Math 205A - Fall 2015 Homework #4 Solutions

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1 Math 25A - Fall 25 Homework #4 Solutions Problem : Let f L and µ(t) = m{x : f(x) > t} the distribution function of f. Show that: (i) µ(t) t f L (). (ii) f L () = t µ(t)dt. (iii) For any increasing differentiable function φ with φ() = : φ( f(x) )dx = φ (λ)µ(λ)dλ. (i) f L () f t = t f = t µ(t). (ii) Follows from (iii). (iii) One can show that Fubini s theores still true for non-negative functions that are measurable with resect to the roduct measure, without ariori knowledge on the integrability of the function. The roof follows essentially the same lines as that of the standard Fubini theorem (chater 5 from the notes). Alying this to φ (t) [, f(x) ] (t)dtdx, we get : φ (t) [, f(x) ] (t)dt dx = and for the left hand side φ (t) [, f(x) ] (t)dx dt = f(x) φ (t)dt dx = φ( f(x) )dx, φ (t) [, f(x) ] (t)dx dt = φ (t)µ(t)dt. Problem 2: Show that for f f L, < and φ L, φ, φ =, if φ t (x) = t φ(x/t), then lim φ t f f =. t The crucial ste is to rove that f g f g : f(x y)g(y) dy f(x y) g(y) g(y) therefore ( f g = g f(x ) g g dy f(x ) g g ) ( dx = g = f(x ) g g, ) f(x y) g(y) dy dx () f(x y) g(y) dx dy = g f g = g f. (2) Now if the result is true for ste functions, then since they are dense in L, there exists a sequence f n of such L functions with f n f. ut then φ t f f φ t f φ t f n + φ t f n f n + f n f φ t f n f + φ t f n f n + f n f f n f + φ t f n f n + f n f. Given ɛ, choose n such that f n f < ɛ/2. In the limit, the 2 nd term vanishes, therefore lim φ t f f ɛ. t So the final ste is to rove the claim for ste functions. y triangle inequality, it s enough to consider f = [a,b].

2 ( φ t f f = φ( x y ( )f(y)dy f(x)) dx = φ(z)f(x tz)dz f(x)) dx t t ( = φ(z)(f(x tz) f(x))dz) dx = φ(z) f(x tz) f(x) dz dx. the same roof as for (2) above Now given ɛ, ick R s.t. z R φ(z) ɛ 2(b a). Note that f(x tz) f(x) = f(x tz) f(x) {, }, and f(x tz) f(x) 2(b a). Using these, for t < b a R : φ(z) f(x tz) f(x) dz dx ɛ + φ(z) f(x tz) f(x) dx dz + φ(z) f(x tz) f(x) dx dz. ut R [ R,] φ(z) f(x tz) f(x) dx dz = R φ(z) a+tz a dx + b+tz b [,R] dx dz 2Rt and similarly the other integral. Hence, lim φ t f f ɛ. This finishes the roof. R φ(z)dz 2Rt as t, FYI: The first inequality is a secial case of Young s inequality: f g r f g q for + q = r +. Problem 3: Let <, f L (X, dµ), and = X f n dµ for n N. Prove that lim n + = f. Clearly + f, so lim n+ f. On the other hand, using Hölder: ( ) n = f n dµ f n n+ X n+ = f n+ n+ n dµ n+ = n+ n+ n+ = + n Now given < ɛ < f, using roblem : + ( f ɛ) n+ µ({x : f f ɛ}), so taking the limit, because µ({x: f f ɛ}) + lim lim n ( f ɛ) n n ( µ({x : f f ɛ}) ) n+ = f ɛ, ( ) n+ n+. > and does not deend on n. Since this is true for every ɛ, sending ɛ we get that lim n+ f as well. Problem 4: Let {, x [, ) φ (t) =, x [, 2), extend it eriodically to all of R, and define φ n (t) = φ (2 n t), n N. Assume that c n 2 < and show that the series c n φ n (t) converges for almost every t. ecause of the -eriodicity of φ n, n, it s enough to rove the statement for a.e. t [, ]. Let s n (t) = c j φ j (t). We want to show that {s n } is almost everywhere Cauchy (hence a.e. convergent). So let E be the bad set where this doesn t haen: E = {t : s n (t) s N (t) m } m N n N+ }{{} E N,/m If we show that m( N E N,ɛ) = for all ɛ >, then m(e) m( N E N,/m) =. Now m( N E N,ɛ) m(e N,ɛ ) 2

3 for all N, so m( N E N,ɛ) lim N,ɛ). On the other hand, N m(e N,ɛ ) = m({t : su n N+ s n (t) s N (t) ɛ}) = m({t : su max k(t) s N (t) ɛ}) n N+ N+ k n = m( {t : max k(t) s N (t) ɛ}) n N+ N+ k n = lim k(t) s N (t) ɛ}) n N+ k n because the sets are increasing Let T (t) N + be the first index k for which s k (t) s N (t) ɛ. Then m({t : max k(t) s N (t) ɛ}) = N+ k n m({t : N + T (t) n}) Hence, = ɛ 2 {t:n+ T (t) n} dt = ɛ 2 n s T (t) s N 2 {N+ T (t) n} dt s T (t) s N ɛ k=n+ = ɛ 2 n k=n+ = ɛ 2 n = ɛ 2 n = ɛ 2 = ɛ 2 s k s N 2 {T (t)=k} dt i=n+ k=n+ i,j=n+ k=n+ { i=n+ k=n+ i=n+ i=n+ c i φ i (t) 2 {T (t)=k} dt c i c j φ i (t) φ j (t) {T (t)=k} dt c i 2 φ i (t) 2 }{{} {T (t)=k} + = c i 2 {T (t)=k} dt c i 2 {N+ T (t) n} dt ɛ 2 n i=n+ c i c j φ i (t) φ j (t) {T (t)=k} } i j } {{ } = (easy) c i 2. m( N E N,ɛ) lim m(e N,ɛ) = lim lim m({ max s k(t) s N (t) ɛ}) N N n N+ k n lim lim N n ɛ 2 c i 2 = ɛ 2 lim c i 2 N i=n+ i=n+ = because c i 2 <. Therefore m(e) = and {s n (t)} converges for almost every t. 3

4 Problem 5: Let Mf(x) = su δ> m((y, δ)) (y,δ) f(y) dy with the suremum taken over all balls (y, δ) that contain x. Show that: (i) There exists c = c(n) > such that c m({x : Mf(x) > }) f(x) dx. (ii) If f L, then Mf L. (iii) If f L and f, there exist C, R > such that Mf(x) C x n for x R. Hence, m({x : Mf(x) > }) C. (i) If f / L, the inequality would be satisfied for any c, so assume f L. Define K := {x : Mf(x) > }. For x K, there exist δ x > and y x with x y x < δ x, and f(y) >. δ x s are bounded, because otherwise, for a sequence {x n } with δ xn, > f f > (y xn, δ xn ). (y x,δ x) (y xn,δ xn ) The balls (x, 2δ x ) contain (y x, δ x ) and (x, 2δ x ) = 2 n (y x, δ x ). Alying esicovitch theorem to {(x, 2δ x ) : x K }, we get disjoint balls {(x i,m, 2δ xi,m )} i with K N(n) m= i (x i,m, 2δ xi,m ). Then m(k ) (x i,m, 2δ xi,m ) = 2 n (y i,m, δ xi,m ) 2 n f 2n 2 n N(n) f = f. m (y i,m,δ xi,m ) (ii) Note that this can only be true for >, otherwise it would contradict (iii). Setting f = {x: f(x) /2} (x)f(x), K {x : Mf (x) > /2}, because for x K, there exists a ball containing x such that < f = f + f f + 2 = 2 < f Mf (x). Then { f >/2} { f /2} m(k ) m({ Mf > /2}) c From roblem (ii), Mf L = m(k ), d c (iii) Choose M s.t. = c = c 2 f /2 f /2 2 f(x) { f(x) /2} (x) dx d = c (2 f(x) ) M f f /2. For x 2M = R, Mf(x) x +M f. f(x) dx d f(x) dx = c2 f L <. x +M f 2 f(x) f x +M 2. ut x + M x + x /2 = 3 x /2, so x +M (3/2) n x n = Mf(x) C x n. 2 f(x) d dx 4

5 For < C/(2 ) this imlies C m(k ) m({ x R : x n > }) = (, (C/)/n ) (, R) = ((C/) ) C 2 = C. In deed, the result can not be true for all f L : for examle, for f = [,] (x), { if x, Mf(x) = 2 x + if x >. ut m({x : Mf(x) > 2}) = can not be bigger than any C /2. Note that this is also an examle of an L function with Mf / L. Problem 6: Let Mf(x) = su δ> m((x, δ)) (x,δ) f(y) dy with the suremum taken over all balls (x, δ). Show that: (i) If f L and f, there exist C, R > such that Mf(x) C x n for x R. Hence, m({x : Mf(x) > }) C. (ii) Mf(x) Mf(x) 2 n Mf(x). (iii) f(x) Mf(x) at every Lebesgue oint if f L ( ). (i) Follows from (ii) and roblem 5. (ii) M takes a suremum over a subset of balls that M does, hence Mf Mf. On the other hand, to every (y, δ) considered in Mf(x), corresonds a (x, 2δ) in Mf(x). The second inequality follows from (x, 2δ) = 2 n (y, δ). (iii) Let x be a Lebesgue oint. Then given ɛ >, there exists r > s.t. f(y) f(x) dy < ɛ, so f(x) = f(x) dy f(y) f(x) dy + f(y) dy ɛ + Mf(x). Letting ɛ we get f(x) Mf(x). 5