Download
1 / 188
E N D
3.5.1 Definition and Basic Properties of Functions of Bounded Variation We will expand on the first part of Section 3.5 of Folland’s text, which covers functions of bounded variation on the real line and related topics. We begin with functions defined on finite closed intervals in R (note that Folland’s ap- proach and notation is slightly different, as he begins with functions defined on R and uses TF(x) instead of our V [f;a,b]). Definition 1. Let f : [a,b] → C be given. Given any finite partition Γ = {a = x0< ··· < xn= b} of [a,b], set n X SΓ = |f(xi) − f(xi−1)|. i=1 The variation of f over [a,b] is V [f;a,b] = sup?SΓ: Γ is a partition of [a,b]?. The function f has bounded variation on [a,b] if V [f;a,b] < ∞. We set BV[a,b] = ?f : [a,b] → C : f has bounded variation on [a,b]?. Note that in this definition we are considering f to be defined at all points, and not just to be an equivalence class of functions that are equal a.e. The idea of the variation of f is that is represents the total vertical distance traveled by a particle that moves along the graph of f from (a,f(a)) to (b,f(b)). Exercise 2. (a) Show that if f : [a,b] → C, then V [f;a,b] ≥ |f(b) − f(a)|. (b) We say that a real-valued function f : [a,b] → R is monotone increasing if a ≤ x ≤ y ≤ b implies f(x) ≤ f(y). Show that every real-valued, monotone increasing function f on [a,b] has bounded variation and that, in this case, V [f;a,b] = f(b) − f(a). These notes follow and expand on the text “Real Analysis: Modern Techniques and their Applications,” 2nd ed., by G. Folland. Additional material is based on the text “Measure and Integral,” by R. L. Wheeden and A. Zygmund. 1
2 3.5 FUNCTIONS OF BOUNDED VARIATION Exercise 3. (a) Show that the Dirichlet functionχQhas unbounded variation on any finite interval. (b) Define f : [−1,1] → R by ( xsin1 0, x, x 6= 0, x = 0. f(x) = Show that f is continuous, but has unbounded variation on [−1,1]. (c) Construct a continuous, piecewise linear function on [0,1] that has unbounded varia- tion. The space BV[a,b] is sometimes defined to consist of only real-valued functions of bounded variation. However, in that case all the definitions and results extend equally to complex- valued functions. In other words, we could just as well have defined bounded variation for real-valued functions, and then declared a complex-valued function to have bounded variation if its real and imaginary parts have bounded variation. Exercise 4. Given f : [a,b] → C, write the real and imaginary parts as f = fr+ ifi. Show that f ∈ BV[a,b] if and only if fr, fi∈ BV[a,b]. For functions on the domain R we make the following definition. Definition 5. The variation of a function f : R → C is V [f;R] = sup V [f;a,b]. a<b We say that f has bounded variation if V [f;R] < ∞, and we define ?f : R → C : f has bounded variation on R?. Lipschitz functions are examples of functions on [a,b] that have bounded variation. BV(R) = Definition 6. A function f : [a,b] → C is Lipschitz on [a,b] if there exists a constant C > 0 such that |f(x) − f(y)| ≤ C |x − y|, We define Lip[a,b] = ?f : [a,b] → C : f is Lipschitz?. Exercise 7. Prove the following. (a) If f is Lipschitz on [a,b], then f is uniformly continuous and has bounded variation, with V [f;a,b] ≤ C (b − a). x,y ∈ [a,b]. (b) A Lipschitz function need not be differentiable. (c) If f is differentiable on [a,b] and f0is bounded on [a,b], then f is Lipschitz, and we can take C = kf0k∞. In particular, if f, f0are both continuous on [a,b], then f is Lipschitz.
3.5 FUNCTIONS OF BOUNDED VARIATION 3 (d) Extend the definition of Lipschitz functions to functions on the domain R instead of [a,b]. Show that continuity of f, f0on R (i.e., f ∈ C1(R)) need not imply that f is Lipschitz. What further hypotheses need be imposed in order to conclude that f is Lipschitz? We will need the following basic exercises on variation. Exercise 8. Let f : [a,b] → C be given. (a) Show that if Γ0is a refinement of Γ, then SΓ≤ SΓ0. (b) Show that if [a0,b0] ⊆ [a,b], then V [f;a0,b0] ≤ V [f;a,b]. (c) Show that if a < c < b, then V [f;a,b] = V [f;a,c] + V [f;c,b]. 3.5.2 The Jordan Decomposition We will prove that every real-valued function of bounded variation can be written as a difference of two monotone increasing functions. Given a real number x, recall the notations x+= max{x,0} and Note that x = x+− x−, while |x| = x++ x−. x−= −min{x,0}. Definition 9. Let f : [a,b] → R be given. Given a partition Γ = {a = x0< ··· < xn= b} of [a,b], define n n ?f(xi) − f(xi−1)?+ ?f(xi) − f(xi−1)?−. Γis the sum of the negative terms. X X S+ S− Γ= and Γ= i=1 i=1 Thus S+ The positive variation of f on [a,b] is Γis the sum of the positive terms of SΓ, and S− V+[f;a,b] = sup?S+ V−[f;a,b] = sup?S− Γ: Γ is a partition of [a,b]?, Γ: Γ is a partition of [a,b]?. and the negative variation is Exercise 10. Show that for any partition Γ we have S+ Γ+ S− Γ= SΓ while S+ Γ− S− Γ= f(b) − f(a). Exercise 11. Show that V+(x) = V+[f;a,x], V−(x) = V−[f;a,x], V (x) = V [f;a,x], are each increasing functions of x ∈ [a,b]. Also observe that 0 ≤ V−(x),V+(x) ≤ V (x) for x ∈ [a,b].
4 3.5 FUNCTIONS OF BOUNDED VARIATION Lemma 12. For any f : [a,b] → R, we have V+[f;a,b] + V−[f;a,b] = V [f;a,b]. Further, if any one of V [f;a,b], V+[f;a,b], or V−[f;a,b] is finite, then they are all finite, and in this case we also have V+[f;a,b] − V−[f;a,b] = f(b) − f(a). Proof. Note that, even if these quantities are infinite, we have V+[f;a,b] = sup S+ Γ≤ sup SΓ = V [f;a,b] Γ Γ and similarly V−[f;a,b] ≤ V [f;a,b]. In particular, if V [f;a,b] is finite, then so are V+[f;a,b] and V−[f;a,b]. On the other hand, (S+ Γ+ S− V [f;a,b] = sup SΓ = sup Γ) Γ Γ S+ S− Γ ≤ sup Γ+ sup Γ Γ = V+[f;a,b] + V−[f;a,b]. In particular, if both V+[f;a,b] and V−[f;a,b] are finite, then so is V [f;a,b]. Further, for every partition Γ, we have by Exercise 10 that S+ fixed and finite constant C = f(b) − f(a). Hence, even if they are infinite, Γ= S− Γ+ C, where C is the V+[f;a,b] = sup{S+ Γ: all partitions Γ} = sup{S− Γ+ C : all partitions Γ} = V−[f;a,b] + C. In particular, V+[f;a,b] is finite if and only if V−[f;a,b] is finite. The above work establishes that if any one of V [f;a,b], V+[f;a,b], or V−[f;a,b] if finite, then so are the other two. Exercise: Show that we can find partitions Γk, where each Γk+1is a refinement of Γk, such that lim k→∞S− Γk= V−[f;a,b]. Then since we have that S+ have k→∞S+ Further, since S− Γk= S− Γk+ C where C = f(b) − f(a) is a finite constant, we k→∞(S− Γk+ C) = V−[f;a,b] + C = V+[f;a,b]. lim Γk= lim Γk+ S+ Γk= SΓk, we have V+[f;a,b] + V−[f;a,b] = lim k→∞S+ k→∞S− k→∞(S+ Γk+ S− Γk+ lim Γk= lim Γk) ≤ V [f;a,b]. Finally, in the case that V [f;a,b] is finite, we have finite limits and therefore can write V+[f;a,b] − V−[f;a,b] = lim k→∞S+ k→∞S− k→∞(S+ Γk− S− Γk− lim Γk= lim Γk) = f(b) − f(a). ?
3.5 FUNCTIONS OF BOUNDED VARIATION 5 Exercise 13. Show that, when they are finite, 1 2 1 2 V+[f;a,b] = V−[f;a,b] = ?V [f;a,b] +f(b) − f(a)? ?V [f;a,b] − f(b) +f(a)?. and Now we can give the main result of this part, characterizing real-valued functions of bounded variation as the difference of two monotone increasing functions. Theorem 14 (Jordan Decomposition). If f : [a,b] → R is given, then the following state- ments are equivalent. (a) f ∈ BV[a,b]. (b) There exist monotone increasing functions f1, f2: [a,b] → R such that f = f1− f2. Proof. (a) ⇒ (b). For x ∈ [a,b], the functions V+[f;a,x] and V−[f;a,x] are monotonically increasing with x. Furthermore, by Lemma 12 we have V+[f;a,x] − V−[f;a,x] = f(x) − f(a). Therefore f = f1− f2where f1= V+[f;a,x] + f(a) and f2= V−[f;a,x] are monotonically increasing functions. ? A complex-valued function f : [a,b] → C will therefore have bounded variation if we can write f = (fr 1− fr 2) + i(fi 1− fi 2) where fr 1, fr 2, fi 1, fi 2are monotone increasing. Exercise 15. Show that f ∈ BV(R) if and only if we can write f = f1−f2where f1, f2are bounded and monotone increasing functions on R. 3.5.3 Variation for Continuous and Differentiable Functions Definition 16. Given any finite partition Γ = {a = x0< ··· < xn= b} of [a,b], we declare the mesh size of the partition to be |Γ| = max{xk− xk−1: k = 1,...,n}. Theorem 17. If f : [a,b] → R is continuous, then V [f;a,b] = lim |Γ|→0SΓ, i.e., for every ε > 0, there exists a δ > 0 such that for any partition Γ of [a,b], |Γ| < δ =⇒ V [f;a,b] − ε < SΓ≤ V [f;a,b].
6 3.5 FUNCTIONS OF BOUNDED VARIATION Proof. Choose any ε > 0. Then by definition of V [f;a,b], there must exist a partition Γ0= {a = x0 0< ··· < x0 n= b} such that V [f;a,b] −ε ≤ SΓ0 ≤ V [f;a,b]. 2 Since f is continuous on the compact domain [a,b], it is uniformly continuous, and therefore we can find an η > 0 such that for x, y ∈ [a,b] we have ε |x − y| < η =⇒ |f(x) − f(y)| < 4(n + 1). Now let δ = min{η, x0 1− x0 0, ..., x0 n− x0 n−1}. Suppose that Γ = {a = x0< ··· < xm= b} is any partition with |Γ| < δ. Let I = {k ∈ {1,...,m} : (xk−1,xk) contains some x0 j} and J = {k ∈ {1,...,m} : (xk−1,xk) contains no x0 j}. Note that since |Γ| < δ, no interval [xk−1,xk) can contain more than more point x0 fore, given k ∈ I we can let ¯ xkdenote that unique x0 we conclude that I can contain at most n+1 elements, since that is how many x0 Now, we have that j. There- jthat is contained in (xk−1,xk). Further, jthere are. X X SΓ = |f(xk) − f(xk−1)| + |f(xk) − f(xk−1)|. k∈I k∈J Let Γ0= Γ ∪ Γ0, and observe that Γ0is a refinement of both Γ and Γ0. Furthermore, by definition of I and J, we have ? ? X X SΓ0= |f(xk) − f(¯ xk)| + |f(¯ xk) − f(xk−1)| + |f(xk) − f(xk−1)| k∈I k∈J = ΣI+ ΣJ. Now, |xk− ¯ xk| < |xk− xk−1| < η and similarly |¯ xk− xk−1| < η, so ε ε |f(xk) − f(¯ xk)| < and |f(¯ xk) − f(xk−1)| < 4(n + 1). 4(n + 1) Since I can contain at most n + 1 elements, we therefore have ε ε 2. X ΣI ≤ ≤ 2(n + 1) k∈I
3.5 FUNCTIONS OF BOUNDED VARIATION 7 Consequently, SΓ ≥ ΣJ = SΓ0− ΣI ≥ SΓ0−ε 2 ≥ SΓ0 −ε since Γ0is a refinement of Γ0 2 ≥ V [f;a,b] −ε 2−ε 2, which completes the proof. ? We will improve on the next theorem later. Specifically, instead of requiring that f be differentiable with a continuous derivative, we will only need f to be absolutely continuous. Absolutely continuous functions need only be differentiable almost everywhere (although that alone is not sufficient to imply absolute continuity). Theorem 18. If f : [a,b] → R is differentiable and f0is continuous on [a,b], then Zb Zb Zb |f0|, V+[f;a,b] = (f0)+, V−[f;a,b] = (f0)−. V [f;a,b] = a a a Proof. Given any particular partition Γ = {a = x0< ··· < xm= b}, by the Mean-Value Theorem we can find points ξk∈ (xk−1,xk) such that f(xk) − f(xk−1) = f0(ξk)(xk− xk−1), so m m X X |f0(ξk)|(xk− xk−1). SΓ = |f(xk) − f(xk−1| = k=1 k=1 Note that this is a Riemann sum forR|f0|. Since f is continuous, we therefore have lim |Γ|→0 k=1 m Z X |f0(ξk)|(xk− xk−1) = |f0|. |Γ|→0SΓ = lim On the other hand, Theorem 17 implies that that |Γ|→0SΓ = V [f;a,b], lim so we conclude that V [f;a,b] =R|f0|. Exercise: Use Exercise 13 to extend to the positive and negative variations. ? Exercise 19. Give an example that shows that the conclusion of Theorem 18 can fail if f has a single removable discontinuity and is continuous at all other points in [a,b].
8 3.5 FUNCTIONS OF BOUNDED VARIATION 3.5.3 Monotonicity and Discontinuities Definition 20. Let f : [a,b] → R be given and let x ∈ [a,b] be fixed. If they exist, we define f(x−) = lim y→x−f(y) and f(x+) = lim y→x+f(y). (a) If f is not continuous at x, then we say that f has a removable discontinuity at x f(x−) and f(x+) both exist and f(x−) = f(x+). (b) If f is not continuous at x, then we say that f has a jump discontinuity at x if f(x−) and f(x+) exist but are not equal. Removable and jump discontinuities are often referred to collectively as discontinuities of the first kind. Lemma 21. Suppose f : [a,b] → R is monotone increasing. (a) f(x+) exists for all x ∈ [a,b), and f(x−) exists for all x ∈ (a,b]. (b) f has at most countably many discontinuities, and they are all jump discontinuities. Proof. (a) This follows from the fact that f is increasing and bounded. (b) Since f is increasing and f(x+) and f(x−) exist, any point of discontinuity of f must be a jump discontinuity. Further, since f is bounded and increasing, given any fixed k ∈ N, the set of x such that f(x+) − f(x−) ≥ 1 k must be finite. Since every jump discontinuity must satisfy this inequality for some k ∈ N, we conclude that there can be at most countably many jump discontinuities. ? Exercise 22. Give an example of a monotone increasing function f : [a,b] → R that has an infinite number of discontinuities. Exercise 23. Show that the conclusions of Lemma 21 also hold for monotone increasing functions f : R → R that are bounded, and that we further have in this case that f(−∞) = limx→−∞f(x) and f(∞) = limx→∞f(x) exist. 3.5.4 Monotonicity and Differentiability In this part we will prove that monotone increasing functions are differentiable a.e. We will formulate the result for monotone increasing functions on R, but it also applies to monotone functions f : [a,b] → R since any such function can be extended to be a monotone function on R (e.g., define f(x) = f(a) for x < a and f(x) = f(b) for x > b). Theorem 24. Suppose that f : R → R is monotone increasing. (a) The function g(x) = f(x+) is defined everywhere, is monotone increasing, is right- continuous, and is differentiable a.e. (b) f is differentiable a.e., f0is measurable, f0≥ 0 a.e., and f0= g0a.e.
3.5 FUNCTIONS OF BOUNDED VARIATION 9 Proof. (a) Exercise: Show that g exists and is monotone increasing and right-continuous. Consequently, our task is to show that g is differentiable almost everywhere. Since g is monotone increasing and right-continuous, there exists a corresponding Lebesgue–Stieltjes measure µg(see Section 1.5). As Lebesgue–Stieltjes measures enjoy many of the same regu- larity properties as Lebesgue measure, we actually have that µgis a regular Borel measure on R. By a theorem from the end of Section 3.4, since µgis regular, if we let µg= k dx + λ be its Lebesgue–Radon–Nikodym decomposition with respect to Lebesgue measure dx, then we have g(x + h) − g(x) h h→0+ A similar argument applies for the case h → 0−, so we conclude that g is differentiable a.e., with g0(x) = k(x) for almost every x. µg(x,x + h] |(x,x + h]| lim h→0+ = lim = k(x) a.e. (b) Let H = g − f. Since f has at most countably many discontinuities, we have that H(x) 6= 0 for at most countably many x, say x1,x2,.... Further, since we always have f ≤ g, we know that H(xj) = g(xj) − f(xj) = f(xj+) − f(xj) > 0 for each j. Define µ = X so H is a finite or countable sum of point mass measures and hence is a Borel measure on R. We claim that µ is locally finite, i.e., finite on compact sets. This follows from the fact that X = X ≤ f(N) − f(−N) < ∞. H(xj)δj, j µ(−N,N) = H(xj) |xj|<N ?f(xj+) − f(xj)? |xj|<N By a theorem from Section 1.5, every locally finite Borel measure on R is a Lebesgue–Stieltjes measure. Therefore µ is a regular Borel measure. Further, µ ⊥ dx since if we set E = {xj} then we have |E| = 0 = µ(EC). Therefore the Lebesgue–Radon–Nikodym decomposition of µ with respect to Lebesgue measure is µ = 0dx + µ. Hence, X H(xj) ???? ????≤ H(x + h) − H(x) h H(x + h) + H(x) |h| x−2|h|<|xj|<x+2|h| ≤ |h| µ(x − 2|h|,x + 2|h|] |(x − 2|h|,x + 2|h|]| = 4 → 0 as h → 0,
10 3.5 FUNCTIONS OF BOUNDED VARIATION where the convergence to zero again follows from a theorem at the end of Section 3.4. ? Since every function of bounded variation is a difference of two bounded monotone in- creasing functions, we obtain the following facts about functions in BV(R). Corollary 25. Suppose that f ∈ BV(R). (a) f(x+) and f(x−) exist for all x ∈ R, as do f(−∞) and f(∞). (b) f has at most countably many discontinuities, each of which is either a removable discontinuity or a jump discontinuity. (c) The function g(x) = f(x+) is right-continuous and has bounded variation on R. (d) f and g are differentiable almost everywhere, and f0= g0a.e. Further, we also have the following inequality for monotone increasing functions on inter- vals (a,b). Compare this inequality to the Fundamental Theorem of Calculus. Theorem 26. If f : [a,b] → R is monotone increasing, then Zb f0≤ f(b) − f(a). 0 ≤ (1) a Proof. Extend f by setting f(x) = f(a) for x ≤ a and f(x) = f(b) for x ≥ b. We have by Theorem 24 that f is differentiable a.e. Therefore, the functions f(x + 1/k) − f(x) 1/k = k?f(x +1 k) − f(x)? fk(x) = converge pointwise to f0(x) a.e. as k → ∞. By Fatou’s Lemma, we therefore have Zb Zb f0≤ liminf fk k→∞ a a Zb+1/k Zb+1/k Zb+1/k Zb Za+1/k Za+1/k = k f − k f a+1/k a = k f − k f b a ≤ k f(b) − k f(a) (since f is increasing) b a = f(b) − f(a). ? Exercise 27. Give an example showing that strict inequality can hold in equation (1). Exercise 28. Extend Theorem 26 to functions f ∈ BV(R).
3.5 FUNCTIONS OF BOUNDED VARIATION 11 The next exercise extends some of the preceding facts from monotone increasing functions to functions of bounded variation. Exercise 29. Show that if f ∈ BV[a,b], then f0(x) exists for a.e. x, and furthermore f0∈ L1[a,b]. Formulate and prove an analogous result for f ∈ BV(R). 3.5.5 The Derivative of the Variation In this part we will show that if f has bounded variation and we set V (x) = V [f;a,x], then we have V0(x) = |f0(x)| a.e. First, we need the following useful lemma, due to Fubini. Lemma 30. For k ∈ N, let fkbe monotone increasing functions on [a,b]. If ∞ X s(x) = fk(x) converges ∀x ∈ [a,b], k=1 then s is differentiable a.e., and ∞ X s0(x) = f0 k(x) a.e. k=1 Proof. For each N ∈ N, define N ∞ X X sN(x) = fk(x) and rN(x) = fk(x). k=1 k=N+1 By hypothesis, the series defining rN(x) converges for each x, and we have s = sN+ rN. Further, both sN and rN are increasing on [a,b], and hence each is differentiable except possibly on some set ZN ⊆ [a,b] with measure zero. The set Z = ∪ZN also has measure zero, so s is differentiable a.e., with s0(x) = s0 N(x) + r0 N(x), x / ∈ Z. By Theorem 26, f0 k, s0 n, r0 n≥ 0 a.e. Consequently, N X f0 k(x) = s0 N(x) ≤ s0 N(x) + r0 N(x) = s0(x) 0 ≤ a.e. k=1 Since this is true for every N, we therefore have that ∞ X f0 k(x) ≤ s0(x) a.e. (2) k=1 Our goal is to show that we actually have equality holding in equation (2) except for a set of measure zero.
12 3.5 FUNCTIONS OF BOUNDED VARIATION Since sN(x) → s(x) everywhere, we have that rN(x) → 0 for every x. Therefore we can choose Njsuch that 1 2j 1 2j, rNj(a) < and rNj(b) < j ∈ N. Combining this with the fact that rN(a) ≤ rN(b), we have ∞ X ?rNj(b) − rNj(a)? 0 ≤ < ∞. (3) j=1 Hence, ∞ Zb X Zb r0 0 ≤ Nj a j=1 ∞ X ∞ X r0 since r0 = Nj≥ 0 a.e. Nj a j=1 ?rNj(b) − rNj(a)? by equation (3). ≤ by Theorem 26 j=1 < ∞ Thus the functionP∞ j=1r0 Nj, which is nonnegative a.e., is integrable, and hence must be finite a.e.: ∞ X r0 0 ≤ Nj(x) < ∞ a.e. j=1 Consequently, we have convergence of a sequence of partial sums: Nj X ? ? s0(x) − f0 ?s0(x) − s0 j→∞r0 Nj(x)? lim j→∞ k(x) = lim j→∞ = lim Nj(x) = 0 a.e.. k=1 However, since each term f0 partial sums converges: kis nonnegative a.e., this implies that that the full sequence of N ? ? X s0(x) − f0 lim N→∞ k(x) = 0 a.e. k=1 This gives us our desired equality. ? Now we can give the main result of this part. Theorem 31. If f ∈ BV[a,b] and V (x) = V [f;a,x], then V is differentiable a.e., and V0(x) = |f0(x)| a.e. x ∈ [a,b].
3.5 FUNCTIONS OF BOUNDED VARIATION 13 Proof. By definition, V (b) is the supremum of all sums SΓ over all partitions Γ of [a,b]. Therefore, we can choose a sequence of partitions ?a = xk 0< xk 1< ··· < xk mk= b? k ∈ N, Γk = such that 0 ≤ V (b) − SΓk< 2−k, where mk X |f(xk j) − f(xk SΓk= j−1)|. j=1 Choose scalars ck jin such a way that f(x) + ck if x ∈ [xk j−1,xk j] and f(xk j) ≥ f(xk j, j−1), fk(x) = −f(x) + ck if x ∈ [xk j−1,xk j] and f(xk j) < f(xk j, j−1), is well-defined at each point x = xk of j and k that j, and satisfies fk(a) = 0. Then we have for each choice fk(xk j) − fk(xk j−1) = |f(xk j) − f(xk j−1)|. Consequently, mk X ?fk(xk j) − fk(xk j−1)? SΓk= = fk(b) − fk(a) = fk(b). j=1 In particular, we have for each k ∈ N that 0 ≤ V (b) − fk(b) ≤ V (b) − SΓk< 2−k. We claim now that for each fixed k, the function V (x) − fk(x) is an increasing function of x. To see this, suppose that a ≤ x < y ≤ b. If there is a single j such that x, y ∈ [xk then fk(y) − fk(x) = |f(y) − f(x)| ≤ V (y) − V (x). On the other hand, if x ∈ [xk j−1,xk j], j−1,xk j] and y ∈ [xk ‘−1,xk ‘] with j < ‘, then ‘−1 X ‘−1 X ?fk(y) − fk(xk ?fk(xk i) − fk(xk ?fk(xk ‘−1)? i−1)? j) − fk(x)? fk(y) − fk(x) = + + i=j+1 ?V (y) − V (xk = V (y) − V (x). ?V (xk i) − V (xk ?V (xk ‘−1)? i−1)? j) − V (x)? ≤ + + i=j+1 In any case, we obtain fk(y) − fk(x) ≤ V (y) − V (x), and therefore V (x) − fk(x) ≤ V (y) − fk(y). Hence V (x) − fk(x) is indeed increasing with x.
14 3.5 FUNCTIONS OF BOUNDED VARIATION Therefore, for x ∈ [a,b], 0 = V (a) − fk(a) ≤ V (x) − fk(x) ≤ V (b) − fk(b) < 2−k. Consequently, ∞ ∞ X X 2−k< ∞. ?V (x) − fk(x)? 0 ≤ ≤ k=1 k=1 Lemma 30 therefore implies that the series ∞ X ?V0(x) − f0 k(x) → V0(x) for a.e. x. But since V is increasing we k(x)? k=1 converges for almost every x. Hence f0 have V0(x) ≥ 0 a.e., so |f0(x)| = |f0 k(x)| → |V0(x)| = V0(x) a.e., and therefore |f0(x)| = V0(x) a.e. ? Additional Problems Problem 32. If f, g ∈ BV[a,b], then αf + βg ∈ BV[a,b] for all α, β ∈ C (so BV[a,b] is a vector space), and fg ∈ BV[a,b]. If |g(x)| ≥ ε > 0 for all x ∈ [a,b] then f/g ∈ BV[a,b]. Problem 33. Set f(x) = x2sin(1/x) and g(x) = x2sin(1/x2) for x 6= 0, and f(0) = g(0) = 0. Show that f and g are differentiable everywhere, but f ∈ BV[−1,1] while g / ∈ BV[−1,1]. Hint: Show that f is differentiable everywhere and f0is bounded, so f is Lipschitz. To show g does not have bounded variation, choose a particular appropriate partition, similar to Exercise 3. Problem 34. Let {fk}k∈N be a sequence of functions of bounded variation on [a,b]. If V [fk;a,b] ≤ M < ∞ for all k and if fk→ f pointwise on [a,b], show that f is of bounded variation and that V [f;a,b] ≤ M. Give an example of a pointwise convergent sequence of functions of bounded variation whose limit is not of bounded variation. Problem 35. Let E ⊆ R be measurable, and suppose that f : E → R is Lipschitz on E, i.e., |f(x) − f(y)| ≤ C |x − y| for all x, y ∈ E. Prove that if A ⊆ E, then |f(A)|e ≤ C |A|e. (4) Note that even if A is measurable, it need not be true that f(A) is measurable, which is why we must use exterior Lebesgue measure in (4).
CONTENTS UNIT I : THE RIEMANN-STIELTJES INTEGRAL 1.0Introduction 1 1.1 Unit Objectives 2 1.2 Riemann-Stieltjes integral 2 1.3 Existence and properties 4 1.4 Integration and Differentiation 23 1.5 Fundamental Theorem of the Integral Calculus 25 1.6 Integration of Vector –Valued Functions 30 1.7Rectifiable Curves 32 1.8References 34 UNIT II: SEQUENCE AND SERIES OF FUNCTIONS 2.0Introduction 35 2.1 Unit Objectives 35 2.2 Sequence and Series of Functions 35 2.3 Pointwise and Uniform Convergence 36 2.4 Cauchy Criterion for Uniform Convergence 45 2.5 Test for Uniform Convergence 46 2.6 Uniform Convergence and Continuity 58 2.7 Uniform Convergence and Integration 72 2.8 Uniform Convergence and Differentiation 77 2.9 Weierstrass’s Approximation Theorem 81 2.10 References 86 UNIT III : POWER SERIES AND FUNCTION OF SEVERAL VARIABLES 3.0Introduction 87 3.1 Unit Objectives 87 3.2 Power Series 88 3.3 Function of several variables 104 3.4References 129
UNIT IV: TAYLOR THEOREM 4.0 Introduction 130 4.1 Unit Objectives 130 4.2 Taylor Theorem 131 4.3 Explicit and Implicit Functions 134 4.4 Higher Order Differentials 138 4.5 Change of Variables 141 4.6 Extreme Values of Explicit Functions 148 4.7 Stationary Values of Implicit Functions 151 4.8 Lagrange Multipliers Method 153 4.9 Jacobian and its Properties 158 4.10 References 173
U UN NI IT T– –I I THE RIEMANN-STIELTJES INTEGRAL Structure 1.0 Introduction 1.1 Unit Objectives 1.2 Riemann-Stieltjes integral 1.2.1 Definitions and Notations Partition P, P* finer than P, Common refinement, Norm (or Mesh) Lower and Upper Riemann-Stieltjes Sums and Integrals Riemann-Stieltjes integral 1.3 Existence and properties 1.3.1 Characterization of upper and lower Stieltjes sums and upper and lower Stieltjes integrals 1.3.2 Integrability of continuous and monotonic functions along with properties of Riemann- Stieltjes integrals 1.3.3 Riemann-Stieltjes integral as limit of sums. 1.4 Integration and Differentiation 1.5 Fundamental Theorem of the Integral Calculus 1.5.1 Theorem on Integration by parts or Partial Integration Formula 1.5.2 Mean Value Theorems for Riemann-Stieltjes Integrals. First Mean Value Theorem for Riemann-Stieltjes Integral Second Mean Value Theorem for Riemann-Stieltjes Integral 1.5.3 Change of variables 1.6 Integration of Vector –Valued Functions 1.6.1 Fundamental theorem of integral calculus for vector valued function 1.7Rectifiable Curves 1.8References 1.0Introduction In this unit, we will deal with the Riemann-Stieltjes integral and study its existence and properties. The Riemann-Stieltjes integral is a generalization of Riemann integral named after Bernhard Riemann and Thomas Joannes Stieltjes. The reason for introducing this concept is to get a more unified approach to the theory of random variables. Fundamental Theorem of the Integral Calculus is discussed later on.
2 The Riemann-Stieltjes Integral 1.1 Unit Objectives After going through this unit, one will be able to define Riemann-Stieltjes integral and characterize its properties. recognize Riemann-Stieltjes integral as a limit of sums. know about Fundamental Theorem of the Integral Calculus and Mean Value Theorems . understand the concept of Rectifiable Curves 1.2 Riemann-Stieltjes integral We have already studied the Riemann integrals in our undergraduate level studies in Mathematics. Now we consider a more general concept than that of Riemann. This concept is known as Riemann-Stieltjes integral which involve two functions f and . In what follows, we shall consider only real-valued functions. 1.2.1 Definitions and Notations Definition1. Let [ , ] set of points a b be a given interval. By a partition (or subdivision) P of [ , ] a b , we mean a finite P { , ,........, x x x } 0 1 n such that . a x x x ........ x x b 0 1 2 n 1 n * * P P Definition 2. A partition P of [ , ] a b is said to be finer than P (or a refinement of P) if P i.e. P P . , that is , * * if every point of P is a point of * Definition 3. The 1P and 2P be two partitions of an interval [ , ] P P P a b . Then a partition P is called their * common refinement of 1P and 2P if . 1 2 Definition 4. The length of the largest subinterval of a partition of [ , ] a b is called the P { , ,...., x x x } 0 1 n Norm (or Mesh) of P. We denote norm of P by P . Thus P max x max{ x x : i 1,2,...., } n i i i 1 * * P P P P We notice that if , then . Thus refinement of a partition decreases its norm. Definition 5. Lower and Upper Riemann-Stieltjes Sums and Integrals Let f be a bounded real function defined on a closed interval[ , ] partition P of [ , ] a b , we put lub ( ) i M f x glb ( ) f x a b . Corresponding to each ( x x x ) i 1 i . i m ( x x x ) i 1 i
Mathematical Analysis 3 and ( ) b a b since ( ) a Let be monotonically increasing function on [ , ] are finite. a b . Then is bounded on [ , ] Corresponding to each partition P={x0,x1,…………,xn } of [a, b], we put . ( ) x ( x ) i i i 1 i . The monotonicity of implies that 0 For any real valued bounded function f on [ , ] a b , we take n ( , , ) L P f m i i i 1 n , ( , , ) U P f M i i i 1 L P f and ( , , ) U P f are respectively M are bounds of f defined above. The sums ( , , ) where called Lower Stieltjes sum and Upper Stieltjes sum corresponding to the partition P. We further define i m and i b fd lub ( , , ) L P f a b fd glb ( , , ) U P f , a b b fd fd where lub and glb are taken over all possible partitions P of [ , ] a b . Then and are a a respectively called Lower integral and Upper integrals of f with respect to . b fd If the lower and upper integrals are equal, then their common value, denoted by , is called the a Riemann-Stieltjes integral of f with respect to , over [ , ] integrable with respect to , in the Riemann sense and we write a b and in that case we say that f is ( ) f . The functions f and are known as the integrand and the integrator respectively. In the special case, when ( ) x x , the Riemann-Stieltjes integral reduces to Riemann-integral. In such b f b f and f respectively in place of ( , , ), L P f ( , , ), U P f a case, we write ( , ), L P f ( , ), U P f , a a b b ( ) fd fd f , and . a a
4 The Riemann-Stieltjes Integral f a andb and does not depend on the symbol fd depends only on , , Clearly, the numerical value of x. In fact x is a “dummy variable” and may be replaced by any other convenient symbol. 1.3 Existence and properties 1.3.1 In this section, we shall study characterization of upper and lower Stieltjes sums, and upper and lower Stieltjes integrals. refinement of the partition increases the lower The next theorem shows that for increasing function , sums and decreases the upper sums. P is a refinement of P, f is bounded real valued function on monotonically increasing function defined on , , , , f P U f P U . b and is a, Theorem1. If b a, . Then, , , L P , f L P , f be a partition of b P a, Proof. Let having one more point. . Further, let P be a refinement of P { x , x ,......... .......... , n x } 0 1 Let x be such that point in the sub-interval that is [ x , x ] i 1 i . Then, let P { x , x ,......... .., x , x , x ,......... ......, x } 0 1 i 1 i n g.l.b. of f in m x , x i i 1 i * w g.l.b. of f in ix 1, x 1 . *, w g.l.b of f in x x 2 i Obviously, . m w ; m w i 1 i 2 Then, L P f , , m m ............ m m ............ m 1 1 2 2 i 1 i 1 i i n n x x x L P , , f m m ............ m w x w 1 1 2 2 i 1 i 1 1 i 1 2 i m m .......... ......... i 1 i 1 n n Thus, P L x x x x , f , L P , f , w x w m x x 1 i 1 2 i i i i 1 w m x x w m x 1 i i 1 2 i i ; Now, w m 0 w m 0 2 i 1 i
Mathematical Analysis 5 Also, is monotonically increasing function and x x i , , , , f P L f P L Similarly, , , U f P U . So, x x x 1 i i x x 0 i 1 0 , , L P , f L P , f 0 , . P , f If P contains more points, then similar process holds and so the result follows. 2 P of b 1P and a, Theorem 2. For any two partitions on b a, , , 1 f P L , let f be a bounded real valued function defined b a, , then and is monotonically increasing function defined on , , 2 f P U . Proof. Let P be the common refinement of have ( , , ) L P f 1P and 2 P , that is, . Then, using Theorem 1, we P P P 1 2 . ( , , ) L P f ( , , ) U P f U P f ( , , ) 1 2 x L m a f M Remark 1. If m . Then, U , b b a , P , f P , f M . Proof . By hypothesis m m M M i i m m M M i i i i i i n n n n m m M M i i i i i i i 1 i 1 i 1 i 1 b a b a b a, m L P f , , U P f , , M . Theorem 3. If f is bounded real valued function defined on on b a, . Then, and is monotonic function defined b b a fd fd a . Proof . Let b denotes the set of all partition of b b , P , a a, P , P P a . For , we know that 1 2
6 The Riemann-Stieltjes Integral 1 L P f , , U P f , , 1 2 b , : 2 P fixed, it follows from 1 that , } P a b . U P f is an upper 1 2, , P P a This holds for each , keeping { L P f , , P bound of the set 1 1 b x a . f d But least upper bound of this set is b f x d . . .{ lub L P f , , : P P a b , } i.e, 1 1 a Since, least upper bound any upper bound b f x d 2 U P f 1, , a b b , . So, it follows from 2 that d x 2 P P a f This holds for each is a lower bound of the set a U P , f , : P P a , b . 2 2 b But greatest lower bound of this set is . fd a b fd . . . g lb U P f , , : P P a b , i.e, 2 2 a Since, any lower bound greatest lower bound. b b a a fd fd So, . Example 1. Let ( ) x x and define f on [0,1] by 1, 0, x x Q Q ( ) f x Then for every partition P of [0,1], we have i m , Therefore M , because every subinterval contain both rational and irrational number. 0 1 [ x , x ] i i 1 i
Mathematical Analysis 7 n m x ( , , ) L P f i i i 1 0 n ( , , ) U P f M x i i i 1 n 1 0 ( x x ) x x 1 i i 1 n 0 i 1 Hence, in this case fd fd . Theorem 4. Let is monotonically increasing on partition P of b a, such that , , f P U b , there exists a ( ) 0 a, then iff for any f , L P , f Proof.The condition is necessary: Let f be integrable on b on b a, f a, i.e, , b b b 1 fd fd fd so that a a a 0 Let be given. b fd L P f P sup{ , , : Since is a partition of [a,b]} a 1P of b a, So, by definition of l.u.b., there exists a partition such that b b L P f fd fd 1, , (by (1)) 2 2 a a b L P f , , fd 1 2 a b 2 L P f , , fd 1 2 a Again,
8 The Riemann-Stieltjes Integral b fd inf{ U P f , , : P Since is a partition of [a,b]}. a 2 P of b a, By the definition of g.l.b., there exists a partition such that b b U P f , , fd fd 2 2 2 a a b 3 U P f , , fd 2 2 a P P P 1P and 2 P , so that 4 Let be the common refinement of 1 2 U P f , , U P f , , 2 And 5 L P f 1, , L P f , , Now, we have b U P f , , U P f , , fd (by (3)) 2 2 a L P , f , (by (2)) 1 2 2 L P f 1, , L P f (by (5)) , , U P f , , L P f , , or U P f , , L P f , , . The condition is sufficient: be any number. Let P be a partition of b 0 a, Let such that 6 U P f , , L P f , , Since lower integral condition exceed the upper integral. b b So, a fd fd . a
Mathematical Analysis 9 b b 7 fd fd 0 a a Now, we know that b b L P f , , fd fd U P f , , a a b b 8 fd fd U P f , , L P f , , a a From (7) and (8) , we have b b a a 0 fd fd b b The non – negative number a being less than every positive number must be zero, fd fd a b b a fd fd 0 i.e, a b b a a fd fd . 1.3.2 In this section, we shall discuss integrability of continuous and monotonic functions along with properties of Riemann-Stieltjes integrals. Theorem 1. If f is continuous on [ , ] a b , then ( ) (i) f there corresponds a (ii) to every 0 0 such that b n ( ) i f t fd i 1 i a a b with P for every partition P of [ , ] and for all . t x , ] x [ i i 1 i
10 The Riemann-Stieltjes Integral and select Proof. (i) Let 0 such that 0 [ ( ) b ( )] a (1) which is possible by monotonicity of on [ , ] a b . Also f is continuous on compact set [ , ] a b . Hence f is uniformly continuous on [ , ] a b . Therefore there exists a 0 such that whenever x t ( ) f x f t ( ) for all , (2). x [ , ] a b t [ , ] a b Choose a partition P with P . Then (2) implies M m ( i 1,2,......, ) n i i Hence n n ( , , ) U P f ( , , ) L P f M m i i i i i 1 i 1 n n ( M m ) i i i i i 1 i 1 n [ ( ) x ( x )] i i i 1 i 1 [ ( ) b ( )] a . , ( ) which is necessary and sufficient condition for . f (ii) We have n ( , , ) L P f ( ) i f t ( , , ) U P f i i 1 and b ( , , ) L P f fd ( , , ) U P f a there exists such that for all partition P with P ( ) Since , for each 0 0 , we have f ( , , ) U P f ( , , ) L P f Thus b n ( ) i f t fd ( , , ) U P f ( , , ) L P f i 1 i a
Mathematical Analysis 11 b n fd Thus for continuous functions f , exists and is equal to . lim ( ) i f t i P 0 i 1 a a b and if is both monotonic and continuous on [ , ] Theorem 2. If f is monotonic on [ , ] ( ) f . a b , then Proof. Let be a given positive number. For any positive integer n, choose a partition P of [ , ] that a b such ( ) b ( ) a . ( i 1,2,......, ) n i n This is possible since is continuous and monotonic on [ , ] bounds ( ) a and ( ) b . It is sufficient to prove the result for monotonically increasing function f , the proof for monotonically decreasing function being analogous. The bounds of f in a b and so assumes every value between its are then [ x , x ] i 1 i , , . m f x ( ) M ( ) f x i 1,2,......, n i i 1 i i Hence n ( , , ) U P f ( , , ) L P f ( M m ) i i i i 1 n ( ) b ( ) a ( M m ) i i n i 1 n ( ) b ( ) a [ ( ) f x f x ( )] i i 1 n i 1 ( ) b ( )[ ( ) a f b ( )] f a n for large n. ( ) Hence . f Example 1. Let f be a function defined by and ( ) f x . * * * for , f x ( ) 1 0 x x a x b b fd Suppose is increasing on [ , ] ( ) * 0 a b and is continuous at x . Then over [ , ] a b and . f a . Since is continuous at * * Solution. Let to each be a partition of [ , ] a b and let x , P { , x x ,......, x 0 } x x 0 1 n such that i there exists 0
12 The Riemann-Stieltjes Integral * * x x ( ) x ( ) x whenever 2 Again since is an increasing function, * * ( ) x ( ) x for 0 x x 2 and * * ( ) x ( ) x for 0 x x 2 Then for a partition P of [ , ] a b , ( ) x ( x ) i i i 1 * * ( ) x ( x ) ( x ) ( x ) 1 i i . 2 2 * 0, t x n i ( ) i f t Therefore i * , t x i 1 i i that is, n ( ) i f t 0 i i 1 Hence b n lim ( ) i f t fd 0 i 0 P 1 i a b fd ( ) 0 and so and . f a ( ) ( ) ( ) Theorem 3. Let and on [ , ] a b , then and f f ( f f ) 1 2 1 2 b b b ( f f d ) f d f d 1 2 1 2 a a a ' ' " i " i , ] x Proof. Let be any partition of [ , ] respectively in the subinterval a b . Suppose further that m and { , ,......, x x } P a x b M m M , , , 0 1 n i i M m are the bounds of then f and . If , , , f f f [ x , x ] , [ x i i 1 2 1 2 i 1 i 1 2 i 1 i )] [ ( ) f ( )] [ ( f ) f ( f 1 2 2 2 1 1 2 1
Mathematical Analysis 13 ( ) ( ) f ( ) f f ( ) f 1 2 1 1 2 2 2 1 ' ' " i " i ( M m ) ( M m ) i i Therefore, since this hold for all , we have , [ x , ] x 1 2 i 1 i ' ' " i " i (1) M m ( M m ) ( M m ) i i i i ( ) Since , there exists a partition 1P and 2 P of [ , ] a b such that , f f 1 2 , ) , ) U P f ( , ( , L P f 1 1 1 1 2 (2) , ) , ) U P f ( , L P f ( , 2 2 2 2 2 These inequalities hold if 1P and 2 P are replaced by their common refinement P. Thus using (1), we have for , f f f 1 2 n ( , , ) U P f ( , , ) L P f ( M m ) i i i i 1 n n ' ' " i " i ( M m ) ( M m ) i i i i i 1 i 1 (using (2)) 2 2 . ( ) Hence . f f f 1 2 Further, we note that ' " i ' " i m m m M M M i i i i i Multiplying by and adding for we get i 1,2,......, , n , ) , ) ( , L P f ( , L P f ( , , ) L P f ( , , ) U P f 1 2 , ) , ) , ) (3) U P f ( , U P f ( , U P f ( , 1 1 2 Also b , ) U P f ( , f d (4) 1 1 2 a b , ) U P f ( , f d (5) 2 2 2 a
14 The Riemann-Stieltjes Integral Combining (3), (4) and (5), we have b , ) , ) fd ( , , ) U P f U P f ( , U P f ( , 1 2 a b b f d f d 1 2 2 2 a a Since is arbitrary positive number, we have b b b fd f d f d (6) 1 2 a a a Proceeding with in place of 1f and 2f respectively, we have ( f ),( f ) 1 2 b b b ( f d ) ( f d ) ( f d ) 1 2 a a a or b b b fd f d f d (7) 1 2 a a a Now (6) and (7) yield b b b b fd ( f f d ) f d f d . 1 2 1 2 a a a a ( ) ( ) Theorem 4.If and then and f f f ( ) b b b fd ( ) fd fd . a a a ( ) ( ) Proof. Since and , there exist partitions 1P and 2 P such that f f ( , , ) U P f ( , , ) L P f 1 1 2 ( , , ) U P f ( , , ) L P f 2 2 2 These inequalities hold if 1P and 2 P are replaced by their common refinement P. Also [ ( ) )] [ ( ) ( ) x ( x x ( x )] 1 1 i i i i i i Hence, if M and i m are bounds of f in , [ x , x ] 1 i i i
Mathematical Analysis 15 n ) ( ( , ,( U P f )) ( , ,( L P f )) ( M m ) i i i i i 1 n n ( M m ) ( M m ) i i i i i i i 1 i 1 . 2 2 Hence . f ( ) Further b ( , , ) U P f fd 2 a b ( , , ) U P f fd 2 a and ( , , U P f ) M M i i i i Also, then b fd ( ) ( , , U P f ) ( , , ) U P f ( , , ) U P f a b b fd fd 2 2 a a b b fd fd a a Since is arbitrary positive number, therefore b b b fd ( ) fd fd . a a a , this inequality is reversed and hence Replacing f by f b b b fd ( ) fd fd . a a a ( ) ( ) ( ) Theorem 5. If [ , ] a b and on [ , ] a b , then on [ , ] a c and on [ , ] c b where c is a point of f f f b c b fd fd fd . a a c
16 The Riemann-Stieltjes Integral ( ) Proof. Since , there exists a partition P such that f . , 0 ( , , ) U P f ( , , ) L P f * * Let P be a refinement of P such that . Then P P { } c * * ( , , ) L P f L P ( , , ) f U P ( , , ) f ( , , ) U P f which yields * (1) U P ( , , ) f ( *, , ) L P f ( , , ) U P f ( , , ) L P f * Let 2 P denote the sets of points of partitions of [ , ] a c and [ , ] 1P and P between[ , ] a c , [ , ] c b respectively. Then 1P and 2 P are * c b and . Also P P P 1 2 * (2) U P ( , , ) f ( , , ) U P f U P f ( , , ) 1 2 and * (3) L P ( , , ) f ( , , ) L P f L P f ( , , ) 1 2 Then (1), (2) and (3) imply that ( , , )] [ ( L P f , , )] * * U P ( , , ) f L P ( , , ) f [ ( , , ) U P f U P f , , ) L P f ( 1 1 2 2 Since each of and is non-negative, it follows that ( , , ) U P f ( , , ) L P f U P f ( , , ) L P f ( , , ) 1 1 2 2 ( , , ) U P f ( , , ) L P f 1 1 and U P f ( , , ) L P f ( , , ) 2 2 Hence f is integrable on [ , ] a c and [ , ] c b . Taking inf for all partitions, the relation (2) yields b c b fd fd fd (4) a a c But since f is integrable on [ , ] a c and [ , ] c b , we have b c b fd fd fd (5) a a c The relation (3) similarly yields
Mathematical Analysis 17 b c b fd fd fd (6) a a c Hence (5) and (6) imply that b c b fd fd fd . a a c ( ) Theorem 6. If , then f b b ( ) and ( cf d ) c fd (i) , for every constant c, cf a a ( ) f x K (ii) If in addition on [ , ] a b , then b [ ( ) fd K b ( )] a . a ( ) Proof.(i) Let and let g . Then f cf n n ' ( , , ) U P g M cM i i i i i 1 i 1 n c M i i i 1 ( , , ) cU P f Similarly ( , , ) L P g ( , , ) cL P f , a partition P such that for every , ( ) Since 0 f ( , , ) U P f ( , , ) L P f c Hence ( , , ) U P g ( , , ) L P g [ ( , , ) c U P f ( , , )] L P f . . cc ( ) Hence . g cf b ( , , ) U P f fd Further, since , 2 a
18 The Riemann-Stieltjes Integral b gd ( , , ) U P g ( , , ) cU P f a b fd c 2 a Since is arbitrary b b gd c fd a a , we get Replacing f by f b b gd c fd a a b b Hence ( cf d ) c fd . a a ( ) (ii) If M and m are bounds of on [ , ] a b , then it follows that f b [ ( ) m [ ( ) b ( )] a fd M b ( )] a for b a (1). a In fact, if a b , then (1) is trivial. If b a , then for any partition P, we have n [ ( ) m b ( )] a m ( , , ) L P f i i i 1 b fd a ( , , ) U P f M i i [ ( ) M b ( )] a which yields b [ ( ) [ ( ) m b ( )] a fd M b ( )] a (2) a ( ) f x K Since for all , we have x ( , ) a b K ( ) f x K so if m and M are the bounds of f in ( , ) a b ,
Mathematical Analysis 19 for all . K m ( ) f x M K x ( , ) a b If b a , then ( ) and we have by (2) b ( ) a 0 b [ ( ) [ ( ) m K b ( )] a b ( )] a fd a [ ( ) [ ( ) M b ( )] a K b ( )] a Hence b [ ( ) fd K b ( )] a . a ( ) a b . , is continuous on [ , [ ( )] f x Theorem 7. Suppose [ , ] a b . Then on[ , ] a b , m m M and ( ) on f on [ , ] f M ] h x ( ) h . Since is continuous on closed and bounded interval [ , ] m M . Therefore there exists Proof. Let continuous on [ , m M , it is uniformly 0 ] such that and 0 if s t ( ) s ( ) t , , . s t [ , m M ] ( ) Since , there is a partition of [ , ] a b such that P { , ,........, x x x } f 0 1 n 2 (1). ( , , ) U P f ( , , ) L P f * i * i Let 1,2,……,n into two classes: M m and m be the lub, glb of ( ) f x and ( ) x respectively in . Divide the number , [ x , x ] M , 1 i i i i if i A M m i i and if . i B M m i i lub ( ) , our choice of implies that k t . Also, for i * i * i * i * i For i , where , A B M m M m 2 k . Hence, using (1), we have t [ , m M ] 2 (2) ( M m ) i i i i i B i B so that . Then we have i i B * i * i * i * i ( , , ) U P h ( , , ) L P h ( M m ) ( M m ) i i i A i B [ ( ) ( )] 2 a b k [[ ( ) ( )] 2 ] a b k
20 Since was arbitrary, The Riemann-Stieltjes Integral , * * . ( , , ) U P h ( , , ) L P h 0 ( ) f Hence . h ( ) f ( ) ( ) ( ), Theorem 8. If and on [ , ] a b , then and f g fg b b fd f d . a a [ ( )] f x ( ) Proof. Let be defined by 1.3.2). Also 2 2 on [ , ] a b . Then by Theorem7(in section ( ) t t ( ) h x f 1[( 4 2 2 fg f g ) ( f g ) ] . ( ) ( ) ( ) ( ) 2 2 Since , , , . Then and and so their ( f g ) ( f g ) f g f g f g difference multiplied by 1 ( ) ( ) 4 also belong to proving that . fg ( ) c so that If we take ( ) f t f , again Theorem 7 implies that . We choose 1 c fd 0 Then c fd fd cfd f d becausecf f . 1.3.3. Riemann-Stieltjes integral as limit of sums. In this section, we shall show that Riemann- Stieltjes integral fd can be considered as the limit of a sequence of sums in which M m involved in , i i fd the definition of are replaced by the values of f . Definition 1. Let be a partition of [ , ] a b and let points 1 t be such P { a x x , ,........, x b } , ,......,n t t 0 1 2 n that . Then the sum t [ x , ] x 1 i i i n ( , , ) S P f ( ) i f t i i 1 is called a Riemann-Stieltjes sum of f with respect to .
Mathematical Analysis 21 Definition 2. We say that lim ( , , ) S P f A P 0 , there exists a such that P If for every 0 0 implies . ( , , ) S P f A ( , , ) S P f lim ( ) Theorem 1. If exists, then and f P 0 b ( , , ) S P f fd lim . 0 P a ( , , ) S P f there exists a lim Proof. Suppose exists and is equal to A. Then given 0 0 such P 0 that P implies ( , , ) S P f A 2 or A ( , , ) S P f A (1) 2 2 If we choose partition P satisfying P and if we allow the points it to range over S P f obtained in this way, the relation (1) gives , taking lub [ x , x ] 1 i i and glb of the numbers ( , , ) A ( , , ) L P f ( , , ) U P f A 2 2 and so ( , , ) U P f ( , , ) L P f 2 2 ( ) Hence . Further f A ( , , ) L P f fd ( , , ) U P f A 2 2 which yields A fd A 2 2 or fd A ( , , ) S P f lim . P 0
22 The Riemann-Stieltjes Integral Theorem 2. If (i) f is continuous, then b ( , , ) S P f fd lim 0 P a and is continuous on [ , ] ( ) (ii) a b , then f b lim ( , , ) S P f fd . 0 P a Proof. Part (i) is already proved in Theorem 1(ii) of section 1.3.2 of this unit. . Then there exists a partition , be continuous and ( ) * (ii) Let P such that 0 f * U P , , ) f fd ( (1) 4 such that for any partition P of [ , ] Now, being uniformly continuous, there exists , we have a b with 0 1 P 1 ( ) x x ( ) for all i 1 i i i Mn 4 where n is the number of intervals into which P* divides [ , ] ( , , ) U P f . Those a b . Consider the sum * intervals of P which contains a point of P in their interior contribute no more than: to U(P*,f, ) (2) n M ( 1) 4 Mn n M ( 1)max . i 4 Then (1) and (2) yield ( , , ) U P f fd (3) 2 P for all P with . 1 such that Similarly, we can show that there exists a 0 2 ( , , ) L P f fd (4) 2 P for all P with . 2 , it follows that (2) and (3) hold for every P such that P Taking . min{ , } 1 2 Since ( , , ) L P f ( , , ) S P f ( , , ) U P f
Mathematical Analysis 23 (3) and (4) yield ( , , ) S P f fd 2 and > ( , , ) S P f fd 2 Hence ( , , ) S P f fd 2 and so P for all P such that lim ( , , ) S P f fd P 0 This completes the proof of the theorem. The Abel’s Transformation (Partial Summation Formula) for sequences reads as follows: Let and be sequences and let na nb A , ........ ( 0) A a a a n 0 1 n 1 then q q 1 n n a b A b ( b ) A b A b n n n 1 q q p 1 p n p n p 1.4 Integration and Differentiation. In this section, we show that integration and differentiation are inverse operations. f Definition 1. If on , then the function F defined by [ , ] a b t F t ( ) ( ) f x dx , t [ , ] a b a is called the “Integral Function” of the function f. f Theorem 1. If on , then the integral function F of is continuous on . [ , ] a b f [ , ] a b Proof. We have t F t ( ) ( ) f x dx a f Since , it is bounded and therefore there exists a number M such that for all x in , [ , ] a b ( ) f x M .
24 The Riemann-Stieltjes Integral Let be any positive number and c be any point of . Then [ , ] a b c h c F c ( ) f x dx F c h ( ) f x dx ( ) ( ) , a a Therefore c h c F c ( h ) F c ( ) ( ) f x dx ( ) f x dx a a c h ( ) f x dx c M h if . h M F c h ( ) F c ( ) Thus implies . Hence F is continuous at any point ( c h ) c c [ , ] a b M and is so continuous in the interval . [ , ] a b Theorem 2. If is continuous on , then the integral function F is differentiable and f [ , ] a b F x , . ( ) ( ) f x x [ , ] a b 0 0 Proof. Let be continuous at in . Then there exists for every such that 0x f [ , ] a b 0 0 f t ( ) ( ) f x (1) 0 . Let t x whenever and , then x s x t x a s t b 0 0 0 0 t F t ( ) F s s ( ) 1 ( ) f x ( ) f x dx ( ) f x 0 0 t t s s t t 1 1 ( ) f x dx ( ) f x dx 0 t s t s s s t t 1 1 [ ( ) f x ( )] f x dx [ ( ) f x ( )] f x dx , 0 0 t s t s s s (using (1)). F x Hence ( ) ( ) . This completes the proof of the theorem. f x 0 0 ' Definition 2. A derivable function F such that Primitive of . f is equal to a given function in is called F f [ , ] a b
Mathematical Analysis 25 Thus the above theorem asserts that “Every continuous function possesses a Primitive, viz the integral f t ( ) f x dx function .” a Furthermore, the continuity of a function is not necessary for the existence of primitive. In other words, the function possessing primitive is not necessary continuous. For example, consider the function defined by [0,1] on f 1 x 1 x 2 sin x cos , x 0 ( ) f x 0, x 0 It has primitive 1 x 2 x sin , x 0 F x ( ) 0, x 0 '( ) x Clearly ( ) f x but is not continuous at , i.e., is not continuous in . F x ( ) f x f [0,1] 0 1.5 Fundamental Theorem of the Integral Calculus Theorem 1 (Fundamental Theorem of the Integral Calculus). If f ∈ R on differential function F on such that [ , ] a b and if there is a [ , ] a b F f , then b ( ) f x dx F b ( ) F a ( ) . a Proof. Let P be a partition of Lagrange’s Mean Value Theorem, we have and choose , such that . Then, by it ( x t x [ , ] a b i 1,2,......, ) n i i i 1 ( ) i F f ( ) F x ( ) ( ) ( ) ( ) f t (Since ). F x x x F t x x i i i i i i i 1 1 1 Further n F b ( ) F a ( ) [ ( ) F x F x ( )] i i 1 i 1 n ( )( i f t x x ) i i 1 i 1 n ( ) i f t x i i 1 b P ( ) f x dx 0 and the last sum tends to as , by theorem 1 of section 1.3.3, taking . Hence ( ) x x a
26 The Riemann-Stieltjes Integral b ( ) f x dx F b ( ) F a ( ) . a This completes the proof of the theorem. ( ) x dx The next theorem tells us that the symbols can be replaced by in the Riemann-Stieltjes d ( ) x b ( ) f x d ( ) x integral . This is the situation in which Riemann-Stieltjes integral reduces to Riemann a integral. f ( ) Theorem 2. If and on , then and [ , ] a b f b b ( ) ( ) f x fd x dx a a f f Proof. Since , , it follows that their product f . Let be given. Choose M such 0 f M that . Since and , using Theorem 2(ii) of section 1.3.3 for integrator as x, we have ( ) i t (1). ( ) i f t x f i P if and and x t x 1 i i i 1 ( ) i t (2). x i P if and . Letting vary in (2), we have x t x it 2 i i i 1 ( ) s (3). x i i P if and . From (2) and (3) it follows that x s x 2 i i i 1 ( ) i t ( ) s x x i i i ( ) i t ( ) s x x i i i 2 or ( ) i t ( ) s x 2 (4). i i P if and , . x t x x s x 2 i i i i i i 1 1 P min{ , } Now choose a partition P so that Theorem, and choose . By Mean Value [ , ] x t x 1 2 i i i 1
Mathematical Analysis 27 ( )( i s ( ) x ( ) ) x x x i i i i i 1 1 ( ) s x i i Then, we have ( )] i t ( ) ( ) i f t ( )[ ( ) i f t ( ) i f t t x s x (5). i i i i i Thus, by (1) and (4), it follows that ( ) i t ( ) s ( )] i t ( ) i f t ( ) i f t ( )[ i f t f x f x i i i i (1 2 2 M M ) Hence b ( ) ( ) f x lim ( ) i f t x x dx i P 0 a or b b ( ) ( ) f x fd x dx . a a 2 2 2 2 2 x dx [ ] x dx Example 1. Evaluate (i) , (ii) . 0 0 Solution. We know that b b ( ) ( ) f x fd x dx a a Therefore 2 2 2 2 2 2 3 x dx x (2 ) x dx 2 x dx 0 0 0 2 4 x . 2 8 4 0 and 2 2 2 [ ] x dx [ ]2 x xdx 0 0 1 2 [ ]2 x xdx [ ]2 x xdx 0 1 2 2 2 x 0 2 0 2 xdx 2 1 1
28 The Riemann-Stieltjes Integral =0+3 = 3. We now establish a connection between the integrand and the integrator in a Riemann-Stieltjes integral. We shall show that existence of implies the existence of fd df . We recall that Abel’s transformation (Partial Summation Formula) for sequences reads as follows: A “Let and be two sequences and let . Then ...... ( 0) na nb A a a a n n 0 1 1 q q 1 (*).” n n a b A b ( b ) q q A b A b n n n 1 p 1 p n p n p ( ) 1.5.1 Theorem (Integration by parts). If on , then on and f [ , ] a b ( ) f [ , ] a b ( ) x df x ( ) ( ) f b ( ) ( ) f a ( ) f x d ( ) x b a ( ) (Due to analogy with (*), the above expression is also known as Partial Integration Formula). a Proof. Let and take { , ,......, x x b } be a partition of . Choose t , ,......,n t t [ , ] such that P a x b t x t x [ , ] a b t t n i i i 0 1 1 a b 2 1 , . Suppose Q is the partition { , ,......, } of . By 0t nt n 1 1 2 1 partial summation, we have 1 n n ( )[ ( ) i f t ( ) ( ) f b ( ) ( ) f a ( , , ) S P f x ( x )] b a ( x )[ ( ) f t f t ( )] 1 1 1 i i i i i 1 1 i i ( ) ( ) f b ( ) ( ) f a ( , , ) S Q b a f P Q 0 0 since . If , , then t x t i i i 1 1 ( , , ) S Q ( , , ) S P f fd f df and . Hence ( ) ( ) f b ( ) ( ) f a fd b a df . 1.5.2 Mean Value Theorems for Riemann-Stieltjes Integrals. In this section, we establish Mean Value Theorems which are used to get estimate value of an integral rather than its exact value. Theorem 1.5.2(a). (First Mean Value Theorem for Riemann-Stieltjes Integral). If and real valued and be is monotonically increasing on that is continuous [ , ] a b f , then there exists a point x in such [ , ] a b b ( )[ ( ) f x fd b ( )] a . a become constant Proof. If and so , the theorem holds trivially, both sides being 0 in that case ( ). Hence we assume that ( ) ( ) a ( ) a ( ) b d . Let b 0 , . M lub ( ) f x m glb ( ) f x a x b
Mathematical Analysis 29 Then m ( ) f x M or [ ( ) m [ ( ) b ( )] a fd M b ( )] a Hence there exists some c satisfying such that m c M b [ ( ) c fd b ( )] a a Since is continuous, there is a point such that and so we have f x [ , ] a b ( ) f x c b ( )[ ( ) f x ( ) f x d ( ) x b ( )] a a This completes the proof of the theorem. Theorem 1.5.2(b) (Second Mean Value Theorem for Riemann-Stieltjes Integral). Let monotonic and be real and continuous. Then there is a point be f such that x [ , ] a b b ( )[ ( ) f a ( )[ ( ) f b fd x ( )] a b ( )] x a Proof. By Partial Integration Formula, we have b b ( ) ( ) f b ( ) ( ) f a fd b a df a a The use of First Mean Value Theorem for Riemann-Stieltjes Integral yields that there is x in that such [ , ] a b b ( )[ ( ) x f b df ( )] f a a Hence, for some , we have x [ , ] a b b ( ) ( ) f b ( ) ( ) f a ( )[ ( ) x f b fd b a ( )] f a a ( )[ ( ) f a ( )[ ( ) f b x ( )] a b ( )] x which proves the theorem. 1.5.3 We discuss now change of variable. In this direction we prove the following result. be continuous on is strictly increasing on [ , ] Theorem 1. Let and . If , where f [ , ] a b , then , a ( ) b
30 The Riemann-Stieltjes Integral b ( ( )) ( ) f x dx f y d ( ) y a b ( ) f x dx (this corresponds to change of variable in by taking ). x ( ) y a Proof. Since is strictly monotonically increasing, it is invertible and so 1( ) a 1( ) b , . Let be any partition of and be the { , ,......, x x } { , ,......, } P a x b Q y y y [ , ] a b n n 0 1 0 1 [ , ] 1( ) x corresponding partition of , where . Then y i i x x x i i i 1 ( ) y ( ) y i i 1 i . ( ) d [ ( )] Let for any , , where . Putting , we have c x d y c ( ) g y f y i i i i i i n (1) ( , ) S P f ( ) f c x i i i 1 ( ( )) d f i i i ( ) g d i i i ( , , ) S Q g b P ( , ) S P f ( ) f x dx 0 Continuity of implies that as and continuity of implies that f g a Q 0 ( , , ) S Q g ( ) g y d as . Q P P 0 0 0 Since uniform continuity of we have on implies that as . Hence letting in (1), [ , ] a b b ( ( )) ( ) f x dx ( ) g y d f y d ( ) y a This completes the proof of the theorem. 1.6 Integration of Vector –Valued Functions. Let and let [ , ] a b 1 2 ( , ,......, f f f be real valued functions defined on into . [ , ] a b R , ,......, f f f k 1 2 k be the corresponding mapping of ) f k
Mathematical Analysis ( ) f 31 ( ) Let be monotonically increasing function on and then the integral of . If for , we say that if [ , ] a b i k 1,2,......, is defined as f b b b b ,........, fd ( f d , f d f d ) . 1 2 k a a a a b b is the point in fd f d k Thus whose ith coordinate is . R i a a ( ) ( ) It can be shown that if , , f g then b b b ( f g d ) fd gd (i) a b a a c b fd fd fd (ii) , . a c b a a c and (iii) if , , then ( ) ( ) ( ) f f f 1 2 1 2 b b b fd ( ) fd fd 1 2 1 2 a a a To prove these results, we have to apply earlier results to each coordinate of theorem of integral calculus holds for vector valued function . Also, fundamental f . We have f ( ) k Theorem 1. If and F map into , if if , then f [ , ] a b f F f b f t dt ( ) F b ( ) F a ( ) a ( ) R k Theorem 2. If , then [ , ] a b maps ( ) R into and if for some monotonically increasing function on R f [ , ] a b f f and b b fd f d . a a Proof. Let . ( ,......, f ) f f k 1 Then 2 2 1/2 ) k f ( f ...... f 1
32 The Riemann-Stieltjes Integral ( ) R ( ) R ( ) R 2x 2 2 2 Since each continuous function of x, the square root function is continuous on R , the function and so their sum . Since is a if if f ...... f 1 k for every real M. Therefore M [0, ] f ( ) . f d y Now, let ( , y y ,.... ) , where , then y y i i k 1 2 fd y and 2 f d 2 i y y y i i i y f d ( ) i i But, by Schwarz inequality y f t ( ) y f t ( ) , ( a t b ) i i Then 2 (1) f d y y y y 0 y If , then the result follows. If , then divide (1) by and get 0 f d y b b fd f d or . a a 1.7 Rectifiable Curves. The aim of this section is to consider application of results studied in this chapter to geometry. k k Definition 1. A continuous mapping of an interval into is called a curve in . R R [ , ] a b :[ , ] a b k If is continuous and one-to-one, then it is called an arc. R :[ , ] a b k If for a curve , R ( ) a ( ) b but ( ) t ( ) t 1 2 for every other pair of distinct points in , then the curve is called a simple closed curve. , t t [ , ] a b 1 2 k Definition 2. Let be a map. If is a partition of , then :[ , ] a b { , ,...., x x } f R P x [ , ] a b n 0 1 n , ( , , ) V f a b lub ( ) f x f x ( ) i i 1 i 1
Mathematical Analysis 33 where the lub is taken over all possible partitions of function is said to be of bounded variation on f , is called total variation of . ( , , ) V f a b on . The [ , ] a b f [ , ] a b if [ , ] a b :[ , ] a b V k Definition 3. A curve rectifiable curve is called rectifiable if is of bounded variation. The length of a . Thus length of rectifiable curve ( , , ) a b R is defined as total variation of , i.e., n ( ) x for the partition ( .... ) . a x x x b lub ( x ) n 0 1 i i 1 i 1 The ith term ( ) x ( ) k ( x ) in this sum is the distance in between the points and . ( ) R ix ix i i 1 1 n ( ) x Further is the length of a polygon whose vertices are at the points ( x ) i i 1 i 1 ), ( ),...., ( x ( . As the norm of our partition tends to zero, then those polygons approach the more and more closely. ) x x n 0 1 range of k Theorem 1. Let be a curve in . If is continuous on , then is rectifiable and has length R [ , ] a b b ( ) t dt . a ( , , ) a b V Proof. It is sufficient to show that . So, let { ,...., x } be a partition of . x [ , ] a b n 0 Using Fundamental Theorem of Calculus for vector valued function, we have x n n i ( ) t dt ( ) x ( ) x i i 1 i 1 i 1 x i 1 x n i ( ) t dt i 1 x i 1 b ( ) t dt a Thus ( , , ) a b V (1). . To prove the reverse inequality, let there exists 0 be a positive number. Since is uniformly continuous on , [ , ] a b such that ( ) s ( ) t s t , if . If mesh (norm) of the partition P is less than and , then we have x t x i i 1 ( ) t ( ) x , i
34 The Riemann-Stieltjes Integral so that x i ( ) t dt ( ) x x x i i i x i 1 x i [ ( ) ( ) x ( )] t dt t i x i 1 x x i i ( ) t dt [ ( ) ( )] t dt x i x x i 1 i 1 ( ) x ( x ) x i i 1 i Adding these inequalities for , we get i 1,2,...., n b n ( ) t dt ( ) x b a ( x ) 2 ( ) i i 1 i 1 a ( , , ) a b V b a 2 ( ) Since is arbitrary, it follows that b ( ) t dt ( , , ) a b V (2). a Combining (1) and (2), we have b ( ) t dt ( , , ) a b V a b ( ) t dt Hence the length of is . a 1.8 References 1.Walter Rudin, Principles of Mathematical Analysis (3rd edition) McGraw-Hill, Kogakusha, 1976, International Student Edition. 2.T. M. Apostol, Mathematical Analysis, Narosa Publishing House, New Delhi, 1974. 3.H.L. Royden, Real Analysis, Macmillan Pub. Co., Inc. 4th Edition, New York, 1993. 4.G. De Barra, Measure Theory and Integration, Wiley Eastern Limited, 1981. 5.R.R. Goldberg, Methods of Real Analysis, Oxford & IBH Pub. Co. Pvt. Ltd, 1976. 6.R. G. Bartle, The Elements of Real Analysis, Wiley International Edition, 2011. 7.S.C. Malik and Savita Arora, Mathematical Analysis, New Age International Limited, New Delhi, 2012.
