REAL ANALYSIS I HOMEWORK 5

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1 REAL ANALYSIS I HOMEWORK 5 CİHAN BAHRAN The questions are from Stein an Shakarchi s text, Chapter Suppose ϕ is an integrable function on R with R ϕ(x)x = 1. Let K δ(x) = δ ϕ(x/δ), δ > 0. (a) Prove that {K δ } δ>0 is a family of goo kernels. (b) Assume in aition that ϕ is boune an supporte in a boune set. Verify that {K δ } δ>0 is an approximation to the ientity. (c) Show that Theorem 2.3 (convergence in the L 1 -norm) hols for goo kernels as well. (a) First, by Exercise 2.16, for every δ > 0 ϕ(x/δ)x = δ R R ϕ(x)x = δ hence R K δ(x)x = 1. Secon, since ϕ is integrable, so is ϕ. Hence A = R ϕ is a finite constant inepenent of δ. Since K δ (x) = δ ϕ(x/δ), By the same reasoning above, we get K δ (x) x = ϕ = A. R R Thir, for every η > 0, letting E η = {x R : x η}, we have K δ (x) x = K δ (x) χ Eη (x)x x η R = ϕ(x) χ Eη (δx)x R = ϕ(x) x 0 as δ 0 by Proposition part (i). x η/δ (b) By assumption, there exists r > 1 such that ϕ vanishes outsie the ball centere at the origin with raius r. So x +1 δ K δ (x) = x +1 ϕ(x/δ) δ+1 x = +1 ϕ(x/δ). δ Since ϕ(x/δ) = 0 if x/δ > r we get that for every x R, x +1 δ K δ (x) r +1 ϕ(x/δ) r +1 M 1

2 REAL ANALYSIS I HOMEWORK 5 2 for some constant M, since ϕ is boune by assumption. observe that A oes not epen on δ an we have K δ (x) Aδ x +1 Writing A = r +1 M, we which is property (iii ) in the efinition of approximations to the ientity. We also have K δ (x) = δ ϕ(x/δ) δ M δ Mr +1 = δ A for each δ > 0, which verifies property (ii ). (c) Let {K δ } δ>0 be a family of goo kernels whose L 1 -norms are uniformly boune by A. We shortly write for the L 1 -norm L 1 (R ). Consier the subset L = {f L 1 (R ) : (f K δ ) f 0 as δ 0} of L 1 (R ). We want to show L = L 1 (R ). We o this in two steps: First we show that L is close in L 1 (R ) an then we show that continuous functions with compact support lie in L. Since such functions are ense in L 1 (R ), showing these are enough. To show L is close in L 1 (R ), let f L. We show f must lie in L. Given > 0, choose g L such that f g. Since g L, we can choose δ 2(A+1) 0 > 0 such that (g K δ ) g whenever 0 < δ δ 2 0. Thus for 0 < δ δ 0, we have (f K δ ) f = (f K δ ) (g K δ ) + (g K δ ) g + g f This shows that f L. (f K δ ) (g K δ ) + g K δ ) g + g f = (f g) K δ + (g K δ ) g + g f f g K δ + (g K δ ) g + g f (A + 1) f g + (g K δ ) g =. Now let s show that if f is a continuous function with compact support E, then f L. WLOG we may assume f > 0. By using property (i) of goo kernels, the triangle inequality an Tonelli s theorem, we have (f K δ ) f = (f K δ )(x) f(x) x E = f(x y)k δ (y)y f(x) x E R = f(x y)k δ (y)y f(x)k δ (y)y x E R R = (f(x y) f(x))k δ (y)y x E R f(x y) f(x) K δ (y) yx E R = f(x y) f(x) K δ (y) xy R E Å ã = K δ (y) f(x y) f(x) x y. R E Since f is uniformly continuous, given > 0 there exists η > 0 such that y < η implies f(x y) f(x). An by property (iii) of goo kernels, there exists δ 2Am(E) 0 > 0

3 REAL ANALYSIS I HOMEWORK 5 3 such that for 0 < δ < δ 0, we have y η K δ(y). Thus for 0 < δ < δ 2 f 0, we have Å ã (f K δ ) f K δ (y) f(x y) f(x) x y Hence f L. + y <η y η y <η 2A K δ (y) =. E Å E Ç ã f(x y) f(x) x y å K δ (y) E 2Am(E) x y + K δ (y) + 2 f K δ (y) y R y η 2. Suppose {K δ } is a family of kernels that satisfies: (i) K δ (x) Aδ for all δ > 0. (ii) K δ (x) Aδ/ x +1 for all δ > 0. (iii) R K δ(x)x = 0 for all δ > 0. y η K δ (y) 2 f y Thus K δ satisfies conitions (ii ) an (iii ) of approximations to the ientity, but the average value of K δ is 0 instea of 1. Show that if f is integrable on R, then (f K δ )(x) 0 for a.e. x, as δ 0. Let {L δ } δ>0 be an approximation to the ientity (we know such a family exist by Exercise 1, for instance) which satisfies the conitions (ii ) an (iii ) with the inepenent constant B. Then (i) R (K δ + L δ )(x)x = R K δ(x)x + R L δ(x)x = = 1. (ii) (K δ + L δ )(x) K δ (x) + L δ (x) Aδ/ x +1 + Bδ x +1 = (A + B)δ/ x +1 for all δ > 0. (iii) (K δ + L δ )(x) Aδ/ x +1 + Bδ/ x +1 = (A + B)δ/ x +1. Thus the family {K δ + L δ } δ>0 is also an approximation to the ientity. So by applying Theorem 2.1 twice, we have for x R Z where Z is a null set an (f L δ )(x) f(x) as δ 0 (f K δ )(x) + (f L δ )(x) = (f K δ + f L δ )(x) = (f (K δ + L δ ))(x) f(x) as δ 0 for x R Z where Z is a null set (the istrubitivity of the convolution over aition is immeiate by efinition). Thus (f K δ )(x) 0 as δ 0 for x R (Z Z ). Since Z Z is a null set, we are one.

4 REAL ANALYSIS I HOMEWORK Suppose 0 is a point of Lebesgue ensity of the set E R. Show that for each of the iniviual conitions below there is an infinite sequence of points x n E, with x n 0, an x n 0 as n. (a) The sequence also satisfies x n E for all n. (b) In aition, 2x n belongs to E for all n. Generalize. The assumption yiels m(( r, r) E) lim = 1. ( ) r 0 + 2r Fix a nonzero real number t such that t 1. We will show that there is an infinite sequence of points {x n } E {0} converging to zero such that x n /t E for every n. So the case t = 1 yiels (a) an the case t = 1/2 yiels (b). For any subset S of R, let s write ts for the set {tx : x S}. We claim that there exists R > 0 such that 0 < r < R implies m(( r, r) E te) > 0. For the moment let s assume the claim an see how the result we want follows. Pick any r 1 < R. Then there exists x 1 0 such that x 1 ( r 1, r 1 ) E te (since this intersection has positive measure). So x 1 E an x 1 /t E. Then pick r 2 > 0 such that r 2 < x 1 /2. Then r 2 < r 1 < R, so there exists x 2 0 such that x 2 ( r 2, r 2 ) E te. So x 2 E, x 2 /t E an x 2 x 1. Going on like this inuctively, we obtain a sequence {x n } where the x n s are istinct, nonzero an x n, x n /t E with x n < x n 1 /2 (note that this implies x n 0 as n since x n x 1 /2 n 1 ). Now we prove the claim: For any r > 0, consier the subsets ( r, r) E an ( r, r) te of ( r, r). Note that ( r, r) = t ( r/ t, r/ t ), so because t 1. So Since 1 t +1 m(( r, r) te) = m(t ( r/ t, r/ t ) te) = m(t (( r/ t, r/ t ) E)) = t m(( r/ t, r/ t ) E) t m(( r, r) E) m(( r, r) te) + m(( r, r) E) ( t + 1) m(( r, r) E). < 1, by ( ) there exists R > 0 such that 0 < r < R implies m(( r, r) E) 2r So combining by the above inequality, we get > 1 t + 1. m(( r, r) te) + m(( r, r) E) > 2r = m(( r, r)) whenever 0 < r < R. This forces the intersection of the subsets ( r, r) E an ( r, r) te of ( r, r) to have positive measure, by the aitivity of the Lebesgue measure.

5 REAL ANALYSIS I HOMEWORK Prove that if f is integrable on R, an f is nonzero on a set of positive measure, then f (x) c, for some c > 0 an all x 1. x Conclue that f is not integrable on R. Then, show that the weak type estimate m({x : f (x) > α}) c/α for all α > 0 whenever f = 1, is best possible in the following sense: if f is supporte in the unit ball with f = 1, then m({x : f (x) > α}) c /α for some c > 0 an all sufficiently small α. [Hint: For the first part, use the fact that B f > 0 for some ball B.] For every n N, let E n = {x R : f(x) > 1/n}. Then since f is positive on a set of positive measure, there exists n N such that m(e n ) > 0. Then by Corollary 1.5, E n contains a point of ensity of itself, say x 0. This ensures that x 0 is containe in some ball B, say of raius r, such that m(b E n ) > 0. Hence f f m(b E n). B B E n n So there exists > 0 such that B f >. Now given x R, B is containe in a ball B x which is centere at x with raius x x 0 + r. Therefore by the efinition of f, we have f (x) 1 f. m(b x ) B x Since B x is containe in a cube with sie length 2( x x 0 + r), we have m(b x ) 2 ( x x 0 + r) 2 ( x + x 0 + r). By the binomial expansion, if x 1, ( x + x 0 + r) A x for some constant A (note that A epens on x 0 an r but not on x). Thus f (x) 1 f 1 f C x B x C x B C x where C = 2 A. Since an C are inepenent of x, we obtain the esire result setting c = C > 0. Since the function 1 is not integrable on the set E = {x R : x 1} (this can be x checke by applying Tonelli an noting that 1/x is not integrable on [1, )) an the nonnegative function f c beats on E, we conclue that f is also not integrable on x E, an hence not integrable on R. By what we have shown in the first part, for every α > 0 there is an inclusion x R : 1 x < (c/α) 1/ = x R : x 1 an c x > α x R : f (x) > α. Thus m Ä x R : f (x) > α ä m Ä x R : 1 x < (c/α) 1/ ä = K(c/α 1)

6 REAL ANALYSIS I HOMEWORK 5 6 where K is the volume of the unit ball in R. In particular, whenever α < c/2, we have c α > c/2 so m Ä x R : f (x) > α ä Kc 2α Thus writing c = Kc, for sufficiently small α we get the esire inequality. 2 (Note that if a function is supporte in the unit ball with L 1 -norm equal to 1, then it clearly satisfies the more general hypotheses state in the beginning of the question. So what we have shown hols for such functions) 7. Using Corollary 1.5, prove that if a measurable subset E of [0, 1] satisfies m(e I) αm(i) for some α > 0 an all intervals I in [0, 1], then E has measure 1. See also Exercise 28 in Chapter 1. Fix x 0 (0, 1). Then for every interval I [0, 1] containing x 0, by assumption we have m(i E) m(i) α. Therefore m(i E) inf m(i) : x 0 I [0, 1] α > 0. But recall that by the proof of Corollary 1.5, there exists a null set Z such that for every x ((0, 1) E) Z = (0, 1) (E Z), m(i E) lim m(i) 0 m(i) x I (0,1) m(i E) = lim m(i) 0 m(i) x I = 0. Thus x 0 / (0, 1) (E Z). Since x 0 (0, 1) was arbitrary, we obtain (0, 1) E Z ; so 1 m(e Z) m(e) + m(z) = m(e). On the other han, m(e) 1 since E [0, 1]. Thus m(e) = Suppose A is a Lebesgue measurable set in R with m(a) > 0. Does there exist a sequence {s n } n=1 such that the complement n=1 (A + s n ) in R has measure zero? [Hint: For every > 0, fin an interval I of length l such that m(a I ) (1 )m(i ). Consier k= (A + t k ), with t k = kl. Then vary.] We first prove a finite measure analogue of the statement an then generalize. Let K be a finite close interval of the form [ N, N]. We claim that there exists a sequence {s n } n N such that the ifference K n N(A + s n ) has measure zero. To get an epsilon of room, fix > 0. Now A has a point of Lebesgue ensity x, that is, m(a I) lim m(i) 0 m(i) x I = 1

7 REAL ANALYSIS I HOMEWORK 5 7 where the limit is over intervals aroun x. So if I is a sufficiently small interval aroun x, then we have m(a I) m(i) > 1 2m(K) m(a I) > m(i) m(i A) = m(i) m(a I) < 2m(K) m(i). 2m(K) m(i) Write l for the length of I. Consier the sets of the form I + ql where q Z, which are mutually isjoint. Since K is a finite interval, there exists a finite subset F Z such that K (I + ql) Z 2K where Z is a finite set, namely the en-points of the intervals I + ql. Now m F m(i A) < m((i A) + ql) < m((i + ql) (A + ql)) < (I + ql) (A + ql) < 2m(K) F m(i) m(i + ql) 2m(K) m(i + ql) 2m(K) 2m(K) m By shrinking the set on the left han sie above, we get m (I + ql) (A + ql). Then since Z has measure zero, we obtain m (A + ql). K (I + ql). So we have shown that given > 0, there exists a finite set S of real numbers (the ql s above) such that m(k s S(A + s)). So for every n N there exists a finite set S n R such that m(k s S n (A + s)) 1/n. Then S = n N S n is a countable subset of real numbers such that ( ) m (A + s) = 0. This proves our claim. K s S Now by what we have just prove, for every N N there exists a countable set S N R such that m [ N, N] (A + s) = 0. s S N

8 R REAL ANALYSIS I HOMEWORK 5 8 So writing S = N N S N, 1 since ( ) (A + s) = [ N, N] (A + s) s S N N N N s S N [ N, N] (A + s), N N s S N we have m(r s S(A + s)) = Construct an increasing function on R whose set of iscontinuities is precisely Q. Disclaimer: I in t come up with the function myself. I i a google search to get the summation iea. Let {q n } n N be an enumeration of Q. Since for any sequence {b n } of 0 s an 1 s the series n N 2 n b n is convergent, the function f : R R x n N 2 n χ [qn, )(x) is well-efine. Fix x 0 R. Since R is Hausorff, given N N, there exists an interval I containing x 0 such that I {q 1,..., q N } {x 0 }. Note that we can t say the intersection is empty, maybe x 0 is one of the q i s. Suppose x 0 is irrational. This time we can say that given N N, then there exists an interval I containing x 0 such that I {q 1,..., q N } =. Then if x I, we have x q n if an only if x 0 q n for n N. In other wors, χ [qn, )(x) = χ [qn, )(x 0 ) for n N; hence Thus for every x I, Therefore f is continuous at x 0. f(x) f(x 0 ) = n>n 2 n Ä χ [qn, )(x) χ [qn, )(x 0 ) ä. f(x) f(x 0 ) n>n 2 n χ[qn, )(x) χ [qn, )(x 0 ) n>n 2 n = 2 N. Now suppose x 0 is rational. Then x 0 = q N for some N. So pick an interval I aroun q N such that I {q 1,..., q N, q N+1 } = {q N }. Let {q kn } be a subsequence of {q n } which lies in I whose terms are smaller than q N but converges to q N. So q kn q r if an only if q N q r for r {1,..., N 1, N + 1} an q kn < q N. So we have f(q N ) f(q kn ) = 2 r Ä χ [qr, )(q N ) χ [qr, )(q kn ) ä r N = 2 N + 2 r Ä χ [qr, )(q N ) χ [qr, )(q kn ) ä 2 N + r>n+1 r>n+1 2 r ( 1) = 2 N 2 N 1 = 2 N 1 1 This is the thir time in this question that I am using the letter S to enote a set of real numbers. First it was a finite set, then it was a countable union of finite sets an now it is a countable union of countable sets. Hopefully it s not confusing.

9 REAL ANALYSIS I HOMEWORK 5 9 hence f(q kn ) f(q N ) as n even though q kn q N. Thus f is not continuous at q N = x 0. Hence we conclue that the set of iscontinuities of f is precisely Q. To see that f is increasing, for each x R write S x = {n N : x q n }. Clearly, if x y then S x S y. Moreover as rationals are ense, if x < y we have S x S y. Since by efinition f(x) = n S x 2 n, if x < y then f(x) < f(y), that is, f is strictly increasing.