexive. Proof. Let α H. For each y H let l y H be defined by l y, x = (y, x). Let A(y) = α, l y. Then A H. Hence there exists a H such that A(y) = (a, y) for all y H. Hence α, l y = (a, y) = (y, a) = l y, a = Ja, l y. Since y l y is onto, we thus have α = Ja. Hence H = J(H) and H is reflexive. Let T : H H be an operator on the Hilbert space H. We define the Hilbert space adjoint T * : H H as follows: For each fixed y H, set f(x) = (T x, y). Then it is easily seen that f H. Hence by the Riesz representation theorem, there exists a unique z H such that (T x, y) = (x, z) for all x H; then define T y = z. So T : H H satisfies We say T B(H) is self-adjoint if T = T. We also have the following. (T x, y) = (x, T *y) for all x, y H. Theorem Let H be a Hilbert space and T B(H). Then T * B(H) and T = T *. Moreover, (T ) = T and N (T ) = R(T ), R(T ) = N (T ). Proof. Clearly T is linear. Moreover, T y 2 = (T y, T y) = (T T y, y) T T y y T T y y ; hence T y T y for all y H, i.e., T B(H) and T T. Note that (x, T y) = (T y, x) = (y, T x) = (T x, y) = (x, (T ) y) x, y H, which proves (T ) = T, and hence T = (T ) T, proving T = T. Moreover, note that x N (T ) T x = 0 y H (T x, y) = (x, T y) = 0 x R(T ) ; hence N (T ) = R(T ). Furthermore, for all x H and y N (T ), we have (T x, y) = (x, T y) = 0 and so T x N (T ), which proves R(T ) N (T ) and hence R(T ) N (T ). To prove the opposite inclusion, by contradiction, suppose x N (T ), but x / R(T ). Let x = y + z, where y R(T ) and 0 z R(T ). Then z R(T ) and z = x y N (T ), so z / N (T ). Let w = T z 0. Thus (T w, z) = (w, T z) = w 2 0, which contradicts z R(T ) Weak Convergence and Compact Operators Weak Convergence. Let X be a normed space. A sequence x n X is said to be weakly convergent to an element x X, written x n x, if f, x n f, x for all f X. Theorem Let {x n } be a sequence in X. (a) Weak limits are unique. (b) If x n x, then x n x. (c) If x n x, then {x n } is bounded and x lim inf x n. Proof. To prove (a), suppose that x and y are both weak limits of the sequence {x n } and set z = x y. Then f, z = 0 for every f X and by Corollary 1.13, z = 0. To prove (b), let f X and note that x n x implies f, x n f, x since f is continuous. To prove (c), assume x n x and consider the sequence {Jx n } of elements of X, where J : X X is the canonical operator defined above. For each f X, sup Jx n (f) = sup f, x n <

12 8 1. Preliminaries (since f, x n converges). By the Banach-Steinhaus Theorem (Theorem 1.9), there exists a constant c such that x n = Jx n c, which implies {x n } is bounded. Finally, for f X f, x = lim f, x n lim inf f x n = f lim inf x n which implies the desired inequality since x = sup f =1 f, x. We note that in a Hilbert space H, the Riesz representation theorem implies that x n x means (x n, y) (x, y) for all y H. Moreover, we have This follows from the estimate (x n, y n ) (x, y) if x n x, y n y. (x, y) (x n, y n ) = (x x n, y) (x n, y n y) (x x n, y) + x n y y n and the fact that x n is bounded. Let X be a normed space. A sequence {f n } in X is said to be weakly* convergent to an element f X, written f n f, if fn, x f, x for all x X. Theorem (Banach-Alaoglu) Let X be a separable normed vector space. Then the closed unit ball in X is weakly* compact in the sense that any sequence {f n } X with f n X 1 has a subsequence which converges weakly* to an f X with f X Compact Operators. Let X and Y be normed spaces. An operator T : X Y is said to be compact if it maps bounded sets in X into relatively compact sets in Y, i.e., if for every bounded sequence {x n } in X, {T x n } has a subsequence which converges to some element of Y. Since relatively compact sets are bounded, it follows that a compact operator is bounded. On the other hand, since bounded sets in finite-dimensional spaces are relatively compact, it follows that a bounded operator with finite dimensional range is compact. It can be shown that the identity map I : X X ( Ix = x ) is compact iff X is finite-dimensional. Finally we note that the operator ST is compact if (a) T : X Y is compact and S : Y Z is continuous or (b) T is bounded and S is compact. One of the main methods of proving the compactness of certain operators is based upon the Ascoli theorem. Let be a subset of the normed space X. A set S C() is said to be equicontinuous if for each ε > 0 there exists a δ > 0 such that f(x) f(y) < ε for all x, y with x y < δ and for all f S. Theorem (Ascoli) Let be a relatively compact subset of a normed space X and let S C(). Then S is relatively compact if it is bounded and equicontinuous. Remark. In other words, every bounded equicontinuous sequence of functions has a uniformly convergent subsequence. Theorem Let X and Y be Banach spaces. If T n : X Y are linear and compact for n 1 and if lim n T n T = 0, then T is linear and compact. Thus, linear compact operators form a closed subspace of B(X, Y ). Proof. To prove the compactness of T, let {x n } be a sequence in X with M = sup n x n <. Let A 1 denote an infinite set of integers such the sequence {T 1 x n } n A1 converges. For k 2 let A k A k 1 denote an infinite set of integers such that the sequence {T k x n } n Ak

13 1.4. Spectral Theory for Compact Linear Operators 9 converges. Choose n 1 A 1 and n k A k, n k > n k 1 for k 2. Choose ε > 0. Let k be such that T T k M < ε/4 and note that T x ni T x nj (T T k )(x ni x nj ) + T k x ni T k x nj < ε/2 + T k x ni T k x nj. Since {T k x ni } i=1 converges, {T x n i } i=1 is a Cauchy sequence. Theorem Let X and Y be Banach spaces and T B(X, Y ) be compact. If x n x in X, then T x n T x in Y. Proof. It suffices to prove that every subsequence of {T x n } has a subsequence converging to T x. We only prove this for the whole sequence {T x n }. Since {x n } is bounded and T is compact, we assume T x nj y. Let g Y ; then g T X. Since x nj x, we have (g T )(x nj ) (g T )(x); that is, g(t x nj ) g(t x). Since T x nj y, we have g(t x nj ) g(y). This proves that g(y) = g(t x) for all g Y. Hence y = T x; so T x nj T x. Theorem Let H be a Hilbert space. If T : H H is linear and compact, then T * is compact. Proof. Let {x n } be a sequence in H satisfying x n m. The sequence {T *x n } is therefore bounded, since T * is bounded. Since T is compact, by passing to a subsequence if necessary, we may assume that the sequence {T T *x n } converges. But then T *(x n x m ) 2 = (x n x m, T T *(x n x m )) 2m T T *(x n x m ) 0 as m, n. Since H is complete, the sequence {T *x n } is convergent and hence T * is compact Spectral Theory for Compact Linear Operators Fredholm Alternative. Theorem (Fredholm Alternative) Let T : H H be a compact linear operator on the Hilbert space H. Then (a) dim N (I T ) = dim N (I T ) <. (b) R(I T ) is closed; hence R(I T ) = N (I T ). (c) N (I T ) = {0} R(I T ) = H. Proof. 1. We prove dim N (I T ) <. Suppose for the contrary that dim N (I T ) =. Then there exists an infinite orthonormal set {u k } in N (I T ); hence T u k = u k for all k = 1, 2,. Now u k u l 2 = u k 2 2(u k, u l ) + u l 2 = 2 for all k l, and so T u k T u l = u k u l = 2 for k l. So {T u k } cannot have a convergent subsequence, which contradicts the compactness of T. 2. We claim there exists a constant γ > 0 such that (1.6) u T u γ u u N (I T ). Indeed, if not, suppose that u k T u k < 1/k for all k = 1, 2, and some u k N (I T ) with u k = 1. By compactness, assume T u kj v in H. Hence u kj = u kj T u kj +T u kj v; so v = 1 and T u kj T v = v. Thus v N (I T ), which implies (u kj, v) = 0 for all j = 1, 2,, and thus v 2 = (v, v) = lim j (u kj, v) = 0, a contradiction to v = 1.

14 10 1. Preliminaries 3. We prove R(I T ) is closed. To show this, let {v k } R(I T ) and v k v in H. Let w k H be such that v k = w k T w k, and let u k N (I T ) be the projection of w k onto N (I T ). Then u k T u k = w k T w k = v k and by (1.6), v k v l = (u k u l ) T (u k u l ) γ u k u l, which implies {u k } is Cauchy and hence u k u in H. Hence u T u = v R(I T ). So R(I T ) is closed. This, combined with the general result Theorem 1.16, proves that Hence (b) is proved. R(I T ) = N (I T ). 4. Assume N (I T ) = {0}. We prove R(I T ) = H. Suppose, for the contrary, that H 1 = R(I T ) = (I T )(H) H. Then H 1 is a proper closed subspace of H. Since I T is one-to-one, H 2 = (I T )H 1 is a proper closed subspace of H 1. Let H k = (I T ) k (H). Then H k+1 is a proper closed subspace of H k. Choose u k H k+1 H k and u k = 1. For all k > l, we have T u k T u l = u k (I T )u k + (I T )u l u l, where u l H l+1, u k H k H l+1, (I T )u k (I T )(H k ) = H k+1 H l+1 and (I T )u l H l+1 and hence T u k T u l 2 = u k (I T )u k + (I T )u l 2 + u l 2 1. So {T u n } cannot have a convergent subsequence, which contradicts the compactness of T. 5. Assume R(I T ) = H. To prove N (I T ) = {0}, we have, by (b), N (I T ) = {0}. Since T is compact, by Step 4 above, R(I T ) = H. Since (T ) = T, by Theorem 1.16, we have N (I T ) = R(I T ) = {0}. Combining Steps 4 and 5, we have proved (c). 6. We prove dim N (I T ) dim R(I T ). By Step 1, m = dim N (I T ) <. Suppose, for the contrary, that m < dim R(I T ). Let {x 1,, x m } be a basis of N (I T ), {y 1,, y m } be linearly independent in R(I T ), and let y 0 R(I T ) be such that y 0 = 1 and (y 0, y i ) = 0 for all i = 1, 2,, m. Let A be the linear operator from N (I T ) onto span{y 1,, y m } such that Ax i = y i for each i = 1, 2,, m. Let F = AP : H R(I T ), where P : H N (I T ) is the orthonormal projection. Let T = T + F. Then T is also compact. Note that x N (I T ) x = T x (I T )x = F x R(I T ) span{y 1,, y m } = {0} F x = 0 x N (I T ) x = 0, which proves N (I T ) = {0}. By Step 4, R(I T ) = H. This would imply y 0 = x T x F x for some x H and hence, 1 = (y 0, y 0 ) = (y 0, (I T )x) (y 0, F x) = 0, a contradiction. 7. Finally, since T is also compact, by Step 6, we have Hence, by Theorem 1.16, it follows that dim N (I T ) dim R(I T ). dim N (I T ) dim R(I T ) = dim N (I T ) dim R(I T ) = dim N (I T ). This proves (a).

15 1.4. Spectral Theory for Compact Linear Operators Spectrum of Compact Operators. A subset S of a Hilbert space H is said to be an orthonormal set if each element of S has norm 1 and if every pair of distinct elements in S is orthogonal. It easily follows that an orthonormal set is linearly independent. An orthonormal set S is said to be complete if x = φ S (x, φ)φ for all x H. It can be shown that (x, φ) 0 for at most countably many φ S. This series is called the Fourier series for x with respect to the orthonormal set {φ}. Let {φ i } i=1 be a countable orthonormal set in H. Upon expanding x N n=1 (x, φ n)φ n 2, we arrive at Bessel s inequality: (x, φ n ) 2 x 2. n=1 Let T : D(T ) H H be a linear operator on the real Hilbert space H. The set ρ(t ) of all scalars λ R for which (T λi) 1 B(H) is called the resolvent set of T. The operator R(λ) = (T λi) 1 is called the resolvent of T. The set σ(t ) = R \ ρ(t ) is called the spectrum of T. It can be shown that ρ(t ) is an open set and σ(t ) is a closed set. The set of λ R for which N (T λi) is called the point spectrum of T and is denoted by σ p (T ). The elements of σ p (T ) are called the eigenvalues of T and the nonzero members of N (T λi) are called the eigenvectors (or eigenfunctions if X is a function space) of T. If T is linear and compact and λ 0, then by the Fredholm alternative, either λ σ p (T ) or λ ρ(t ). Moreover, if H is infinite-dimensional, then 0 ρ(t ); otherwise, T 1 B(H) and T 1 T = I is compact. As a consequence, σ(t ) consists of the nonzero eigenvalues of T together with the point 0. The next result asserts that σ p (T ) is either finite or a sequence converging to zero. Theorem Let T : X X be a compact linear operator on the normed space X. Then for each r > 0 there exist at most finitely many λ σ p (T ) for which λ > r. Proof. Suppose there exists a sequence {λ n } of distinct λ n σ p (T ) with λ n > r. Let u n be an eigenvector to λ n. Define Y n = span{u 1,, u n }. Then Y n is a strictly increasing sequence of subspaces of X. By the Riesz s lemma below, choose unit vectors y n Y n such that dist(y n, Y n 1 ) > 1/2 for each n = 1, 2,. Then, for n < m, T y n T y m = (T λ n )y n + λ n y n (T λ m )y m λ m y m. But (T λ n )y n + λ n y n (T λ m )y m Y m 1 (here the fact that y n is a linear combination of eigenvectors u 1,, u n is used), therefore, by the choice of y m, we have T y n T y m > λ m dist(y m, Y m 1 ) > r/2 and hence {T y n } cannot have a convergent subsequence, a contradiction to the compactness of T. Lemma (Riesz s lemma) Let X be a Banach space and Y be a proper closed subspace of X. Then, for each ɛ > 0, there exists x X such that x = 1 and dist(x, Y ) 1 ɛ. Proof. Let u X \ Y ; then R = dist(u, Y ) > 0. For δ > 0 let y Y be such that y u < R + δ. Let x = (u y)/ u y, so x = 1 and dist(x, Y ) = inf x v = inf u v Y v Y u y y u y v = inf w Y u u y y u y w u y = inf z Y u z = u y R u y > R R + δ > 1 ɛ,

16 12 1. Preliminaries if δ > 0 is chosen sufficiently small. Remark. Using Riesz s lemma, many conclusions of the Fredholm Alternative Theorem still hold for compact linear operators on Banach spaces Self-Adjoint Compact Operators. The next result implies that a self-adjoint compact operator on a Hilbert space has at least one eigenvalue. On the other hand, an arbitrary bounded, linear, self-adjoint operator need not have any eigenvalues. As an example, let T : L 2 (0, 1) L 2 (0, 1) be defined by T u(x) = xu(x). Theorem Suppose T B(H) is self-adjoint, i.e., (T x, y) = (x, T y) for all x, y H. Then T = sup (T x, x). x =1 Moreover, if H {0}, then there exists a real number λ σ(t ) such that λ = T. If λ σ p (T ), then in absolute value λ is the largest eigenvalue of T. Proof. Clearly m sup x =1 (T x, x) T. x, y H To show T m, observe that for all 2(T x, y) + 2(T y, x) = (T (x + y), x + y) (T (x y), x y) m( x + y 2 + x y 2 ) = 2m( x 2 + y 2 ) where the last step follows from the paralleogram law. Hence, if T x 0 and y = ( x / T x )T x, then 2 x T x = (T x, y) + (y, T x) m( x 2 + y 2 ) = 2m x 2 which implies T x m x. Since this is also valid when T x = 0, we have T m. To prove the moreover part, choose x n H such that x n = 1 and T = lim n (T x n, x n ). By renaming a subsequence of {x n }, we may assume that (T x n, x n ) converge to some real number λ with λ = T. Observe that (T λ)x n 2 = T x n 2 2λ(T x n, x n ) + λ 2 x n 2 2λ 2 2λ(T x n, x n ) 0. We now claim that λ σ(t ). Otherwise, we arrive at the contradiction 1 = x n = (T λ) 1 (T λ)x n (T λ) 1 (T λ)x n 0. Finally, we note that if T φ = µφ, with φ = 1, then µ = (T φ, φ) T which implies the last assertion of the theorem. Lemma Suppose T B(H) is self-adjoint and Y is a subspace of H such that T (Y ) Y. Then T (Y ) Y. Proof. Let z Y. For each y Y, since T y Y, (T z, y) = (z, T y) = 0, we have T z Y. Thus T (Y ) Y. Finally we have the following result.

17 1.4. Spectral Theory for Compact Linear Operators 13 Theorem Let H be a separable Hilbert space and suppose T : H H is linear, self-adjoint and compact. Then there exists a countable complete orthonormal set in H consisting of eigenvectors of T. More precisely, let {λ j } N j=1 with λ 1 = T λ 2 λ n be the distinct nonzero eigenvalues of T ; if N = then λ n 0. Let N j = N (T λ j I) be the λ j - eigenspace. Then d j = dim N j < and let {φ j,k : k = 1, 2,, d j } be the orthonormal basis of N j. Let N 0 = N (T ) be the 0-eigenspace and let {ψ k } with k = 0, 1,, N be an orthonormal basis of N 0 (since N 0 is separable, N = 0 if dim N 0 = 0, and N = if dim N 0 = ). Then, for all x H, and x = d N j (φ j,k, x)φ j,k + j=1 k=1 T x = N (ψ k, x)ψ k k=0 d N j λ j (φ j,k, x)φ j,k. j=1 k=1 When N = the sum is taken as infinite series. Proof. It is easy to see that N j Ni if i j. Hence {N j : j = 0, 1, N} is an orthonormal set. Let Z = N j=0 N j and X = Z in H. We have T (X) X. Hence T (X ) X. Then T X is also linear, compact and self-adjoint on X. Suppose X {0}. Then, by Theorem 1.26 above, there exists an eigenvalue µ of T X on X such that µ = T X. Let w X be an eigenvector to this eigenvalue. This eigenvector must be an eigenvector of T on H; however, all eigenvectors of T are in X. Therefore, X = {0} and thus X = H.

18

19 Chapter 2 Sobolev Spaces This chapter is devoted to the study of the necessary Sobolev function spaces which permit a modern approach to partial differential equations Weak Derivatives and Sobolev Spaces Weak Derivatives. Let be a nonempty open set in R n. Suppose u C m () and ϕ C0 m (). Then by integration by parts (2.1) ud α ϕdx = ( 1) α vϕdx, α m where v = D α u. Motivated by (2.1), we now enlarge the class of functions for which the notion of derivative can introduced. Let u L 1 loc (). A function v L1 loc () is called the αth weak derivative of u if it satisfies (2.2) ud α ϕdx = ( 1) α vϕdx for all ϕ C α 0 (). It can be easily shown that the weak derivative is unique. Thus we write v = D α u to indicate that v is the α th weak derivative of u. If a function u has an ordinary α th derivative lying in L 1 loc (), then it is clearly the αth weak derivative. In contrast to the corresponding classical derivative, the weak derivative D α u is defined globally on all of by (2.2). However, in every subregion the function D α u will also be the weak derivative of u. It suffices to note that (2.2) holds for every function ϕ C α 0 ( ), and extended outside by assigning to it the value zero. In particular, the weak derivative (if it exists) of a function u having compact support in has itself compact support in and thus belongs to L 1 (). We also note that in contrast to the classical derivative, the weak derivative D α u is defined at once for order α without assuming the existence of corresponding derivatives of lower orders. In fact, the derivatives of lower orders may not exist as we will see in a forthcoming exercise. However, it can be shown that if all weak derivatives exist of a certain order, then all lower order weak derivatives exist. 15

20 16 2. Sobolev Spaces Example 2.1. (a) The function u(x) = x 1 has in the ball = B(0, 1) weak derivatives u x1 = sgn x 1, u xi = 0, i = 2,..., n. In fact, we apply formula (2.2) as follows: For any ϕ C0 1() x 1 ϕ x1 dx = x 1 ϕ x1 dx + x 1 ϕ x1 dx where + = (x 1 > 0), = (x 1 < 0). Since x 1 ϕ = 0 on and also for x 1 = 0, an application of the divergence theorem yields x 1 ϕ x1 dx = ϕdx + ϕdx = (sgn x 1 )ϕdx. + Hence x 1 x1 = sgn x 1. Similarly, since for i 2 x 1 ϕ xi dx = ( x 1 ϕ) xi dx = 0ϕdx x 1 xi = 0 for i = 2,..., n. Note that the function x 1 has no classical derivative with respect to x 1 in. (b) By the above computation, the function u(x) = x has a weak derivative u (x) = sgn x on the interval = ( 1, 1). On the other hand, sgn x does not have a weak derivative on due to the discontinuity at x = 0. (c) Let = B(0, 1/2) R 2 and define u(x) = ln(ln(2/r)), x, where r = x = (x x2 2 )1/2. Then u L () because of the singularity at the origin. However, we will show that u has weak first partial derivatives. First of all u L 2 (), for u 2 dx = 2π 1/2 0 0 r[ln(ln(2/r))] 2 dr dθ and a simple application of L hopitals rule shows that the integrand is bounded and thus the integral is finite. Similarly, it is easy to check that the classical partial derivative u x1 = cos θ r ln(2/r), where x 1 = r cos θ also belongs to L 2 (). Now we show that the defining equation for the weak derivative is met. Let ε = {x : ε < r < 1/2} and choose ϕ C0 1 (). Then by the divergence theorem and the absolute continuity of integrals uϕ x1 dx = lim uϕ x1 dx = lim ε 0 ε ε 0 [ ] u x1 ϕdx + uϕn 1 ds ε r=ε where n = (n 1, n 2 ) is the unit outward normal to ε on r = ε. But (ds = εdθ) 2π uϕn 1 ds u ϕ εdθ 2πεc ln(ln(2/ε)) 0 as ε 0. Thus r=ε 0 uϕ x1 dx = u x1 ϕdx. The same analysis applies to u x2. Thus u has weak first partial derivatives given by the classical derivatives which are defined on \{0}.

21 2.1. Weak Derivatives and Sobolev Spaces Sobolev Spaces. For p 1 and k a nonnegative integer, we let W k,p () = {u : u L p (), D α u L p (), 0 < α k} where D α u denotes the α th weak derivative. When k = 0, W k,p () will mean L p (). It is clear that W k,p () is a vector space. A norm on W k,p () is introduced by defining { ( α k (2.3) u k,p = u W k,p () = Dα u p dx) 1/p if 1 p <, α k Dα u L () if p =. The space W k,p () is known as a Sobolev space of order k and power p. We define the space W k,p 0 () to be the closure of the space C0 k () with respect to the norm k,p. As we shall see shortly, W k,p () W k,p 0 () for k 1. (Unless = R n.) Remark 2.1. The spaces W k,2 () and W k,2 0 () are special since they become a Hilbert space under the inner product (u, v) k,2 = (u, v) W k,2 () = D α ud α vdx. α k Since we shall be dealing mostly with these spaces in the sequel, we introduce the special notation: H k () = W k,2 (), H k 0 () = W k,2 0 (). Theorem 2.2. W k,p () is a Banach space under the norm (2.3). If 1 < p <, it is reflexive, and if 1 p <, it is separable. Proof. 1. We first prove that W k,p () is complete with respect to the norm (2.3). We prove this for 1 p < ; the case p = is similar. Let {u n } be a Cauchy sequence of elements in W k,p (), i.e., u n u m p k,p = D α u n D α u m p dx 0 as m, n. α k Then for any α, α k, when m, n D α u n D α u m p dx 0 and, in particular, when α = 0 u n u m p dx 0. Since L p () is complete, it follows that there are functions u α L p (), α k such that D α u n u α (in L p ()). Since each u n (x) has weak derivatives (up to order k) belonging to L p (), a simple limit argument shows that u α is the α th weak derivative of u 0. In fact, ud α ϕdx u n D α ϕdx = ( 1) α ϕd α u n dx ( 1) α u α ϕdx. Hence u 0 W k,p () and u n u 0 k,p 0 as n. This proves the completeness of W k,p (); hence it is a Banach space. 2. Consider the map T : W 1,p () (L p ()) n+1 defined by T u = (u, D 1 u,..., D n u).

22 18 2. Sobolev Spaces If we endow the latter space with the norm n+1 v = ( v i p p) 1/p i=1 for v = (v 1,..., v n+1 ) (L p ()) n+1, then T is a (linear) isometry. Now (L p ()) n+1 is reflexive for 1 < p < and separable for 1 p <. Since W 1,p () is complete, its image under the isometry T is a closed subspace of (L p ()) n+1 which inherits the corresponding properties as does W 1,p () (see Theorem??). Similarly, we can handle the case k 2. The following result is of independent importance. We omit the proof. Theorem 2.3. u n u in W k,p () if and only if D α u n D α u in L p () for all α k. Example 2.4. Let be a bounded open connected set in R n. Divide into N open disjoint subsets 1, 2,..., N. Suppose the function u : R has the following properties: (i) u is continuous on. (ii) For some i, D i u is continuous on 1, 2,..., N, and can be extended continuously to 1, 2,..., N, respectively. (iii) The surfaces of discontinuity are such that the divergence theorem applies. Define w i (x) = D i u(x) if x N i=1 i. Otherwise, w i can be arbitrary. We now claim that w i L p () is a weak partial derivative of u on. Indeed, for all ϕ C0 1 (), the divergence theorem yields ud i ϕdx = ud i ϕdx j j = ( ) uϕn i ds ϕd i udx j j j = ϕd i udx. Note that the boundary terms either vanish, since ϕ has compact support, or cancel out along the common boundaries, since u is continuous and the outer normals have opposite directions. Similarly, if u C k ( ) and has piecewise continuous derivatives in of order k + 1, then u W k+1,p (). Remark 2.2. More generally, by using a partition of unity argument, we can show the following: If O is a collection of nonempty open sets whose union is and if u L 1 loc () is such that for some multi-index α, the α th weak derivative of u exists on each member of O, then the α th weak derivative of u exists on. Exercise 2.3. (a) Consider the function u(x) = sgnx 1 + sgnx 2 in the ball B(0, 1) R 2. Show that the weak derivative u x1 does not exist, yet the weak derivative u x1 x 2 does exist. (b) Let be the hemisphere of radius R < 1 in R n : r 2 x 2 i R 2, x n 0 (n 3). i=1 Show that u = (r (n/2 1) ln r) 1 H 1 ().

23 2.2. Approximations and Extensions 19 (c) Let B = B(0, 1) be the open unit ball in R n, and let u(x) = x α, x B. For what values of α, n, p does u belong to W 1,p (B)? 2.2. Approximations and Extensions Mollifiers. Let x R n and let B(x, h) denote the open ball with center at x and radius h. For each h > 0, let ω h (x) C (R n ) satisfy ω h (x) 0; ω h (x) = 0 for x h ω h (x)dx = ω h (x)dx = 1. R n B(0,h) Such functions are called mollifiers. For example, let { k exp [( x 2 1) 1 ], x < 1, ω(x) = 0, x 1, where k > 0 is chosen so that R n ω(x) dx = 1. Then, a family of mollifiers can be taken as ω h (x) = h n ω(x/h) for h > 0. Let be a nonempty open set in R n and let u L 1 (). We set u = 0 outside. Define for each h > 0 the mollified function u h (x) = ω h (x y)u(y)dy where ω h is a mollifier. Remark 2.4. There are two other forms in which u h can be represented, namely (2.4) u h (x) = ω h (x y)u(y)dy = ω h (x y)u(y)dy R n B(x,h) the latter equality being valid since ω h vanishes outside the (open) ball B(x, h). Thus the values of u h (x) depend only on the values of u on the ball B(x, h). In particular, if dist(x, supp(u)) h, then u h (x) = 0. Theorem 2.5. Let be a nonempty open set in R n. Then (a) u h C (R n ). (b) If supp(u) is a compact subset of, then u h C0 small. () for all h sufficiently Proof. Since u is integrable and ω h C, the Lebesgue theorem on differentiating integrals implies that for α < D α u h (x) = u(y)d α ω h (x y)dy i.e., u h C (R n ). Statement (b) follows from the remark preceding the theorem. With respect to a bounded set we construct another set (h) as follows: with each point x as center, draw a ball of radius h; the union of these balls is then (h). Clearly (h). Moreover, u h can be different from zero only in (h).

24 20 2. Sobolev Spaces Corollary 2.6. Let be a nonempty bounded open set in R n and let h > 0 be any number. Then there exists a function η C (R n ) such that 0 η(x) 1; η(x) = 1, x (h) ; η(x) = 0, x ( (3h) ) c. Such a function is called a cut-off function for. Proof. Let χ(x) be the characteristic function of the set (2h) : χ(x) = 1 for x (2h), χ(x) = 0 for x (2h) and set η(x) χ h (x) = ω h (x y)χ(y)dy. R n Then η(x) = ω h (x y)dy C (R n ), (2h) 0 η(x) ω h (x y)dy = 1, R n and { B(x,h) η(x) = ω h (x y)χ(y)dy = ω h(x y)dy = 1, x (h), 0, x ( (3h) ) c. B(x,h) In particular, we note that if, there is a function η C0 () such that η(x) = 1 for x, and 0 η(x) 1 in. Henceforth, the notation means that, are open sets, is bounded, and that. We need the following well-known result. Theorem 2.7. (Partition of Unity) Assume R n is bounded and N i=1 i, where each i is open. Then there exist C functions ψ i (x) (i = 1,..., N) such that (a) 0 ψ i (x) 1 (b) ψ i has its support in i (c) N i=1 ψ i(x) = 1 for every x Approximation Theorems. Lemma 2.8. Let be a nonempty bounded open set in R n. Then every u L p () is p-mean continuous, i.e., u(x + z) u(x) p dx 0 as z 0. Proof. Choose a > 0 large enough so that is strictly contained in the ball B(0, a). Then the function { u(x) if x, U(x) = 0 if x B(0, 2a) \ belongs to L p (B(0, 2a)). For ε > 0, there is a function Ū C( B(0, 2a)) which satisfies the inequality U Ū L p (B(0,2a)) < ε/3. By multiplying Ū by an appropriate cut-off function, it can be assumed that Ū(x) = 0 for x B(0, 2a) \ B(0, a). Therefore for z a, U(x + z) Ū(x + z) L p (B(0,2a)) = U(x) Ū(x) L p (B(0,a)) ε/3.

25 2.2. Approximations and Extensions 21 Since function Ū is uniformly continuous in B(0, 2a), there is a 0 < δ < a such that Ū(x + z) Ū(x) L p (B(0,2a)) ε/3 whenever z < δ. Hence for z < δ we easily see that u(x + z) u(x) L p () = U(x + z) U(x) L p (B(0,2a)) ε. Theorem 2.9. Let be a nonempty open set in R n. If u L p () (1 p < ), then (a) u h p u p (b) u h u p 0 as h 0. If u C k ( ) and is compact, then, for all, (c) u h u C k ( ) 0 as h 0. Proof. 1. If 1 < p <, let q = p/(p 1). Then ω h = ω 1/p h ω1/q h and Hölder s inequality implies ( ) p/q u h (x) p ω h (x y) u(y) p dy ω h (x y)dy ω h (x y) u(y) p dy which obviously holds also for p = 1. An application of Fubini s Theorem gives ( ) u h (x) p dx ω h (x y)dx u(y) p dy u(y) p dy which implies (a). 2. To prove (b), let ω(x) = h n ω h (hx). Then ω(x) C (R n ) and satisfies ω(x) 0; ω(x) = 0 for x 1 ω(x)dx = ω(x)dx = 1. R n B(0,1) Using the change of variable z = (x y)/h we have u h (x) u(x) = [u(y) u(x)]ω h (x y)dy = B(x,h) B(0,1) Hence by Hölder s inequality u h (x) u(x) p d and so by Fubini s Theorem u h (x) u(x) p dx d B(0,1) B(0,1) [u(x hz) u(x)]ω(z)dz. u(x hz) u(x) p dz ( u(x hz) u(x) p dx)dz. The right-hand side goes to zero as h 0 since every u L p () is p-mean continuous. 3. We now prove (c) for k = 0. Let, be such that. Let h 0 be the shortest distance between and. Take h < h 0. Then u h (x) u(x) = [u(y) u(x)]ω h (x y)dy. B(x,h)

26 22 2. Sobolev Spaces If x, then in the above integral y. Now u is uniformly continuous in and ω h 0, and therefore for an arbitrary ε > 0 we have u h (x) u(x) ε ω h (x y)dy = ε provided h is sufficiently small. exercise. B(x,h) The case k 1 is handled similarly and is left as an Remark 2.5. In (c) of the theorem above, we cannot replace by. Let u 1 for x [0, 1] and consider u h (x) = 1 0 ω h(x y)dy, where ω h (y) = ω h ( y). Now h h ω h(y)dy = 1 and so u h (0) = 1/2 for all h < 1. Thus u h (0) 1/2 1 = u(0). Moreover, for x (0, 1) and h sufficiently small, (x h, x + h) (0, 1) and so u h (x) = x+h x h ω h(x y)dy = 1 which implies u h (x) 1 for all x (0, 1). Corollary Let be a nonempty open set in R n. Then C 0 () is dense in Lp () for all 1 p <. Proof. Suppose first that is bounded and let. For a given u L p () set { u(x), x v(x) = 0, x \. Then u v p dx = u p dx. \ By the absolute continuity of integrals, we can choose so that the integral on the right is arbitrarily small, i.e., u v p < ε/2. Since supp(v) is a compact subset of, Theorems 2.5(b) and 2.9(b) imply that for h sufficiently small, v h (x) C0 () with v v h p < ε/2, and therefore u v h p < ε. If is unbounded, choose a ball B large enough so that u p dx < ε/2 \ where = B, and repeat the proof just given. We now consider the following local approximation theorem. Theorem Let be a nonempty open set in R n and suppose u, v L 1 loc (). Then v = D α u iff there exists a sequence of C α () functions {u h } with u h u L 1 (S) 0, D α u h v L 1 (S) 0 as h 0, for all compact sets S. Proof. 1. (Necessity) Suppose v = D α u. Let S, and choose d > 0 small enough so that the sets S (d/2), S (d) satisfy. For x R n define u h (x) = ω h (x y)u(y)dy, v h (x) = ω h (x y)v(y)dy. Clearly, u h, v h C (R n ) for h > 0. Moreover, from Theorem 2.9 we have u h u L 1 (S) u h u L 1 ( ) 0. Now we note that if x and 0 < h < d/2, then ω h (x y) C0 ( ). Thus by Theorem 2.5 and the definition of weak derivative, D α u h (x) = u(y)dx α ω h (x y)dy = ( 1) α u(y)dy α ω h (x y)dy = ω h (x y) v(y)dy = v h (x).

27 2.2. Approximations and Extensions 23 Thus, D α u h v L 1 (S) (Sufficiency) Choose ϕ C α 0 () and consider a compact set S supp(ϕ). Then as h ud α ϕdx u h D α ϕdx = ( 1) α ϕd α u h dx ( 1) α vϕdx S which is the claim. S Theorem Let be a domain in R n. If u L 1 loc () has a weak derivative Dα u = 0 whenever α = 1, then u =const. a.e. in. Proof. Let. Then for x and with u h as in Theorem 2.11, D α u h (x) = (D α u) h (x) = 0 for all h sufficiently small. Thus u h = const = c(h) in for such h. Since u h u L 1 ( ) = c(h) u L 1 ( ) 0 as h 0, it follows that c(h 1 ) c(h 2 ) L 1 ( ) = c(h 1 ) c(h 2 ) mes( ) 0 as h 1, h 2 0. Consequently, c(h) = u h converges uniformly and thus in L 1 ( ) to some constant. Hence u = const (a.e.) in and therefore also in, by virtue of it being connected. We now note some properties of W k,p () which follow easily from the results of this and the previous section. (a) If and if u W k,p (), then u W k,p ( ). (b) If u W k,p () and a(x) k, <, then au W k,p (). In this case any weak derivative D α (au) is computed according to the usual rule of differentiating the product of functions. (c) If u W k,p () and u h is its mollified function, then for any compact set S, u h u W k,p (S) 0 as h 0. If in addition, u has compact support in, then u h u k,p 0 as h 0. More generally, we have the following global approximation theorems. Serrin H = W. The proofs make use of a partition of unity argument.) S S (See Meyers and Theorem Assume is bounded and let u W k,p (), 1 p <. Then there exist functions u m C () W k,p () such that u m u in W k,p (). In other words, C () W k,p () is dense in W k,p (). Theorem Assume is bounded and C 1. Let u W k,p (), 1 p <. Then there exist functions u m C ( ) such that In other words, C ( ) is dense in W k,p (). u m u in W k,p (). Exercise 2.6. Prove the product rule for weak derivatives: D i (uv) = (D i u)v + u(d i v) where u, D i u are locally L p (), v, D i v are locally L q () (p > 1, 1/p + 1/q = 1). Exercise 2.7. (a) If u W k,p 0 () and v C k ( ), prove that uv W k,p 0 (). (b) If u W k,p () and v C0 k k,p (), prove that uv W0 ().

28 24 2. Sobolev Spaces Chain Rules. Theorem Let be an open set in R n. Let f C 1 (R), f (s) M for all s R and suppose u has a weak derivative D α u for α = 1. Then the composite function f u has a weak derivative D α (f u) = f (u)d α u. Moreover, if f(0) = 0 and if u W 1,p (), then f u W 1,p (). Proof. 1. According to Theorem 2.11, there exists a sequence {u h } C 1 () such that u h u L 1 ( ) 0, D α u h D α u L 1 ( ) 0 as h 0, where. Thus f(u h ) f(u) dx sup f u h u dx 0 as h 0 f (u h )D α u h f (u)d α u dx sup f D α u h D α u dx + f (u h ) f (u) D α u dx. Since u h u L 1 ( ) 0, there exists a subsequence of {u h }, which we call {u h } again, which converges a.e. in to u. Moreover, since f is continuous, {f (u h )} converges to f (u) a.e. in. Hence the last integral tends to zero by the dominated convergence theorem. Consequently, the sequences {f(u h )}, {f (u h )D α u h } tend to f(u), f (u)d α u respectively, and the first conclusion follows by an application of Theorem 2.11 again. 2. If f(0) = 0, the mean value theorem implies f(s) M s for all s R. Thus, f(u(x)) M u(x) for all x and so f u L p () if u L p (). Similarly, f (u(x))d α u L p () if u W 1,p (), which shows that f u W 1,p (). Corollary Let be a bounded open set in R n. If u has an α th weak derivative D α u, α = 1, then so does u and D α u if u > 0 D α u = 0 if u = 0 D α u if u < 0 i.e., D α u = (sgn u)d α u for u 0. In particular, if u W 1,p (), then u W 1,p (). Proof. The positive and negative parts of u are defined by u + = max{u, 0}, u = min{u, 0}. If we can show that D α u + exists and that { D α u + D = α u if u > 0 0 if u 0 then the result for u follows easily from the relations u = u + u and u = ( u) +. Thus, for h > 0 define { (u f h (u) = 2 + h 2 ) 1 2 h if u > 0 0 if u 0. Clearly f h C 1 (R) and f h is bounded on R. By Theorem 2.15, f h(u) has a weak derivative, and for any ϕ C0 1 () f h (u)d α ϕdx = D α (f h (u))ϕdx = ϕ udα u dx. u>0 (u 2 + h 2 ) 1 2

29 2.2. Approximations and Extensions 25 Upon letting h 0, it follows that f h (u) u +, and so by the dominating convergence theorem u + D α ϕdx = ϕd α udx = vϕdx u>0 where v = { D α u if u > 0 0 if u 0 which establishes the desired result for u +. The next result extends the result on u, u + and u. Theorem Let f : R R be Lipschitz continuous with f(0) = 0. Then if is a bounded open set in R n, 1 < p < and u W 1,p 0 (), we have f u W 1,p 0 (). Proof. Given u W 1,p 0 (), let u n C 1 0 () with u n u 1,p 0 and define v n = f u n. Since u n has compact support and f(0) = 0, v n has compact support. Also v n is Lipschitz continuous, for v n (x) v n (y) = f(u n (x)) f(u n (y)) c u n (x) u n (y) c n x y. Hence v n L p (). Since v n is absolutely continuous on any line segment in, its partial derivatives (which exist almost everywhere) coincide almost everywhere with the weak derivatives. Moreover, we see from above that v n / x i c n for 1 i n, and as is bounded, v n / x i L p (). Thus v n W 1,p () and has compact support, which implies v n W 1,p 0 (). From the relation v n (x) f(u(x)) c u n (x) u(x) it follows that v n f u p 0. Furthermore, if e i is the standard ith basis vector in R n, we have and so v n (x + he i ) v n (x) h lim sup v n n c u n(x + he i ) u n (x) h x i p c lim sup n u n x i p. But, { u n / x i } is a convergent sequence in L p () and therefore { v n / x i } is bounded in L p () for each 1 i n. Since v n 1,p is bounded and W 1,p 0 () is reflexive, a subsequence of {v n } converges weakly in W 1,p (), and thus weakly in L p () to some element of W 1,p 0 (). Thus, f u W 1,p 0 (). Corollary Let u W 1,p 0 (). Then u, u +, u W 1,p 0 (). Proof. We apply the preceding theorem with f(t) = t. Thus u W 1,p 0 (). Now u + = ( u + u)/2 and u = (u u )/2. Thus u +, u W 1,p 0 ().