Real Variables # 7

Brandon Behring

Real Variables # 7

Exercise 16 (a) Assume F is a function of bounded variation on [b, a]. Then, using the Jordan decomposition we can write it as the sum of the positive and negative variations

F (x) = PF (a, x) − NF (a, x) + F (a).

Since PF and NF are monotonic increasing they are diﬀerentiable almost everywhere, thus a.e.

0 0 0 F (x) = PF (a, x) − NF (a, x) and since both derivatives are positive

0 0 0 |F (x)| = |PF (a, x) − NF (a, x)| 0 0 ≤ |PF (a, x)| + |NF (a, x)| 0 0 = PF (a, x) + NF (a, x).

Integrating using Corallary 3.7 we have

Z b Z b 0 0 0 |F (x)|dx ≤ PF (a, x) + NF (a, x)dx a a ≤ PF (a, b) − PF (a, a) + NF (a, b) − NF (a, a)

= PF (a, b) + NF (a, b) = TF (a, b)

R x 0 (b) We are given that TF (a, x) = a |F (t)|dt. By theorem 3.11, we know that the total variation is an absolutely continuous function of x. Take a partition a ≤ x ≤ y ≤ b, then we have

|F (y) − F (x)| < TF (a, y) − TF (a, x).

1 Thus, F is absolutely continuous. (⇐) Assume F is absolutely continuous. Then for any partition {a = t0, t1, ··· ,TN = b} of [a, b]. Then the variation is bounded by

As this is true for any partition, we take the supremum to find that R b 0 TF (a, b) ≤ a |F (t)|dt. Since a function that is absolute continuous is also of bounded variation we have using part (a) that TF (a, b) = R b 0 a |F (t)|dt. See also Proposition 4.2 for a different proof. 17 Let us take the case d = 1. The proof for general d is no different- just 2 replace m(Br) = Vdr . Let us split our integral into |x| < and |x| ≥ and use the two properties of an approximation to the identity, (i) K(x) ≤ A/δ for all δ > 0 and 2 (ii) K(x) ≤ Bδ/|x| for all δ > 0 and x ∈ R . Using Corollary 1.10 in R d+1 Chapter 2 ( |x|≥ dx/|x| ≤ C/) and that the ess sup norm is bounded Z |(f ∗ K)(x)| = | f(x − y)K(y)dy| Z R ≤ |f(x − y)||K(y)|dy R Z + Z = |f(x − y)||K(y)|dy + |f(x − y)||K(y)|dy − |x|≥ A Z + Z |f(x − y)| ≤ |f(x − y)||dy + B 2 dy − |x|≥ |x| A Z x+ Z 1 ≤ |f(y)||dy + B|f|∞ 2 dy x− |x|≥ |x| A Z x+ ≤ |f(y)||dy + K x− C Z x+ ≤ |f(y)||dy x−

2 In the last step we assume f is non-zero somewhere on the ball- otherwise the problem is trivial. However,

1 Z 1 Z x+ f ∗(x) = sup |f(y)|dy = sup |f(y)|dy x∈B m(B) B >0 2 x− and we are done. 21 For all that follows let F be absolutely continuous and increasing on [a, b] with F (a) = A and F (b) = B and f is any measurable function on [A, B].

(a) Since f is measurable, there exists a series of simple functions on 0 0 [A, B] such that ρn → f. Then ρn(F (x))F (x) → f(F (x))F (x) is also a measurable function. R 0 (b) Lemma 1: m(O) = F −1(O) F (x)dx. Proof: O is an arbitrary open set and can be written as a countable union of open intervals O = ∪n(An,Bn) and since F is increasing −1 F (O) = ∪n(an, bn) where F (an) = An and F (bn) = Bn. Using that F is absolutely continuous and applying the fundamental theorem we have

Z Z F 0(x)dx = F 0(x)dx −1 F (O) ∪n(an,bn) X Z bn = F 0(x)dx n an X = [F (bn) − F (an)] n X = (Bn − An) = m(O). n

Lemma 2: Let N be a subset of [A, B] with mA = 0, then m(F −1(A)∩ {F 0(x) > 0}) = 0 Proof: Since A has measure 0 we can cover it by an open set O such that A ⊂ O and for any > 0 we have m(O) < . However, from our R 0 previous result (noting x:F 0(x)=0 f (x)dx = 0) Z Z F 0(x) = F 0(x) = m(O) < . F −1(O)∩{F 0(x)>0} F −1(O)

But since A ⊂ O and that this is true for all > 0 we have Z F 0(x) = 0. F −1(A)∩{F 0(x)>0}

Since F 0(x) > 0 we must have m(F −1(A) ∩ {F 0(x) > 0}) = 0.

3 Lemma 3: Let H = {x : F 0(x) > 0} and E be a measurable subset of [A, B] then H = F −1(E) ∩ {F 0(x) > 0} is measurable. Proof: This is problem 3.20(c).

Since H is measurable, there exists a Gδ ,G, such that m(G − E) = R 0 0. Our previous result this tells us that F −1(G) ∩{F (x) > 0} = R 0 F −1(E) ∩{F (x) > 0} so m(E) = m(G) Z = F 0(x)dx F −1(G)∩{F 0(x)>0} Z = F 0(x)dx H Z b 0 = χF −1(E)F (x)dx a Z b 0 = χE (F (x)) F (x)dx a

We have now shown that the chain rule applies for a characteristic function Z b Z b 0 χE(t)dt = mE = χE (F (x)) F (x)dx. a a Summing over characteristic functions we easily see that it applies for simple functions. Taking a series of simple functions converging to f and using part (a) and using the monotone convergence theorem we have that it applies for nonnegative functions. Decomposing f to negative and positive parts we see that it applies for a general measurable function.

25 (a) Every null set is measurable, thus E is measurable, thus by the def- inition of measurability for any > 0 there exists an open set O such that E ⊂ O where m(O − E) < . Noting that m(O − E) = −n m(O)−m(E) = m(O) we have by choosing = 2 we have On such −n that m(On) < 2 . P∞ Now, let fN (x) = n=1 χOn(x). Noting that the monotonic convergence theorem allows us to interchange the limit and the integral (as

χOn(x) ≥ we have

∞ Z 1 Z X f(y)dy = χ dx m(B) On(x) B B n=1 ∞ 1 X Z = dx m(B) n=1 B∩On(x)

4 ∞ X m(B ∩ On(x)) = m(B) n=1

For suﬃciently large n On becomes closer and closer to a set of a measure so m(B ∩ On) → m(B). We have that tail end of our series P∞ P goes as n=N m(B)/m(B) = n 1 which diverges. R x (b) Construct an absolutely continuous function F (x) = −∞ f(t)dt. This is clearly increasing as f(t) is positive. In 1-dimension part (a) tells us over asymmetric balls |x − h/2| < h/2 and |x + h/2| < h/2 we have 1 Z x+h lim inf = dx = ∞ h→0+ h x and 1 Z x lim inf = dx = ∞. h→0+ h x−h The derivates of F are

f(x + h) − f(x) 1 Z x+h D+(F )(x) = lim inf = lim inf f(t)dt h→0+ h h→0+ h x and f(x) − f(x − h) 1 Z x D−(F )(x) = lim inf = lim inf f(t)dt h→0+ h h→0+ h x−h

so we are done. Problem 4 (a) We have for any h > 0 and y > x + h that

|ρ(b) − ρ(x)| |ρ(x + h) − ρ(x)| ≤ h · |b − x|

Taking h → 0+ we have ρ(x + h) − ρ(x) so ρ is right continuous. We also have |ρ(b) − ρ(y|) |ρ(y) − ρ(y − h)| ≤ h · |b − y|

Taking h → 0+ we have ρ(y) − ρ(y − h) = 0 so ρ is left continuous. (b) Letting a + h = a0 and b − h = b0 we have

ρ(a0) − ρ(a) ρ(y) − ρ(x) ρ(b) − ρ(b0) M = ≤ ≤ = M 1 a0 − a y − x b − b0 2

then |f(y) − f(x)| < M|y − x|. Since f is Lipschitz it is absolutely continuous.

5 (c) Whenever x < y, h > 0 and x + h < y we have

ρ(x + h) − ρ(x) ρ(y) − ρ(x) ρ(y) − ρ(y − h) ≤ ≤ , h y − x h

so taking lim inf as h → 0+ we have

ρ(y) − ρ(x) D ρ(x) ≤ ≤ D ρ(y), + y − x −

So ρ(x) is an increasing function on (a, b). A monotone function can only be discontinuous at a countable number of points. Since it is absolutely continuous from part (b) we have for [x, y] ∈ (a, b)

Z y ρ0(t)dt = ρ(y) − ρ(x). x R x (d) Let ρ(x) = c ψ(t)dt and ψ be an increasing function in (a, b) with c ∈ (ab). Then ρ0(x) = ψ(x) is also an increasing function. Deﬁne for a < x < y < b a function on θ ∈ [0, 1] by

χ(θ) = ρ(1 − θ)x + θy) − ((1 − θ)ρ(x) + θρ(y))

and χ0(θ) = (y − x)ψ((1 − θ)x + θy) − (ρ(y) − ρ(x))

As ψ is increasing, so is χ0(θ) because ψ moves from x to y as θ moves from 0 to 1. If we can show χ(θ) ≤ 0 we are done. Let us ﬁnd the absolute maximum of χ(θ) on [0, 1]. We have χ(0) = χ(1) = 0 so if it occurs at the endpoints we are done. Assume it occurs at z ∈ (0, 1), then χ0(z) ≤ 0 and since χ0 is increasing χ0 ≤ 0 on [0, z] and χ is decreasing on [0, z] so χ ≤ 0 and the maximum value is nonpositive. This argument was taken from Royden (3rd edition).