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Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L p Spaces Yung-Hsiang Huang * 2018/04/11 6.1 Basic Theory of L p Spaces 1. When does equality hold in Minkowski’s inequality? Proof. (a) For p = 1, since we know the triangle inequality |(f + g)(x)|≤|f (x)| + |g(x)| is always true, to characteriz the case when the equality holds in Minkowski’s inequality is equivalent to seeking the condition for the equality |(f + g)(x)| = |f (x)| + |g(x)| for all x, that is f (x) g(x) 0 for a.e. x. (We consider the complex-valued functions here.) (b) For 1 <p< , in its proof, we know the equality holds in Minkowski inequality if and only if the ones in triangle and H¨ older’s inequalities hold, that is, if and only if f (x) g(x) 0 for a.e. x, and there are constants α, β 0 such that |f (x)| p = α|f (x)+ g(x)| p and |g(x)| p = β |f (x)+ g(x)| p for a.e. x. Combining them, we know the equality holds in Minkowski if and only if f (x)= γg(x) for a.e. x, for some γ 0. (c) For p = . Define the set A f = {{x n }} where {x n } is a sequence in X with |f (x n )|→kf k . We claim that kf + gk = kf k + kgk iff either one of the functions is 0 a.e., or A f A g is nonempty and for some {x n }∈ A f A g , where f (x n ) a and g(x n ) b, there is a λ> 0 such that b = λa. () If one of them is 0 a.e., then it’s done. If not, then kf k + kgk = lim n |f (x n )| + lim n |g(x n )| = |a| + |b| = |a + b| = lim n |f (x n )+ g(x n )|≤kf + gk . Combining with the Minkowski inequality, we have the desired equality. * Department of Math., National Taiwan University. Email: [email protected] 1

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  • Real Analysis, 2nd Edition, G.B.Folland

    Chapter 6 Lp Spaces

    Yung-Hsiang Huang

    2018/04/11

    6.1 Basic Theory of Lp Spaces

    1. When does equality hold in Minkowskis inequality?

    Proof. (a) For p = 1, since we know the triangle inequality |(f + g)(x)| |f(x)| + |g(x)|

    is always true, to characteriz the case when the equality holds in Minkowskis inequality is

    equivalent to seeking the condition for the equality |(f + g)(x)| = |f(x)|+ |g(x)| for all x, that

    is f(x)g(x) 0 for a.e. x. (We consider the complex-valued functions here.)

    (b) For 1 < p < , in its proof, we know the equality holds in Minkowski inequality if and

    only if the ones in triangle and Holders inequalities hold, that is, if and only if f(x)g(x) 0

    for a.e. x, and there are constants , 0 such that |f(x)|p = |f(x) + g(x)|p and |g(x)|p =

    |f(x) + g(x)|p for a.e. x. Combining them, we know the equality holds in Minkowski if and

    only if f(x) = g(x) for a.e. x, for some 0.

    (c) For p =. Define the set Af = {{xn}} where {xn} is a sequence in X with |f(xn)| f.

    We claim that f + g = f + g iff either one of the functions is 0 a.e., or Af Agis nonempty and for some {xn} Af Ag, where f(xn) a and g(xn) b, there is a > 0

    such that b = a.

    () If one of them is 0 a.e., then its done. If not, then

    f + g = limn|f(xn)|+ lim

    n|g(xn)| = |a|+ |b| = |a+ b| = lim

    n|f(xn) + g(xn)| f + g.

    Combining with the Minkowski inequality, we have the desired equality.

    Department of Math., National Taiwan University. Email: [email protected]

    1

  • () We assume both functions are not zero, so f and g > 0. Since there exists xnsuch that |(f + g)(xn)| f + g = f + g. By a contradiction argument, we see

    |f(xn)| f and |g(xn)| g, that is, {xn} Af Ag. If f(xn) f > 0 and

    g(xn) g < 0, then |f(xn) + g(xn)| |f g| < f + g = f + g, a

    contradiction. Similarly, its impossible that f(xn) f and g(xn) g. Hence we get

    the desired result with = gf .

    2. Prove Theorem 6.8.

    Proof. (a),(b) and (c)() are obvious, we also omit (d) since its proved in many textbooks.

    (c)(): Given m N there exists Mm N such that for all n Mm, fn f < 1m . Then

    for such (m,n), there is Xmn M such that (Xmn ) = 0 and |fn f | < 1m on X \Xmn . Take

    Ec = m=1 n=Mm Xmn

    which has zero measure and fn f uniformly on E.

    (e) is proved by partition the given range [f, f) equally into 2k disjoint closed-open

    subintervals, and for each k N, consider the proper simple function fk corresponding to the

    sets {{x : f

    j

    2k |f(x)| < f

    j + 1

    2k}, {x : f = |f(x)|}

    }2k1j=2k

    .

    3. If 1 p < r , Lp Lr is a Banach space with norm f = fp + fr, and if

    p < q < r, the inclusion map Lp Lr Lq is continuous.

    Proof. fq fpf1r ff1, where is defined by q1 = p1 + (1 )r1.

    4. If 1 p < r , Lp+Lr is a Banach space with norm f = inf{gp+hr : f = g+h},

    and if p < q < r, the inclusion map Lq Lp + Lr is continuous.

    Remark 1. (2018.04.10) See Sergey Astashkin and Lech Maligrandas paper (arXiv:1804.03469),

    Lp+Lq and LpLq are not isomorphic for all 1 p, q , p 6= q and the references therein.

    Proof. Since p < q < r,

    f f{|f |>1}p + f{|f |1}r f{|f |>1}q + f{|f |1}q = fq.

    2

  • There is a related set equality due to Alvarez [1]:

    Theorem 2. Let Lpq := Lp + Lq. Then (a) Lpq + L

    rs = L

    min(p,r)max(q,s) (b) L

    pq Lrs = Luv , where

    u1 = p1 + r1; v1 = q1 + s1 (c) Lpq(Lp0q0

    + Lp1q1 ) = Lpq Lp0q0 + L

    pq Lp1q1 .

    We prove this theorem in additional exercise 6.1.3.

    Remark 3. Lpq is a special case of Orlicz spaces. Its duality result to LpLq and generalization

    are given in Stein-Shakarchi [18, Exercise 1.24, 1.26 and Problem 1.5].

    5. Suppose 0 < p < q < . Then Lp 6 Lq iff X contains sets of arbitrarily small positive

    measure, and Lq 6 Lp iff X contains sets of arbitrarily large finite measure. What about the

    case q =

    Proof. Since the first and second assertions are dual to each other, we only prove the first one.

    () By assumption, there is a sequence of sets {Fn} with 0 < (Fn+1) < 31(Fn) for all

    n N and (F1) < 1/3. Let En = Fn \ n+1Fj, then we obtain a sequence of disjoint sets

    En with 0 < 21(Fn) < (En) < 3

    n. Finally we consider the function f =anEn with

    an = (En)1/q which is in Lp but not in Lq.

    () Assume there is c > 0 such that any set with positive measure has measure c. Take

    f Lp and consider the sets An = {x : n |f(x)| < n + 1}. By Chebyshevs inequality, for

    n 1, either (An) c or (An) = 0. Since f Lp, only finitely many n such that (An) c.

    So f Lq and thus Lp Lq, which is a contradiction.

    For the case q =, L 6 Lp iff (X) =. This is easy to prove.

    On the other hand, Lp 6 L iff X contains sets of arbitrarily small positive measure. The

    proof for () is the same as above except taking an = (En)1/(p+1) now. The proof for ()

    is the same.

    Remark 4. A. Villani [19] simplified Romeros previous work [14] and show that the followings

    are equivalent: Let 0 < p < q . (i) Lp() Lq() for some p < q; (ii) Lp() Lq() for

    all p < q; (iii) inf{(E) : (E) > 0} > 0. In the case q < p, condition (iii) is replaced by (iii)

    sup{(E) : (E)

  • Proof. (i) If p1 < , then consider f(x) = x1/p1(0, 12

    )(x) + x1/p0(2,)(x). If p1 = , then

    consider f(x) = | log x|(0, 12

    )(x) + x1/p0(2,)(x).

    (ii) If p1 0, so that f Lq for q (0, p).

    Proof. (a) If log |f |q L1, then this is Jensens inequality. See Exercise 3.42.

    If log |f |q 6 L1, then| log |f |q| = and

    | log |f |q| =

    log |f |q{|f |>1}

    log |f |q{|f |1} |f |q{|f |>1}

    log |f |q{|f |1}

    Since f Lq,

    log |f |q{|f |1} = . Therefore

    log |f | = which implies the inequality.

    (b) Since log(x) x 1 for all x > 0. Let x = fqq, we yields the first inequality. The second

    assertion is a consequence of monotone convergence theorem.

    (c) By (a), fq exp(

    log |f |) for all q (0, p), and then lim infq0 fq exp(

    log |f |). On

    the other hand, from (b) we know lim supq0 log fq lim supq0(|f |q 1)/q =

    log |f |.

    Taking the exponential of both sides, we have lim supq0 fq exp(

    log |f |).

    9. Proof. The first assertion is Chebyshevs inequality. To prove the second one, we notices the

    convergence in measure implies convergence a.e., up to a subsequence, and then the limit

    function |f | g. This implies the desired result by LDCT and the fact |fn f | 2g.

    Remark 5. The first assertion is trivial true for p =, but the second one is false.

    10. Suppose fn, f Lp, fn f a.e. for some p [1,). Then fnp fp fn fp 0.

    Proof. = part is easy (we dont need to assume pointwise convergence here). = part

    is an application of Fatous lemma to the functions 2p|fn|p + 2p|f |p |fn f |p 0. (Another

    method is through Egoroffs theorem and Fatous lemma, see Rudin [15, Exercise 3.17(b)].)

    4

  • Remark 6. Rudin [15] mentions that the proof seems to be discovered by Novinger [13], where

    an application to F.Rieszs theorem on mean convergence of Hp functions to their boundary

    function is mentioned. See Duren [7, Theorem 2.6 and 2.2] and Rudin [15, Theorem 17.11].

    If fn f a.e., by Fatous lemma, we always have fp lim inf fnp. What can be said

    about fp. There is a more general and quantitative result:

    Theorem 7 (Brezis-Lieb lemma[4]). Suppose fn f a.e. and fnp C < for all n and

    for some 0 < p 0 , since |f(x) z|

    |f(x) zn|+ |zn z|, by choosing n large such that |zn z| < , we have

    {x : |f(x) zn| < } {|f(x) z| < 2}.

    Therefore 0 < ({x : |f(x) zn| < }) ({|f(x) z| < 2}) and hence z Rf .

    (b) By definition and proof by contradiction, f sup{|w| : w Rf}. The compactness

    follows from (a) and Bolzano-Weierstrauss. So there is z Rf , such that |z| = sup{|w| : w

    Rf}. If |z| < f, then |z| < 12(|z|+ f).

    Hence the set {x : |f(x) 12(|z| + f)| < 0} has zero measure for some 0 > 0, but this

    implies {x : |f(x) 12(|z| + f)| < } has zero measure for all 0 < < 0. In particular,

    choosing 1 > > 0 such that 2(|z|+ f) < 0. Since

    {x : |f(x)| > 1 2

    (|z|+ f)} {x : |f(x)1

    2(|z|+ f)|

    12

    (|z|+ f)}) = 0,

    which contradicts the definition of f since 0 < 12 (|z| + f) < f. Therefore,

    max{|w| : w Rf} = |z| = f.

    Remark 9. See Rudin [15, Exercise 3.19] and additional exercise 6.1.7 for an extended discus-

    sion.

    12. Proof. Let (X,, ) be the measure space. We assume 1 p

  • Now we assume there exists A such that 0 < (A) < . If (ii) for every E ,

    (EA) = 0, then dim(Lp) = 1, that is, every function equals a.e. to a multiple of A. The

    inner product is defined by (A, A) = (A)2.

    If (iii) there is a set C with 0 < (C) 0. We may assume B := C \A has

    positive measure. Note A2p = (A)2/p, B2p = (B)2/p, A + B2p = ((A) + (B))2/p

    and AB2p = ((A) +(B))2/p, so 2(A2p + B2p) = 2((A)2/p +(B)2/p) which is not

    equal to 2((A) + (B))2/p = A + B2p + A B2p except p = 2.

    For p = , in case (iii), A + B2 + A B2 = 2 6= 4 = 2(A2 + B2). In

    case (ii), the situation is the same as above, so it possesses inner product structure. In case

    (i), if (X) = 0, then its done. For (X) = , if there are proper subset A of X such that

    (A) = (X\A) =, then we see there is no inner product structure as above with B = X\A;

    if for every set A with (A) =, we always have (X \A) = 0, then Lp has dimension 1, since

    for each f L, if f takes two distinct values on disjoint sets A,B, then either (A) = 0 or

    (B) = 0. That is, f = X , for some scalar .

    13. Proof. The case 1 p (g )Ep, which

    means T > g . Since > 0 is arbitrary, g T.

    15. This is Vitali Convergence Theorem. Necessity of the hypotheses and further discussions are

    given in Rudin [15, Exercise 6.10-11]. For example, in finite measure space, Vitali implies

    LDCT, and there is some case Vitali works but LDCT cant be applied. The details are in

    additional exercise 6.1.8.

    Proof. (Necessity) Use Chebyshevs inequality to prove (i). Completeness of Lp implies the

    existence of limit function, which controls the integrability of fn for the large n, which is the

    central idea to prove (ii) and (iii).

    6

  • (Sufficiency) Given > 0, let E be the set in condition (iii). Take

    Anm := {|fm(x) fn(x)|

    (E)}.

    By (i), we know if n,m large enough, then (Anm) < , where is the modulus of uniform

    integrability. Hence, for all large n,mAnm

    |fn fm|p d < .

    Therefore, for all large enough n,mX

    |fn fm|p d =Ec|fn fm|p d+

    Anm

    |fn fm|p d+E\Anm

    |fn fm|p d

    2+ + (E)

    (E),

    which implies the desired result.

    16. Proof. The completeness is a revision of the proof of Theorem 6.6, we omit it. The triangle

    inequality is a consequence of the inequality (a + b)p ap + bp, a, b 0, whose proof is

    straightforward.

    Remark 11. There are several properties for 0 < p < 1; the first is in finite measure space,

    the behavior of q as q 0, answered in exercise 8(c); the second is that the dual space of

    Lp([0, 1],L ,m) is {0}, that is, its NOT a locally convex space; the last is the reversed Holder

    and Minkowski inequalities, see Jones [10, p.252-253] and Rudin [16, p.37].

    Theorem 12. Let (X,M , ) be a measure space, 0 < p < 1 and f, g be two nonnegative and

    measurable functions. Then X

    fg d (fp d

    )1/p(gpd)1/p

    ,

    and

    f + gp fp + gp.

    6.2 The Dual of Lp

    7

  • 17. Proof. First, we note that Mq(|g|) 8Mq(g) since for each f =n

    k=1(ak + ibk)Ek , the

    set of simple functions that vanish outside a set of finite measure

    |f |g||2 = |

    nk=1

    ak

    Ek

    |g|+ ibkEk

    |g||2 =( nk=1

    ak

    Ek

    |g|)2

    +( nk=1

    bk

    Ek

    |g|)2

    ( nk=1

    ak

    Ek

    |Re(g)|+ |Im(g)|)2

    +( nk=1

    bk

    Ek

    |Re(g)|+ |Im(g)|)2

    =( nk=1

    ak(

    E1k

    Re(g)E2k

    Re(g) +

    E3k

    Im(g)E4k

    Im(g)))2

    +( )2

    2( nk=1

    ak

    E1k

    Re(g) akE2k

    Re(g))2

    + 2( nk=1

    ak

    E3k

    Im(g) akE4k

    Im(g))2

    + 2( )2

    + 2( )2

    2Mq(g) + 2Mq(g) + 2Mq(g) + 2Mq(g),

    where Ek = E1k E2k = E3k E4k are separated by the sign of Re(g) and Im(g) respectively and

    the last inequality follows from the definition of Mq(g) (by taking fr =n

    k=1 akE1k akE2kand fi =

    nk=1 akE3k akE4k).

    Given > 0. Let E := {x : |g(x)| > }. If (E) = , then for each n N there is En E

    such that n < (En)

  • Since (1 g)1A 0 -a.e., there exists an increasing sequence of simple functions n 0

    converging to A(1 g)1 -a.e. Since A(1 g)1 < -a.e., n L2() for all n. For all

    E M , by monotone convergence theorem, we have

    a(E) = (AE) =E

    A1 g

    (1g) d = limn

    E

    n(1g) d = limn

    E

    ng d =

    E

    A1 g

    g d,

    which is the desired result.

    19. Proof. Note that |n(f)| f, which means n(l) 1 for all n. Then Banach-Alaoglus

    theorem implies the sequence {n} has a weak cluster point . If (f) =

    j f(j)g(j) for some

    g l1, then by taking fi l with fi(j) = ij, the Kronecker delta, we see g 0 since

    g(i) = (fi) = limn n(fi) = 0 for each i. But this leads a contradiction that 0 = (1) =

    limn n(1) = 1, where 1(j) = 1 for all j N.

    20. Suppose supn fnp 0 such thatE|g|q < whenever (E) < , (ii) A X

    such that (A) < andX\A |g|

    q < , and (iii) B A such that (A \ B) < and

    fn f uniformly on B.)

    (b) The result of (a) is false in general for p = 1

    Proof. (a) We omit it since every step is given in the hint and should be verified easily. A more

    precise result about fn and f is Brezis-Lieb lemma, which is proved in additional exercise

    6.1.5.

    Note that Banach-Alaoglu theorem implies the following similar and important result: (for Lp

    case, we can prove it by Cantor diagonal process without Banach-Alaoglu, see my solution file

    for Stein-Shakarchi [18, Exercise 1.12(c)])

    Theorem 14. Let X be a reflexive Banach space. If fnX M

  • 21. If 1 < p

  • 23. See Exercise 1.16 for the notion of locally measurable sets.

    Proof. (a) If 0 < (E) < , then 0 = (E E) > 0, a contradiction. If is semifinite, then

    for every locally null set E with (E) = , there exist F E with 0 < (F ) < , but

    0 = (E F ) = (F ) > 0, a contradiction.

    (b) If f = 0, then for each 1n there exist an (0,1n) such that {|f | > an} is locally null.

    Since {|f | > an} {|f | > 1n}, {|f | >1n} is locally null. Given F M with (F ) < , we

    have 0 (F {f 6= 0}) = (F n{|f | > 1n})

    n (F {|f | >1n}) = 0, that is, {f 6= 0}

    is locally null.

    Given f, g L and then given > 0. Note that

    {|f+g| > f+g+} {|f |+|g| > f+g+} {|f | > f+/2}{|g| > g+/2}.

    11

  • Since the latter set is locally null, {|f + g| > f + g + } is locally null. Then f + g

    f + g + for all > 0 and therefore f + g f + g.

    Given 0 6= C and f L, since {|f | > ||f + } = {|f | > f + ||} is locally null

    for each > 0 and {|f | > ||f } = {|f | > f ||} is not locally null for each > 0,

    we see f = ||f. Consequently, is a norm on L.

    Given {fn} L with

    n fn < . We are going to show

    n fn L. More precisely,

    n fn

    n fn 0{|n

    fn| >n

    fn + }{

    n

    |fn| >n

    fn + } n

    {|fn| > fn +

    2n

    },

    where each indexed-set in the last set is locally null. Therefore, (L, ) is complete.

    The last assertion is due to = here, which is a consequence of (a).

    24. In contrast to Proposition 6.13 and Theorem 6.14, we do NOT need to assume is semifinite

    or Sg = {g 6= 0} is -finite here. Also see additional exercise 6.2.2.

    Proof. (i) Given f L1 with f1 = 1. Since, for each n N, {|f | > 1n} has finite measure

    and the set {|g| > g} is locally null, the intersection of these two sets are null. Therefore

    |f(x)g(x)| |f(x)|g a.e.. Hence sup{|fg| : f1 = 1} g.

    Given > 0, since {|g| > g } is not locally null, there is a measurable set A with

    finite measure such that B = A {|g| > g } has nonzero finite measure. Consider

    h = sgn(g)B((B))1, then h1 = 1 and

    sup{|fg| : f1 = 1} |

    hg| = ((B))1

    B

    |g| > g .

    Therefore sup{|fg| : f1 = 1} g.

    (ii) Now we consider the counterpart of Theorem 6.14: Given > 0. If {|g| > M(g) + } is

    not locally null, then there is a set A with finite measure, such that B = A{|g| > M(g)+ }

    has nonzero finite measure. But for h = sgn(g)B((B))1, h1 = 1 and hence

    M(g) |hg| = ((B))1

    B

    |g| > M(g) + ,

    which is a contradiction. So {|g| > M(g) + } is locally null for each > 0. Therefore

    g M(g)

  • Now for decomposable , let {Fa}aA be the decomposition of . Let ga L(Fa) be so

    L1(Fa) = ga and define g = ga on each Fa. Then g L since for each R, g1((,)) =

    ag1a (,)) is measurable by property (iv) in the definition of decomposable measur and

    since g = supa ga supa L1(Fa) .

    Since for each f L1, the set En = {|f | > 1n} has finite measure and (En) =

    a (En Fa).

    Therefore for each k N only finite many index a such that (EnFa) > 1k , and only countably

    many indices a such that (En Fa) > 0. Furthermore, only countably many indices a such

    that ({|f | > 0} Fa) > 0. We call this index set I and thus,

    (f) =iI

    (fFi) =iI

    Fi

    fgi =

    fg,

    where we apply LDCT in the last equality.

    Remark 18. The definition of decomposable measure we used here should be precisely called

    almost decomposable due to (iii) (E) =(EF ) is required to be true for finite measure

    set E only. (cf: Bogachev [2, 1.12.131-132] and Fremlin [8, Vol. 2].)

    A subtle difference is the following facts:

    is decomposable is semifinite : is almost decomposable.

    () is straightforward. The counterexample for (:) is recorded in additional exercise 6.2.6.

    From Bogachev [2, 1.12.134], we know is semifinite ; is decomposable.

    Example 19. I learned these examples from Professor Nico Spronks Notes

    (i) The first is an application of Exercise 25 which is not a consequence of Theorem 6.15.

    Let I be an uncountable set, and M be the -algebra of sets of countable or co-countable. is

    the counting measure. Then the collection M of all locally measurable sets is P (I), the power

    set of I and we can define the saturation of on M (cf: Exercise 1.16).

    Note that L1(I,M , ) = L1(I, M , ) and {{i}}iI is a decomposition of for . The theorem

    we just proved implies that (L1) = L.

    (ii) The second example sits outside Theorem 6.15 and Exercise 25!

    Let X = {f, i} and consider : P (X) [0,] be the unique measure satisfying ({f}) = 1

    and ({i}) = . Note that {i} is locally null. Then L is the span of {f}. On the other

    hand L1 is also the span of {f}. Then by definition, (L1) = R = L.

    But the existence of infinite atom {i} means that we cannot partition X into subsets of finite

    measure, that is, is neither semifinite nor decomposable!

    13

    http://www.math.uwaterloo.ca/~nspronk/math451/Lone_dual.pdf

  • Remark 20. Further results related to Exercise 23-25 may also be confirmed with Cohn [5,

    Chapter 3].

    6.3 Some Useful Inequalities

    26. The case p = 1 is a consequence of Fubini-Tonellis Theorem. The case p = is apparent.

    27. (Hilberts inequality, a special case of Schurs lemma, Theorem 6.20, our proof here is essential

    the same as the one given there.)

    Proof. By Minkowskis inequality,( 0

    |

    0

    f(y)

    x+ ydy|p dx

    )1/p=(

    0

    |

    0

    f(xz)

    1 + zdz|p dx

    )1/p

    0

    ( 0

    |f(xz)|p dx)1/p 1

    1 + zdz

    = fp

    0

    z1/p(1 + z)1 dz = fp csc(/p).

    To compute the last integral, one use the residue theorem in one complex variable.

    28. This exercise extends Corollary 6.20 with T = J1. Further generalizations are given in the

    exercise 29 and 44.

    Proof. For p = , it is straightforward. For 1 < p < , using change of variables and

    Minkowskis inequality, we have( 0

    1x

    x0

    f(t)(x t)1 dtp dx)1/p (

    0

    ( 10

    |f(xw)|(1 w)1 dw)pdx)1/p

    1

    0

    ( 0

    |f(xw)|p(1 w)pp dx)1/p

    dw = fp 1

    0

    (1 w)1w1/p dw = fp()(1 p1)( + 1 p1)

    .

    A counterexample for p = 1 is (0,1)(t). Note that 0

    x x

    0

    (x t)1(0,1)(t) dt dx

    =

    1

    x 1

    0

    (x t)1 dt dx+ 1

    0

    x x

    0

    (x t)1 dt dx

    =

    1

    x(x)1 dx+ 1,

    where x (x 1, x) for all x (1,) by mean value theorem. We consider > 1 and

    0 < 1 separately. For > 1, the integrand is larger than x(x1)1 = (1 1x)(x1)1

    which is not integrable on (2,). For 1, the integrand is larger than xx1 = x1 which

    is not integrable on (1,). So J((0,1)(t)) 6 L1(0,).

    14

  • Remark 21. Is the constant (1p1)

    (+1p1) sharpest ? When does the equality hold? See Rudin

    [15, Exercise 3.14] for the Hardys inequality, that is, the case = 1.

    Remark 22. A similar operator to fractional integral operator is Riesz and Bessel potential

    operators, which is strongly related to Sobolev embedding theorem. See Stein [17, Chapter V]

    and Ziemer [20, Chapter 2]. Also see exercise 44-45.

    29. (Hardys inequality)

    Proof. Again, we prove this without using Theorem 6.20. By Minkowskis inequality,( 0

    xr1[ x

    0

    h(y) dy]pdx)1/p

    =(

    0

    xpr1[ 1

    0

    h(xz) dz]pdx)1/p

    1

    0

    ( 0

    xpr1(h(xz))p dx)1/p

    dz =

    10

    ( 0

    ypr1h(y)p dy)1/p

    zrp1 dz

    =p

    r

    ( 0

    ypr1h(y)p dy)1/p

    .

    ( 0

    xr1[

    x

    h(y) dy]pdx)1/p

    =(

    0

    xp+r1[

    1

    h(xz) dz]pdx)1/p

    1

    ( 0

    xp+r1(h(xz))p dx)1/p

    dz =

    1

    ( 0

    yp+r1h(y)p dy)1/p

    zrp1 dz

    =p

    r

    ( 0

    yp+r1h(y)p dy)1/p

    .

    30. Proof. (a)Since all terms in the integrand is nonnegative, by Fubini-Tonelli, we can interchange

    the order of integrations freely. K(xy)f(x)g(y) dx dy =

    K(z)f(

    z

    y)g(y)y1 dz dy =

    f(z

    y)g(y)y1 dy K(z) dz

    (

    [f(

    z

    y)]pyp dy)1/p(

    g(y)q dy)1/qK(z) dz

    =

    (

    [f(w)]p(w/z)pz/w2 dw)1/pK(z) dz(

    g(y)q dy)1/q

    =

    z1+

    1pK(z) dz(

    [f(w)]pwp2 dw)1/p(

    g(y)q dy)1/q.

    (b) is proved by (a) with p = q = 2 and g(y) =k(zy)f(z) dz.

    31. Proof. (Sketch) Mathematical induction and apply the original Holders inequality to rr.

    32. (Hilbert-Schmidt integral operator.)

    15

  • Proof. By Tonelli-Fubinis theorem and Holders inequality, we know for -a.e. x, the integralY

    |K(x, y)||f(y)| d(y) (

    Y

    |K(x, y)|2 d(y))1/2fL2() 0.

    Since for any x (0,)

    |Tf(x) T (f(0,n))(x)| x1/p x

    0

    |f(t) f(0,n)(t)| dt f f(0,n)q,

    we can choose a large N such that Tf T (f(0,N)) < . Since for x > N

    |T (f(0,N))(x)| x1/p N

    0

    |f(t)| dt,

    the left hand side is less than if x is sufficient large. Combine these inequalities and since is

    arbitrary, Tf C0(0,). The boundedness of T is a consequence of Holders inequality.

    34. Proof. Since f is absolutely continuous on any [, 1], we know for any [x, y] (0, 1),

    |f(x) f(y)| = | yx

    f (s)ds| ( y

    x

    s1p

    pp1

    ) p1p( y

    x

    s|f (s)|p)1/p

    (i) If p > 2, then by the above calculation, for any sequence xn 0, {f(xn)} forms a Cauchy

    sequence. Therefore limx0 f(x) exists.

    (ii) If p = 2, given > 0 and pick y = y() (0, 12) such that

    y0t|f (t)|2 dt < 2, then for any

    x (0, y),

    0 |f(x)|| log x|1/2

    =|f(y)

    yxf (t)dt|

    | log x|1/2 |f(y)|| log x|1/2

    +| log y log x|1/2| log x|1/2

    .

    16

  • Hence there exist = () (0, y) such that for any x (0, ), the last term is less than 2.

    (iii) If p (1, 2), given > 0 and pick y = y() (0, 12) such that

    y0t|f (t)|p dt < p, then for

    any x (0, y),

    0 |f(x)|x1(2/p)

    =|f(y)

    yxf (t)dt|

    x1(2/p) |f(y)|x1(2/p)

    + Cp|y

    p2p1 x

    p2p1 |

    p1p

    x1(2/p).

    Hence there exist = () (0, y) such that for any x (0, ), the last term is less than 2.

    (iv)If p = 1, given > 0 and pick y = y() (0, 12) such that

    y0t|f (t)| dt < , then for any

    x (0, y),

    0 |f(x)|x1

    =|f(y)

    yxf (t)dt|

    x1 |f(y)|

    x1+x1

    x1.

    Hence there exist = () (0, y) such that for any x (0, ), the last term is less than 2.

    From(ii)-(iv), we see the second and third assertions hold.

    6.4 Distribution Functions and Weak Lp

    35. For any measurable f and g we have [cf ]p = |c|[f ]p and [f + g]p 2([f ]pp + [g]pp)1p ; hence

    weak Lp is a vector space. Moreover, the balls {g : [g f ]p < r} (r > 0, f weak

    Lp) generate a topology on weak Lp that makes weak Lp into a topological vector

    space.

    Proof. The first assertion follows from definition; the second assertion is through the fact

    {|f + g| > } {|f |+ |g| > } {|f | > 2} {|g| >

    2}.

    The addition (x,y) 7 x+y is continuous under the given topology by definition and the second

    assertion.

    The scalar multiplication (a, x) 7 ax is continuous by definition and the above two assertions.

    Remark 24. Normability of weak Lp, see Grafakos [9, p.14-15] and additional exercise 6.4.1.

    36. If f weak Lp and ({x : f(x) 6= 0}) < , then f Lq for all q < p. On the other

    hand, if f (weak Lp) L, then f Lq for all q > p.

    Proof. (i) By Proposition 6.24 and assumption, we see q p < 0 and

    1

    q

    |f |q d =

    10

    +

    1

    q1({x : |f | > }) d M 1

    0

    q1 d + C

    1

    q1p d

  • for some positive constants M,C.

    (ii) Similarly, we see q p > 0 and

    1

    q

    |f |q d =

    f0

    +

    f

    q1({x : |f | > }) d C f

    0

    qp1 d 2k} and Fk = {x : 2k < f(x) 2k+1}. Prove that

    f Lp()

    k=

    2kp(Fk) 0 such that{|f |} |f |

    p < . Thus p(f () f ()) {

  • 40. If f is a measurable function on X, its decreasing rearrangement is the function

    f (0,) [0,] defined by

    f (t) = inf{ : f () t} (where inf =).

    (a) f is decreasing. If f (t) 0, If f (t) = 0, then t1/pf (t) [f ]p. If f (t) > 0 then by definition, for all

    0 < < f (t), we have

    t(f (t) )p < f (f (t) )(f (t) )p [f ]pp.

    Taking p-th root and letting 0, we have t1/pf (t) [f ]p. Since t > 0 is arbitrary,

    supt>0

    t1/pf (t) [f ]p.

    On the other hand, given > 0 and assume f () > 0 without loss of generality. Since for any

    0 < < f (), we have f(f () ) > . Therefore

    supt>0

    t1/pf (t) (f () )1/pf (f () ) > (f () )1/p.

    Letting 0, we have supt>0 t1/pf (t) [f ]p.

    19

  • Remark 27. See Lieb-Loss [11, Chapter 3-4] and Dibenedetto [6, Chapte 9] for further dis-

    cussions and applications on higher-dimensional rearrangement and Rieszs rearrangement in-

    equality.

    6.5 Interpolation of Lp Spaces

    41. Suppose 1 < p and 1p

    + 1q

    = 1. If T is a bounded linear operator on Lp such

    that

    (Tf)g =f(Tg) for all f, g Lp Lq, then T extends uniquely to a bounded

    linear operator on Lr for all r [p, q] (if p < q) or [q, p] if q < p). We should assume

    is semifinite when p =, see the errata sheet provided by Prof. Folland.

    Proof. (i) If q < p < or p < q < , then we define T on Lp Lq the same as the original

    bounded linear map on Lp. Given f LpLq and g LpLq, we have, by Holders inequality,

    |

    (Tf)g| = |f(Tg)| fqTppgp, (1)

    where Tpp is the operator norm of T on Lp. Since Lp Lq is dense in Lp, for any g Lp,

    we choose gn Lp Lq such that gn g in Lp, then the sequence {

    (Tf)gn} is Cauchy in C

    and we define (Tf)g = lim

    n

    (Tf)gn.

    This definition is independent of the choice of gn since the union of two approximation se-

    quences is still an approximation sequence and the corresponding sequence is still Cauchy in

    C. Therefore we know (1) holds for any g Lp with gp = 1, and hence

    Tfq fqTpp

    for any f Lp Lq. Since Lp Lq is dense in Lq, this inequality holds for any f Lq. So T is

    bounded on Lq. The desired boundedness of T on Lr follows from Riesz-Thorin theorem.

    (ii) If p = , then q = 1. First, we show T maps L L1 L1 continuously by Theorem

    6.14: Let be the set of simple functions that vanish outside a set of finite measure. Given

    f L L1 and g L L1, then fg L1 and we have, by Holders inequality, for any

    g = 1

    |

    (Tf)g| = |f(Tg)| f1T, (2)

    where T is the operator norm of T on L. Therefore

    M1(Tf) := sup{|

    (Tf)g| : g = 1} f1T.

    20

  • By Theorem 6.14 and Exercise 17, we have Tf L1 and Tf Tf1. Since L1 L

    is dense in L1, we can extend T to L1 continuously with norm less than T. The desired

    boundedness of T on Lr follows from Riesz-Thorin again.

    The uniqueness part is due to the fact Lp Lq is dense in Lr which is proved by considering

    cut-off in domain and range simultaneous.

    Remark 28. Here I emphasis that case (i) can also be proved by the method given in (ii), but

    the density argument in (i) may not be suitable for case (ii).

    42. Prove the Marcinkiewicz theorem in the case p0 = p1.

    Proof. X

    |Tf |q = q

    0

    sq1({|Tf | > s}) ds

    = q

    fp0

    sq1({|Tf | > s}) ds+ q fp

    sq1({|Tf | > s}) ds

    q fp

    0

    sq1(C0fp

    s)q0 ds+ q

    fp

    sq1(C1fp

    s)q1 ds

    = q( Cq00q q0

    +Cq11q1 q

    )fqp

    43. Let H be the Hardy-Littlewood maximal operator on R. Compute H(0,1) explicitly.

    Show that it is in Lp for all p > 1 and in weak L1 but not in L1, and that its Lp

    norm tends to like 1p1 as p 1

    +, although (0,1)p = 1 for all p.

    Proof. By direct computations,

    H(0,1)(x) =

    1

    2(1x) for x 0,

    1 for x (0, 1),1

    2xfor x 1.

    (3)

    For t > 0,

    m({H(0,1)(x) > t}) =

    0 for t > 1,

    1 for t [12, 1],

    1t 1 for t (0, 1

    2),

    (4)

    which is easy to see the weak (1,1) estimate with constant 1 and strong (p,p) estimate with

    constant 1(p1)2p1 + 1.

    21

  • 44. Let I be the fractional integration operator of Exercise 61 in 2.6. If 0 < < 1,

    1 < p < 1, and 1

    r= 1

    p , then I is weak type (1, 11) and strong type (p, r) with

    respect to Lebesgue measure on (0,). Also see the last paragraph in section 6.6.

    Proof. This exercise is essentially the same as the next one. We change our in this problem

    to and put the following parameters in the next exercise: n = 1, = 1 , f = g(0,) and

    K(x) := |x|(0,). Using the results there, we have the desired weak type estimate.

    45. If 0 < < n, define an operator T on functions on Rn by

    Tf(x) =

    |x y|f(y) dy.

    Then T os weak type (1,n

    ) and strong type (p, r) with respect to Lebesgue measure

    on Rn, where 1 < p < nn and

    1r

    = 1p n

    n.

    Proof. Given f Lp, we may assume fp = 1.

    Fix > 0 to be determined later, we decompose

    K(x) := |x| = K(x){|x|} +K(x){|x|>}

    =: K1(x) +K(x).

    Then Tf = K1 f +K f . The first term is an Lp function since its the convolution of an

    L1 function K1 and Lp function f and similarly, the second term is an L function.

    Since for each > 0

    m({|K f | > 2}) m({|K1 f | > }) +m({|K f | > })

    Note that m({|K1 f | > }) K1fpp

    p (K1

    p1

    )p = c1(

    n

    )p and Kp = c2n/r. Choose

    such that c2n/r = and therefore K f , that is, m({|K f | > }) = 0. Then

    m({|K f | > 2}) c1(n

    )p = c3

    r

    The special case for p = 1 gives the weak (1, n1) estimate. The strong (p,r) estimate follows

    from Marcinkiewicz interpolation theorem.

    Remark 29. There are examples for the non-strong type at end point p = 1 and non-embedding

    phenomena at p = nn given in Stein [17, Section V.1]. For simplicity, we construct one-

    dimensional case only, but they can extend to higher-dimensional case easily.

    22

  • (i) Take f = (12, 12

    ) . Then f1 = 1 and we see (Tf)(x) (|x| + 1

    2

    ) 0 for all |x| 1.

    Hence

    =|x|1

    (2|x|)1 |x|1

    (|x|+ 1

    2

    ) 1

    (Tf)(x) 1 .(ii) Let f(x) = |x|1(log 1/|x|)(1)(1+) for |x| 1

    2and 0 for |x| > 1

    2, where 0 < 1. Then

    f L1

    1 (R). But

    (Tf)(0) =

    |x| 1

    2

    |x|1(log 1|x|

    )(1)(1+) dx =,

    as long as ( 1)(1 + ) 1. Therefore Tf 6 L.

    Acknowledgement

    The author would like to thank Joonyong Choi for his reminder that the proof for problem 17 has

    to consider En (not E) and to consider complex-valued g (not only real-valued). (2017.08.20)

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    24