Notes on Integrable Functions and the Riesz Representation Theorem Math 8445, Winter 06, Professor J. Segert. f(x) = f + (x) + f (x).

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1 References: Notes on Integrable Functions and the Riesz Representation Theorem Math 8445, Winter 06, Professor J. Segert Evans, Partial Differential Equations, Appendix 3 Reed and Simon, Functional Analysis, Chapter 1 Rudin Integrable and Square-Integrable Functions Let R be a measurable set. Let f : R be a measurable function. There exist unique nonnegative measurable functions f + : R and f : R such that f(x) = f + (x) f (x), f(x) = f + (x) + f (x). Definition: A measurable function f : R is integrable (or absolutely integrable, or summable) if both of the following integrals are finite (nonnegative real numbers): f + dx <, f dx <. The one-norm of an integrable function is the finite nonnegative real number f 1 = f dx = f + dx + f dx 0. The integral of an integrable function f is bounded above and below: f 1 f dx f 1. Definition: A measurable function f : R is square-integrable (or square-summable) if the following integral is finite (nonnegative real number): f 2 dx <. The two-norm of a square-integrable function is the nonnegative real number (here y 1 2 nonnegative square root) ( ) 1 ( ) 1 f 2 = f 2 2 dx = f 2 2 dx 0. = y 0 is the Exercise: Suppose f and g are square integrable. Prove the Cauchy inequality f g 1 f 2 g 2. The Cauchy inequality has a very important corollary: Corollary: The product f g of two square-integrable functions has finite integral: fg 1 f g dx fg 1. Exercise: Supposing that has finite Lebesgue measure, use the Cauchy inequality to show that every square-integrable function is integrable. Show that this is not true if has infinite Lebesgue measure. 1

2 We say that f : R is locally integrable (or locally summable) if the restriction to every subset Y of finite measure µ(y ) < is integrable. If has finite measure, then locally integrable is the same as integrable. But if has infinite measure, then a function may be locally integrable without being integrable. For example, consider the constant function f = 1 on = R. Vanishing Almost Everywhere Let f : R be a measurable function. We say that f vanishes everywhere if the preimage of {0} is the entire set, f 1 ({0}) =. Equivalently, f vanishes everywhere if the preimage of the complement R {0} is the empty set, f 1 (R {0}) =. We have often used the following fact and its generalization to higher dimensions, for example in the discussion of the Dirichlet principle: Lemma: Suppose = U is the closure of an open and bounded subset U R, and f : R is a continuous function. Then f vanishes everywhere if and only if f 1 = 0. The result remains true is the one-norm if replaced by the two-norm. We want a similar result that holds even for measurable functions which are not continuous. Note that the empty set has zero Lebesgue measure, µ( ) = 0, but there are also lots of non-emtpy sets with zero measure. In general, one says that a given property holds almost everywhere if it holds on some set of zero measure. In particular: Definition: A measurable function f : R vanishes almost everywhere if the preimage of R {0} has zero Lebesgue measure, µ(f 1 (R {0})) = 0. Exercise: Suppose f : R integrable. Show that f 1 = 0 if and only if f vanishes almost everywhere. Exercise: Suppose f : R is square-integrable. Show that f 2 = 0 if and only if f vanishes almost everywhere. MAIN RESULT: Return to the Riesz Representation Theorem We now consider an open bounded interval U = (a, b) R and its closure = U = [a, b]. Let V = C([a, b]) be the vector space of continuous functions f : R, with the inner product (f, g) = f g dx. We have already seen that this inner product space is not complete, so the strong form of the Riesz representation theorem does not apply. In particular, it is easy to construct a Cauchy sequence f n of continuous functions that does not converge to any continuous function. We now have the ingredients to fix this problem, although the fix introduces some complications. The following observation was already used implicitly in defining the inner product on V, let us state it explicitly: Lemma: C([a, b]) is a subspace of L 2 ([a, b]). Proof: We are claiming that a continuous function f : [a, b] R is square-integrable. Now [a, b] is a bounded and closed subset or R, so it is compact. A continuous function on a compact set attains a maximum and minimum, so f is bounded. Now [a, b] has finite measure µ([a, b]) = b a. A bounded measurable function on a set of finite measure is integrable, and square integrable. QED 2

3 We are now ready to state the main result. It says that the Riesz representation theorem works if we accept some modifications: Theorem: (Main Result) Let φ : C([a, b]) R be a bounded linear map. Then there exists a squareintegrable function q : [a, b] R such that φ(f) = (q, f) = [a,b] q f dx for every f C([a, b]). The function q is unique up to addition of a function that vanishes almost everywhere. In particular, we have to weaken continuity to square-integrability, and we have to accept a certain degree of non-uniqueness. Exercise: Let [a, b] = [0, 2]. Consider the bounded linear map φ : C([0, 2]) R defined by φ(f) = Find a square-integrable function q as described by the Theorem. Check that q cannot be continuous. [1,2] Continued... f dx. 3

4 Overview of the Technical Details This section outlines the ingredients used in the proof of the Main Result. This material is not strictly needed to understand and apply the main result. Definition: Suppose f and g are measurable functions. If the difference f g vanishes almost everywhere, we will say that f and g are equal almost everywhere, and will denote this by f g. Exercise: Show that the set of integrable functions is a vector space, and similarly for the set of squareintegrable functions. (Note that the set of all functions f : R is a vector space, we need to check that these subsets are vector subspaces.) The Lebesgue space L 1 Exercise: Show that is an equivalence relation on the set of integrable functions. Definition: The set of equivalence classes is called L 1 (). An element f L 1 is an integrable function defined up to addition of a function vanishing almost everywhere. It does not make sense to ask the value of f L 1 at a point x, since this can change if one adds a function vanishing almost everywhere (but not vanishing at the point x.) However, adding such a function does not affect the values of either of the two integrals f 1 = f dx, f dx. So both of these integrals are well-defined for an element f L 1, even though f(x) is not. The following Theorem and Remark are not needed for our applications, and may be skipped: Theorem: L 1 () with 1 is a complete normed space (real Banach space). Remark: f 1 does not satisfy the parallelogram identity, so it is not the norm associated to any inner product. The Lebesgue space L 2 Exercise: Show that is an equivalence relation on the set of square-integrable functions. Definition: The set of equivalence classes is called L 2 (). An element f L 2 is a square-integrable function defined up to addition of a function vanishing almost everywhere. It does not make sense to ask the value of f L 2 at a point x, since this can change if one adds a function vanishing almost everywhere (but not vanishing at the point x.) However, adding such a function does not affect the values of the integral ( f 2 ) 2 = f 2 dx = f 2 dx. So this integral is well-defined for an element f L 2, even though f(x) is not. In fact this norm is associated to an inner product. The integral (f, g) = f g dx. is well-defined for elements f, g L 2 (and is finite by the Cauchy inequality). Proposition: (f, g) is an inner product on the vector space L 2 (). The associated norm is of course (f, f) = f 2 dx = f 2. 4

5 The following two non-trivial theorems are essential: Theorem: (see Rudin Theorem 11.42) L 2 () with the inner product (f, g) is a complete inner-product space (real Hilbert space). This means that every Cauchy sequence f n L 2 () converges to some f L 2 () (and this f is unique). Now specialize to the closed interval = [a, b] R. Clearly this is a measurable set. Theorem: (Rudin Theorem 11.38) C([a, b]) is a dense subset of L 2 ([a, b]). All the technical details we need from Lebesgue theory are contained in the following statement, which combines the two previous theorems: Proposition: Every Cauchy sequence f n C([a, b]) has a unique limit f L 2 ([a, b]). Equivalently, L 2 ([a, b]) is a completion of C([a, b]) with respect to the inner product (f, g). Combining with the Riesz Representation Theorem, the Main Result follows. The End 5