
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 jF (x)j = jPF (a; x) − NF (a; x)j 0 0 ≤ jPF (a; x)j + jNF (a; x)j 0 0 = PF (a; x) + NF (a; x): Integrating using Corallary 3.7 we have Z b Z b 0 0 0 jF (x)jdx ≤ 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 jF (t)jdt. 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 jF (y) − F (x)j < TF (a; y) − TF (a; x): Since TF (a; x) is absolutely continuous we can choose δ > 0 such that P P for any intervals [xi; yi] we have i jyi−xij < δ implies jTF (a; yi)− TF (a; xi)j < . Thus we must also have X X jF (yi) − F (xi)j ≤ jTF (a; y) − TF (a; x)j < . i i 1 Thus, F is absolutely continuous. (() Assume F is absolutely continuous. Then for any partition fa = t0; t1; ··· ;TN = bg of [a; b]. Then the variation is bounded by N X VF (b; a) = jF (tj) − F (tj−1)j j=1 N X Z tj−1 = j F 0(t)dtj j=1 tj N X Z tj−1 Z b ≤ jF 0(t)jdt = jF 0(t)jdt j=1 tj a As this is true for any partition, we take the supremum to find that R b 0 TF (a; b) ≤ a jF (t)jdt. 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 jF (t)jdt. 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 jxj < and jxj ≥ 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=jxj for all δ > 0 and x 2 R . Using Corollary 1.10 in R d+1 Chapter 2 ( jx|≥ dx=jxj ≤ C/) and that the ess sup norm is bounded Z j(f ∗ K)(x)j = j f(x − y)K(y)dyj Z R ≤ jf(x − y)jjK(y)jdy R Z + Z = jf(x − y)jjK(y)jdy + jf(x − y)jjK(y)jdy − jx|≥ A Z + Z jf(x − y)j ≤ jf(x − y)jjdy + B 2 dy − jx|≥ jxj A Z x+ Z 1 ≤ jf(y)jjdy + Bjfj1 2 dy x− jx|≥ jxj A Z x+ ≤ jf(y)jjdy + K x− C Z x+ ≤ jf(y)jjdy 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 jf(y)jdy = sup jf(y)jdy x2B 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 theo- rem 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)\ fF 0(x) > 0g) = 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)\fF 0(x)>0g 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)\fF 0(x)>0g Since F 0(x) > 0 we must have m(F −1(A) \ fF 0(x) > 0g) = 0. 3 Lemma 3: Let H = fx : F 0(x) > 0g and E be a measurable subset of [A; B] then H = F −1(E) \ fF 0(x) > 0g 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) \fF (x) > 0g = R 0 F −1(E) \fF (x) > 0g so m(E) = m(G) Z = F 0(x)dx F −1(G)\fF 0(x)>0g 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 . P1 Now, let fN (x) = n=1 χOn(x). Noting that the monotonic conver- gence theorem allows us to interchange the limit and the integral (as χOn(x) ≥ we have 1 Z 1 Z X f(y)dy = χ dx m(B) On(x) B B n=1 1 1 X Z = dx m(B) n=1 B\On(x) 4 1 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 P1 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 jx − h=2j < h=2 and jx + h=2j < h=2 we have 1 Z x+h lim inf = dx = 1 h!0+ h x and 1 Z x lim inf = dx = 1: 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 jρ(b) − ρ(x)j jρ(x + h) − ρ(x)j ≤ h · jb − xj Taking h ! 0+ we have ρ(x + h) − ρ(x) so ρ is right continuous. We also have jρ(b) − ρ(yj) jρ(y) − ρ(y − h)j ≤ h · jb − yj 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 jf(y) − f(x)j < Mjy − xj. 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).
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