209: Honors Analysis in Rn Functions of bounded variation and absolutely continuous functions
1 Nondecreasing functions
Definition 1 A function f :[a, b] → R is nondecreasing if x ≤ y ⇒ f(x) ≤ f(y) holds for all x, y ∈ [a, b].
Proposition 1 If f is nondecreasing in [a, b] and x0 ∈ (a, b) then
lim f(x) and lim f(x) x→x0, x
Proof Exercise.
Definition 2 The function f is sait to be continuous from the right at x0 ∈ (a, b) if f(x0 + 0) = f(x0). It is said to be continuous from the left if f(x0 − 0) = f(x0). A function is said to have a jump discontinuity at x0 if the limits f(x0 ± 0) exist and if |f(x0 + 0) − f(x0 − 0)| > 0.
Let x1, x2 ... be a sequence in [a, b] and let hj > 0 be a sequence of positive P∞ numbers such that n=1 hn < ∞. The nondecreasing function X f(x) = hn
xn
Proposition 2 Let f be nondecreasing on [a, b]. Then it is measurable and bounded and hence it is Lebesgue integrable.
Proof Exercise. (Hint: the preimage of (−∞, α) is an interval.)
1 Proposition 3 Let f :[a, b] → R be nondecreasing. Then all the discon- tinuities of f are jump discontinuities. There are at most countably many such.
Proof Exercise. (Hint: The range of the function is bounded, and the total length of the range is larger than the sum of the absolute values of the jumps, so for each fixed length, there are at most finitely many jumps larger than that length.)
Exercise 1 A jump function associated to {xn}{hn} is continuous from the left and jumps hn = f(xn + 0) − f(xn − 0).
Definition 3 Let x ∈ (a, b). We define the derivative numbers by
f(y)−f(x) f(y)−f(x) (Dlf)(x) = lim infy
When all the derivative numbers are equal then we say that the function is differentiable at x and denote the common value by f 0(x).
Proposition 4 Let f be continuous in [a, b]. If one of the derivative numbers is greater or equal to zero everywhere in (a, b), then f is nondecreasing in [a, b]
Proof. We consider Dr for instance. Assume first Dr > 0 on (a, b). If a ≤ x < y ≤ b is such that max[x,y] f is achieved at some w < y, then for z ∈ (w, y) we have f(z) − f(w) ≤ 0 z − w contradicting the assumption. Thus f must be nondecreasing. If Drf ≥ 0 we take f(x) + x and then let → 0.
Theorem 1 Let f be nondecreasing on [a, b]. Then f 0 exists almost every- where (with respect to Lebesgue measure), f 0 is measurable, nonnegative, and
Z b f 0(t)dt ≤ f(b) − f(a). a
2 Proof. Let
Eαβ = {x ∈ [a, b] | Drf(x) ≤ α < β ≤ Drf(x)}.
Let U be an open set in [a, b] such that Eαβ ⊂ U. Let
F = {[x, x + h] | a ≤ x < x + h ≤ b, [x, x + h] ⊂ U, f(x + h) − f(x) ≤ αh}
Then F is a fine cover of Eαβ and by Corollary 3 of Vitali’s theorem (or Corollary 4 of Besicovich’s thm), there exist [xj, xj + hj] = Ij ∈ F, disjoint, j = 1, 2,... so that λ(Eαβ \ ∪Ij) = 0 where λ is Lebesgue outer measure. Let S = Eαβ ∩ (∪Ij). Let
I = {[x, x + h] | ∃j, [x, x + h] ⊂ Ij, f(x + h) − f(x) ≥ βh}
Then I is a fine cover of S \ ∪{xj + hj}. Therefore, there exist disjoint [yk, yk + gk] = Jk ∈ I such that
λ(S \ ∪Jk) = 0
Now P P β λ(Jk) ≤ (f(yk + gk) − f(yk)) P k k P ≤ j (f(xj + hj) − f(xj)) ≤ α j λ(Ij) ≤ αλ(U) We deduce that P λ(Eαβ) ≤ λ(S) ≤ k λ(Jk) α ≤ β λ(U) α Because U is arbitrary we deduce λ(Eαβ) ≤ β λ(Eαβ), so Eαβ is negligible.
This implies that Drf = Drf almost everywhere. Similar arguments show that all four derivative numbers are equal almost everywhere. In order to prove the rest, condider the functions 1 g (x) = n f(x + ) − f(x) n n where we take the extension of f to the right by f(x) = f(b) for x ≥ b. Then 0 gn are nonnegative and gn → f almost everywhere in [a, b]. By Fatou’s lemma Z b Z b 0 f (t)dt ≤ lim inf gn(t)dt = f(b) − f(a) a n→∞ a
3 and this concludes the proof. There exist nondecreasing functions f such that f 0 = 0 almost every- where and f(b) − f(a) > 0. Simple step functions are examples of such functions. More interesting are Cantor functions, which are nondecreasing and continuous, with f(b) − f(a) > 0 and f 0(t) = 0 almost everywhere. We consider the interval I = [0, 1] and we construct first Cantor’s set. This is done inductively by removing the open middle third interval. We set R1 the open interval (1/3, 2/3), R2 = (1/9, 2/9) ∪ (1/3, 2/3) ∪ (7/9, 8/9), and so on. Rn ⊂ Rn+1 is a sequence of open sets. Their complements are denoted Cn. n −n Each Cn is closed, it is formed with 2 disjoint closed intervals of length 3 n each. The Lebesgue measure of Cn is therefore λ(Cn) = (2/3) . The intersec- tion C = ∩Cn is not empty because Cn+1 ⊂ Cn are non-empty compact sets. The set C is the Cantor “middle thirds” set; its Lebesgue measure is zero. Its complement R = ∪Rn has measure 1 and is open. Now we construct (many) continuous, nondecreasing functions f such that f 0(t) = 0 for all t ∈ R, and such that f(0) = 0, f(1) = 1. This is done inductively as follows. Start with f0(t) = t. We are going to describe suggestively Rn as “resting times” and we are going to implement consistent “travel plans” based on the preceding travel plan. So, for instance, we can decide that we are going to follow exactly the preceding travel plan for a third of the time, rest for a third of the time, and catch up in the last third. If we travel at constant speeds, this produces the function f1(t) = t for t ≤ 1/3, f1(t) = 1/3, for t ∈ R1, f1(t) = 2t − 1 on [2/3, 1]. Inductively, we find functions fn+1(t) ≤ fn(t) with fn(0) = 0, 0 fn(1) = 1, fn(t) piece-wise linear, fn continuous, nondecreasing, fn(t) = 0 for t ∈ Rn. Moreover fn+1(t) = fn(t) for t = Rn. The nonzero speeds of the n-th traveler are 1, 2,... 2n. The distance between successive travelers n+1 obeys supt∈[0,1] |fn+1(t) − fn(t)| ≤ (2/3) . Because these numbers form a convergent series, the sequence fn is Cauchy in C([0, 1]), which means that it is uniformly convergent to a function f. The function f is clearly non- decreasing, continuous (as uniform limit of continuous functions), f(0) = 0, f(1) = 1 and f 0(t) = 0 for t ∈ R because f is constant in each interval that comprises R. Other travel plans work as well, as long as we keep the rule that the resting time of the nth traveler is Rn, and that the next traveler always rests with his predecessor. For instance we may say that we want to keep the speeds of the travelers the same, when they are not resting. Starting from the same f0 we obtain then f1(t) = (3/2)t for t ≤ 1/3, f1(t) = 1/2 on R1 and f1(t) = (3/2)t − (1/2) for t ∈ [2/3, 1]. The n-th traveler has nonzero speed
4 n R t n (3/2) and is fn(t) = 0 gn(s)ds with gn = (3/2) 1Cn . The maximal distance 1 −n between succsessive travelers is now supt∈[0,1] |fn+1(t) − fn(t)| = 6 2 . The functions fn are nondecreasing, continuous, fn+1 = fn on Rn, fn(0) = 0, 0 fn(1) = 1, fn(t) = 0 for t ∈ Rn. We obtain a different Cantor function in the limit. (The rest places are different, although they rest at the same times).
2 BV
We consider finite collections of closed intervals Ij = [lj, rj], included in a fixed interval [a, b], with j = 1,...N, a ≤ rj ≤ lj+1 < rj+1 ≤ b for 1 ≤ j ≤ N − 1. We use the notation
I([a, b]) = {{I1,...IN } | N ∈ N} to represent the set of all such collections. A partition is an element P ∈ N I([a, b]) such that [a, b] = ∪j=1Ij. We denote by P([a, b]) the set of all partitions. We’ll write I, P when the interval [a, b] is clear from the context.
Definition 4 A function f :[a, b] → R is said to be of bounded variation, f ∈ BV ([a, b]) if
N b X Va (f) = sup |f(rj) − f(lj)| P ∈I([a,b]) j=1
b is finite. Va (f) is called the total variation of f on [a, b].
Clearly, the total variation remains the same if we replace I by P.
Proposition 5 Let c ∈ [a, b]. Then, for any f :[a, b] → R
b c b Va (f) = Va (f) + Vc (f)
c Proof. Clearly, because I([a, c]) ∪ I([c, b]) ⊂ I([a, b]), we have Va (f) + b b Vc (f) ≤ Va (f). If, on the other hand P ∈ I([a, b]) with P = {I1,...,IN }, we consider J = max{j ≤ N | rj < c} and then, if c ∈ (lJ+1, rJ+1) we replace IJ+1 by two intervals, [lJ+1, c] and [c, rJ+1]. If J = N, or if c ≤ lJ+1 we leave the intervals as they are. We construct P1 ∈ I([a, c]) by listing all the intervals Ij with j ≤ J and adding, if neccesary, [lJ+1, c]; we construct P2 ∈ I([c, b]) by listing all intervals Ij with j ≥ J + 1 and replacing, if necessary,
5 IJ+1 by [c, rJ+1]. Because of the triangle inequality |f(rJ+1) − f(lJ+1)| ≤ |f(c) − f(lJ+1)| + |f(rJ+1) − f(c)| we have PN |f(r ) − f(l )| ≤ P |f(r ) − f(l )| + P |f(r ) − f(l )| j=1 j j Ij ∈P1 j j Ij ∈P2 j j c b ≤ Va (f) + Vc (f), b c b and because P was arbitrary, we deduce Va (f) ≤ Va (f) + Vc (f). Proposition 6 Let f ∈ BV ([a, b]). Then there exist two nondecreasing functions g and h in [a, b] such that f = g − h. Conversely, if f = g − h with g, h nondecreasing, then f ∈ BV ([a, b]). Proof. If f = g − h with g, h nondecreasing then f ∈ BV ([a, b]) obviously. x If f ∈ BV ([a, b]), let g(x) = Va (f). This is clearly nondecreasing. We have x to show that h(x) = f − Va (f) is nondecreasing. Let x1 ≤ x2. We have obviously x1 x2 Va (f) + f(x2) − f(x1) ≤ Va (f) and this is the same as h(x1) ≤ h(x2). Corollary 1 If f ∈ BV ([a, b]) then f 0 exists almost everywhere and f 0 ∈ L1([a, b]). Proof Indeed, if f = g − h with g, h nondecreasing then f 0 = g0 − h0 almost 0 0 0 R b 0 everywhere and |f | ≤ g + h , a |f (t)|dt ≤ g(b) + h(b) − g(a) − f(a) < ∞.
3 AC
Definition 5 We say that the function f :[a, b] → R is absolutely continu- ous, f ∈ AC([a, b]) is for every > 0 there exists δ > 0 such that, for every PN PN P ∈ I([a, b]) with j=1(rj − lj) ≤ δ we have j=1 |f(rj) − f(lj)| ≤ . Proposition 7 AC([a, b]) ⊂ C([a, b]) ∩ BV ([a, b]). Proof. C([a, b]) is the space of continuous functions in [a, b]. It is obvious from the definition that absolutely continuous functions are continuous. In order to show that absolutely continuous functions are functions of bounded variation, we take the δ corresponding to = 1 in the definition. We find that the total variation of f on any interval of length δ is at most 1. Because b−a we can write [a, b] as the union of at most N = [ δ ] + 1 closed intervals, and b in view of Proposition 5 we have that Va (f) ≤ N.
6 Proposition 8 If f ∈ AC([a, b]) then f 0 exists almost everywhere, and is x integrable. The function Va (f) is absolutely continuous.
Proof The first part follows from Proposition 7 and Corollary 1. the second part is an easy exercise using the definition and Proposition 5.
Lemma 1 Let f ∈ AC([a, b]) and assume f 0(t) = 0 almost everywhere. Then f is constant.
Proof. Let r ≤ b and let E ⊂ [a, r] be a set of measure r − a such that f 0(t) = 0 for t ∈ E. Let > 0 and let
(y − x) F = [x, y] | [x, y] ⊂ [a, r], |f(x) − f(y)| ≤ 2(r − a)
By the corollary to Vitali’s covering lemma, there exist a countable disjoint P subfamily Ij = [x−j, yj] of F included in [a, b] such that j(yj −xj) = r−a. Taking δ > 0 from the definition of absolute continuity corresponding to /2, PN there exists N large enough so that j=1(yj − xj) ≥ r − a − δ. Suppose we numbered the intervals from left to right, so that yj ≤ xj+1 and we introduce PN y0 = a and xN+1 = r. Then j=0(xj+1 − yj) ≤ δ. Therefore
|f(r) − f(a)| = PN (f(x ) − f(y )) + PN (f(y ) − f(x )) j=0 j+1 j j=1 j j ≤ 2 + 2(r−a) (r − a).
In view of the fact that is arbitrary, it follows that f(r) = f(a). Lemma 2 Let g be bounded in [a, b]. Consider
Z x f(x) = f(a) + g(t)dt a with some constant f(a). Then f ∈ AC([a, b]) and f 0 = g almost everywhere.
Proof. The fact that f ∈ AC([a, b]) follows from the fact that g ∈ L1([a, b]). The fact that f 0 exists almost everywhere follows from Proposition 8. Let
1 Z x+ n gn(x) = n g(t)dt x
7 Note that supx |gn(x)| ≤ M if supx |g(x)| ≤ M, and also
1 f(x + n ) − f(x) 0 gn(x) = 1 → f (x) a.e. n Therefore Z r Z r 0 gn(t)dt → f (t)dt 0 0 follows from the LDC theorem, but
1 R r R t+ n a n t g(s)ds dt = 1 R r+ n R s n g(s) 1 dt ds a s− n 1 R r+ n R r = a g(s)ds → a g(s)ds = f(r) − f(a). We obtained thus Z x Z x f 0(t)dt = g(t)dt 0 a for all x ∈ [a, b]. This implies that f 0 = g almost everywhere, by the Lebesgue differentiation theorem.
Theorem 2 Let h ∈ L1([a, b]). Consider
Z x f(x) = f(a) + h(t)dt a with some constant f(a). Then f ∈ AC([a, b]) and f 0 = h holds almost everywhere in [a, b].
Proof.WLOG h ≥ 0 a.e. (Indeed h = h+ − h−, with nonnegative h± ∈ 1 L ([a, b])). Define hn = max{h, n}. Then, by Lemma 2 Z x fn(x) = f(a) + hn(t)dt a
0 are in AC([a, b]) and obey fn(t) = hn(t) a.e. By the Lebegue monotone convergence theorem, we deduce that Z x Z x fn(x) = f(a) + hn(t)dt → f(a) + h(t)dt = f(x) a a
8 0 so limn fn(x) = f(x) holds everywhere. Also fn(x) = hn(x) converges almost R x everywhere to h(x) by construction. The function gn(x) = a (h − hn)(t)dt is a nondecreasing function of x and therefore it is differentiable almost ev- erywhere and its derivative is nonnegative. Therefore, because
f(x) = gn(x) + fn(x) it follows that 0 0 f (x) ≥ fn(x) holds almost everywhere. (The function f ∈ AC([a, b]) obviously, so it is differentiable a.e.) Passing to the limit, we deduce f 0(x) ≥ h(x) almost everywhere. Integrating we deduce Z b Z b f 0(t)dt ≤ h(t)dt = f(b) − f(a) a a R b 0 but, because f is nondecreasing we have, in view of Theorem 1 a f (t)dt ≤ f(b) − f(a) so we have Z b (f 0(t) − h(t))dt = 0. a 0 0 Now, because f − h ≥ 0 a.e., it follows that f = h a.e. Theorem 3 Let f :[a, b] → R. the following are equivalent: (a) f ∈ AC([a, b]) (b) The function f is differentiable almost everywhere, f 0 ∈ L1([a, b]) and Z x f(x) = f(a) + f 0(t)dt a holds for all x ∈ [a, b]. Proof In view of Theorem 2 we need only to prove (a) ⇒ (b). Because of Proposition 8 we know that f 0 exists almost everywhere and is in L1([a, b]). Then we can define Z x g(x) = f(a) + f 0(t)dt a and from Theorem 2 we know that g0 = f 0 holds almost everywhere. Then g − f is an absolutely continuous function whose derivative vanishes almost everywhere. By Lemma 1 this must be a constant function; the constant is zero because G(a) = f(a).
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