i. v = 0 if and only if v 0. iii. v + w v + w. (This is the Triangle Inequality.)

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1 Definition A (real) normed vector space is a real vector space V, equipped with a function called a norm, denoted by, provided that for all v and w in V and for all α R the real number v 0, and i. v = 0 if and only if v 0. ii. α v = α v. iii. v + w v + w. (This is the Triangle Inequality.) The norm that we will use is called the L 1 -norm, named after Henri Lebesgue. Definition Let (, A, µ) be a measure space. We define the L 1 -norm of an integrable function f by f 1 = f dµ and we note that for each f L(, A, µ) we have f 1 <. Exercise Show that the L 1 -norm on the vector space L(R, L, l) satisfies properties (2) and (3) of Definition but does not satisfy the first property. The difficulty is that an integrable function f can satisfy f = 0 without it being true that f is the (identically) zero function. We can remedy this defect and turn L(, A, µ) into a normed vector space L 1 (, A, µ) as follows. Definition We call two functions f and g in L(, A, µ) equivalent, denoted by f g, provided f g 1 = 0. We define L 1 (, A, µ) = L(, A, µ)/ = L(, A, µ)/{f L(, A, µ) f 1 = 0} the quotient space of the vector space L(, A, µ) modulo the subspace of all functions having L 1 -norm equal to zero. The equivalence class of an integrable function f is commonly denoted [f]. Exercise If f L(, A, µ), prove that f 1 = 0 if and only if f = 0 almost everywhere. 78

2 When dealing with L 1 (, A, µ) it is very common to go back and forth between the considering functions and the equivalence classes of functions. It is not common to denote the integral of an equivalence class, for example. It is easy to check that if g [f] then f dµ = g dµ, so that it does not matter which member of an equivalence class one integrates. Subtleties arise however of the following kind. Sometimes one speaks of some f L 1 (R) being a continuous function. What this actually means is that there exists a continuous function which can be selected for use as a representative of that equivalence class. But other representatives of [f] need not be continuous even though f C(R). Still if f is continuous then it is common to speak of f L 1 (R) being continuous, even though it is only the individual function f that is continuous and not every member of its equivalence class. And strictly speaking, we ought not to write f L 1 (R) because it is really [f] L 1 (R). Still everyone writes in the simple, loose way, but bearing in mind that the notation f L 1 (R) is just a common abuse of notation for the sake of simplicity of expression. Definition A set S L 1 (, A, µ) is said to be dense in L 1 (, A, µ) provided that for each f L 1 (, A, µ) and for each ǫ > 0 there exists s S such that s f 1 < ǫ. Theorem The space S 0 of special simple functions is dense in the space L 1 (, A, µ). The reader should note the conventional abuse of language in the statement of this theorem: S 0 is not actually contained in L 1 (, A, µ) because the latter space is a set of equivalence classes of function - not of functions themselves. Thus it is really not the set of special simple functions but the set of equivalence classes of those functions that is dense in L 1! However, we will abuse language in this way for simplicity of expression and it should not cause difficulty if the reader remains alert to it. Proof. Let f L 1 (, A, µ) and write f = f + f as in Definition We know there exists a function φ + S 0 such that 0 φ + f + and also such that φ + dµ > f + dµ ǫ 2. We define φ similarly, let φ = φ + φ, and apply the triangle inequality as follows: f φ f + φ + + f φ < ǫ. 79

3 In the case of the real line with Lebesgue measure, there is a very special subset of S 0 called the step functions. Definition Define a step-function on an interval [a, b] R as follows. We call σ a step-function if there exists a partition a = x 0 < x 1 < < x n = b of [a, b] into finitely many contiguous closed intervals such that σ(x) = c i, a constant, for all x (x i 1, x i ), i = 1,..., n. Thus σ is constant on each open interval (x i 1, x i ). The values of σ at x 0, x 1,...x n are arbitrary. Let S denote the family of all step functions on the real line. Exercise On the real line R, prove that the vector space S is dense in both S 0 and L 1 (R). (Hint: If A is a measurable set, prove that the indicator function 1 A is the limit of a sequence of step functions: 1 A σ n 1 0 as n. You can use the result of Exercise ) Exercise Let f L 1 (R) and define the Fourier transform f by f(α) = R f(x) sin αxdl(x) for each α R. Prove that f(α) 0 as n. This statement is known as the Riemann-Lebesgue lemma. (Hint: first treat the special case in which f is a step function.) Exercise If f L 1 (R), define f t L 1 (R) by f t (x) = f(x + t), the translate of f by t. For each t R define φ(t) to be the linear transformation φ(t) : L 1 (R) L 1 (R) given by φ(t) : f f t for each t R and for each f L 1 (R). a. Prove that φ(t)f is continuous at 0, meaning that for each fixed f in L 1 (R) we have φ(t)f φ(0)f 1 0 as t 0. Prove also continuity at each value of t R. In words, this exercise says that the mapping t φ(t)f is a continuous mapping from R L 1 (R). 80

4 b. Prove that φ is a homomorphism from the additive group of real numbers into the group L (L 1 (R)) of linear transformations of the vector space L 1 (R) into itself. The group L (L 1 (R)) is equipped with the operation of composition. 1 c. Fix t R and show that φ(t)f 1 = f 1 for each f L 1 (R). (For those who know what is meant by the norm of a linear transformation, this says that φ(t) is a bounded hence continuous linear mapping of L 1 (R). 2 ) d. Fix any t > 0, no matter how small. Show that there exists a function f L 1 (R) such that f 1 = 1 yet φ(t)f f 1 = 2. Thus φ(t) φ(0) 0 as t 0, using the concept of the norm of a linear transformation on a normed linear space. 3 Definition In a normed linear space V, a sequence of vectors v n is called a Cauchy sequence provided that for each ǫ > 0 there exists an N N such that for all m and n greater than or equal to N we have v n v m < ǫ. A normed linear space is called complete provided that for each Cauchy sequence v n in V there exists v V such that v n v, which means that v n v 0. A complete normed real linear space is called a real Banach space, and a complete normed complex linear space is called a Banach space. Exercise Give an example of a Cauchy sequence of functions f n L 1 [0, 1] such that there does not exist any point x [0, 1] for which f n (x) converges. Prove that your example has these properties. 1 This is a purely algebraic question concerning the group operations. 2 This part is only for those students who know how to put a norm on the vector space of continuous linear transformations of a normed linear space. See for example the texts Advanced Calculus: An Introduction to Linear Analysis ([10]) and Functional Analysis ([11]). 3 The point here is that continuity in analysis is a very delicate issue indeed. It is very much a matter of what is the mapping, what is the domain (and its norm or topology), and what is the range (and its norm or topology). 81

5 Theorem Let (, A, µ) be a measure space. Then L 1 (, A, µ) is a complete normed linear space. Proof. Let f n be a Cauchy sequence in L 1 (, A, µ). We must show there exists f L 1 (, A, µ) such that f n f 1 0 as n. The main difficulty in the proof is to find a suitable function f in L 1 (, A, µ). The reason this is challenging is that a Cauchy sequence in L 1 (, A, µ) need not converge pointwise at any point x, as shown by Exercise In order to remedy this difficulty, we note that since f n is Cauchy there exists an increasing sequence of natural numbers n k N such that if n and m are greater than or equal to n k then In particular f n f m 1 < 1 4 k (5.3) f nk f nk+1 1 < 1 4 k for each k N. Note also that the set { A k = x f nk f nk+1 1 } 2 k is measurable since f is A-measurable. Furthermore, 1 2 kµ(a k) f nk f nk+1 dµ < 1 4 k so that µ(a k ) < 1 2 k for each k N. Next we define the lim sup A k of the sequence of sets A k by N = lim sup A k = p=1 k=p and we note that N is the set of all points x that lie in infinitely many of the sets A k. Furthermore ( ) 1 µ A k 2 = 1 k 2 0 p 1 k=p k=p as p. It follows that µ(n) = 0, so that N is an (A, µ)-null-set. We will show that the sequence f nk (x) converges for each x \ N. 82 A k

6 If x / N then there exists p N such that k p implies f nk f nk+1 < 1 2 k. Therefore if k and l are greater than or equal to p, repeated application of the triangle inequality tells us that f nk (x) f nl (x) < 1 2 p 1 so that f nk (x) is a Cauchy sequence of real numbers. Thus the function given by f(x) = lim k f nk (x) exists almost everywhere on and is measurable. 4 And if we replace n by n k and m by n l and let l in Equation 5.3 we find that f nk f 1 = f nk f dµ = lim inf l f nk f nl dµ lim inf l f nk f nl dµ 1 4 k by Fatou s theorem. Thus f nk f L 1 () and it follows that f L 1 () and that f nk f in the L 1 -norm. Equation 5.3 tells us also that if n n k then we have f n f nk 1 < 1 4 k. It follows that if n n k we have f n f 1 f n f nk 1 + f nk f 1 < 2 4 k 0 as k. Hence f n f 1 0 as n and L 1 () is complete. Exercise Denote [0, 1] Q = {q k k N}, the countable set of all rational numbers in [0, 1]. Let f k : [0, 1] R be defined by f k (x) = 1 x qk for each k N. Note that the graph of f k has a vertical asymptote at x = q k. (a) Show that f k L 1 [0, 1] and find an upper bound for f k 1 that is independent of k. 4 If (, A, µ) happens to be a complete measure space, then it is clear that f is measurable. If the measure space is not complete, we can define f to be constant on the null-set N of non-convergence pointwise, and again f will be measurable. 83

7 (b) Let S n = n k=1 for each n N. Prove that S n is a Cauchy sequence in the L 1 -norm, and thus that S = lim n S n L 1 [0, 1]. (c) Prove for each x [0, 1] that S n (x) either diverges to infinity or else converges, and that lim n S n (x) < for almost all x. (d) Show that S 1 is improperly Riemann integrable in the sense of elementary calculus. Show that S n may be considered improperly Riemann integrable in a plausible sense, that should be explained. Show that S(x) is infinite at each rational point x E. Can you find any reasonable way to describe S as improperly Riemann integrable? 5 Exercise Suppose f n L 1 (, A, µ) and f n 1 0 as n. Prove that f n 0 in measure as well. f k 2 k 5.6 Complex Valued Functions What we have learned thus far about L 1 (, A, µ) can be extended easily to complex valued functions. (This should not be confused with complex analysis, in which the independent variable x is replaced by a complex independent variable z.) In contexts in which it would not be obvious in what set the values of a function lie, it is common to write such symbols as L 1 (, R) or L 1 (, C) to indicate (with the last entry) what type of value the function has. In more advanced subjects, such as harmonic analysis on Lie groups and representation theory, one learns about L 1 (, H), in which the functions take their values in a complex Hilbert space H rather than in the real or complex field. If f : C we write f(x) = u(x) + iv(x) where the real part of f(x) is denoted by u(x) = Rf(x) and the imaginary part is denoted by v(x) = If(x), both of which are real valued. Note that the complex modulus f(x) = (Rf) 2 + (If) 2. 5 This exercise is intended to show that Lebesgue integration obviates the need for a concept of improper integration and goes very much farther than the concept of improper integration used for the Riemann integral. 84