Functional Analysis. Thierry De Pauw

Transcription

1 Functional Analysis Thierry De Pauw

2 The drawing on the front is c Courtney Gibbons brownsharpie.courtneygibbons.org 1

3 Contents 1 Preliminaries Countability Metric Spaces Norms Subspaces and Cartesian products Examples of Normed Spaces Finite Dimensional Spaces Young, Hölder and Minkoswki Sequence Spaces Spaces of Continuous Functions Spaces of bounded functions Other Norms on Spaces of Continuous Functions Lebesgue Spaces Modes of convergence beyond normed spaces Bounded Operators Linear Mappings between Normed Spaces Spaces of Linear Mappings Some Continuous Embeddings Isomorphisms Completeness Complete Metric Spaces Completion Examples and nonexamples of Banach Spaces Finite Dimensional Spaces Convergence of Series in Normed Spaces Schauder basis

4 CONTENTS 3 5 Dense Subspaces and Separability Dense Subspaces Piecewise Affine Functions Form a Dense Subspace of C[a, b] Simple Measurable Functions Form a Dense Subspace of L p C c (U) is dense in L p (U) Separable and Nonseparable Banach Spaces Dual Spaces Elementary properties The dual of l n p (K) and the dual of l p (K) The dual of L p Hahn-Banach Theorem: Analytic Form The Hahn-Banach Theorem: Geometric Form The dual of C c (U) Convolution Averages Weierstrass Theorem Comment Fubini and Tonelli Support of a member of L p (R m ) Convolution Product Approximate identity Smoothing

5 Chapter 1 Preliminaries 1.1 Countability Definition. Let S be a set. We say that: (A) S is finite if it is either empty or there exists n N and a bijection {1,..., n} S; (B) S is countably infinite if there exists a bijection N S; (C) S is countable if it is either finite or countably infinite; (D) S is uncountable if none of the above occurs. Note that if S is nonempty and finite then the integer n appearing in (A) is uniquely determined by S. It is called the cardinality of S and denoted card S. We agree that card = 0. We explicitely notice that if there exists a bijection between two sets S and S then S is countable iff S is countable. We start we a few examples of countably infinite sets Example. The set S of even positive integers is countably infinite because N S : n 2n is a bijection. The set Z is countably infinite because the map f : N Z defined as follows is a bijection: { m if n = 2m f(n) = m 1 if n = 1 + 2m Example. We next show that N N is countably infinite. In order to do so we first partition N N into the diagonal lines S m = N N {(p, q) : p + q = m}, m N, and we enumerate each such S m. Notice each S m is a finite set of cardinality m + 1, thus we need m + 1 integers to enumerate S m. Therefore in order to enumerate N N it suffices to partition N into successive

6 1.1 Countability 5 blocks of length m + 1 and to use the m th block to enumerate S m. Notice the first block (i.e. block number 0) is {0}, the second block (i.e. block number 1) is {1, 2}, the third block is {3, 4, 5} etc. Observe that the m th block starts at the integer m j=1 j = m(m+1) 2. These remarks make it plain that f : N N N defined as follows is a bijection: f(n) = ( n ) m(m + 1) m(m + 3), n 2 2 where m is such that n {m(m + 1)/2,..., m(m + 2)/2 1} Proposition. Each subset of a countable set is countable as well. Proof. Let S be countable and S S. If S is empty or finite there is nothing to prove. We may therefore assume that both S and S are countably infinite. Let f : N S be a bijection and define A = f 1 (S ). Observe that f A S is a bijection. Thus it suffices to establish the existence of a bijection g : N A. We define inductively subsets of A denoted A 0, A 1, A 2,..., and integers a 0, a 1, a 2,... as follows. A 0 = A and a 0 = min A 0. Next A n+1 = A n \ {a n } and a n+1 = min A n+1. We notice that none of the A n is empty for otherwise A would be finite. The map N A : n a n is a bijection Proposition. Let S be a nonempty set. The following are equivalent: (A) S is countable; (B) There exists a surjection N S; (C) There exists an injection S N. Proof. (A) (B) If S is finite then there is a bijection {1,..., n} S, n = card S, which can be arbitrarily extended to a surjection N S. Otherwise there is a bijection N S, thus a surjection as well. (B) (C) Let f : N S be a surjection. Thus for each ξ S the set f 1 ({ξ}) and we let g(ξ) = min f 1 ({ξ}). It is clear that g is injective. (C) (A) Let f : S N be an injection. Then f : S f(s) is a bijection and f(s) is countable according to Proposition Proposition. Let m 2 and let S j, j = 1,..., m, be countable sets. The Cartesian products m j=1 S j is countable. Proof. We recall that there exists a bijection between m j=1 S j and ( m 1 j=1 S j) S m. Proceeding inductively it is thus sufficient to establish the conclusion for m = 2. Let f j : N S j, j = 1, 2, be surjective (Proposition 1.1.5) and define f : N N S 1 S 2 : (n 1, n 2 ) (f 1 (n 1 ), f 2 (n 2 )). Since f is surjective and N N is countable (Example 1.1.3) there exists a surjection N S 1 S Proposition. Let J be a countable set and let S j, j J, be countable as well. It follows that S = j J S j is countable.

7 1.2 Metric Spaces 6 Proof. For each j J we choose a surjection f j : N S j, and we define f : J N S : (j, n) f j (n). This is clearly surjective and since J N is countable (Proposition 1.1.6) we infer that so is S Example. Here we observe that Q is countable. To prove it we define a surjection f : Z N Q by f(m, n) = m/n. Since Z is countable (Example 1.1.2) and N is countable, so is Z N (Proposition 1.1.6). Thus there exists a surjection N Q which shows that Q is countable according to Proposition It follows from Proposition that Q n is countable, for every n N Example. Here we show, following G. Cantor, that P(N) (the set consisting of subsets of N) is uncountable. We idenfity P(N) with F (N; {0, 1}), the collection of mappings N {0, 1}. Assume if possible that there exists a surjection N F (N; {0, 1}) : n f n. Define f : N {0, 1} as follows. f(n) = 0 if f n (n) = 1 and f(n) = 1 if f n (n) = 0. By its very definition we see that f does not coincide with any f n, a contradiction Example. If a < b are reals then [a, b] is uncountable. Suppose if possible that there exists a surjection N [a, b] : n x n. Define inductively a sequence of closed subintervals I n [a, b] with nonmepty interiors and such that I n+1 I n, x n I n and diam I n 0 as n. By compactness n I n, and by the condition imposed on diam I n this set is a singleton. Let x [a, b] be such that {x} = n I n. If x = x n for some n then x I n, a contradiction. 1.2 Metric Spaces This section is a reminder (meaning that the reader is expected to be familiar with the material contained here, and able to provide proofs for each statement). A metric space consists in a pair (X, d) where X is a set and d : X X R + verifies the following axioms. (A) For every x, y X: d(x, y) = 0 iff x = y; (B) (Symmetry) For every x, y X: d(x, y) = d(y, x); (C) (Triangle inequality) For every x, y, z X: d(x, z) d(x, y) + d(y, x). The function d is called a metric and sometimes d(x, y) is called the distance between x and y. Thus d implements a way of deciding whether two points x, y X are close to each other or not. Beware, though, that the definition allows for metrics spaces where points are either equal or not so close, as the following example shows Example (Discrete metric). Let X be any set and let d(x, y) = 1 whenever x y, x, y X, and of course d(x, y) = 0 when x = y. It is not hard to check that (X, d) is a metric space. The following canonical example is probably closest to the reader s initial intuition.

8 1.2 Metric Spaces Example. Consider X = R and let d(x, y) = x y. It follows that (X, d) is a metric space Example. Let X = C. For x C define x = R(x) 2 + I(x) 2. This is called the modulus of x. One checks that d(x, y) = x y defines a metric on C. See also the next two sections! The concept of a metric makes it possible to define the notion of convergence of a sequence {x k } in X Definition. A sequence {x k } in X converges to x X whenever the following holds. For every ε > 0 there exists k 0 N such that d(x, x k ) ε for every k k 0. This is equivalent to asking that the sequence {d(x, x k )} converges to 0 in the metric space of Example If a sequence {x k } converges to some x X and converges to some y X as well, then x = y as follows from axiom (A) 1. We say that the limit of a sequence is unique. We call x the limit of the sequence {x k } and we write lim k x k = x or x k x as k. A sequence which admits a limit will be termed a convergent sequence. The notion of convergence of a sequence leads to the definition of continuous mappings Definition. Let (X, d X ) and (Y, d Y ) be two metric spaces, and let f : X Y be a map. We say that f is continuous at x whenever the following holds. For every sequence {x k } in X converging to x, the sequence {f(x k )} in Y converges to f(x). We say that f is continuous if it is continuous at every x X Example. Let 0 < α < 1. If X = R and d α (x, y) = x y α, x, y R, then (R, d α ) is a metric space. Letting d denote the (standard) distance on R introduced in Example 1.2.2, one checks that the identity map id : (R, d) (R, d α ) is continuous, together with its inverse. In other words the metrics d and d α on R have exactly the same convergent sequences. 2 The notions of convergence of a sequence and continuity of a map can be conveniently phrased in the slightly more abstract language of open and closed sets. We now turn to introducing the relevant vocabulary. Let (X, d) be a metric space. Given x X and r 0 we define the open ball centred at x of radius r as follows: B(x, r) = X {y : d(x, y) < r}. 1 More precisely we refer here to the statement that d(x, y) = 0 implies x = y. This seemingly obvious axiom will cause some trouble when we will study Lebesgue spaces. 2 Yet they are not equivalent. We say two metrics d 1 and d 2 on a set X are equivalent if there exists C 1 such that C 1 d 1 (x, y) d 2 (x, y) Cd 1 (x, y) for every x, y X.

9 1.2 Metric Spaces 8 Similarly we dfine the closed ball centred at x of radius r as follows: B[x, r] = X {y : d(x, y) r}. In a general metric space one should be warned that some of our intuitions built upon the real line have become entirely wrong. For instance the following can occur: (a) If B X is a ball (either open or closed) then neither its center nor its radius need be unique; (b) A set B X is sometimes both an open ball and a closed ball; (c) In some metric spaces (called ultrametric spaces) if two closed balls intersect then their intersection coincides with one of them. A set U X is open whenever the following holds: For every x U there exists r > 0 such that B(x, r) X. In this definition the open ball can be equivalently replaced by a closed ball. The links between the notion of an open set and that of a convergent sequence is a follows. A sequence {x k } in X converges to some x X iff for every open set U X containing x there exists an integer k 0 such that x k U for every k k 0. We say that a sequence {x k } is eventually in a set S X if there exists an integer k 0 such that x k S for every k k 0. With this vocabulary we have: A set U X is open iff each sequence {x k } in X converging to an element of U is eventually in U. A set C X is closed whenever the following holds: For every sequence {x k } in C, if {x k } converges to some x X then x C. It is not so hard to check that C is closed iff X \ C is open. The link between these notions and that of a continuous map is as follows Proposition. Let (X, d X ) and (Y, d Y ) be two metric spaces and let f : X Y be a map. The following are equivalent: (A) f is continuous; (B) For every open set U Y in Y, f 1 (U) is open in X; (C) For every closed set C Y in Y, f 1 (C) is closed in X; (D) For every subset S X one has f(clos S) Clos f(s). Beware that the direct image of an open (resp. closed) set by a continuous map need not be open (resp. closed). Open balls are open sets, and closed balls are closed sets. An arbitrary union of open sets is open as well, and a finite intersection of open sets is open as well. All these are easily verified directly from the definitions. Consequently, an arbitrary intersection of closed sets is closed, and a finite union of closed sets is closed too. The interior of a set S X consists of all those points x S with the following property: There exists r > 0 such that B(x, r) S. The interior of S is denoted Int S. One checks that S 1 S 2 implies Int S 1 Int S 2, and that Int(S 1 S 2 ) = (Int S 1 ) (Int S 2 ). The Closure of a set S X consists of all points x X which are the limit of a sequence {x k } contained in S. The closure

10 1.3 Norms 9 of S is denoted Clos S. Note that S 1 S 2 implies Clos S 1 Clos S 2 and that Clos(S 1 S 2 ) = (Clos S 1 ) (Clos S 2 ). The important notions of completeness, separability, and compactness will be reminded in relevant subsequent sections. 1.3 Norms We will consider linear spaces over the field K = R of real numbers or over the field K = C of complex numbers Definition. A normed space is a linear space X over the field K equipped with a norm, i.e. a real valued function subject to the following requirements: : X R + (A) For every x X : x = 0 iff x = 0; (B) (Homogeneity) For every x X and every λ K : λx = λ x ; (C) (Triangle inequality) For every x, y X : x + y x + y. A normed space is usually denoted as (X, ). If we must insist we call (X, ) a real normed space (resp. a complex normed space) when X is a linear space over the field R (resp. over the field C). In item (B) of the definition of a normed space the symbol λ denotes the absolute valued of λ when K = R and the modulus of λ when K = C Remark. With a normed linear space (X, ) we associate a metric space (X, d) by letting d(x, y) = x y. Thus the following metric notions make sense: open and closed subsets of X, balls in X, convergent sequences {x k } in X, separability, completeness and compactness of subsets of X Proposition. Let (X, ) be a normed linear space. For every x, y X one has x y x y. Proof. Notice that x = x y + y x y + y, according to the triangle inequality. Thus x y x y. Similarly, y y x + x, therefore y x y x = x y. The proof is complete Definition. Let (X, ) be a normed space. A subset S X is bounded if it is contained in a ball, i.e. there exists 0 < r < such that S B X (0, r). Equivalently sup{ x : x S} <. A sequence {x k } in X is called bounded if sup k x k <. For instance if {x k } is convergent then it is bounded. Indeed, letting x denote its limit, there exists an integer k 0 such that x x k 1 whenever k k 0. For such k the preceding proposition implies that x k 1 + x. Thus sup k x k 1 + x + max{ x k : k = 0,..., k 0 }.

11 1.4 Subspaces and Cartesian products Subspaces and Cartesian products Here we briefly discuss two ways of obtaining new normed spaces from old ones. If (X, ) is a normed space and X X is a linear subspace of X then the restriction of the norm to X, still denoted, defines on X a structure of a normed space. This is the only structure of a normed space we will ever consider on a linear subspace of a normed space. Let (X 1, 1 ) and (X 2, 2 ) be two normed spaces (over the same field K). We consider X = X 1 X 2 as a linear space over K (with the operations defined as (x 1, x 2 ) + (y 1, y 2 ) = (x 1 + y 1, x 2 + y 2 ) and λ(x 1, x 2 ) = (λx 1, λx 2 )). For x = (x 1, x 2 ) X we define x = max{ x 1, x 2 }. It is not hard to check that (X, ) is a normed space. A sequence x k = (x 1,k, x 2,k ) in X 1 X 2 converges to some x = (x 1, x 2 ) iff {x 1,k } converges to x 1 and {x 2,k } converges to x 2. Consequently if S 1 X 1 and S 2 X 2 then Clos(S 1 S 2 ) = (Clos S 1 ) (Clos S 2 ). By induction on n one can then define a structure of a normed space of a Cartesian product X 1... X n Proposition. Let (X, ) be a normed linear space. The following mappings are continuous: X X X : (x, y) x + y and K X : (λ, x) λx. Proof. Assume that {x k } converges to x and that {y k } converges to y. The triangle inequality implies that (x k +y k ) (x+y) = (x k x)+(y k y) x k x + y k y 0 as k which exactly means that the sequence {x k + y k } converges to x + y. Similarly, assume that {λ k } converges to λ in K and that {x k } converges to x in K. It follows that λ k x k λx = λ k x k λx k + λx k λx (λ k λ)x k + λ(x k x) λ k λ sup x k + λ x k x k 0 as k. The following is easy and often useful Proposition. Let (X, ) be a normed space and Y X a linear subspace of X. It follows that Clos Y is a linear subspace of X as well.

12 1.4 Subspaces and Cartesian products 11 Proof. The mapping X X X : (x, y) x + y maps Y Y into Y, thus also Clos(Y Y ) into Clos Y because it is continuous. Since Clos(Y Y ) = (Clos Y ) (Clos Y ), it follows that Clos Y is stable under addition. The analogous argument shows that Clos Y is stable under multiplication by a scalar.

13 Chapter 2 Examples of Normed Spaces 2.1 Finite Dimensional Spaces The first example of real normed space is simply X = R equipped with the norm x = x, the absolute value of x R. The first example of a complex normed space is simply X = C equipped with the norm x = x, the modulus of x C. Next we may consider X = K n as a linear space over the field K. An element x K n is a finite sequence, of length n, of elements of K, denoted x = (x(1),..., x(n)) Example. The first norm we consider on X = K n is the following: x 1 = n x(j). j=1 The corresponding normed space is usually denoted l n 1 (K) Example. Another usual norm is: x = max{ x(j) : j = 1,..., n}. The corresponding normed space is denoted l n (K). Verifying that these are indeed norms on K n consists in routine calculations. Checking that the following are norms is, however, somewhat more involved. Let 1 p < and define n x p = x(j) p j=1 1 p.

14 2.2 Young, Hölder and Minkoswki Young, Hölder and Minkoswki We recall that a function f : R R is convex if f(λξ + (1 λ)η) λf(ξ) + (1 λ)f(η) whenever 0 λ 1 and ξ, η R. In case f is twice continuously differentiable, f is convex if and only if f (ξ) 0 for every ξ R. Given a real number 1 < p < we call conjugate of p the real number q characterized by the relation p 1 + q 1 = 1. Notice that 1 < q < and that the conjugate of q is p therefore we simply say that p and q are conjugate numbers. Notice the conjugate of 2 is 2 itself. For later use we extend the definition to saying that 1 and are conjugate as well. Letting p = q = 2 in the next inequality we obtain ab a2 2 + b2 2 for all a, b R +. This is easily seen to hold because 0 (a b) 2 = a 2 + b 2 2ab. We now turn to the case of general conjugate numbers p and q Proposition (Young s inequality). Let 1 < p < and 1 < q < be conjugate numbers. For every a, b R + one has ab ap p + bq q. Proof. If a = 0 or b = 0 the inequality is obviously verified, thus we can assume a 0 b. The exponential function R R : ξ exp(ξ) is convex. Consequently, ab = exp (log a + log b) ( 1 = exp p log ap + 1 ) log bq q 1 p exp (log ap ) + 1 q exp (log bq ) = ap p + bq q. Given x K n and 1 < p < we define n x p := x(j) p j=1 1 p. (2.1) Our goal is to show this is a norm on K n. We observe that axioms (A) and (B) are indeed verified. Axiom (C) is the content of the Minkowski inequality below Proposition (Hölder s inequality). Let 1 < p < and 1 < q < be conjugate numbers. For every x, y K n one has n x(j)y(j) x p y q j=1

15 2.2 Young, Hölder and Minkoswki 14 Proof. We shall first prove Hölder s inequality under the additional assumption that x p = 1 = y q. In that case we infer from Young s inequality that n n ( ) x(j) p x(j)y(j) + y(j) q p q j=1 j=1 = 1 p x p p + 1 q y q q = 1 = x p y q. In the case when x 0 y we put x := x 1 p x and ỹ := y 1 q y so that x p = 1 and ỹ q = 1. Thus the previous case applies to x and ỹ: x 1 p y 1 q n x(j)y(j) = j=1 n x(j)ỹ(j) 1, and multiplying both sides by x p y q we obtain the required inequality. Finally if x = 0 or y = 0 then the conlusion of the proposition is readily verified. If p and q are conjugate real numbers then one checks that p(q 1) = q and q(p 1) = p Proposition (Minkowski s inequality). Let 1 p. For every x, y K n one has x + y p x p + y p. Proof. For p = 1 and p = this was observed respectively in Example and Example We now assume that 1 < p < and we let 1 < q < be conjugate to p. If x + y = 0 the result is obvious and we henceforth assume that x + y 0. We infer from Hölder s inequality that n x + y p p = x(j) + y(j) p j=1 j=1 n ( x(j) + y(j) ) x(j) + y(j) p 1 j=1 n ( x p + y p ) x(j) + y(j) (p 1)q j=1 n = ( x p + y p ) x(j) + y(j) p j=1 = ( x p + y p ) x + y p 1 p. 1 q 1 p (p 1) Dividing both sides by x + y p 1 yields the required inequality.

16 2.3 Sequence Spaces Example. Let 1 < p <. Equation (2.1) defines a norm on X = K n : The triangle inequality in this context is Minkowski s inequality, Proposition The corresponding normed space is denoted l n p (K). 2.3 Sequence Spaces Let 1 p. Here we intend to generalize the examples l n p (K) to the case when n =, i.e. for infinite sequences of elements of K. Thus an element of the underlying linear space over K is a sequence x : N K and its norm should be defined as 1 x lp = x(j) p j N p (when p ). Observe that in order that lp be a norm the above series is assumed to at least converge Definition. Let x : N K be a sequence of elements of K and 1 p <. Define x lp = x(j) p j N 1 p [0, + ] as well as x l = sup{ x(j) : j N}. Next we associate with each 1 p the set l p (K) = K N {x : x lp < }. Our goal is to show that l p (K) is a linear space and that lp We shall perform both tasks at once Proposition. Let 1 p. l p (K) is a normed space. is a norm. Proof. We need to show conditions (A), (B) and (C) of Definition are verified. We start with the case where 1 p <. (A) It is clear that x l p R + and that x l p = 0 if and only if x = 0. (B) If x l p (K) and λ K then it is readily checked that λx l p (K) and λx lp = λ x lp. (C) Thus it remains to show that if x, y l p (K) then x + y l p (K) and x + y l p x l p + y l p. For fixed n N we associate with x and y the following elements of K n : x n := (x(1),..., x(n)) and y n = (y(1),..., y(n)). It

17 2.3 Sequence Spaces 16 follows from Minkowski s inequality applied to x n and y n that n x(j) + y(j) p j=1 1 p = x n + y n p x n p + y n p n = x(j) p j=1 x(j) p j N = x l p + y l p. 1 p 1 p n + y(j) p j=1 + y(j) p j N Letting n in this expression shows at once that x + y l p (K) and that x + y l p x l p + y l p. Next we treat the case p =. Conditions (A) and (B) of Definition are easily verified in this case as well. It remains to establish the triangle inequality. Let x, y l (K). For each j N we have x(j) + y(j) x(j) + y(j) x l + y l. Taking the supremum over j N yields the result. We now define a subspace of l (K) Definition. We let c 0 (K) = l (K) { } x : lim x(j) = 0. j Proposition. c 0 (K) is a closed linear subspace of l (K). Proof. We first note that c 0 (K) is a linear space. This means that if x, y : N K are sequences in K converging to zero, and λ K, then λx and x+y are sequences in K converging to zero as well. Next we turn to showing that c 0 (K) is closed in l (K). Let {x k } be a sequence in c 0 (K) that converges to some x l (K), i.e. x x k l 0 as k. We ought to show that x c 0 (K). Let ε > 0. Choose k 0 N such that x x k l < ε whenever k k 0. Since x k0 c 0 (K) there exists j 0 N such that x k0 (j) < ε whenever j j 0. For such j observe that x(j) x(j) x k0 (j) + x k0 (j) x x k0 l + x k0 (j) 2ε. Since ε is arbitrary we infer that lim j x(j) = 0. To end this section we want to explicitely notice that the l p (K) are infinite dimensional normed spaces. 1 p 1 p

18 2.4 Spaces of Continuous Functions Definition. By an infinite dimensional linear space we mean one that is not finite dimensional, i.e. one that does not admit an algebraic basis consisting of finitely many elements. Let X be a linear space over K. Recall that a family e 1,..., e n spans X if each x X can be written as x = n j=1 α je j for some α j K, j = 1,..., n, and that the family e 1,..., e n is linearly independent if the relation n j=1 α je j = 0 implies α 1 =... = α n = 0. We say e 1,..., e n is an algebraic basis if it both spans X and is linearly independent. The cardinality n of such basis, if it exists, does not depend on the particular basis but only on the space X. In order to check that a linear space X is not finite dimensional it suffice to exhibit for each n N a linearly independent family of elements of X of cardinality n. Indeed if X were finite dimensional then every linearly independent family (of, say, n elements) could be completed in an algebraic basis (of more than n elements) Proposition. Let 1 p. l p (K) is infinite dimensional. Proof. We define e k l p (K), k N, by the relation { 1 if j=k e k (j) = 0 otherwise, k N. Since clearly each family e 0,..., e n, n N, is linearly independent, l p (K) is infinite dimensional Remark. It is natural to ask whether the family e 0, e 1, e 2,... is in some sense a basis of l p (K). This will be touched upon in... In case p = 2 a thourough discussion appears in Chapter Spaces of Continuous Functions We start this section by observing that the sum, and mutliplication by a scalar, of K valued functions are well-defined. Suppose indeed that S is a set and u, v : S K are two K valued functions defined on S. The sum u + v is a K valued function on S defined pointwise : (u + v)(ξ) = u(ξ) + v(ξ), ξ S. Similarly if λ K then λu is a K valued function on S defined as (λu)(ξ) = λu(ξ), ξ S. Of course these definitions are made possible owing to the fact that the operations sum and multiplication by a scalar exist in the range K. Now let us look at the particular case where S = [a, b] R is a closed interval. We will consider the collection consisting of all continuous functions [a, b] K.

19 2.4 Spaces of Continuous Functions Definition. We denote by C([a, b]; K) the collection of all continuous functions [a, b] K. When K = R we sometimes simply write C[a, b]. This is a linear space because if u, v : [a, b] K are continuous and λ K then u + v and λu are continuous as well. It goes without saying that the zero element of C([a, b]; K) if the function that vanishes identically, [a, b] K : ξ 0. Owing to the compactness of [a, b] and the continuity of u we infer that u is bounded, i.e. there exists M R + such that u(ξ) M for every ξ [a, b]. Thus for each u C([a, b]; K) the following is finite: u = sup{ u(ξ) : ξ [a, b]}. In fact since u achieves its supremum on [a, b] we see that u = max{ u(ξ) : ξ [a, b]}. (2.2) Proposition. (C([a, b]; K), ) is a normed space. Proof. We ought to check that conditions (A), (B) and (C) of Definition are verified. (A) If u = 0 then u = 0. Reciprocally if u C([a, b]; K) and u then, according to (2.2), u vanishes identically on [a, b], thus so does u. (B) This is simply because λu = max{ λu(ξ) : ξ [a, b]} = max{ λ u(ξ) : ξ [a, b]} (C) Let u, v C([a, b]; K) and ξ [a, b]. Notice that = λ max{ u(ξ) : ξ [a, b]} = λ u. (u + v)(ξ) = u(ξ) + v(ξ) u(ξ) + v(ξ) u + v. Taking the supremum over ξ [a, b] we obtain u + v u + v Remark. The notion of convergence in (C([a, b]; K), ) is that of uniform convergence of a sequence of functions: If {u k } is a sequence in C([a, b]; K) that converges to u C([a, b]; K) with respect to the norm, i.e. u u k 0 as k, then lim k sup{ u(ξ) u k (ξ) : ξ [a, b]} = 0. There are other notions of convergence of sequences {u k } in C([a, b]; K) but these may not correspond with uniform convergence. For instance we say that a sequence {u k } converges pointwise to u C([a, b]; K) if the following holds: For every ξ [a, b] one has lim k u k (ξ) = u(ξ). We recall that the uniform convergence of a sequence {u k } implies its pointwise convergence, but the converse does not hold. Here is an example of a sequence {u k } in C[0, 1] that converges

20 2.4 Spaces of Continuous Functions 19 pointwise to 0 but does not converge uniformly (thus it has no limit in the normed space (C[0, 1], )). kξ if 0 ξ 1 k u k (ξ) = k ( ) ξ 2 k if 1 k < ξ 2 k 0 if 2 k ξ 1. Other notions of convergence in C([a, b]; K) will be considered in Section... We now turn to showing that C([a, b]; K) is an infinite dimensional space. For that purpose we introduce the following vocabulary Definition. Let u C([a, b]; K). We define its support to be supp u = Clos ([a, b] {ξ : u(ξ) 0}). For instance in case u k are the functions defined at the end of Remark then supp u k = [0, 2k 1 ] Lemma. Let a c < d b be real numbers. There exists u C([a, b]; K) such that supp u = [c, d] and u ( ) c+d 2 0. Proof. Let ξ 0 = (c + d)/2 be the center of the inerval [c, d] and define { (d c)/2 ξ ξ 0 if c ξ d u(ξ) = 0 otherwise. It is clear that u is continuous and that {u 0} = (c, d), so that supp u = [c, d]. Notice the lemma fails if c = d Proposition. Let a < b be real numbers. The linear space C([a, b]; K) is infinite dimensional. Proof. Suppose if possible that C([a, b]; K) is finite dimensional and let n be its dimension. Choose n + 1 nonoverlapping closed intervals 1 with nonempty interiors, I j [a, b], j = 1,..., n + 1, and for each I j choose u j C[a, b] according to Lemma so that supp u j = I j. For instance I 1 = [a, a + (b a)/(n + 1)],..., I n+1 = [b (b a)/(n + 1), b]. Let ξ j be the center of I j. We claim that u 1,..., u n+1 are linearly independent. Indeed if λ 1,..., λ n+1 R and 0 = n+1 j=1 λ ju j then each λ j = 0: For fixed j we simply note that n+1 0 = λ j u j (ξ j ) = λ j u j (ξ j ), j=1 because ξ j I j = supp u j when j j, thus u j (ξ j ) = 0. As u j (ξ j ) 0 we infer that λ j = 0. Since j is arbitrary the family u 1,..., u n+1 is linearly independent. Therefore the dimension of C([a, b]; K) is at least n + 1, a contradiction. 1 By I 1 and I 2 being nonoverlapping we mean that Int I 1 Int I 2 =

21 2.4 Spaces of Continuous Functions Remark. If a = b then C([a, b]; K) is linearly isomorphic to K. Indeed in that [a, b] = {a}, each function u : {a} K is continuous and uniquely determined by the scalar u(a) K. Thus C({a}; K) K : u u(a) is a linear bijection. We next extend our construction of a normed space of continuous functions in a somewhat larger generality. We wish to replace the domain [a, b] with more general domains, say some subsets K of R m. As each subset K R m inherits its structure of a metric space from the ambient R m it makes sense to speak of continuous functions u : K K. If we want to define a norm u the same way we did when K = [a, b] we need to know that u is bounded on K. One way to enforce this is by requiring that K be compact. For instance K may a bounded closed interval in R, the Cantor set, a closed ball in R m, a Cartesian product of bounded closed intervals in R m, a Cartesian product of Cantor sets in R m, or simply a finite subset of R m Definition. Let K R m be a nonempty compact set. We let C(K; K) denote the collection of continuous functions u : K K and for each u C(K; K) we let u = sup{ u(ξ) : ξ K} = max{ u(ξ) : ξ K}. In case K = R we simply write C(K) instead of C(K; R). The following is proved in the exact same way as Proposition Proposition. Let K R m be compact and nonempty. It follows that (C(K; K), ) is a normed space. The notion of support of a continuous function makes sense in a similar way as before Definition. Let K R m be a nonempty compact set. The support of u C(K; K) is defined as supp u = Clos(K {ξ : u(ξ) 0}). In analogy with Lemma we have the following Lemma. Let K R m be a compact nonempty set. For each ξ 0 K and r > 0 there exists u C(K; K) such that u(ξ 0 ) 0 and supp u = K B[ξ 0, r]. Proof. It suffices to consider { r ξ ξ 0 2 if ξ K B[ξ 0, r] u(ξ) = 0 otherwise. We now generalize Proposition and the Remark thereafter.

22 2.4 Spaces of Continuous Functions Proposition. Let K R m be a compact nonempty. The following alternative holds: (A) If K is infinite then C(K; K) is infinite dimensional; (B) If K is finite then C(K; K) is linearly isomorphic with K n where n = card K. Proof. (A) Assume if possible that C(K; K) is finite dimensional and let n be its dimension. Let ξ 1,..., ξ n+1 be distinct n + 1 elements of K. Put r := 1 3 min { ξ j ξ k 2 : j, k = 1,..., n + 1 and j k}. Let u j C(K; K) be associated with ξ j and r > 0 in Lemma If λ 1,..., λ n+1 are scalars such that 0 = n+1 j=1 u j then, owing to the fact that the u j have pairwise disjoint supports, we infer that 0 = λ j u j (ξ j ), and in turn λ j, for each j = 1,..., n + 1. Thus u 1,..., u n+1 is linearly independent, a contradiction. (B) Let n = card K and enumerate the points of K as K = {ξ 1,..., ξ n }. It follows that C(K; K) K n : u (u(ξ 1 ),..., u(ξ n )) is a linear bijection. What if we wish to define a space of continuous functions u : U K where now U R m is a nonmepty open subset of R m? That would again make a linear space but we would run into trouble when trying to define a norm u. Indeed there is no reason why u should be bounded. For instance if U = R then u(ξ) = ξ is continuous and unbounded, or if U = (0, 1) and u(ξ) = 1/ξ then u is continuous and unbounded as well. Thus we must impose an extra condition on u Definition. Let U R m be a nonempty open set. We say that a continuous function u : U K has compact support, or compactly supported in U, provided there exists a compact set K U such that u(ξ) = 0 for every ξ U \ K. We define its support to be supp u = Clos(U {ξ : u(ξ) 0}). The collection of continuous compactly supported functions u : U K is denoted C c (U; K). If K = R we write C c (U) instead of C c (U; R) Remark. Assume U is bounded (so that Bdry U ). If K is a compact subset of U then of course K Bdry U =. In fact there exists δ > 0 such that ξ ζ 2 δ for every ξ K and every ζ Bdry U. In order to prove this we consider the function f : K R + : ξ dist(ξ, Bdry U)

23 2.4 Spaces of Continuous Functions 22 defined by the formula dist(ξ, Bdry U) = inf{ ξ ζ 2 : ζ Bdry U}. We claim this function is continuous. Indeed let ξ, ξ K and ε > 0. Choose ζ Bdry U so that ξ ζ 2 < ε + dist(ξ, Bdry U) and observe that dist(ξ, Bdry U) dist(ξ, Bdry U) ξ ζ 2 ( ξ ζ 2 ε) ξ ξ 2 +ε, according to Proposition Since ε is arbitrary and ξ and ξ can be swapped we conclude that dist(ξ, Bdry U) dist(ξ, Bdry U) ξ ξ 2. The continuity of ξ dist(ξ, U) follows at once. Since K is compact this function achieves its minimum on K at, say, ξ K. Assume if possible that dist(ξ, Bdry U) = 0. Thus there would exist a sequence {ζ k } in Bdry U such that ξ ζ k 2 0 as k. Since U is bounded, so is Bdry U. As Bdry U is closed as well, it is compact, thus {ζ k } has a subsequence {ζ j(k) } converging to some ζ Bdry U. Therefore ξ ζ 2 = lim k ξ ζ j(k) 2 = 0 so that ξ = ζ Bdry U, a contradiction. We summarize what we can prove using the techniques developed so far Proposition. Let U R m be an open nonempty set. The following hold. (A) For every u C c (U; K), supp u is a compact subset of U and if ξ U \ supp u then u(ξ) = 0; (B) For every u, v C c (U; K) and λ K one has supp(u + v) (supp u) (supp v) and supp(λu) supp u ; (C) C c (U; K) is a linear space; (D) (C c (U; K), ) is a normed space where (E) C c (U; K) is infinite dimensional. u = sup{ u(ξ) : ξ U} ; (2.3) Proof. (A) Notice supp u is closed by definition. If u C c (U; K) then there exists a compact K U such that u(ξ) = 0 if ξ U \ K. Thus supp u K and the first conclusion follows. If u(ξ) 0 then clearly ξ supp u, which is the second conclusion.

24 2.5 Spaces of bounded functions 23 (B) If ξ U and u(ξ) = 0 = v(ξ) then (u + v)(ξ) = 0. Thus and in turn {u + v 0} {u 0} {v 0} supp(u + v) = Clos{u + v 0} Clos ({u 0} {v 0}) = Clos{u 0} Clos{v 0} = (supp u) (supp v). The second inclusion is even easier to prove. (C) This is an immediate consequence of (B) and the fact that (supp u) (supp v) is a compact subset of U because so are supp u and supp v. (D) Note that if u C c (U; K) then u is bounded. Indeed u is bounded on the compact set supp u and vanishes outside of supp u according to (A). The fact that is a norm is proved exactly as in Proposition (E) Since U is an infinite set one can mimic the proof of Proposition (A). With the same notations as there (with K replaced by U), the real r should be subjected to the extra requirement that r 1 2 min{dist(ξ j, Bdry U) : j = 1,..., n + 1} to guarantee that supp u j U for all j = 1,..., n + 1. Other than that, the argument runs the same way. 2.5 Spaces of bounded functions Here we generalize the space l (K). Recall that this is the space of functions N K which happen to be bounded. We shall now replace the domain N with an arbitrary set S. Thus we consider the functions u : S K. Addition of two such functions is defined pointwise again, as well as multiplication by an element of K. One checks that the sum of two bounded functions is bounded as well, as is a scalar multiple of a bounded function. Thus the collection B(S) defined below is a linear space over K Definition. Let S be a set. We let B(S) denote the linear space of all bounded functions S K, and for u B(S) we define u = sup{ u(ξ) : ξ S} Proposition. (B(S), ) is a normed space. Proof. Similar to that of Proposition In the special case when S happens to be a metric space as well, we will consider the subspace of B(S) consisting of those bounded continuous functions u : S K Definition. Let S be a metric space. We define C b (S) = B(S) {u : u is continuous}.

25 2.6 Other Norms on Spaces of Continuous Functions 24 Notice carefully that if S is an arbitrary metric space then continuous functions need not be bounded, so that C b (S) does not contain all continuous functions. If S is compact, however, then C b (S) is exactly the space of all continuous functions S K Proposition. Let S be a metric space. The subspace C b (S) is closed in (B(S), ). Proof. This amounts to showing that the uniform limit of a sequence of continuous functions is continuous too. Let {u k } be a sequence in C b (S) and u B(S) such that u u k 0 as k. Let ξ S and let {ξ j } be a sequence in S converging to ξ. We must show that u(ξ j ) u(ξ) as j. Let ε > 0 and choose an integer k such that u u k ε. Observe that for every j u(ξ) u(ξ j ) u(ξ) u k (ξ) + u k (ξ) u k (ξ j ) + u k (ξ j ) u(ξ j ) 2 u u k + u k (ξ) u k (ξ j ) 2ε + u k (ξ) u k (ξ j ). Since u k is continuous, letting j in the above inequality yields lim sup u(ξ) u(ξ j ) 2ε. j As ε is as small as we wish the proof is complete. 2.6 Other Norms on Spaces of Continuous Functions We will subsequently define new norms on C([a, b]; K) and C c (U; K). We first briefly expose our intent. Since C([a, b]; K) is infinite dimensional its complexity might be hard to grasp and we may want to approximate the elements of C([a, b]; K) by those belonging to some finite dimensional subspace Z C([a, b]; K) that we understand better. For instance we may choose Z to consist of the polynomial functions of degree at most 357. Thus our problem is to approximate an arbitrary continuous u C([a, b]; K) by a polynomial of degree 357 or less. One attempt may be to choose a polynomial v Z closest to u. In order to specify the problem we must tell in what sense we mean to approximate u by a polynomial v, i.e. in what sense the word closest in the last sentence was meant. Since C([a, b]; K) is equipped with a norm we may try to look for v Z minimizing the quantity u v, i.e. uniformly close (as close as possible) to u. This process raises several questions: (1) Does a best approximant v always exist? (2) If yes, is it unique? (3) If yes, how does it depend on v? For instance is the map C([a, b]; K) Z : u v linear? Continuous?

26 2.6 Other Norms on Spaces of Continuous Functions 25 Answering anyone of these questions is by no means entirely obvious. In fact the answers vary according to the norm being minimized in the definition of the approximant v and it turns out that the maximum norm is not necessarily the best theoretical choice. Instead of trying to have v approximate u uniformly on its domain [a, b], we may try to arrange for v to approximate u on average on the domain [a, b]. For instance we may want to minimize the quantity b v u, corresponding to a the norm u C[a, b]. u L1 = b a u, Definition. Let 1 p < and u C[a, b]. Define u Lp = ( b a u p ) 1 p Since u is continuous so is u p and therefore the above integral exists in the sense of Riemann. Note in particular that u Lp R + is finite. One of the main topics of Chapter?? is to give positive answers to questions (1), (2) and (3) above when C([a, b]; K) is equipped with L Proposition. Let 1 p < and u C[a, b]. It follows that u = 0 iff u Lp = 0. Proof. Notice that 0 Lp = 0. Next we must show that if u C[a, b] and u Lp = 0 then u = 0. We proceed by contraposition. Thus u 0. There exists ξ 0 [a, b] such that u(ξ 0 ) 0. Since u is continuous there exists δ > 0 such that u(ξ) u(ξ 0 ) /2 whenever ξ [a, b] [ξ 0 δ, ξ 0 + δ]. Therefore b ( ) p u p L p = u p u p u(ξ0 ) δ > 0. 2 a [a,b] [ξ 0 δ,ξ 0+δ] Proposition (Hölder inequality). Let 1 < p < and 1 < q < be conjugate exponents, and let u, v C[a, b]. The following holds: b a uv u Lp v Lq. Proof. If u = 0 or v = 0 the conclusion is immediate. We henceforth assume that u 0 v and hence u Lp 0 v Lp, according to Proposition Corresponding to each ξ [a, b] we apply Young s inequality with K = R, a = u(ξ) u 1 L p and b = v(ξ) v 1 L q : u(ξ)v(ξ) u Lp v Lq u(ξ) p p u p + v(ξ) q L p q v q. L q

27 2.7 Lebesgue Spaces 26 Integrating over ξ [a, b] we obtain at once 1 u Lp v Lq b a u(ξ)v(ξ) dξ b b a u(ξ) p dξ p u p a + v(ξ) q dξ L p q v q = 1 L q p + 1 q = Proposition. (C[a, b], Lp ) is a normed space. Proof. (A) This is Proposition (B) This follows from the homogeneity of the integral: λu Lp = ( b a ) 1 ( p ) 1 b p λu p = λ u p = λ u Lp. a (C) It remains only to check the triangle inequality. Let u, v C[a, b] and write w = u+v. Observe that w p = w p 1 u+w p 1 v, let 1 < q < be conjugate to p, and apply Hölder inequality (Proposition 2.6.3) twice to obtain b a u + v p = b a b w p w p 1 u + b a a w p 1 v w p 1 Lq u Lp + w p 1 Lq v Lp. (2.4) Finally notice that w p 1 Lq = ( b a w (p 1)q ) 1 q = ( b a w p ) 1 q. Dividing both sides of (2.4) by w p 1 Lq yields the triangle inequality. 2.7 Lebesgue Spaces Let S be a set, A a σ-algebra of subsets of S, and µ a measure on A. This means that A is a collection of subsets of S subject to the following requirements: (1) A; (2) If A A then X \ A A; (3) If {A k } is a sequence in A then k A k A. Notice that also X A and k A k A whenever {A k } is a sequence in A. By a measure on A we mean µ : A [0, ] such that

28 2.7 Lebesgue Spaces 27 (1) µ( ) = 0; (2) µ( k A k ) = k µ(a k) for every disjointed sequence {A k } in A. By a disjointed sequence {A k } we mean that A k A k = whenever k k. Notice that µ( k A k ) doesn t depend on a particular numbering of {A k }, and neither does k µ(a k) because µ(a k ) 0 for all k. An A-measurable function u : S K is one such that u 1 (O) A for every open set O K. One further says that u is simple if its range is finite, u(s) = {λ 1,..., λ κ }. Thus u = κ λ k 1 Ak k=1 where A k = u 1 {λ k } A. In case u 0 we define the integral of such u by the formula κ udµ = λ k µ(a k ). S k=1 In case u 0 is A-measurable we let { } udµ = sup vdµ : v is simple and v u. S S If udµ < we say that u is µ-summable. Finally, in general, we decompose S u = u + u where u + = max{u, 0} and u = min{u, 0} (we observe that u + and u are A-measurable) and we say that u is µ-summable if both u + and u are. In that case we put udµ = u + dµ u dµ. S S S In the present and next sections we shall use only elementary properties of measure and integration, which we now list Proposition. Let (S, A, µ) be a measure space and {A k } a sequence in A. The following hold. (A) If {A k } is increasing then µ( k A k ) = lim k µ(a k ); (B) If {A k } is decreasing and µ(a 1 ) < (in particular if µ(s) < ) then µ( k A k ) = lim k µ(a k ) Proposition. Let (S, A, µ) be a measure space, let u, v : S K be A-measurable functions, and let λ K. The following hold. (A) If both u abd v are µ-summable then so is u + v and (u + v)dµ = S S udµ + S vdµ; (B) If u is µ-summable then so is λu and S (λu)dµ = λ S udµ; (C) If 0 u then 0 S udµ;

29 2.7 Lebesgue Spaces 28 (D) If u is µ-summable then udµ = 0 iff u = 0 µ-almost everywhere, i.e. S µ(s {ξ : u(ξ) = 0}) = 0. Conceivably the single most important measure considered in these notes is stated to exist in the following theorem, a proof of which should be offered in a class devoted to measure and integration. The Borel σ-algebra in R m is the smallest one containing all open subsets of R m Theorem. Let B denote the Borel σ-algebra of R m. There exists a unique measure L m defined on B with the following properties: (A) L m (h + B) = L m (B) for every B B and every h R m ; (B) L m ([0, 1] m ) = 1. This measure L m is called the Lebesgue measure in R m Definition. Let u : S K be A-measurable. (A) Let 1 p <. We say that u is p-summable if S f p dµ <. In that case we define ( ) 1 u Lp = u p p dµ. The collection of p-summable functions S K is denote L p (S, A, µ). (B) We say that u is essentially bounded if there exists t > 0 such that S {ξ : u(ξ) t} is µ-null. In that case we define S u L = inf R + {t : µ(s {ξ : u(ξ) > t}) = 0}. The set of essentially bounded functions S K is denoted L (S, A, µ). Notice that if u L p (S, A, µ) then u Lp < Proposition. Let 1 p. L p (S, A, µ) is a linear space over K. Proof. Let u, v L p (S, A, µ) and λ K. It is clear that λu L p (S, A, µ): In case p this follows from the homogeneity of the integral, whereas in case p = it suffices to notice that { u > t} = { λu > λ t}. It remains to show that u + v L p (S, A, µ). Assume p. Since the function R + R + : s s p is convex we infer that for every ξ S the following holds: ( ) p u(ξ) + v(ξ) ( u(ξ) p + v(ξ) p ). Therefore u + v p 2 p 1 ( u p + v p ) and it follows from the Comparison Theorem that u + v is p-summable. In case p = we select t u, t v R + such that the sets S { u > t u } and S { u > t v } are µ-null. Thus their union is µ-null as well. Since it contains S { u + v > t u + t v } the proof is complete. Let us now examine whether Lp is a norm.

30 2.7 Lebesgue Spaces Proposition. Let 1 p. The following holds: (A) For every u L p (S, A, µ): u Lp = 0 iff u = 0 µ-almost everywhere. Proof. Assume u = 0 µ-almost everywhere. If 1 p < then u p = 0 µ- almost everywhere as well, thus S u p dµ = 0 and therefore u Lp = 0. If p = then taking t = 0 in the definition of u L shows this number equals zero. The other way round, if 1 p < and S u p dµ = 0 then u p = 0 µ-almost everywhere, and so does u. Finally, suppose p = and u L = 0. Observe that {u 0} = k { u > k 1 } and since µ{ u > k 1 } = 0 for every k N we infer that µ{u 0} = 0 as well Remark. Note that Lp is not a norm on L p (S, A, µ) since the relation u Lp = 0 does not imply u = 0. In order to establish the triangle inequality we first state and prove the Hölder inequality Proposition (Hölder inequality). Let 1 p and 1 q be conjugate exponents. For every u L p (S, A, µ) and every v L q (S, A, µ) one has uv L 1 (S, A, µ) and uv dµ u Lp v Lq. S Proof. The proof is exactly similar to that of Proposition We nevertheless spell it out. If u = 0 µ-almost everywhere or v = 0 µ-almost everywhere the conclusion is immediate. We henceforth assume u Lp 0 v Lp (Proposition 2.7.6) and we start with the case where 1 < p < and 1 < q <. Corresponding to each ξ S we apply Young s inequality with a = u(ξ) u 1 L p and b = v(ξ) v 1 L q : u(ξ)v(ξ) u Lp v Lq u(ξ) p p u p + v(ξ) q L p q v q. L q Integrating over ξ S we obtain at once 1 S u(ξ)v(ξ) dµ(ξ) u(ξ) p dµ(ξ) u Lp v Lq p u p S + v(ξ) q dµ(ξ) L p q v q = 1 L q p +1 q = 1. S Finally we assume p = 1 and q =. For each t R t such that µ{ v > t} = 0 we see that uv t u µ-almost everywhere and thus uv dµ S t S u dµ. Taking the infimum over all such t we obtain at once uv dµ S v L S u dµ Proposition. Let 1 p. The following hold: (B) For every u L p (S, A, µ) and every λ K : λu Lp = λ u Lp ; (C) For every u, v L p (S, A, µ) : u + v Lp u Lp + v Lp.

31 2.7 Lebesgue Spaces 30 Proof. We need to recall that A-measurability is preserved under multiplication by a scalar as well as under addition. (B) For 1 p < this follows from the homogeneity of the integral. For p = and λ 0 we simply notice that λu L = inf R + {t : µ{ λu > t} = 0} = inf R + {t : µ{ u > λ 1 t} = 0} = inf R + { λ s : µ{ u > s} = 0} = λ. ( inf R + {s : µ{ u > s} = 0} ) = λ u L. For λ = 0 the conclusion is trivially verified. (C) For 1 p < the mimicking the proof of Proposition 2.6.4(C) is left to the reader. We only need to realize that indeed w = u + v is p-summable as was established in Proposition We now turn to proving the triangle inequality in case p =. Let ε > 0 and choose t u, t v R + such that µ{ u > t u } = 0 and t u < ε + u L as well as µ{ v > t v } = 0 and t v < ε + v L. The triangle inequality in K implies that { u + v > t u + t v } { u > t u } { v > t v } and therefore µ{ u + v > t u + t v } = 0. Consequently, u + v L t u + t v < 2ε + u L + v L. Since ε is arbitrary the proof of (C) is complete. It is of course a bummer that (L p (S, A, µ), Lp ) is not a normed space simply because it fails half of condition (A) in the definition of a norm. The way to fix this is to consider instead of L p (S, A, µ) a quotient space of the latter where each function u : S K that vanishes almost everywhere is identified with zero. Formally we let Z = {u : S K : u is A-measurable and µ{u 0} = 0}. Notice that Z is a linear subspace of each L p (S, A, µ) so that the following definition makes sense Definition. We define the Lebesgue space L p (S, A, µ) to be the quotient space L p (S, A, µ)/z. For each u L p (S, A, µ) we let u Lp = u Lp where u u is some representative of u. Notice that the definition of u Lp doesn t depend on the choice of a representative of u. Indeed if u, u u then u u Z which means that u = u µ-almost everywhere and therefore S u p dµ = S u p dµ. In case p = we observe that u u L = 0 according to Proposition and u L u L u u L = 0 (which is a consequence of the triangle inequality Proposition 2.7.9(C) as in Proposition 1.3.3). Thus we obtain the following.