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Real Analysis, 2nd Edition, G.B.Folland
Chapter 6 Lp Spaces
YungHsiang 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 complexvalued 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 = limnf(xn)+ lim
ng(xn) = a+ b = a+ b = lim
nf(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 closedopen
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 SteinShakarchi [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 2pfnp + 2pf 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 (BrezisLieb 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 BolzanoWeierstrauss. 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 6.1011]. 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 fmp d < .
Therefore, for all large enough n,mX
fn fmp d =Ecfn fmp d+
Anm
fn fmp d+E\Anm
fn fmp 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.252253] 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 g2 = 
nk=1
ak
Ek
g+ ibkEk
g2 =( 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 BanachAlaoglus
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 thatEgq < 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 BrezisLieb lemma, which is proved in additional exercise
6.1.5.
Note that BanachAlaoglu theorem implies the following similar and important result: (for Lp
case, we can prove it by Cantor diagonal process without BanachAlaoglu, see my solution file
for SteinShakarchi [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 indexedset 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, 1.12.131132] 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 cocountable. 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 2325 may also be confirmed with Cohn [5,
Chapter 3].
6.3 Some Useful Inequalities
26. The case p = 1 is a consequence of FubiniTonellis 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+ ydyp dx
)1/p=(
0

0
f(xz)
1 + zdzp 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 4445.
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 FubiniTonelli, 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. (HilbertSchmidt integral operator.)
15

Proof. By TonelliFubinis 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
sf (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
y0tf (t)2 dt < 2, then for any
x (0, y),
0 f(x) log x1/2
=f(y)
yxf (t)dt
 log x1/2 f(y) log x1/2
+ log y log x1/2 log x1/2
.
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
y0tf (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)
+ Cpy
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
y0tf (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.1415] 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 pth 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 LiebLoss [11, Chapter 34] and Dibenedetto [6, Chapte 9] for further dis
cussions and applications on higherdimensional 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 RieszThorin 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 RieszThorin again.
The uniqueness part is due to the fact Lp Lq is dense in Lr which is proved by considering
cutoff 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 HardyLittlewood 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 yf(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 nonstrong type at end point p = 1 and nonembedding
phenomena at p = nn given in Stein [17, Section V.1]. For simplicity, we construct one
dimensional case only, but they can extend to higherdimensional 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
=x1
(2x)1 x1
(x+ 1
2
) 1
(Tf)(x) 1 .(ii) Let f(x) = x1(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
x1(log 1x
)(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 complexvalued g (not only realvalued). (2017.08.20)
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