Exercises not in Rudin:
6.3:0. Say whether the following statement is true or false.
(a) If | f (x) | ≤ | g(x) | for all x∈[a , b], and g∈ (α), then f∈ (α).
6.3:1. The obvious formula for ∫1 dα. (d : 1)
Show (from the definition of the integral) that for any increasing function α on an interval [a, b], one has
∫
ab1 dα = α(b) –α(a). (This should probably have been made part of Theorem 6.12.)6.3:2. Functions whose values are ‘‘mostly’’ zero have integral zero. (d : 2)
(a) Let f : [0,1] → R be the function of 4:R18 (p.100) which takes the value 0 at all irrationals, and the value 1 ⁄ n at a rational number whose expression in lowest terms is m ⁄ n. Show that for every continuous increasing function α on [0,1] one has f∈ (α), and
∫
01f dα = 0. (You may do this by doing part (b) below, if you choose.)(b) Show that the same conclusion is true of any function f on [0,1] with the property that for every ε > 0, the set {x∈[0,1]! | f (x)| ≥ ε} is finite. Show that the function f of part (a) has that property.
6.3:3. An α which does all its increasing on the Cantor set. (d : 4, 3, 2, 1, 3, 1)
Let P denote the Cantor set, and let us define a function α0 on the complement of P in [0,1] as follows. For all x in the segment (1 ⁄ 3,2 ⁄ 3), i.e., the segment one deletes at the first step in constructing ...
Answers to True/False question 6.2:0. (a)F. (b)T. (c) F. (d)T. (e)T. (f )F.
the Cantor set, let α0(x) = 1 ⁄ 2. At all points of the segments (1 ⁄ 9,2 ⁄ 9) and (7 ⁄ 9,8 ⁄ 9), the two segments one deletes at the second stage in the construction of the Cantor set, let α0(x) have the values 1 ⁄ 4 and 3 ⁄ 4 respectively. Similarly, on the 2n –1 segments deleted at the nth stage of the construction of the Cantor set, let α0 have the constant values 1 ⁄ 2n, 3 ⁄ 2n, ... , (2n–1) ⁄ 2n (each one half-way between the values previously assigned on the two surrounding segments; or in the case of the first and last of these segments, half-way between the value at the adjacent previously assigned segment and the value 0, respectively 1). Now
(a) Show that α0 can be extended to a function α on all of [0,1] so as to give a continuous increasing function. (Suggestion: Show that the set on which we have defined α0 is dense in [0,1], and that α0(x –) and α0(x+) are defined and equal for all x∈[0,1], and deduce the result using these facts.)
(The function α can also be constructed as follows: Given x∈[0,1], write it in base 3 notation. If it has any digit ‘‘1’’, change the first such digit to a ‘‘2’’, and all digits after it to ‘‘000...’’. In the resulting string of 0’s and 2’s, change all 2’s to 1’s, and regard the result as an expression for α(x) in base 2. But if you use this description, you must prove it equivalent to the one stated above.)
(b) For α as in part (a), show that if f is a continuous real-valued function on [0,1] which is zero on all points of the Cantor set, then
∫
f dα = 0.(c) Give an example of a function f as in (b) which is not the zero function.
(d) Deduce from (b) that if f and g are two continuous real-valued functions on [0,1] which agree on all points of the Cantor set, then
∫
f dα =∫
g dα.(e) Let f : [0,1] → R be the function such that f (x) = 0 for all x in the Cantor set, and f (x) = 1 for all other x. Show that f∈ (α).
(f ) Why does the result of (e) not contradict the result of (b) above? Why does not it contradict the result of 6:R6 (p.138)?
6.3:4. An integrable function of an integrable function need not be integrable. (d : 2)
Show by example that, in contrast to Theorem 6.11, if f and ϕ are Riemann-integrable functions, their composite ϕ°f need not be. Suggestion: Let f be as in 4:R18 , and choose ϕ to be discontinuous at the real number which f most often approaches. (Examples are even known where f is continuous;
but they are more difficult to describe.)
6.3:5. Condition for an increasing function to be Riemann-Stieltjes integrable. (d : 4. > 4.3:5(a))
Prove that f∈ (α) if f is increasing and is continuous at every point where α is discontinuous.
(Suggestion: Combine the ideas of the proofs of Theorems 6.8 and 6.9, using 4.3:5(a). Incidentally, the idea of the first displayed formula in the proof of Theorem 6.9 is that the numbers ∆αi are all small.
You won’t be able to get a precise formula like that one in your proof, but you will want to use the same idea.)
6.3:6. A condition for mutual integrability of α and β. (d : 4. > 6.2:1)
Show that the three conditions of 6.2:1 are also equivalent to: (iv) For every x∈[a, b), either α(x) = α(x+) or β(x) = β(x+), and for every x∈(a, b], either α(x) = α(x –) or β(x) = β(x –). (Suggestion:
assuming (iv) holds, let M = (α(b) + β(b)) – (α(a) + β(a)). Given ε, find a partition such that in every interval of the partition, either ∆αi or ∆βi is < ε⁄ 2 M. Deduce that ∆αi ∆βi ≤ (∆αi+∆βi) (ε⁄ 2 M ), and sum these inequalities.)
6.3:7. ‘‘Sinceε is arbitrary ...’’ (d : 2) (a) Find the fallacy in the following argument:
‘‘Theorem’’ Every function f : X→ Y between metric spaces is continuous.
Proof Given any ε> 0 and any x∈X, take δ = 1, and consider a point y≠x of X satisfying d (x , y) <δ. Choose any C > d ( f (x), f (y)) ⁄ε. Multiplying this inequality by ε, we get
...
Answer to True/False question 6.3:0. (a)F.
d ( f (x), f (y)) < Cε.
But since ε was arbitrary, we can choose it to make this as small as we like, proving continuity.
(b) The ends of the proofs of Theorem 3.50 (pp.74-75) and Theorem 6.11 (p.127) use a similar argument that since ε is arbitrary, an expression having ε as a factor can be made arbitrarily small, yielding a desired conclusion. Why do those proofs not suffer from the same fallacy as above?
6.4. Basic properties (beginning of Rudin’s section PROPERTIES OF THE INTEGRAL). (pp.128-130)
Relevant exercises in Rudin:
6:R5. Does f2∈ or f3∈ imply f∈ ? (d : 2) 6:R10. Ho¨lder’s inequality. (d : 3, 2, 4, 3)
The only way I can see to do part (a) is quite roundabout: Let up = s, q = t, 1 ⁄ p = a, and hence, by the first displayed equation, 1 ⁄ q = 1 – a. Turn the resulting inequality into an inequality concerning s ⁄ t, and write s ⁄ t = r. The resulting inequality will be true for r = 1; prove it for general r using the Mean Value Theorem, assuming the standard formula for the derivative of xa. (Without the above hint, I would rate that part at least d : 4.)
Hint for part (c) in the case where neither of the integrals on the right is zero: Use a scalar multiplication to reduce to the case where those integrals are 1, and apply (b).
I don’t know what method Rudin had in mind for the case where one or both of those integrals is zero.
It could be proved using results from Chapter 11, but these are not available yet. One method that will work is to approximate f and /or g by functions for which the integrals in question are nonzero, and show that the inequality for those approximating functions implies the same inequality for the limit functions. Another proof of this case of (c) for the special case p = q = 2 is given in 6.4:2 below, and an alternative development of the rest of part (c) for those values of p and q in 6.4:3.
Part (d ), of course, depends on 6:R7 and 6:R8. (I didn’t indicate this above because parts (a)-(c), which form the core of the exercise, do not.)
6:R11. The triangle inequality for the L2 norm. (d : 1. > 6:R10(c), or 6.4:2 and 6.4:3)
The ‘‘Schwarz inequality’’ that Rudin refers to here is not the result of that name which we saw in Theorem 1.35, but the version of that result with integration replacing summation referred to in part (c) of the preceding exercise. As the difficulty-rating indicates, this exercise is easy – assuming the difficult exercise 6:R10 discussed above, or the somewhat easier substitute exercises given below. Incidentally,
|| u ||2 is known as ‘‘the L2norm of u’’; hence the titles I have given this and the next exercise.
6:R12. An integrable function is L2-approximable by a continuous function. (d : 3. > 6:R11) Exercises not in Rudin:
6.4:0. Say whether the following statement is true or false.
(a) If f∈ (α) on an interval [a , b], then f∈ (α) on every subinterval [c , d] ⊆ [a , b].
6.4:1. Integration with respect to α and α–1. (d : 1)
A real-valued function α on a set E of real numbers is said to be strictly increasing if for all x, y∈E, x < y ⇒ α(x) < α(y). Clearly, such a function is one-to-one. If α is a strictly increasing continuous function on an interval [a , b], the Intermediate Value Theorem shows that α is onto [α(a),α(b)], hence is a bijection from [a , b] to [α(a),α(b)], hence it has an inverse function, α–1: [α(a),α(b)] → [a , b].
For such an α, show that for every real-valued function f on [a, b] we have
∫
abf (x) dα(x) =∫
αα(a)(b)f (α–1(x)) dx,in the sense that if either side is defined, so is the other, and they are then equal.
(This result is related to Theorem 6.19; but that is outside of section 6.4, so you can’t use it here.)
6.4:2. Functions that behave like zero under Riemann-Stieltjes integration. (d : 3)
(a) Suppose that f∈ (α) on [a, b], and that
∫
abf (x)2dα(x) = 0. Prove that for all g∈ (α) one also has∫
abf (x) g(x) dα(x) = 0.Suggestion: Fixing g, verify on general principles that for all real numbers t,
∫
ab( f (x) + t g(x))2dα(x) ≥ 0, and that as a function of t, this integral has a local minimum at t = 0.Now expand the integral in terms of the integrals of f2, f g and g2, and draw a conclusion.
(b) Deduce that if f∈ (α) on [a, b], and f (x) ≥ 0 for all x, then the following conditions are equivalent: (i)
∫
abf (x) dα(x) = 0. (ii)∫
abf (x)2dα(x) = 0. (iii)∫
abf (x) g(x) dα(x) = 0 for all g∈ (α).(Hint: Can you write f as the square of a function in (α) ?)
(c) Show that if from part (b) we delete the assumption that f (x) ≥ 0 for all x, then the equivalence still holds, with f (x) replaced by | f (x) | in statement (i), but not in statements (ii) and (iii). (Hint: f = ( | f | + f ) ⁄ 2 – ( | f | – f ) ⁄ 2.)
6.4:3. The Schwarz inequality for integrals. (d : 3. > 6.4:2)
The calculations by which we proved the Schwarz inequality for n-tuples of real numbers in 1.7:2 can be mimicked using integrals instead of sums. Given an increasing function α, the analog of the dot product for f, g∈ (α) is
∫
abf g dα. Use the method of that exercise to obtain a Schwarz inequality for such functions (namely, the bottom display on p.139 with p = q = 2), assuming the integral of the square of each function is nonzero.
For vectors, the case of ‘‘zero norm’’ created no difficulty, because vectors of zero norm were zero;
however, the analogous statement for integrals is not true (cf. 6:R1, p.138). Show, however, that that case can be handled with the help of 6.4:2 above.
6.4:4. Description of (α+β). (d : 2)
Show that if α and β are increasing functions on [a, b], then (α+β) = (α) ∩ (β).
6.4:5. Extending the Riemann-Stieltjes integral to the case of non-increasing α. (d : 3)
The definition of the Riemann-Stieltjes integral
∫
f dα requires that α be an increasing function. In this exercise we will see that, having defined such integrals for increasing α, we can extend the definition in a natural way to the wider class of all functions that can be written as differences of increasing functions. (Exercise 5.2:3 looked at that class of functions; but the result proved there is not needed for this exercise.)Suppose that f and λ are functions on [a, b], and that we can express λ as λ = α1 –α2,
where α1 and α2 are increasing functions, and f is integrable with respect to both α1 and α2. Show that if we define
∫
f dλ =∫
f dα1 –∫
f dα2then
∫
f dλ is well-defined, independent of our choice of decomposition of λ as α1 –α2.Thus, what you must prove is that if λ can also be written λ = β1 – β2, where β1 and β2 are increasing and f is integrable with respect to both β1 and β2, then
∫
f dα1 –∫
f dα2 =∫
f dβ1 –∫
f dβ2.(Suggestion: abbreviating
∫
f dα1 –∫
f dα2 to I (α1,α2), show that I (α1,α2) = I (α1+β2,α2+β2)= I (α2+β1,α2+β2) = I (β1,β2).)
6.4:6. Converse to Theorem 6.12 (c). (d : 2)
Show that if a < c < b, and α is a monotonically increasing function on [a , b], and f is a function on [a , b] such that f∈ (α) both on [a , c] and on [c , b], then f∈ (α) on [a , b].
Thus the equality
∫
acf dα +∫
cbf dα =∫
abf dα,...
Answer to True/False question 6.4:0. (a)T.
which Rudin proves under the hypotheses of Theorem 6.12(c), also holds in this situation.
6.4:7. Integrable functions can have infinitely many big jumps. (d : 1, 3)
Let s1 < s2 < ... < sn < ... be elements of an interval [a , b] such that s1 = a and limn→∞ sn = b.
Let f : [a , b] → [0,1] be defined by the condition that for x∈[si, si+1), f (x) = (x – si) ⁄ (si+1 – si), while f (b) = 0.
(a) At what points x is f discontinuous, and at each of these points, what are the values of f (x), f (x – ), and f (x+ ) if the latter two are defined? (No proof required for this part.)
(b) Assuming the properties noted in part (a), show that f∈ . The above example is generalized in the next exercise.
6.4:8. Integrability piece by piece. (d : 3)
Let s1 < s2 < ... < sn < ... be elements of an interval [a , b] such that s1 = a and limn→∞ sn = b.
Let α be an increasing function on [a , b], and let f be any bounded real-valued function on [a , b].
For each positive integer i, let fi denote the restriction of f to [si, si+1], and αi the restriction of α to that interval.
Prove that f∈ (α) if and only if for every i, fi∈ (αi), and that when this holds, we have
∫
abf dα = Σ∞i = 1( ∫
si si+1
f dα
)
,with the sum on the right absolutely convergent. (The ith summand of that sum could, of course, be written more formally as
∫
si si+1
fidαi. )
6.4:9. Functions with only countably many discontinuities are integrable. (d : 2, 3)
Let us begin by proving explicitly an ‘‘intuitively obvious’’ general fact that we will need.
(a) Suppose an interval [a , b] is covered by finitely many open neighborhoods, Nε
1( p1), ... , Nε
m( pm).
Show that there is a partition x0, ... , xn of [a , b] such that every subinterval [xi–1, xi] is contained in (at least) one of the neighborhoods Nε
i( pj). Show further that in this situation, given any subset S ⊆ {1, ... , m}, if we let T denote the set of indices i∈ {1, ... , n} such that [xi–1, xi] is contained in Nε
1( pj) for some j∈S, then Σi∈T ∆xi ≤ Σj∈S 2εj.
(Suggestion: Let y1 < ... < yk be those boundary points of the neighborhoods Nε
i( pi) – i.e., points of the form pi–εi or pi+εi – that lie in (a , b), arranged in increasing order, and let y0 = a, yk+1 = b. Use the partition y0 < (y0+ y1) ⁄ 2 < (y1+ y2) ⁄ 2 < . . . < (yk –1+ yk) ⁄ 2 < (yk+ yk+1) ⁄ 2 < yk+1, verifying that it has the required properties.)
Now for the interesting result.
(b) Show that a bounded real-valued function f on an interval [a , b] → R which is continuous except at a countable set of points is Riemann-integrable.
(Suggestion: Suppose f continuous at all points except s1, s2, ... si, ... . Given ε, choose a series Σδi of positive real numbers which converges to a sum <ε. Surround each point si by the neighborhood Nδ
i(si), while for each point x where f is continuous, show that there is a neighborhood Nδ(x)(x) such that diam( f (Nδ(x)(x))) < ε. These two sorts of neighborhoods together cover [a , b].
Take a finite subcovering, choose a partition as in part (a), deduce that the total length of the intervals [xi–1, xi] with diam( f ([xi–1, xi])) > ε is small, and complete the proof like that of Theorem 6.11.)
6.4:10. A Riemann-integrable function with uncountably many discontinuities. (d : 3)
Let f : [0,1] → R denote the function which has the value 1 at all points of the Cantor set, and 0 elsewhere. Show that f is discontinuous at uncountably many points, but is Riemann integrable.
Determine its integral.
(Suggestion: Consider the partition of [0,1] into 3n equal intervals, and count the number of such intervals which contain at least one point of the Cantor set. The fact that many of them have just an endpoint in the Cantor set is a minor nuisance, but even counting those, the fraction of intervals containing a point of the Cantor Set behaves well as n→
∞
.)6.4:11. One case of Theorem 6.12(a). (d : 1)
Prove the last equation in Theorem 6.12(a) in the case where c < 0.
(Not hard, but one needs to notice that in proving that equation, the cases c > 0 and c < 0 must be distinguished.)
6.4:12. Theorem 6.11 for a function ϕ of several variables. (d : 3)
Suppose α is a monotonically increasing real-valued function on an interval [a , b], and f1, ... , fk are real-valued functions on [a , b] which each belong to (α). Let us write f : [a , b] → Rk for the function defined by f (x) = ( f1(x), ... , fk(x)). Let K be any compact subset of Rk containing f ( [a , b] ) = { f (x) | x∈[a , b] }, and let ϕ be any continuous real-valued function on K.
Below, you will prove that the function ϕ°f : [a , b] → R defined by (ϕ°f )(x) = ϕ(f (x)) also belongs to (α), generalizing Rudin’s Theorem 6.11.
(a) Given ε > 0, show that there is a δ > 0 such that for p, q∈K one has d ( p, q) < δ ⇒
|ϕ( p) – ϕ(q) | < ε, and a partition P = {x0, ... , xn} of [a , b] such that for j = 1, ... , k, U (P,α, fj) – L (P,α, fj) <δ ε.
(b) For δ and P = {x0, ... , xn} as above, let A ⊆ {1, ... , n} be the set of indices such that for j = 1, ... , k,
supx∈[x
i–1, xi] fj(x) – infx∈[x
i–1, xi] fj(x) < δ⁄0k.
Show that for i∈A and x, y∈[xi–1, xi], one has d (f (x), f (y)) < δ. Deduce from this an upper bound on the sum of the terms of U (P,α,ϕ°f ) – L (P,α,ϕ°f ) indexed by members of A, such that this upper bound approaches 0 as ε→ 0.
(c) Let B = {1, ... , n} – A. Prove an upper bound on Σi∈B∆αi that approaches 0 as ε → 0. (Hint: ε was used in choosing P. Note that k, f and ϕ are fixed in this exercise, so the bound can depend on these.)
(d) Deduce a bound on U (P,α,ϕ°f ) – L (P, α, ϕ°f ) that approaches 0 as ε → 0, and conclude that ϕ°f∈ (α), as claimed.
Remark: Rudin proves Theorem 6.11 only for functions of one variable. He deduces Theorem 6.13(a), that a product of functions in (α) again lies in (α), by a trick that reduces the two-variable multiplication operation to the one-variable squaring operation. The trick is impressive, but that result suggests the question of whether a similar result holds for continuous functions other than multiplication.
The above exercise answers this question.
6.5. Step functions, differentiable α, and change of variables (end of Rudin’s section PROPERTIES OF THE INTEGRAL). (pp.130-133)
Relevant exercises in Rudin: None Exercises not in Rudin:
6.5:0. Say whether each of the following statements is true or false.
(a) There exists an increasing function α on [0,1] such that for every continuous function f on that interval,
∫
01 f dα = Σ 2– nf (1 ⁄ n).(b) There exists an increasing function α on [0,1] such that for every continuous function f on that interval,
∫
01 f dα = Σ (1 ⁄ n) f (2– n).(c) If α increases monotonically on [a , b] and is differentiable, then α′ ∈ on [a , b].
(d) If f∈ on [0,1], then
∫
01 f (x) d x =∫
01⁄2 f (2x) d(2x).6.5:1. Getting the Fundamental Theorem of Calculus from Theorem 6.17. (d : 2. > 6.3:1)
This exercise obtains of one of the main results of the next section from a result in this section.
Let f∈ on [a, b], and suppose there exists a differentiable function F on [a, b] such that F′= f.
(a) Show that if f (x) ≥ 0 for all x∈[a, b], then one can apply Theorem 6.17 with the constant function
1 as the ‘‘f ’’ of that theorem, and F as the α of that theorem, and that 6.3:1 above is applicable to one side of the formula you get. Show the formula that results from applying 6.3:1 to that side.
(b) Deduce that the same formula holds if f can be written f1– f2, where each of f1 and f2 is nonnegative, is Riemann-integrable, and is the derivative of a differentiable function.
(c) Show that any f which is Riemann-integrable and is the derivative of a differentiable function can in fact be written f = f1– f2 as in (b). (Hint: Take f2 to be an appropriate constant function.) Conclude that the formula you got in (a) is applicable to any such f .
6.5:2. Strengthening Theorem 6.16. (d : 2)
Show that Theorem 6.16 (p.130) remains valid if the assumption that f is continuous on [a, b] is replaced by the condition that it is bounded, and is continuous at each of the points sn. Specifically –
We see that no continuity condition is needed until display (24). Say why that display is valid under the condition stated above. Then modify the remainder of the proof (in which Rudin uses integrability of f with respect to α2, which under his assumptions follows from Theorem 6.8) to conclude that (23) holds with the upper, respectively the lower integral in place of the left-hand side. Conclude that f∈ (α) and that (23) holds.
6.5:3. Weaker conditions for integrability. (d : 3. > 6.5:2)
Let J(x) be defined like I(x) at the bottom of p.129, except that J(0) = 1. (Equivalently, J(x) = 1 – I(– x) for all x.)
(a) Show that every increasing function α(x) on an interval [a, b] can be written α(x) = Σ1∞c
nI(x – sn) + Σ1∞d
nJ(x – sn) + β(x),
where Σcn and Σdn are convergent series of nonnegative real numbers, (sn) is a sequence of distinct points of [a, b], and β is a continuous increasing function on [a , b].
(Hint: If α is discontinuous at infinitely many points, can you write the set of these points as { sn ! n = 1, 2, ... } ? If so, what should the cn and dn be to make the above equation plausible? Once you have found these, to prove that α(x) – ΣcnI(x – sn) – ΣdnJ(x – sn) is increasing, approximate this function using finite partial sums in place of the infinite sums, and prove each of the resulting functions increasing. Finally, verify continuity.)
(b) Taking for granted that the version of Theorem 6.16 described in 6.5:2 is also valid with J in place of I, deduce that Theorem 6.9 remains valid if the assumption that α is continuous on [a, b] is replaced by the assumption that it is continuous at every point where f is discontinuous.
(The proof of the result gotten by replacing ‘‘I ’’ with ‘‘J ’’ in 6.5:2 is virtually identical to the original proof of that exercise. It can also be gotten using the next exercise.)
6.5:4. Integration with respect to d ( –α( – x)). (d : 2)
Suppose f∈ (α) on [a, b]. We would like to be able to apply Theorem 6.19 in the case ϕ(x) = – x, and deduce that
∫
– a– bf (– x) dα( – x) =∫
abf (x) dα(x). Unfortunately, α( – x) is not an increasing function of x, and – a is not < – b, so the left-hand integral has not been defined. However, –α( – x) is an increasing function, so
(a) Show that
∫
– b– af (– x) d ( –α( – x)) =∫
abf (x) dα(x).(Remark: Rudin doesn’t like showing the variable in integrations. To satisfy his preference, we could let ρ : R → R (ρ = rho, for ‘‘reversal’’) be defined by ρ(x) = – x, and write the above equation
∫
( f°ρ) d (ρ°α°ρ) =∫
f dα.)(b) From part (a) and Theorem 6.19, deduce that if, in the first line of that theorem we change
‘‘increasing’’ to ‘‘decreasing’’, then we get a formula like (32), but with dβ changed to d( –β).
...
Answers to True/False question 6.5:0. (a)T. (b)F. (c) F. (d)T.
6.6. INTEGRATION AND DIFFERENTIATION (the Fundamental Theorem of Calculus). (pp.133-134)
Relevant exercises in Rudin:
6:R9. Integration by parts for improper integrals. (d : 4. uses definitions in 6:R7, 6:R8) 6:R13. Calculations involving ∫ sin (t2) d t . (d : 3)
Note that in this exercise, parentheses ( ) and brackets [ ] do not denote the ‘‘fractional part’’ and
‘‘integer part’’ functions of 4:R16 . They are simply being used to make very clear what the squaring operation is and isn’t being applied to.
6:R14. Calculations involving ∫ sin (et) d t . (d : 3. > 6:R13)
6:R15. Properties of a function such that ∫ f2d t = 1. (d : 2. > 6:R10(c) or 6.4:3) 6:R16. The Riemann zeta-function. (d : 3. uses definitions in 6:R8)
Before attempting this, you should be sure you are familiar with the greatest-integer function [x ]; e.g., sketch its graph. The improper integrals are to be understood in the sense of 6:R8, p.138 (but the result of that exercise isn’t needed for this one). To evaluate the corresponding integrals on intervals [1, N ] as suggested in Rudin’s hint, compute them for the subintervals where [x ] is constant (except at the endpoints), and use Theorem 6.12(b). In doing the integrations, you can assume the formula for the derivative of xc, but justify the integration formula you get from that formula using a theorem in this chapter; also justify the way you handle the behavior at endpoints of intervals of integration (hint: 6:R1 ).
In the final step of deducing from the values of the integral over intervals [1, N ] the value of the improper integral, remember that you need a limit of
∫
1C... as C approaches +∞
through all real values, not just through the integer values examined so far.Get (b) from (a). (The point of (b) is that it gives a formula for the zeta function that makes sense not only for s∈(1, +
∞
), where the original series converges, but also for s∈(0, 1).)6:R17. Integration by parts for Riemann-Stieltjes integrals. (d : 4)
Rudin’s Hint begins ‘‘Take g real, without loss of generality.’’ What this means is, prove the result for an R-valued function g, then deduce from that the corresponding statement for an Rk-valued function g .
In obtaining the relation Rudin gives at the end of his Hint, you might use Theorem 3.41, or 3.11:1 above.
Exercises not in Rudin:
6.6:0. Say whether each of the following statements is true or false.
(a) If f∈ on [a, b], and for all x∈[a, b] we define F(x) =
∫
axf (t) dt, then F is differentiable and F′ = f.(b) If f is a continuous function on [a, b], and for all x∈[a, b] we define F(x) =
∫
xbf (t) dt, then F is differentiable and F′ = – f.6.6:1. A sort of derivative formula for Riemann-Stieltjes integrals. (d : 2)
On [a, b], let α be a strictly increasing function (as defined in 5.2:1(b)) and f a continuous function, and for x∈[a, b] define F(x) =
∫
axf (t) dα(t). Show that for all x∈[a, b], d F (x) ⁄ dα(x) = f (x), where the left-hand side is defined as limt→x (F(x) – F(t)) ⁄ (α(x) –α(t)), and the equality includes the assertion that this limit exists.6.6:2. Repeated integration reduces to a single integration. (d : 3, 4, 4) (a) Show that if f is a continuous function on [a, b], then
∫
t = ab &(∫
s = at f (s) ds') dt =∫
t = ab (b – t) f (t) dt.Hint: Rewrite the above equation with arbitrary x∈[a, b] in place of b, and name the left-hand side P(x) and the right-hand side Q(x). Show that both P and Q are differentiable functions of x, and have the same derivatives. In figuring out how to differentiate Q, use Theorem 6.12.
One can also get this formula using results about change of order of integration; but Rudin will not treat that subject till Chapter 10.
(b) Find a similar formula for the n-fold iterated integral of f, of which the above is the n = 2 case.
(c) Show that the result of (a) and, if you did it, (b), continues to hold if f is merely assumed Riemann-integrable, but not necessarily continuous. (Hints: To show the right-hand side is differentiable with the correct derivative, first verify that if a function u(x) is bounded, then for any x0, the function (x – x0) u(x) is continuous at x0, while if u(x) is continuous at x0, then (x – x0) u(x) is differentiable at x0. Now suppose you want to find the derivative of the right-hand integral at x0. Rewrite the factor (x – t) as (x – x0) + (x0 – t), and treat each of the resulting integrals in the neighborhood of x0 with the help of the above results.)
6.6:3. The Fundamental Theorem, minus the condition that the derivative be integrable. (d : 2)
Prove the following generalization of Theorem 6.21, p.134: If F is any differentiable function on [a, b], then
– a
∫
bF′(x) d x ≤ F (b) – F (a) ≤
∫
– ba F′(x) d x .6.6:4. A formal product law for d (α β). (d : 3)
Let f be a function on [a, b] and α, β monotonically increasing nonnegative functions on [a, b]
such that f∈ (α)∩ (β), α∈ (β), and β∈ (α). (By 6.2:1, these conditions are slightly redundant.) Prove that
∫
f d(α β) =∫
fα dβ +∫
fβ dα.(If we use this as a formula for evaluating
∫
fα dβ in terms of the other two integrals, it can be thought of as a generalization of integration by parts.)6.6:5. Repeated Riemann-Stieltjes integration. (d : 4)
We shall obtain here a generalization of the result of 6.6:2 in which ds and dt are replaced by more general expressions dα(s) and dβ(t).
(a) Suppose α, β are increasing functions on [a, b] such that α∈ (β), and f∈ (α)∩ (β). Show that
∫
axf (t) dα(t), regarded as a function of x, belongs to (β), and that∫
t = ab &(∫
s = at f (s) dα(s)') dβ(t) =∫
t = ab (β(b) – β(t)) f (t) dα(t).(Suggestion: First consider the case where f is everywhere > 0, since in this case one can easily describe the least and greatest values of
∫
s = at f (s) dα(s) on any interval as the values at the respective ends of the interval; then get the general case by writing any f∈ (α)∩ (β) as a difference of two nonnegative-valued functions in (α)∩ (β). Cf. the method of passing from part 6.5:1(b) to (c).) (b) Show that given continuous functions u and on [a, b], there exists a continuous function w of two variables such that for any f∈ on [a, b], one has∫
t = ax u(t) &(∫
s = at (s) f (s) ds') dt =∫
t = ax w(x, t) f (t) dt.Hint: Prove the case where u and are nonnegative-valued using (a).
6.6:6. A characterization of the Riemann-Stieltjes integral. (d : 2, 4, 4)
Let α be an increasing function on [a, b], and f any bounded function on that interval. Given x1, x2 with a ≤ x1 < x2 ≤ b, let us use ∆F as an abbreviation for F(x2) – F(x1) and ∆α as an abbreviation for α(x2) –α(x1).
(a) Show that if f∈ (α) and we define F(x) =
∫
t = ax f (t) dα(t), then(i) F(a) = 0,
(ii) For all x1, x2 with a ≤ x1 < x2 ≤ b, we have ∆F∈ [(∆α) (inf f (x)), (∆α) (sup f (x))], where the inf and sup are over all x∈[x1, x2].
...
Answers to True/False question 6.6:0. (a)F. (b)T.
(We could write the conclusion of (ii) more suggestively as ∆F ⁄∆α∈[ inf f (x), sup f (x)], were it not that ∆α might be zero for some choices of x1 and x2.)
(b) Show that F is the unique function satisfying (i) and (ii).
(c) Show that for any bounded function f on [a, b], if there is a unique function F on [a, b]
satisfying these conditions, then f∈ (α).
Suggestion: Show that for any bounded f, the upper and lower integrals F+(x) =
∫
– xt = af (t) dα(t) andF–(x) = – t = a
∫
x f (t) dα(t) each satisfy the indicated conditions, and that any function F satisfying those conditions satisfies F–(x) ≤ F(x) ≤ F+(x) for all x∈[a, b].6.6:7. A change-of-variables result. (d : 3. > 6.6:6)
Suppose α is a monotonically increasing real-valued function on [a, b] and f, g are continuous real-valued functions on that interval, with g nonnegative-valued. Prove
∫
t = ax f (t) d&(∫
s = at g(s) dα(s)') =∫
t = ax f (t) g(t) dα(t) .6.7. INTEGRATION OF VECTOR-VALUED FUNCTIONS. (pp.135-136) Relevant exercises in Rudin: None
(But since the concepts of this section are essential to the next, exercises for the next section also test the material in this one.)
Exercises not in Rudin:
6.7:0. Say whether each of the following statements is true or false.
(a) If f is a differentiable Rk-valued function on [a, b] and f′ ∈ , then
∫
abf′d x = f (b) – f (a).(b) If α is a monotonically increasing function on [a, b] and f is a differentiable Rk-valued function on [a, b] such that f′ ∈ (α), then !
∫
abf′dα! ≥∫
ab| f′| dα.6.7:1. Integrability of vector-valued functions expressed in terms of partitions. (d : 2) Let f be a function [a, b]→ Rk, and α: [a, b]→ R an increasing function.
(a) Show that f is integrable with respect to α if and only if for every ε > 0 there exists a partition P = {x0, ... , xn} of [a, b] such that
Σni =1 diam ( f ( [xi–1, xi] ))∆αi < ε. where ‘‘diam’’ denotes the diameter of a set (Definition 3.9, p.52).
In parts (b) and (c) below, assume f is indeed integrable with respect to α, and let ε be a positive real number, and P = {x0, ... , xn} a particular partition making the above inequality hold.
(b) Show that for every choice of points ti∈[xi –1, xi] (i = 1, ... , n), we have
! (Σni =1f (ti)∆αi) – (
∫
abf dα)! < ε.(c) Show that every refinement P* of P also satisfies the inequality of (a), and hence the condition of (b).
6.8. RECTIFIABLE CURVES. (pp.136-137)
Relevant exercises in Rudin:
6:R18. Rectifiability and length of a curve don’t depend on its range alone ... . (d : 1, 1, 3)
Here eix is defined as in Example 5.17. You may assume differentiability of the sine and cosine function, and the standard formulas for their derivatives. The definition Rudin gives for γ3(t) is undefined for t = 0; let γ3(0) = 1. Before beginning the exercise, point out why this choice makes γ3(t) continuous at 0.
6:R19. ... but they are not affected by reparametrization. (d : 2)
Exercises not in Rudin:
6.8:0. Say whether each of the following statements is true or false.
(a) If γ is a rectifiable curve in Rk, then γ is differentiable and γ′ is continuous.
(b) If γ is the curve in the complex plane C defined by γ(t) = cos t + i sin t, with the parameter interval [0, 4π], then Λ(γ) = 4π.
6.8:1. A space-filling curve can’t be rectifiable. (d : 4)
In 7:R14 (p.168) Rudin will construct a ‘‘space-filling curve’’; specifically, a curve in R2 whose image is the whole unit square [0, 1]2. Show that a curve with this property cannot be rectifiable.
One possible approach: Find a constant c such that every curve in R2 whose image contains all points of a solid square of side s must have a sub-curve of length at least c s lying in the interior of that square; then use the fact that the unit square can be broken into n2 squares of side 1 ⁄ n. (A variant of this argument is indicated in the next exercise.)
6.8:2. Busy-body curves have to be long. (d : 4)
Prove that for every positive constant C there exists an ε > 0 such that any curve in R2 whose image has points at distance ≤ ε from each point of the unit square [0, 1]2 must have length ≥C .
Show how the result of the preceding exercise follows from this.
6.8:3. Integrability of γ′ is enough to prove the curve-length formula. (d : 2. > 6.7:1(a))
Suppose γ: [a , b] → Rk is a differentiable curve such that γ′ ∈ . Exercise 6.7:1(a) tells us that given ε> 0 we can find a partition P = {x0, ... , xn} such that
Σni =1 diam ( f ( [xi–1, xi] ))∆xi < ε.
(a) Show that if P is such a partition, and we take any point ti in each interval [xi–1, xi], then Σ |γ(xi) –γ(xi–1) | – |γ′(ti) | ∆xi < ε.
(b) Deduce that Theorem 6.27 remains true if the assumption ‘‘γ′ is continuous’’ is weakened to
‘‘γ′ ∈ ’’.