Lecture Notes. Functional Analysis in Applied Mathematics and Engineering. by Klaus Engel. University of L Aquila Faculty of Engineering

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1 Lecture Notes Functional Analysis in Applied Mathematics and Engineering by Klaus Engel University of L Aquila Faculty of Engineering ( = %7E) (Preliminary Version of October 16, 2012)

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3 Contents Introduction 1 Chapter 1. Normed Vector Spaces 2 1. Vector Spaces 2 2. Metric Spaces 5 3. Normed Vector Spaces 6 4. Continuity and Convergence in Metric Spaces 9 5. Topology in Metric Spaces Completeness and Banach Spaces 14 Chapter 2. Lebesgue Integration and L p -spaces A Reminder on Riemann s Integral The Lebesgue Measure and Integrable Functions The L p -Spaces 30 Chapter 3. Linear Operators Basic Definitions Three Cornerstones of Functional Analysis 37 Chapter 4. Dual Spaces and Weak Topologies Linear Functionals and Dual Spaces The Weak Topology Sobolev Spaces and Distributions 48 Chapter 5. Hilbert Spaces Basic Definitions Orthogonality Fourier Series in Hilbert Spaces Dual Spaces and Adjoint Operators 61 Chapter 6. Spectral Theory for Bounded Linear Operators Basic Definitions Compact Operators Spectral Theorem for Normal Compact Operators 75 Chapter 7. Some Applications 77 Index 78 iii

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5 Introduction Roughly speaking, Functional Analysis is Linear Algebra with the aid of Analysis in infinite dimensional vector spaces. Its development started about 100 years ago from the observation that problems in quite different fields of mathematics enjoy similar features and properties. By omitting unnecessary details FA provides an effective and unified approach to these problems. The advantage of this abstract approach is that it concentrates on the essential facts, so that these facts become clearly visible since the attention is not disturbed by unimportant details. In this respect this abstract method provides the simplest and most economical way for treating certain mathematical problems. Moreover, it helps to find relations between fields which have at a first glance no contact with each other. In the abstract approach, one usually starts from a set of elements and operations satisfying certain axioms. The theory then consists of logical consequences resulting from the axioms and are derived as theorems. These general theorems can then be applied to various special sets satisfying those axioms. In this connection the concept of space is used in a wide and general sense. By choosing different sets of axioms one obtains different types of abstract spaces. This idea goes back to Maurice Fréchet (1906, definition of metric space) and is justified by its great success. 1

6 CHAPTER 1 Normed Vector Spaces In the spirit of the Introduction, we start by pursuing the following Aim 1.1. Generalize the concept of the three dimensional euclidian space (R 3, ) or, more generally, of the n-dimensional space (K n, ), n 1, given by 1 K n := { x = (x 1,..., x n ) x i K 1 i n }, ( n x := x i 2) 1 2 = length of the vector x K n. i=1 Basically, this space has two (independent) structures: firstly a linear (or vector space ) structure which allows to add and subtract elements x, y K n (called vectors ) and to multiply them with numbers α K (called scalars ). Roughly speaking, this encodes how we can move in this space. Secondly, there is a metric structure which allows to measure the distance d(x, y) = x y between two points x and y in the space, or the length (called norm ) x of a vector x. This metric is sufficient to introduce the fundamental concepts of convergence and limit (e.g. of a sequence (x n ) n N K n ) and many further analytical notions based on this like continuity of a function or openness of a set. We already mention that the concept of metric is not sufficient to measure angles between vectors in a space, or even to define the important concept of orthogonality. To do so one needs further refinements which lead to the notions of inner product and Hilbert spaces, see Chapter 5. The introduction of the above abstract concepts (i.e., vector space, metric, inner product) is the quintessence of a long development in the history and mathematics and is justified by its great success in many fields of mathematics. In the sequel of this chapter we will start to introduce in a formal way some of the mentioned concepts and will also derive some first consequences. 1. Vector Spaces Definition 1.2. Let be given a field K (e.g. K = R or C), a set X and two mappings + : X X X, (x, y) x + y (addition) : K X X, (α, y) α x (scalar multiplication) such that x, y, z X, α, β K (i) x + y = y + x (ii) (x + y) + z = x + (y + z) (iii) 0 X such that 0 + x = x (iv) x X such that x + ( x) = 0 (v) (α + β) x = α x + β x (vi) α (x + y) = α x + α y (vii) (α β) x = α (β x) (commutativity), (associativity), ( zero element), ( inverse element), (distributivity), (distributivity), (associativity), 1 Here and in the sequel K indicates always the real numbers R or the complex numbers C. Moreover, A := B or B =: A indicates that A = B by definition. 2

7 1. VECTOR SPACES 3 (viii) 1 x = x (1 = unit element of K). Then X (more precisely (X, K,, +) is called a vector space (over the field K). Remarks 1.3. (i) (iv) (X, +) is an abelian group. If X is a K-vector space, then x X = vectors, α K = scalars. We will only consider vector spaces over R (i.e., real vector spaces) and C (i.e., complex vector spaces) and use as already mentioned the notation K to indicate either of them. Let s consider some Examples 1.4. is a K-vector space. Let S be a set, then X = K n with the operations equipped with the operations (x 1,..., x n ) + (y 1,..., y n ) := (x 1 + y 1,..., x n + y n ) α (x 1,..., x n ) := (α x 1,..., α x n ) F (S) : = {f : S K} = set of all K-valued functions on S (f + g)(s) := f(s) + g(s), (α f)(s) := α f(s) is a K-vector space. Note that K n = F ({1, 2,..., n}). The set X = C([a, b], K) = C[a, b] of all continuous, K-valued functions on [a, b] is a K-vector space with respect to the addidion and scalar multiplication defined in the previous example. Remark 1.5. C[a, b] F ([a, b]), i.e. C[a, b] is a subspace of F ([a, b]) in the following sense. Definition 1.6. Let X be a vector space and let Y X. If (1.1) α x + β y Y }{{} =linear combination x, y Y, α, β K (i.e., Y is closed under addition and scalar multiplication), then Y is called (linear) subspace of X. Remarks 1.7. A subspace of a vector space is always a vector space. Condition (1.1) is equivalent to α x Y and x + y Y x, y Y, α K. Examples 1.8. Y = {(x, 0, z) x, z R} R 3 is a subspace of R 3. If s 0 S, then {f : S K f(s 0 ) = 0} is a subspace of F (S). P := {p : p is a polynomial} C(R) is a subspace of C(R). An arbitrary intersection of subspaces of a vector space X is again a subspace. However, the union of two subspaces in general is not a subspace anymore. Translating a subspace Y of a vector space X by some fixed vector x 0 X we obtain x 0 + Y := {x 0 + y : y Y } = affine subspace. If Y is maximal (i.e. subspaces Z X satisfying Y Z it follows Z = X), then x 0 + Y is called hyperplane.

8 4 1. NORMED VECTOR SPACES Examples 1.9. Every line passing through the origin is a subspace of R 2, every line is a hyperplane in R 2. Every line or plane passing through the origin is a subspace of R 3, every line is an affine subspace of R 3, planes are exactly the hyperplanes in R 3. The solution set S of a linear equation ( ) T f = g 0 for a linear 2 map T : X Y between two vector spaces X and Y and a given g 0 Y is an affine subspace. More precisely, one can represent S = f 0 + ker T where f 0 is a particular solution of ( ) and ker T = {x X : T (x) = 0}, i.e., ker(t ) is the solution set of the corresponding homogeneous equation (with g 0 = 0). Note that this is just an abstract formulation of the well known fact that the general solution of a linear system (or a linear differential equation) is given by the sum of a particular solution and the general solution of the associated homogeneous equation. Next we study how vector spaces can be generated. Definition Let X be a vector space, then x 1,... x n X are called linearly independent if n α k x k = 0 α 1 = α 2 =... = α n = 0; }{{} = linear combination linearly dependent if there exist α 1,..., α n not all = 0, such that n α k x k = 0. More generally a set V X is called linearly independent if every nonempty finite subset of V is linearly independent Examples (1, 0, 0), (1, 1, 0), (1, 1, 1) R 3 are linearly independent while (1, 1), ( 1, 1), (1, 2) R 2 are linear dependent. Let p k (s) = s k, then 3 {p k : k N 0 } C[0, 1] is linearly independent. In fact, any linear combination of the p k, k N 0 being = 0 is a polynomial having infinitely many zeros and hence is zero. Definition If X ia a vector space and x 1,..., x n X, then { n } span{x 1,..., x n } := α k x k α 1,..., α n K is called the subspace generated (or spanned) by x 1,..., x n. If V X (possibly infinite), then we define its span as the set of all (finite!) linear combinations of elements in V, i.e., { n } span V := α k x k x1,..., x n V, α 1,..., α n K, n N. In other words, for a subset V X of a vector space X, span V is the smallest subspace of X containing V. Examples span{(1, 0, 1), (0, 1, 0)} = {(x, y, x) : x, y R} R 3. If p k (s) = s k, then span{p 0,..., p n } = {p : p is a polynomial of degree span{p k : k N 0 } = {p : p is a polynomial}. 2 Linear means that T (αx + βy) = αt (x) + βt (y) for all scalars α, β and all vectors x, y. 3 Here and in the sequel N0 := N {0}. n}, while

9 2. METRIC SPACES 5 Definition If X = span{x 1,..., x n } for n linearly independent vectors x 1,..., x n, then we say that X has dimension dim(x) := n and call {x 1,..., x n } a (Hamel) basis of X. If in X there exists an infinite linearly independent subset, then X is called infinite dimensional and we write dim(x) = +. Remarks If {x 1,..., x n } and {y 1,..., y m } are two bases of X, then necessarily m = n, i.e., the concept of dimension is well defined. Using Zorn s lemma one can show that every vector space {0} has a basis. Examples K is a one-dimensional K-vector space. C considered as a R-vector space has dimension 2. For example C = span{1, i} and 1 and i are linearly independent over R. R considered as a Q-vector space is infinite dimensional. The vectors e 1 := (1, 0, 0), e 2 := (0, 1, 0) and e 3 := (0, 0, 1) R 3 are linearly independent and R 3 = span{e 1, e 2, e 3 }. Hence R 3 is 3-dimensional. More generally, dim(k n ) = n. If S denotes the number of elements of the set S (where S = + if S is infinite), then dim ( F (S) ) = S. As we have seen above, P = {p n : n N 0 } C[0, 1] (p k (s) = s k ) is linearly independent and hence C[0, 1] is infinite dimensional. As we will see next among the finite dimensional K-vector spaces K n is fundamental. To this end we need the following Definition Two vector spaces X 1 and X 2 over the same field K are (algebraically) isomorphic if there exists a bijective map T : X 1 X 2 such that α 1, α 2 K, x 1, x 2 X Now one can show: T (α 1 x 1 + α 2 x 2 ) = α 1 T (x 1 ) + α 2 T (x 2 ) (i.e., T is linear) Proposition Every n-dimensional K-vector space is isomorphic to K n. Simply speaking this means that every finite dimensional vector space X (over the field K) can be identified by K n where n = dim(x). 2. Metric Spaces In the previous section we considered only the linear, i.e., the algebraic structure of K n and derived the notion of a vector space. In this section we introduce the concept of a metric which generalizes the notion of distance between two points from everyday life. Definition Let X a set 4 and consider a map d : X X R such that x, y, z X (i) d(x, y) = 0 iff x = y (identity of indiscernibles), (ii) d(x, y) = d(y, x) (symmetry), (iii) d(x, z) d(x, y) + d(y, z) (triangle inequality). Then X (more precisely the pair (X, d)) is called a metric space. The function d is called metric, distance function or simply distance since it has the following important interpretation 4 Note that we do not need X to be a vector space. d(x, y) = distance between x and y in X.

10 6 1. NORMED VECTOR SPACES We note that d(x, y) is always non-negative, since by the above axioms 2d(x, y) = d(x, y) + d(y, x) d(x, x) = 0. Examples Ignoring mathematical details, for any system of roads and terrains the distance between two locations can be defined as the length of the shortest route connecting those locations. To be a metric (symmetry!) there shouldn t be any one-way roads. The triangle inequality then expresses the fact that detours aren t shortcuts. Any subset X R with the distance function d(x, y) = x y given by the absolute difference is a metric spaces. More generally, any subset X K n with the distance function ( n d(x, y) = x y := x i y i 2) 1 2 i=1 given by the norm of the difference is a metric spaces. If X is an arbitrary non-empty set, then { 0 if x = y, d(x, y) := 1 if x y for x, y X defines a metric, called the discrete metric on X. Many more examples of metric spaces can be found here. The importance of metric spaces stems from the fact that the concept of distance allows to generalize many concepts from calculus like convergence, continuity or open/closed set to metric spaces. In most cases it suffices to replace terms like x y (or x y in higher dimensions) by d(x, y). Before we will elaborate in this direction we combine the algebraic and the metric structures defined previously. 3. Normed Vector Spaces Above we generalized independently the algebraic and the metric structures of K n to obtain the two concepts of vector- and of metric space. However doing so in general these concepts are not compatible. More precisely, in a vector space equipped with a metric d the quantity d(x, 0) represents the distance between x and the origin 0, in other words the length of the vector x. Then it is natural to ask that for example the vectors ±2x have twice the length of x, i.e., d(±2x, 0) = 2d(x, 0) x. This, however, does not follow from the axioms of a metric. Moreover, in a vector space the distance should be shift invariant, i.e., d(x, y) = d(x z, y z) x, y, z which, again, in general not holds in case of an arbitrary metric. In the following definition we will add a homogeneity condition which takes care of these shortcomings. Definition If X is a K-vector space then a map is called a norm on X if x, y X, α K (i) x = 0 x = 0 (ii) x + y x + y (iii) α x = α x : X R X equipped with a norm is called a normed (vector) space. (positive definiteness); (triangle inequality); (positive homogeneity).

11 3. NORMED VECTOR SPACES 7 Returning to the previous considerations, we remark that every normed vector space is also a metric space if we define d(x, y) := x y In particular this implies x = d(x, 0) 0 for all x X. However, not every metric stems from a norm, for example the discrete metric on R is not induced by a norm. Examples The absolute value is a norm on K, hence (K, ) is a one-dimensional normed vector space. More generally, (K n, 2 ) where for x = (x 1,..., x n ) we define ( n ) 1 x 2 := x k 2 2 is a normed space. Here the properties (i) and (iii) of a norm are trivially satisfied. As we will see later in Proposition 5.3 in Chapter 5, the triangle inequality (ii) follows from Cauchy Schwarz s inequality n 2 ( n ) ( n ) x k y k x k 2 y k 2 The 2-norm 2 introduced in the previous example is the natural (euclidian) generalization of length from 3 to n dimensions. However, this is by far not the only example of a norm on K n and also x := max 1 k n x k, ( n x p := x k p) 1 p, (1 p < + ) define norms on K n. The only non-trivial part to verify this is to show the triangle inequality for the p-norm p, 1 p < +, which follows from Minkowski s inequality 5 ( n x k + y k p ) 1 p ( n ) 1 ( x k p p n + y k p ) 1 p If in the last example we (formally) consider n = + we arrive at various sequence spaces. Example Let X := { x = (x 1, x 2, x 3,...) : x k K k N } = space of all K-valued sequences. Since X = F (N) it is clear that X is an infinite dimensional vector space with respect to coordinatewise operations. On X we define the (possibly infinite-valued) maps 6 : X R + {+ }, p : X R + {+ }, x p := x := sup x k, k N (+ x k p) 1 p, (1 p < + ). Considering only elements with finite norm we obtain l p := { x X : x p < + } { bounded sequences if p = +, = p-summable sequences if 1 p < +. 5 This inequality also holds for n = + if the two series on the right-hand-side converge. 6 Here and in the sequel we will use the notation R+ := [0, + ).

12 8 1. NORMED VECTOR SPACES By Minkowski s inequality on page 7 for n = + it follows that (l p, p ) is a normed space 1 p +. Moreover, l 1 l p l 1 < p <. For example ( 1 n ) n N l p p > 1, while (1) n N l p p = +. Example Let X := C[a, b], then for 1 p < + f := sup f(s), s [a,b] ( b f p := f(s) p ds a ) 1 p, (1 p < + ) define norms on X. This is trivial for p = +. For 1 p < condition (i) of the norm follows from the fact that h C[a, b], b a h(s) ds = 0 h(s) = 0 s [a, b]. Finally, the triangle inequality follows from Minkowski s inequality ( b a ) 1 f(s) + g(s) p p ds (cf. Theorem 2.30 on page 30 below). ( b a ) 1 f(s) p ds p + ( b a ) 1 g(s) p p ds As we have seen the same vector space can be equipped with different norms, which (as we will see in the sequel) might have quite different properties. To relate two normed spaces we make the following Definition Two normed vector spaces (X, X ) and (Y, Y ) are called topologically isomorphic if there exists a bijective linear map T : X Y and constants c 1, c 2 > 0 such that c 1 x X T x Y c 2 x X x X. In this case we write X Y. In finite dimensions we have Proposition If (X, X ) is a n-dimensional normed space over K, then X K n, where K n can be equipped with an arbitrary norm. If in the previous definition we can choose c 1 = c 2 = 1 we arrive at the following Definition Two normed vector spaces (X, X ) and (Y, Y ) are called isometrically isomorphic if there exists a bijective linear map T : X Y such that ( ) T x Y = x X x X. Moreover, a map satisfying ( ) is called an isometry. Starting from one or more normed space(s), next we will construct new related ones. The first construction is very simple. Definition A subspace Y of a normed space (X, X ) with the induced norm is called a normed subspace of X. y Y := y X, y Y

13 4. CONTINUITY AND CONVERGENCE IN METRIC SPACES 9 Example Let Y := {p : [0, 1] R p is a real polynomial of degree 5} then Y is a normed subspace of (C[0, 1], ) if it is equipped with the norm Next we consider products of normed spaces. p Y := sup p(s). s [0,1] Proposition Let (X 1, X1 ),..., (X n, Xn ) be normed spaces and define X := X 1 X n := { (x 1,..., x n ) : x k X k 1 k n }. Then each of the following maps : X R +, x := max k Xk, 1 k n p : X R +, ( n ) x p := x k p 1 p X k, (1 p < + ). defines a norm on X, i.e., (X, p ) is a normed space for all 1 p Continuity and Convergence in Metric Spaces Using the concept distance it is quite easy to generalize concepts like convergence, limit or continuity, etc. from R to metric spaces. As already mentioned, in most cases it suffices to replace expressions of the form x y in R (or x y in higher dimensions) by d(x, y). Definition Consider the map F : X Y between two metric spaces (X, d X ) and (Y, d Y ). Then F is said to be continuous in x 0 X, if ε > 0 δ = δ(x 0, ε) > 0 such that d Y ( F (x0 ), F (x) ) < ε x X satisfying d X (x 0, x) < δ; continuous if it is continuous in every x 0 X; uniformly continuous if ε > 0 δ = δ(ε) > 0 such that d Y ( F (x0 ), F (x 1 ) ) < ε x 0, x 1 X satisfying d X (x 0, x 1 ) < δ; 7 Lipschitz continuous if there exists L 0 such that d ( F (x 0 ), F (x 1 ) ) L d(x 0, x 1 ) x 0, x 1 X; in this case L is called Lipschitz constant of F. Examples It is clear that Lipschitz continuity uniform continuity continuity, while the converse even in a one-dimensional normed space is not true. For example the square F : R R, F (s) = s 2 is continuous but not uniformly continuous with respect to the canonical metric induced by the absolute value on R. Moreover, for the same metric the square root G : [0, + ) R, G(s) = s is uniformly continuous but not Lipschitz continuous. Let F : X Y be an arbitrary map between two metric spaces (X, d X ) and (Y, d Y ). Then F is uniformly continuous independently of the metric d Y on Y if on X we choose the discrete metric d X (cf. page 6). Let F : K n K m be linear where we equip K n and K m with the sup-norm. Then F can be represented by a m n-matrix A = (a ij ) m n and for 8 x = (x 1,..., x n ) t, 7 Hence uniform means that δ might depend on ε but not on x0, x 1. 8 Here (...) t denotes the transposed vector.

14 10 1. NORMED VECTOR SPACES x 0 = (x 0 1,..., x0 n) t we obtain F (x) F (x 0 ) = F (x x 0 ) n a 1j (x j x 0 j ) j=1 =. n a mj (x j x 0 j ) j=1 n = max a ij (x j x 0 j) 1 i m j=1 max 1 i m j=1 = q x x 0 n a ij max (x j x 0 j) 1 j m for q := max 1 i m n j=1 a ij = maximal row sum of A. Hence, F is Lipschitz continuous. Remark Since every finite dimensional normed space X is topologically isomorphic to (K n, ) for n = dim(x) and = (see above, page 8), every linear map F : (X, X ) (Y, Y ) defined on a finite dimensional X is uniformly continuous. Next we consider sequences in metric spaces. Definition We say that the sequence (x n ) n N (X, d) converges to x 0 X if In this case we write Remarks As in the scalar case is continuous in x 0 X lim d(x 0, x n ) = 0. n + lim x n = x 0 or x n x 0 as n +. n + the limit of a sequence, if it exists, is unique; sequences can be used to characterize continuity in metric spaces: } F : (X, d X ) (Y, d Y ) lim F (x n) = F (x 0 ) (convergence in Y ) n + (x n ) n N X with lim n + x n = x 0 (in X). If X = K n is equipped with an arbitrary norm and (x k ) k N X, x k = (x k 1,..., xk n), then lim k + xk 1 = x0 1, lim x k = x 0. k + lim k + xk n = x 0 n. Hence convergence in K n in norm is equivalent to convergence in each coordinate. Next we consider series in normed spaces. Definition The series + x k for elements (x n ) n N (X, X ) in a normed space is said to converge to the sum x X if lim n + x n x k }{{} n-th partial sum X = 0.

15 5. TOPOLOGY IN METRIC SPACES Topology in Metric Spaces In real analysis the concepts of open, closed and compact sets are of fundamental importance. Using the notion of convergence of sequences it is easy to generalize them to metric spaces. Definition Let (X, d) be a metric space and A X a subset. Then A is closed if (x n ) n N A and lim x n = x 0 in X implies x 0 A; n + A is open if X \ A is closed; x 0 X is an accumulation point of A if there exists (x n ) n N A \ {x 0 } such that lim x n = x 0 ; n + Ā := A {x 0 X : x 0 is an accumulation point of A} is called the closure of A in X. Remarks A is open x 0 A ε > 0 such that {x X : d(x, x 0 ) < ε} A, i.e., each x 0 A is an interior point of A. Ā is the smallest closed set containing A. By the following proposition it can be represented as Ā = C. A C C closed Proposition Arbitrary unions of open sets are open, arbitrary intersections of closed sets are closed; finite unions of closed sets are closed, finite intersections of open sets are open; Examples In general, arbitrary unions of closed sets are not closed anymore. For example if X = (R, ) and A n := [ 1 n, 1 1 n ], then each A n is closed while n N A n = (0, 1) is not closed anymore. In general, arbitrary intersections of open sets are not open anymore. For example if X = (R, ) and A n := ( 1 n, 1 n ), then each A n is open while n N A n = {0} is not open anymore. Definition A subset A of a metric space X is called dense, if Ā = X. Hence A is dense in X means that any x X can be approximated by elements y A as close as we like, i.e., for every ε > 0 there exists y A such that d(x, y) < ε. Examples Q is dense in (R, ) or, more generally, Q n is dense in (R n, ). Consider the vector space Φ : = { (x k ) k N K : k 0 N such that x k = 0 k N } = space of all finite K-valued sequences. Then (cf. Example 1.23) Φ l p for all 1 p + ; Φ is dense in (l p, p ) 1 p < +. Φ = c 0, where c 0 := { (x k ) k N K : lim x k = 0 } = space of all null sequences. k + Here the first assertion is trivial. The third assertion and the first part of the second assertion follows by truncation of a given sequence x l p while the second part is clear by the fact that for e := (1, 1, 1, 1,...) l we have e x 1 x Φ. By Weierstrass s approximation theorem the space P of all real polynomials is dense in (C([a, b], R), ).

16 12 1. NORMED VECTOR SPACES Definition A metric space is said to be separable, if it contains a countable dense subset S. Examples Every finite dimensional normed space X is separable. (Consider the countable set of all rational linear combinations of elements of a (finite) basis of X). X := (l p, p ) for 1 p + is separable 1 p < +. In fact Φ Q := {(x k ) k N Φ : all x k are rational} is countable and dense in X for 1 p < +. However, l is not separable. (C[a, b], ) is separable. (Consider polynomials with rational coefficients.) Definition A set A (X, d) is called bounded if sup d(x, y) < +. x, y A For a normed space this is equivalent to sup x A x X < +. Next we look at compactness which is one of the most useful concepts of mathematical analysis. Definition Let (X, d) be a metric space and A X a subset. (A i ) i J (here J denotes an arbitrary index set) is called an open cover of A if each A i is open and A i J A i. A is compact if every open cover (A i ) i J has a finite subcover, i.e., if there exists a finite set I J such that A i I A i. A is relatively compact if the closure Ā is compact. The next result relates some of the previous concepts. Proposition Let (X, d) be a metric space and A X. A is compact every sequence in A has a subsequence converging in A (i.e., A is sequentially compact) every infinite subset in A has an accumulation point in A (Bolzano Weierstraß property). If A is compact and C A, then C is relatively compact. In particular, every closed subset of a compact set is compact. Now assume that (X, X ) is a normed space. If A is compact, then it is bounded and closed; the converse holds if and only if dim(x) < + (Theorem of Heine-Borel). Example In a normed space X let U := {x X : x 1} be the closed unit ball. Then U is always bounded and closed. However, U is compact if and only if dim X < +. For example, for 1 p + take X = (l p, p ). Then 9 A := {e k : k N} U where e k := (δ nk ) n N. However, A has no accumulation points since e k e l 1 k l. Continuous maps on compact domains enjoy very strong properties. This fact is one of the reasons of the importance of compact sets in analysis. Proposition Let X, Y be two metric spaces, where X is compact and let F : X Y be continuous. Then F is uniformly continuous. F (X) := {f(x) : x X} is compact in Y. If F is invertible, then F 1 : Y X is continuous. If Y = R, then F achieves its minimum and its maximum (in particular F is bounded). We give a possible application of the previous result to 9 Here δnk denotes the Kronecker delta: δ nk = 1 if n = k, δ nk = 0 if n k.

17 5. TOPOLOGY IN METRIC SPACES 13 Best Approximation. Let (X, d) be a metric space and suppose that a fixed x X is to be approximated by a y 0 C where C is a fixed subset of X. Let δ := dist(x, C) := inf d(x, y) = distance between x and C. y C Clearly, δ depends on both x and C. If there exists y 0 C such that d(x, y 0 ) = δ, i.e., if y 0 C has minimum distance from the given x, then y 0 is called best approximation to x out of C. x δ y 0 C Figure 1. Best approximation y 0 of x in C. Now assume that C is compact in X. Since for every fixed x X the function F : (C, d) R, F (y) := d(x, y) is continuous (proof!) it assumes its minimum in some point y 0 which gives the best approximation of x 0 in C. We give another application. Example By the previous proposition it follows that if K is a compact metric space then f := sup f(s) < + s K defines a norm on C(K) := {f : K K f is continuous}. Relatively compact sets in C(K) are characterized by the following result. Theorem 1.51 (Arzela Ascoli theorem). A set A (C(K), ) is relatively compact if and only if (i) {f(s) : f A} is bounded in K for every s K (i.e. A is pointwise bounded), and (ii) A is equicontinuous. Here a set A C(K) is called equicontinuous, if for every s K and every ε > 0 there exists δ = δ(s) > 0 such that for all f A f(s) f(r) < ε if r K satisfies d(s, r) < δ. 10 Remark The concept of open sets (and consequently closed and compact sets) can be defined in the much more general context of topological spaces. 10 Hence equicontinuous means that δ = δ(ε) might depend on ε but not on f A.

18 14 1. NORMED VECTOR SPACES 6. Completeness and Banach Spaces Note that both (Q, ) and (R, ) are normed spaces, however the latter is much more important. The fundamental difference between these spaces is their completeness: while R is complete Q is not. The completeness of R can be described in various ways (e.g. using the order structure), in a metric space we use the following approach. Definition A sequence (x n ) n N (X, d) in a metric space is called Cauchy sequence if for every ε > 0 there exists n 0 = n 0 (ε) such that d(x n, x m ) < ε n, m n 0. It is easy to show that every convergent sequence in a metric space is Cauchy, however the converse in general does not hold even in a normed space. Example Let X = C[ 1, 1] be equipped with the 1-norm f 1 = 1 1 f(s) ds. Consider the sequence (f n ) n N X defined by { s 1 n if s (0, 1] f n (s) := 0 if s [ 1, 0]. This sequence is Cauchy, since for n > m f n f m 1 = 1 = n Clearly, the pointwise limit of f n is f(s) := 0 [ ( 1 1 ) s 1+ 1 n s n s m ds = m lim f n(s) = n n s1+ 1 ] 1 m m 0 as m, n +. { 1 if s (0, 1] 0 if s [ 1, 0]. which defines a function f / X because of the discontinuity in s = 0. The previous example justifies the following fundamental definition Definition Let (X, d) be a metric space. Then X is said to be complete if every Cauchy sequence in X converges in X; A complete normed space (X, X ) is called Banach space. Examples Every finite-dimensional normed space over K is complete. (Φ, p ), 1 p + is never complete. For example x n := ( 1, 1, 1, , 0,... ) Φ n defines a Cauchy sequence (x n ) n N in (Φ, p ) converging to the infinite sequence ( 1 2 ) n n N l p \ Φ. Since uniform limits of continuous functions are continuous, (C[a, b], ) is complete, hence a Banach space. More generally, if K R n is compact, then (C(K), ) is a Banach space. For each 1 p + the sequence space l p is a Banach space. A subspace Y X of a Banach space (X, X ) is complete if and only if it is closed in X. Every metric space can be embedded in a complete one. More precisely for normed spaces the following holds.

19 6. COMPLETENESS AND BANACH SPACES 15 Proposition Let X be a normed space. Then there exists a Banach space X and a linear isometry T : X X with dense range T (X) = {T (x) : x X} X. The space X is unique up to isometric isomorphy and is called the completion of X. Examples The completion of Q is R, i.e., Q = R. Using similar arguments as in Example 1.54 one can show that X = (C[a, b], p ) (cf. Example 1.24) for 1 p < + is not complete. It is clear that its completion must contain discontinuous functions. One might guess that X = ( C[a, b], p ) coincides with the space of all Riemann integrable functions on [a, b] with finite p-norm. However, this space is not complete either. In Chapter 2 we will introduce a more sophisticated theory of integration in order to obtain the explicit representation ( C[a, b], p ) = L p [a, b] as the space of all measurable, p-integrable 11 functions. The following examples are important for the study of differential equations. More Examples For an open set Ω R n and k N 0 we consider the (vector) space of all real, k-times continuously differentiable functions on Ω, i.e., { C k D α } f exists and can be continuously (Ω) := f : Ω R extended to Ω. α k Here for a multi-index (α 1,..., α n ) N n 0 we define D α α f f := α 1 1 αn n where α := α α n. Then C k (Ω) equipped with the norm f k, := max 0 α k Dα f is complete, hence a Banach space. Let X := C [a, b], the (vector) space of all infinitely differential functions on [a, b]. Then for all k N, 1 p < + ( b k f k,p := D i f(s) ) 1 p ds a i=0 where D i denotes the i-th derivative, defines a norm on X for which X is not complete. In this way we obtain a whole scale of normed spaces (X, k,p ) and since it follows that f k+1,p f k,p f X ( C [a, b], k+1,p ) ( C [a, b], k,p ) with continuous inclusion. The previous example can be generalized to higher dimensions. For an open set Ω R n we consider the (vector) space of all infinitely differentiable functions on Ω having compact support, i.e. { } C supp f is compact and c (Ω) := f : Ω R D α f exists α N n in the sense of Lebesgue p,

20 16 1. NORMED VECTOR SPACES Here the support of f is given by supp f := {s Ω : f(s) 0}. Alternatively, one can say that a function has compact support if it is zero outside of a compact set. Then for all k N 0 and 1 p < + ( ) 1 f k,p := D α f(s) p p ds, Ω α k defines a norm on X for which X is not complete. The completions of (X, k,p ) are called Sobolev spaces, and are very important in the study of partial differential equations. We will come back to these spaces later, see Definition We close this chapter by the following important result which has numerous applications in applied mathematics. First we need a Definition Let X be a metric space and F : X X. If there exists 0 q < 1 such that d ( F (x), F (y) ) q d(x, y) x, y X, then F is called a contraction (with constant q). 12 It is easy to see that every contraction is uniformly continuous. Theorem 1.61 (Banach s fixed point theorem 13 ). Let X be a complete metric space (e.g. a Banach space) and F : X X be a contraction with constant q. Then there exists a unique fixed point x X of F, i.e., a unique x such that F (x ) = x. Moreover, if we choose an arbitrary x 0 X and define (recursively) the sequence (x n ) n N by then lim x n = lim F n (x 0 ) = x and n + n + x n+1 := F (x n ), =x 1 {}}{ d(x n, x ) qn 1 q d( x 0, F (x 0 ) ). Proof. Let m, n N and assume m > n. Then by the triangle inequality it follows that d(x n, x m ) d(x n, x n+1 ) + d(x n+1, x n+2 ) d(x m 1, x m ). Moreover, for every k N we have d(x k, x k+1 ) = d ( F (x k 1 ), F (x k ) ) q d(x k 1, x k ) = q d ( F (x k 2 ), F (x k 1 ) ) q 2 d(x k 2, x k 1 ). = q k 1 d ( F (x 0 ), F (x 1 ) ) q k d(x 0, x 1 ). Using the convergence of the geometric series + k=0 qk = 1 1 q d(x n, x m ) m 1 k=n + q n m n 1 q k d(x 0, x 1 ) = q n k=0 q k d(x 0, x 1 ) = k=0 for 0 q < 1 this implies that q k d(x 0, x 1 ) qn 1 q d(x 0, x 1 ) m > n. 12 In other words: F is a contraction iff it is Lipschitz continuous with Lipschitz constant L = q < also called Contraction Mapping Principle

21 6. COMPLETENESS AND BANACH SPACES 17 Since q n 0 as n + this shows that (x n ) n N is a Cauchy sequence. Since X is complete, lim x n =: x exists and from the continuity of F one obtains n + ( ) F (x ) = F lim x n = lim F (x n) = lim x n+1 = x. n + n + n + Finally, if we suppose that x and y are two different fixed points of F, then from it follows a contradiction. d(x, y ) = d ( F (x ), F (y ) ) q d(x, y ) < d(x, y ) Note that the previous proof is constructive, i.e., not only gives the existence of the fixed point but also a procedure how to construct it. This fact is used in various numerical methods in the form of iteration schemes. Frequently in applications the following generalization is very useful. Corollary If X is a complete metric space and F : X X is such that F k is a contraction for some k N, then F has a unique fixed point. Proof. By Banach s fixed point theorem G := F k has a unique fixed point x = Gx = lim n + Gn (x 0 ) for arbitrary x 0 X. Hence, choosing x 0 := F (x ) we obtain x = lim Gn( F (x ) ) = lim F n k+1 (x ) = lim F ( G n x ) ( ) = F lim n + n + n + n + Gn (x ) = F (x ), i.e., x is a fixed point of F. Since every fixed point of F is also a fixed point of G, this fixed point is unique. As already mentioned above, the previous two results have various applications.

22 18 1. NORMED VECTOR SPACES Example 1.63 (Linear systems). For a given n n-matrix A and a vector b K n we consider the equation ( ) Ax = b, x K n. In order to apply BFT we decompose A as A = B + (A B) for some invertible matrix B. Then ( ) is equivalent to B 1 Ax = x + ( B 1 A Id ) x = B 1 b ( Id B 1 A ) x + B 1 b = x. Hence, introducing the map F : K n K n by F (x) := ( Id B 1 A ) x + B 1 b =: Cx + d equation ( ) is equivalent to the fixed point equation F (x) = x. If we assume that C = (c ij ) n n satisfies n max c ij =: q < 1 (maximal row sum criterion), 1 i n j=1 then by the calculations from the example on page 9 it follows that F is a contraction on (K n, ), i.e., for the metric defined by d(x, y) = x y. Hence the linear system ( ) has a unique solution x. Moreover, this solution is given by x = lim F k (x 0 ) for any initial n + value x 0 K n, i.e., can be obtained as the limit of the iteration scheme 14 x k+1 := Cx k + d, k N 0. We note that there are various ways to choose the invertible matrix B above which leads to different methods like the Jacobi-, Richardson-, Gauß Seidel- or SOR-methods. Example 1.64 (Picard Lindelöf theorem). For a continuous function f : R R R and u 0 R we consider the Cauchy problem { u (t) = f ( u(t), t ), (CP) u(t 0 ) = u 0. Substituting in (CP) t by s and integrating from s = t 0 to s = t we obtain the Volterra integral equation 15 t u(t) = u 0 + f ( u(s), s ) ds, t 0 which is equivalent (proof!) to (CP). Now we choose δ > 0 and consider on the Banach space X := (C[t 0 δ, t 0 + δ], ) the operator F : X X defined by [ F (u) ] (t) := u0 + t t 0 f ( u(s), s ) ds. Since f is continuous, F is well-defined, i.e., F (u) X for all u X. Moreover, u is a solution of (CP) F (u) = u. In order to apply Banach s fixed point theorem, we assume that F is Lipschitz continuous, i.e., that there exists a constant L > 0 such that f(u, t) f(v, t) L u v u, v R, t [t 0 δ, t 0 + δ]. 14 One can show (see an exercise on page 20) that this scheme already converges if λ < 1 for all eigenvalues λ of C. 15 Integral equations had a huge effect on the development and promotion of FA.

23 Using this we conclude for u( ), v( ) X [ F (u) ] (t) [ F (v) ] (t) = Hence [ F 2 (u) ] (t) [ F 2 (v) ] (t) = 6. COMPLETENESS AND BANACH SPACES 19 t t 0 t ( f ( u(s), s ) f ( v(s), s )) ds L u(s) v(s) ds t 0 L t t 0 u v. t t 0 t ( f ( [F (u)](s), s ) f ( [F (v)](s), s )) ds L [F (u)](s) [F (v)](s) t 0 }{{} L s t 0 u v ( t ) L 2 s t 0 ds t 0 ds u v = L 2 t t 0 2 u v. 2 Proceeding in this way one can show by induction that [ F k (u) ] (t) [ F k (v) ] (t) Lk t t 0 k u v k N, t [t 0 δ, t 0 + δ], u, v X. k! This implies that F k (u) F k (v) L k δ k u v k N, u, v X. k! Since L k δ k lim = 0 k + k! this shows that F k for k sufficiently large is a contraction and hence F has a unique fixed point u which gives the unique solution u(t), t [t 0 δ, t 0 + δ] of (CP). Since δ > 0 can be chosen arbitrarily large, the solution exists for all times t R. Exercises Show that the set {1, e x, e 2x } is linearly independent in C[0, 1]. Which of the sets X 1 := {f X : f(0) = 1}, X 2 := {f X : f(1) = 0}, X 3 := {f X : f 0} are subspaces or affine subspaces of C[0, 1]? Let X be a vector space and Y X be a subset of X. Show that the following assertions are equivalent: - Y is a affine subspace of X, - for one/all y 0 Y the set y 0 + Y := {y y 0 : y Y } is a linear subspace of X. Let X be a K-vector space and Y X be a subset of X. Show that the following assertions are equivalent: - Y is a hyperplane, - there exists x 0 X and a linear functional ϕ : X K, ϕ 0, such that Y = x 0 + ker(ϕ), where ker(ϕ) := {x X : ϕ(x) = 0} denotes the kernel of ϕ. Does p(f g) := f(0) g(0) define a metric on C[0, 1]? Let (X, d) be a metric space. Show that for every fixed x X the map F : X R, F (y) := d(x, y) is continuous. Let (X, d) be a metric space. Prove that d : X X R, d(x, y) := d(x, y) 1 + d(x, y)

24 20 1. NORMED VECTOR SPACES defines a new metric on X which satisfies d(x, y) < 1 for all x, y X. Moreover, a sequence (x n ) n N converges to x 0 with respect to d iff it converges with respect to d (i.e., the metrics d and d are equivalent). Let X be a normed space then a map p : X R satisfying the conditions (ii) and (iii) of the definition of a norm (cf. page 6) is called a half-norm on X. Moreover a (finite or infinite) sequence p 1, p 2, p of half-norms on X is called total if p 1 (x) = p 2 (x) = p 3 (x) =... = 0 implies x = 0. Show that for a total sequence p 1, p 2, p of half-norms on X the map d : X X R given by d(x, y) := + defines a metric on X. Let (X, X ) be a normed space. Show that x y x y 1 2 k p k (x y) 1 + p k (x y) x, y X and conclude that the norm : X R is always continuous. Show that the closed unit ball U := {x X : x X 1} in a normed space is always closed and convex 16. For 0 < p < 1 consider the map f p : R 2 R +, f p (x, y) := ( x p + y p) 1 p. Explain why this does not define a norm on R 2. (Hint: Look at f p (x, y) 1 and use the previous exercise.) Draw the unit spheres S p := {x R 2 : x p = 1} in R 2 for 1 p +. What kind of relation between x p1 and x p2 for p 1 < p 2 does this picture imply? Two norms 1 and 2 on a vector space X are called equivalent if there exist c 1, c 2 > 0 such that c 1 x 1 x 2 c 2 x 1 x X. In this case we write 1 2. Show that all norms on a finite dimensional vector space are equivalent. True or false? Let X be a vector space equipped with two norms 1 and 2. Define X k := (X, k ) for k = 1, 2. Then X 1 X Show that the norms 1 and are not equivalent on C[0, 1]. Which relation holds between this two norms? Show that X := l is not separable. (Hint: for a set S = {s 1, s 2, s 3,...} X, where s k = (s k 1, sk 2, sk 3,...) consider x = (x 1, x 2, x 3,...) X defined by { s k k x k := + 1 if sk k 1, 0 else. Find conditions implying that a linear map C = (c ij ) n n : K n K n is a contraction on (K n, 1 ) or (K n, 2 ). Use these conditions in order to solve linear equations of the form Cx + d = x as in the example on page 18. As before, let C = (c ij ) n n : K n K n be a linear map where K n is equipped with an arbitrary norm. Show that there exists k N such that C k is a contraction λ < 1 for all eigenvalues λ of C. Let f C 1 [a, b] satisfy f L. Show that f is Lipschitz continuous with Lipschitz constant L. In particular, f is a contraction if L < A set A in a vector space is called convex if x, y A α x + (1 α) y A α (0, 1).

25 6. COMPLETENESS AND BANACH SPACES 21 Apply in an appropriate way Banach s fixed point theorem to the function F 1 (s) := s to show that f(s) := s 3 s 1 has a unique zero in [1, 2]. Approximate this zero with an error < Explain why one can not argue using F 2 (s) = s 3 1 on R; F 3 (s) = 1 on [1 + δ, + ) for some δ > 0 sufficiently small. s 2 1 Using in an appropriate way Banach s fixed point theorem, calculate the solution of the following Cauchy problem: (CP) { u (t) = u(t), u(0) = 1. Using in an appropriate way Banach s fixed point theorem, prove that the equation u(s) = 1 + s 0 s r u(r) dr, s [0, 1] has a unique solution u( ) C[0, 1]. Moreover, find a power series representation of u(s). Using in an appropriate way Banach s fixed point theorem, show that for every g C[a, b] there exists a unique solution u C[a, b] of the following equations (a) Volterra integral equation u(t) µ t a k(t, s) u(s) ds = g(t), t [a, b], where k C( a,b ) for the triangle a,b := {(t, s) R 2 : t [a, b], a s t} R 2 and µ C. (b) Fredholm integral equation u(t) µ b a k(t, s) u(s) ds = g(t), t [a, b], where k C([a, b] [a, b]) and µ C sufficiently small. Compare the solvability of Volterra s and Fredholm s integral equations in the previous exercise with the solvability of the following linear systems in X = C n, where x = (x 1,..., x n ) t X, b = (b 1,..., b n ) t X, µ C and A = (a ij ) n n is a n n matrix i (a) x i µ a ij x j = b i, 1 i n; (b) x i µ j=1 n a ij x j = b i, 1 i n; j=1 satisfying a ii = 0, 1 i n in case (a). Show that for every M > 0 the set A M := { f C 1 (a, b) : f(a) + f M } is relatively compact in X = ( ) C[a, b],. On the space X = (C[a, b], ) consider the Fredholm integral operator T : X X defined by (T f)(s) := b a k(t, s) f(s) ds for some fixed k C([a, b] [a, b]). Show that the set A := { T f : f X, f 1 } is relatively compact in X.

26 CHAPTER 2 Lebesgue Integration and L p -spaces 1. A Reminder on Riemann s Integral If f : [a, b] R is bounded and P : a = s 0 < s 1 <... < s m = b is a partition of [a, b] we define m i := inf{f(s) : s [s i 1, s i ]} and M i := sup{f(s) : s [s i 1, s i ]} for 1 i m. Then the expressions s(p, f) : = S(P, f) : = m m i (s i s i 1 ) m M i (s i s i 1 ) (lower sum), (upper sum) approximate the area A between the graph of f and the s-axis from below and from above, respectively. M f(s) s(f, P ) M f(s) S(f, P ) m m a = s 0 s 1 s 2 s 3 s 4 s s 5 = b a = s 0 s 1 s 2 s 3 s 4 s s 5 = b Figure 1. Lower sum s(p, f) and upper sum S(P, f) (m = 5). Clearly, the best approximation from below would be the biggest possible one, while the best from above the smallest possible one and coincidence of the two would give the area between the graph of f and the s-axis. However, these best approximations usually do not exist and we are thus led to the following Definition 2.1 (Riemann Integral 1 ). A bounded function f : [a, b] R is called Riemann integrable if sup { s(p, f) : P partition of [a, b] } = inf { S(P, f) : P partition of [a, b] } =: A In this case we set b a f(s) ds := A (Riemann integral). While for many purposes the Riemann integral is sufficient, for our needs it is too restrictive. For example the space of Riemann integrable functions is not closed with respect to pointwise convergence. 1 More precisely, Riemann Darboux integral 22

27 2. THE LEBESGUE MEASURE AND INTEGRABLE FUNCTIONS 23 Example 2.2. Since Q is countable there exists a sequence (r k ) k N Q such that [0, 1] Q = {r k : k N}. Now define { 1 if s {r 1,..., r k }, f k (s) := 0 else. Then all f k are Riemann integrable (having integral = 0) and converge pointwise, i.e., { lim f 1 if s [0, 1] Q, k(s) =: f(s) = k + 0 else. However, the limit function f is not Riemann integrable. In fact, since Q is dense in R, the set {r k : k N} is dense in [0, 1] and hence for every partition P of [0, 1] it follows s(p, f) = 0, S(P, f) = 1. Summing up, the Riemann integral is simple to define, however has drawbacks which make it of few use in advanced mathematics. In the sequel we will construct a better theory of integration which then will be used to define certain very important (complete) spaces of integrable functions. 2. The Lebesgue Measure and Integrable Functions While in Riemann s approach to integration the domain of a function f is partitioned, in Lebesgue s approach the range of the function will be divided. To explain the basic idea suppose that f : D [0, + ) is a non-negative function defined on some domain D R. Then every partition 0 = t 0 < t 1 <... < t m induces a partition of the domain D, i.e., we can decompose 2 { D = m D D i := { } s D : t i f(s) < t i+1 if 0 i m 1 i where i=0 D m := { s D : t m f(s) }. This partition can be used to approximate the area between the graph of f and the s-axis from below by the expression (cf. Figure 2 below). m m t i λ(d i ) = t i λ(d i ), i=0 where λ(d i ) denotes the measure of the set D i. If D i is an interval, then clearly λ(d i ) is f(s) i=1 t 2 t 1 m i=1 t i λ(d i ) t 0 s D 0 D 1 D 2 Figure 2. Decomposition D = m i=0d i of the domain (m = 2). 2 Here denotes the disjoint union of sets.

28 24 2. LEBESGUE INTEGRATION AND L p -SPACES just its length. However, in general the sets D i will be much more complicated and it is by no means clear how to define λ(d i ). Hence, before we pursue this idea to construct a new and better integral we need to define the measure of a set The Outer Measure. For simplicity we considered above domains in R but now we turn to the general case and pose the problem to define the measure of a set D R n. In case D = B = n i=1 [a i, b i ] is a box things are easy and we define its measure as n l(b) := (b i a i ). For an arbitrary set we make the following Definition 2.3. The Lebesgue Outer Measure of a set D R n is { + + λ (D) := inf l(b i ) : D B i }, where each B i denotes a box. i=1 i=1 Note that l(b i ) 0 and hence the sum of the series appearing in the definition of λ (D) is independent of the order of summation, i.e., λ (D) is well-defined. The function λ is defined on the whole power set P(R n ) and takes values in [0, + ] =: [0, + ) {+ }, i.e., λ : P(R n ) [0, + ]. We remark that λ (B) = l(b) for every box B R n. Some further properties of the outer measure are collected in the following Proposition 2.4. (i) λ ( ) = 0. (ii) If D 1 D 2 R n then λ (D 1 ) λ (D 2 ) (i.e., the outer measure is monotone). (iii) For every sequence D k R n, k N one has ( + ) λ D k + λ (D k ) i=1 (countable subadditivity) Example 2.5. We show that every countable set D = {s k : k N} has outer measure λ (D) = 0. To this end choose ε > 0 and a box B k R n such that s k B k and λ (B k ) ε 2 k. Then clearly D + B k and by monotony and subadditivity we conclude λ (D) λ ( + ) B k + λ (B k ) + ε 2 k = ε. Since we can choose ε > 0 arbitrarily small this implies λ (D) = 0 as claimed. Remark 2.6. Not every set having outer measure zero is countable. For example the Cantor set C is an example of an uncountable set having outer measure λ (C) = 0. One would expect in Proposition 2.4.(iii) to have equality if the sets D k are pairwise disjoint. However, in general (2.1) D 1, D 2 R n, D 1 D 2 = λ (D 1 D 2 ) = λ (D 1 ) + λ (D 2 ) i.e., λ is not even finitely additive. Surprisingly, this is not a defect of the function λ but of its domain P(R n ) which is too big. If one restricts λ to a smaller domain Σ, things become much nicer and, as we well see below, one gets even countable additivity.