WHAT IS THE DUAL OF C [0, 1]?

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1 WHAT IS THE DUAL OF C [, ]?. Orientation The space C [, ] consists of the continuous real-valued functions defined on the unit interval. It is a vector space under pointwise addition and scalar multiplication, and is infinite dimensional since x n C [, ] for every n N. The uniform norm is defined on C [, ] as f = sup f (x, f C [, ]. x Exercise.. Prove C [, ] is complete as a metric space under the metric d (f, g = f g. That is, show that if f n f uniformly then f C [, ]. We summarize the facts above by saying C [, ] is a Banach space, i.e. a normed linear space which is complete. It turns out to be useful to consider elements of the dual space (C [, ] which consists of the continuous linear functionals l : C [, ] R. Linearity requires and by continuity we mean l (f + g = l (f + l (g, f, g C [, ] l (λf = λl (f, f C [, ], λ R f n f = l (f n l (f. The convergence on the left is uniform, i.e. f n f. The convergence on the right is that of a sequence of real numbers. If l is linear, continuity at any one f C [, ] implies continuity on all of C [, ]. In particular, a linear functional l : C [, ] R is in the dual space (C [, ] if and only if f n = l (f n. A sufficient condition for continuity is the existence of a positive constant M such that (. l (f M f f C [, ]. If a linear functional satisfies this condition we say it is norm-bounded and we define its norm l to be the smallest such constant M for which (. holds. Suppose now l is continuous at zero, then there is a ball about the origin such that l (f ɛ whenever f δ. Given any f C [, ] we find l ( δ f f ɛ and hence l is norm-bounded with l ɛ/δ. We have proved the following Proposition.2. A linear functional l : C [, ] R belongs to the dual space (C [, ] if and only if it is norm-bounded. Let s see some examples.

2 WHAT IS THE DUAL OF C [, ]? 2 Example.3. Fix x [, ] and define the Dirac mass at x, δ x (f = f (x. This is clearly in the dual space and has δ x =. Example.4. Given a sequence of points x i [, ], i N along with absolutely summable weights a i, define l (f = i a i f (x i. This is linear and has l i a i so it is in the dual space. In fact l = i a i as can be seen by considering f n with f n (x i = sign (a i, i =,..., n. Example.5. The Riemann integral is in the dual space. That is, the mapping f I (f = f dx is linear and has I by the triangle inequality for integration f dx f dx. By choosing f = we see I =. The next example is more complicated and involves defining a different type of integral known as the Lebesgue-Stieljies integral. It is worth understanding well. 2. The main example: Lebesgue-Stieljies integration Given α BV [, ] with α ( = we define an integral via the following recipe. Let = x < x < < x n = be a partition of [, ] and make the sum f (x k+ (α (x k+ α (x k. k= If f C [, ] then as we refine the partition and take the mesh size δ, the sum converges to a number f dα = lim f (x k+ (α (x k+ α (x k. δ k= Indeed if P, P 2 are two partitions they have a common refinement P, and if the mesh sizes δ, δ 2 are small enough then we can write P f (x k+ α k P 2 f (x k+ α k 2ɛVar (α. Now from our definition it is clear that the map f I (f = f dα is linear. Moreover if we define the running total variation of α as { } α (x = sup α (x k+ α (x k, = x < x < < x n = x, we find k= f dα f d α Var (α f

3 WHAT IS THE DUAL OF C [, ]? 3 I Var (α and I (C [, ]. Again this inequality is actually equality, and this is a good exercise to work out before moving on. Exercise 2.. Prove I = Var (α. (Hint: consider non-decreasing α first. Exercise 2.2. Each of the examples from the previous section can be written in the form l (f = f dα for some α BV, α ( =. Work this out. Are the α s uniquely determined? 3. The Riesz theorem A theorem due to Riesz asserts that this last example is generic. Theorem 3. (Riesz. Given l (C [, ] there exists α BV, α ( = l (f = f dα f C [, ]. Exercise 3.2. Check that if α, β BV, α ( = β ( = satisfy f dα = f dβ f C [, ] then α = β except at most countably many points. In fact, if α is non-decreasing prove α (x inf f dα α ( x +. f [,x] For a general α BV argue similarly to conclude α (x + = lim δ α (x + δ is determined uniquely by knowledge of all the integrals { f dα, f C [, ] }. Discussion: If we assume α is right-continuous then it is uniquely determined, and indeed we can assume this at x (, ] without changing anything we ve said already. However, we may not simultaneously assume α is right-continuous at and α ( = if we wish to reproduce the entire dual space. We chose α ( = simply to clean up the definition of the Lebesgue-Stiejies integral. We begin the proof of the theorem by discussing the positive linear functionals, i.e. those linear functionals l with the property that l (f if f. Exercise 3.3. Show every positive linear functional l : C [, ] R is automatically continuous. Given a positive linear functional l define α (x = inf { l (f [,x] f, f C [, ] }, < x and set α ( =. Then α is non-decreasing and it defines a Lebesgue-Stiejies integral. We will prove l (f = f dα f C [, ]. Pick a partition = x < x < < x n = on which f is almost constant, so f (x f (x k ɛ whenever x k x x k+, k =,..., n. Using the definition of α find functions [,xi] ψ i [,xi+], i =,..., n

4 WHAT IS THE DUAL OF C [, ]? 4 satisfying α (x i l (ψ i α (x i + ɛ n. If we take ψ =, ψ n = we may define a partition of unity φ i = ψ i ψ i, i =,..., n well-suited for approximation of f on the mesh. Then n l (f f (x i l (φ i l ɛ and l < since every positive linear functional on C [, ] is automatically continuous. Also n f (x i l (φ i = f (x i+ (α (x i+ α (x i + e (n where l (f i= e (n f ɛ, f dα ( l + f ɛ. Since this holds for all ɛ > we have the result. Given a general l (C [, ] we have to be more sly. First we define an auxiliary linear functional l with the property that the linear functionals l ± l are positive. Then since 2l = ( l + l ( l l we finish by applying the proof above twice and superimposing the results. Now if we expect l = dα then it is natural to go after l = d α. (Remember α does not signify the usual absolute value but rather the running total variation. Therefore we define α (x = inf y>x sup { l (ψi [,x] ψ i [,y, ψ i, ψ i C [, ] along with α ( =. Note α is non-decreasing and finite since l (ψi = l ( ±ψi l whenever ψ i are admissable. Thus α defines a Lebesgue-Stieljies integral l (f = f d α f C [, ], and we claim l (f l (f f C [, ], f. As before, the heart of the proof is in setting up a useful partition of unity. Given f C [, ], begin by finding a partition = x < x < < x n = such that f (x f (x k ɛ whenever x k x x k+, k =,..., n. Then find y k (x k, x k+ and admissable ψ k,i with [,xk ] i ψ k,i [,yk, k =,..., n }, < x α (x k ɛ n + i l (ψ k,i α (x k + ɛ n +.

5 WHAT IS THE DUAL OF C [, ]? 5 From the definition of α it is clear we can pick the ψ k,i inductively to include the previously chosen ψ k,i along with new admissable functions supported on [x k, y k. Indeed if we have ψ k,i already and ψ k,i are proposed, and if we call Ψ k = i ψ k,i, we can break up each proposed function into two pieces ψ k,i = ψ k,i ( [,yk Ψ k + ψk,i Ψ k and make a new proposed family consisting of the individual pieces. This new family is still admissable and gets a larger value in the supremum part of the definition of α. So we may assume the ψ k,i come to us already with support either on [, y k or on [x k, y k, and then we clearly get a better approximation to the supremum by replacing those ψ k,i supported on [, y k by the already chosen ψ k,i. Having built up ψ k,i as described above we are ready to build the partition of unity. Take Ψ =, Ψ n =, and Ψ k = i ψ k,i for k =,..., n, then the relevant partition of unity is given by Note for fixed k =,..., n we have Then since we conclude for f that Φ k = Ψ k Ψ k, k =,..., n. l (Ψ k Ψ k = supp(ψ k,i [x k,y k l (ψ k,i l (Φ k α (x k α (x k + 2ɛ n +. l (f n f (x i l (Φ i l ɛ, l (f f (x i+ l (Φ i+ + l ɛ i= f (x i+ ( α (x i+ α (x i + ( l + 2 ɛ. i= Refining the partition and taking ɛ we conclude l (f l (f whenever f, f C [, ]. Now we know the linear functionals l ± l are positive, hence we may apply the first half of the proof to find α +, α BV with ( l + l (f = f dα + 2 ( l l (f = f dα. 2 Finally we have l (f = f d (α + α, f C [, ] and we re done.