Spaces of functions

11. Spaces of functions

11.1 Introduction Even if we are only interested in S ∗ f , when f ∈ Lp , we shall need some other spaces of functions. Since these spaces are not usually included in a first course of real analysis we give here a brief summary of their definitions and properties. The reader can find a more complete study in the references Hunt [22] and Bennett and Sharpley [3]. We will define the Lorentz spaces Lp,1 (µ) and Lp,∞ (µ). In Theorem 11.6, we prove that Lp,1 (µ) is the smallest rearrangement invariant space of measurable functions such that
χM
=
χM
p . We express these property by saying that it is an atomic space. Then we see Lp,∞ (µ) as its dual space, this dictates our selection of the norm
·
p,∞ . The proof that we presents of the duality Theorem 11.7 may appear unnecessarilly complicated, but it has the merit of getting absolute constants. We presents Marcinkiewicz’s Theorem with special attention to the constant that appear that will have a role in our theorems on Fourier series. We end the chapter studying a class of spaces near L1 (µ) that play a prominent role in the next chapter. We prove that they are atomic spaces, a fact that allows very neat proof in the following chapter.

11.2 Decreasing rearrangement The functions we are considering will be defined on an interval X of R and we will always consider the normalized Lebesgue measure on this interval, µ. Therefore µ is a probability measure µ(X) = 1. Given a measurable function f : X → R we consider its distribution function µf (y) = µ{|f | > y}. µf : [0, +∞) → [0, 1] is a decreasing and right-continuous function. Observe that if (fn ) is a sequence of measurable functions such that |fn | is an increasing sequence converging to |f |, then µfn is increasing and converges to µf .

J.A. de Reyna: LNM 1785, pp. 127–143, 2002. c Springer-Verlag Berlin Heidelberg 2002


128

11. Spaces of functions

Apart from µf we also consider the decreasing rearrangement of f that is defined as f ∗ (t) = m{y > 0 : µf (y) > t}.

If f is a positive simple function, there is a decreasing finite sequence of measurable sets (Aj )nj=1 and positive real numbers sj such that f =

n j=1 sj χAj . (If the non null values that f attains are a1 > a2 > · · · > an , and an+1 = 0 we can take

n Aj = {f > aj+1 } and sj = aj − aj+1 ). Then it is easy to see that f ∗ = j=1 sj χ[0,µ(Aj )) . Proposition 11.1 For every measurable function f : X → [0, +∞] and every measurable set A   µ(A) f dµ ≤ f ∗ (t) dt. 0

A

Proof. Since |f | ≤ |g| implies f ∗ ≤ g ∗ we only need to prove the case where f is a simple function. Then with the above representation we get   µ(A) n ∞   f dµ = sj µ(A ∩ Aj ); f ∗ (t) dt = sj inf{µ(Aj ), µ(A)}. A

0

j=1

j=1




The comparison between these quantities is trivial.

Theorem 11.2 (Hardy and Littlewood) For all measurable functions f and g: X → C we have    1   f ∗ (t)g ∗ (t) dt.  f g dµ ≤ 0

X

Proof. Since (|f |)∗ = f ∗ we can assume that f and g are positive. First assume that f is a simple function and consider its representation as above. We shall have    µ(Aj )  1 n n   f g dµ = sj g dµ ≤ sj g ∗ (t) dt = f ∗ (t)g ∗ (t) dt. j=1

Aj

j=1

0

0

For a general f let (fn ) be an increasing sequence of positive simple functions converging to f . We shall have   1  1 fn g dµ ≤ fn∗ g ∗ dm ≤ f ∗ g ∗ dm. 0

0

And now we can apply the monotone convergence theorem.




11.2 Decreasing rearrangement

129

Proposition 11.3 A measurable function f : X → C is equimeasurable with f ∗ , that is, for every y > 0 we have µ{|f | > y} = m{f ∗ > y}. Proof. Since f ∗ is decreasing m{f ∗ > y} = sup{t : f ∗ (t) > y}. By the same reasoning f ∗ (t) = m{µf (s) > t} = sup{s : µf (s) > t}. Hence we get m{f ∗ > y} = sup{t : there is s > y, with µf (s) > t}. Since µ{f > sn } → µ{f > y} for every decreasing sequence (sn ) converging to y, we get m{f ∗ > y} = sup{t : µf (y) > t} = µf (y).
Proposition 11.4 If ϕ is a nonnegative, Borel measurable function on [0, +∞), we have  1  ϕ(|f |) dµ = ϕ(f ∗ (t)) dt, X

0

for every measurable function f . Proof. Let ν1 be the Borel measure on [0, +∞) image of µ by the function |f |, and let ν1 the Borel measure on [0, +∞) image of m by the function f ∗ . The two functions |f | and f ∗ are equimeasurable m{t ∈ (0, 1) : f ∗ (t) > s} = µ{x ∈ X : |f (x)| > s}. Therefore, for s > 0, we have ν1 (s, +∞) = ν2 (s, +∞). Since ν1 and ν2 are probabilities we have also ν1 {0} = ν2 {0}. Therefore the two image measures are the same. Then since ϕ is Borel measurable and positive we have     1 ϕ(|f |) dµ = ϕ dν1 = ϕ dν2 = ϕ(f ∗ ) dm. X

[0,+∞)

[0,+∞)

0


For every measurable function f : X → C we define for t > 0 f ∗∗ (t) in the following way  1 t ∗ f ∗∗ (t) = f (s) ds. t 0 This function is also the supremum of the mean values of f . Proposition 11.5 For every measurable function # 1  $ f ∗∗ (t) = sup |f | dµ : µ(A) = t . µ(A) A

130

11. Spaces of functions

Proof. By Proposition 11.1, for every measurable set A with µ(A) = t   t |f | dµ ≤ f ∗ (s) ds A

0

Therefore the supremum of the mean values of |f | is less than or equal to f ∗∗ (t). To prove the equality, first assume that there is an y > 0 such that µ{|f | > y} = m{f ∗ > y} = t. Then we apply Theorem 11.4 with ϕ(t) = t χ(y,+∞) (t). If we denote by A the set {|f | > y} we get     t ∗ |f | dµ = ϕ(|f |) dµ = ϕ(f ) dm = f ∗ dm. 0

A

In the other case there is some point y0 such that µ{|f | > y} < t for y > y0 and µ{|f | > y} > t for y < y0 . Then there is a set of positive measure on X where |f | = y0 . Hence f ∗ takes this value on a set of positive measure. Then we can obtain a set A = A0 ∪ A1 where A0 = {|f | > y0 } and A1 ⊂ {|f | = y0 } and such that µ(A) = t. It is easy to see that in this case we also obtain the equality.
Therefore we have (f + g)∗∗ (t) ≤ f ∗∗ (t) + g ∗∗ (t).

11.3 The Lorentz spaces Lp,1 (µ) and Lp,∞ (µ) First we consider the Lorentz spaces Lp,1 (µ). It is the set of those measurable functions f : X → C such that   1 1 1 1/p ∗ dt
f
p,1 = t f (t) = f ∗ (tp ) dt < +∞. p 0 t 0 Every function f ∈ Lp,1 (µ) is in Lp (µ). In the case that p = 1 there is nothing to prove. For p > 1 let q be the conjugate exponent. Then    1  
f
p = sup  f g dµ ≤ f ∗ g ∗ dm.

g q ≤1

0

It is easy to see that for every such g, we have g ∗ (t) ≤ t−1/q . Therefore we get
f
p ≤ p
f
p,1 . It can be proved that the above inequality can be improved, removing the coefficient p. Now we show that
f
p,1 is a norm and Lp,1 (µ) is a Banach space. From the equality (λf )∗ = |λ|f ∗ , we get
λf
p,1 = |λ|
f
p,1 . To prove the triangle inequality we bring in f ∗∗ ,

11.3 The Lorentz spaces Lp,1 (µ) and Lp,∞ (µ)


f + g
p,1 = p−1 =p



1

t−1/q (f + g)∗ dt

0

−1 −1/q

t

131

 1 −1 −1 t(f + g)  + p q ∗∗ 

0

1

t(f + g)∗∗ t−1−1/q dt.

0

Since (f + g) ∈ Lp we know that limt→0+ t1/p (f + g)∗∗ (t) = 0. Hence  1
f + g
p,1 = p−1 (f + g)∗∗ (1) + p−1 q −1 (f + g)∗∗ t−1/q dt. 0

∗∗

We have seen that (f + g)

≤f

∗∗

∗∗

+ g . It follows that
·
p,1 is a norm.

To prove that the space is complete it suffices to prove that for every

∞ ∞ sequence of functions (fj ) with j=1
fj
p,1 < +∞, the series j=1 fj is absolutely convergent almost

∞ everywhere on X, and that the sum S of the series satisfies
S
p,1 ≤ j=1
fj
p,1 . To prove this assertion we can assume that the functions fj ≥ 0 for every j ∈ N. Then the partial sums Sj are increasing, therefore Sj∗ is increasing and converges to S ∗ . Hence  1  1 ∞  ∗ p−1 t−1/q S ∗ (t) dt = lim p−1 t−1/q SN (t) dt ≤
fj
p,1 . N →+∞

0

0

j=1

For a characteristic function χA a simple computation shows that
χA
p,1 is equal to
χA
p . The space Lp,1 (µ) (p > 1) can be defined as the smallest Banach space with this property. This is the content of the following theorem. Theorem 11.6 There is an absolute constant C such that for every f ∈ ∞ Lp,1 (µ) there exist a sequence of measurable sets (Aj )∞ j=1 and numbers (aj )j=1 such that f=

∞  j=1

aj χAj ,


f
p,1 ≤

∞ 

|aj |µ(Aj )1/p ≤ C
f
p,1 .

j=1

Proof. Given the measurable function f , by induction we can define a parp −pj tition of X into measurable sets (Aj )∞ j=1 such that µ(Aj ) = (e − 1)e and for every x ∈ Aj , y ∈ Ak with j < k we have |f (x)| ≤ |f (y)|. Define aj = supx∈Aj |f (x)|.

∞ We can easily check that f ∗ ≥ j=1 aj χIj , were Ij denotes the interval Ij = [e−p(j+1) , e−pj ). Therefore
f
p,1 ≥ (1 − e−1 )

∞  j=1

aj e−j .

132

11. Spaces of functions

By the construction of Aj we have a1 ≤ a2 ≤ · · ·. If we assume that f is not equivalent to 0, there exists some N such that a1 = · · · = aN −1 = 0 and aN > 0. Then we can write f=

∞ 

aj fj ,

fj = a−1 j f χAj ,

where


fj
∞ ≤ 1.

j=N

∞ Since every function with values in [0, 1) can be writen as j=1 2−j χTj , we get ∞ ∞   βj,k χTj,k , Tj,k ⊂ Aj , |βj,k | ≤ 4. fj = k=1

k=1

Therefore we obtain the expression f=

∞ 

aj βj,k χTj,k .

j,k=1

This is the decomposition we are seeking. In fact 

aj |βj,k |µ(Tj,k )1/p ≤ 4

∞ 

aj µ(Aj )1/p = 4(ep − 1)1/p

j=1

j,k

∞ 

aj e−j

j=1

4e ≤
f
p,1 . 1 − e−1
In particular we have proved the density of the simple functions in the space Lp,1 (µ). Remark. I stress the fact that C is an absolute constant in the previous theorem. It follows, for example that the inequality
f
p ≤ p
f
p,1 obtained above, can be improved now. In fact, under the conditions of the theorem,

we have
f
p ≤ |aj |
χAj
p ≤ C
f
p,1 . Now we consider the dual space. We define Lp,∞ (µ) as the set of measurable functions f such that sup0
0
0
For p > 1, these two quantities are equivalent. Indeed if we assume sup t1/p f ∗ (t) = C,

0
then

11.3 The Lorentz spaces Lp,1 (µ) and Lp,∞ (µ)

1 t

f ∗∗ (t) =

 0

t

f ∗ (s) ds ≤

133

Cp −1/p t p−1

(11.1).

It is easy to see that as we have defined it
·
p,∞ is a norm, and Lp,∞ (µ) a vector space. For p = 1, we can see that L1,∞ (µ) is a vector space. It is the weak L1 space. It can be shown that it is not a normed space. Therefore for p = 1 we put
f
1,∞ = sup t1/p f ∗ (t). 0
This is not a norm but a quasi-norm. Now it is straightforward to see that for p > 1 the space Lp,∞ (µ) is a older’s Banach space. Also it is clear that Lp (µ) ⊂ Lp,∞ (µ). In fact, by H¨ inequality,  t 1/p ∗∗ 1/p−1 t f (t) = t f ∗ (s) ds ≤ t1/p−1 t1/q
f ∗
p =
f
p . 0

Therefore
f
p,∞ ≤
f
p . Theorem 11.7 For p > 1, Lq,∞ (µ) is the dual space of Lp,1 (µ), where q is the conjugate exponent to p. Proof. Given g ∈ Lq,∞ (µ), for every f ∈ Lp,1 (µ) we put f =

∞ with j=1 |aj |
χAj
p ≤ C
f
p,1 . Then  ∞      |aj |  f g dµ ≤ X

Aj

j=1

≤

∞ 

|g| dµ ≤

∞ 

j=1

aj χAj ,

|aj |µ(Aj )g ∗∗ (µ(Aj ))

j=1

|aj |µ(Aj )1−1/q
g
q,∞ ≤

j=1

∞

∞ 

|aj |
χAj
p
g
q,∞

j=1

≤ C
f
p,1
g
q,∞ . This allows us to identify every function g ∈ Lq,∞ (µ) with a continuous linear functional defined in Lp,1 (µ). Let u be a continuos linear functional on Lp,1 (µ). Put ν(A) = u(χA ), for every measurable set A. Then ν is an additive set function. Since |ν(A)| ≤
u

χA
p,1 =
u
µ(A)1/p , the function ν is a signed measure and ν  µ. By the Radon-Nikodym Theorem there is a measurable function g, such that  ν(A) = A g dµ, for every measurable set A. This function is in Lq,∞ (µ). In fact we have      g dµ ≤
u
µ(A)1/p . A ∗∗

From this we derive that g (t) ≤
u
t−1/q . Therefore
g
q,∞ ≤
u
.

134

11. Spaces of functions

 It follows that for every simple function ϕ we have  u(ϕ) = ϕg dµ. By the density of the simple functions and the inequality  f g dµ ≤ C
f
p,1
g
q,∞ ,  it follows that u(f ) = f g dµ for every f ∈ Lp,1 (µ). Therefore Lq,∞ (µ) is the dual space. Also, if we denote by
·
•p,∞ the dual norm in the space Lq,∞ (µ),     •
g
q,∞ = sup  f g dµ.

f p,1 ≤1

X

Then we have proved that for some absolute constant C we have
g
q,∞ ≤
g
•q,∞ ≤ C
g
q,∞ .


11.4 Marcinkiewicz Interpolation Theorem To prove this interpolation theorem we shall need the following inequality. Theorem 11.8 (Hardy’s inequality) For every positive real function f defined on (0, +∞), and every 1 ≤ p1 < p < p0 < +∞ we have   t p  p1 1  p1  ∞ 1 1 −1  ∞ − pp p1 −1 1 t f (s)s ds dt ≤ − f (s)p ds , p1 p 0 0 0  +∞  1 ∞     ∞     p1 p p 1 1 1 −1 p t− p0 f (s)s p0 −1 ds dt ≤ f (s)p ds . − p p0 0 t 0 Proof. Both inequalities are proved in the same way. For example, to prove the first one, we apply H¨ older’s inequality to the inner integral 

t

− p1


f (s)s 0

s

1 p1

1 −p

ds ≤



t p

f (s) s 0

1 p1

1 −p

ds

 p1  t p11 − p1 1− p1 1 p1

−

1 p

.

Then, if we denote by I the first term of the inequality, we get   1  1 1−p ∞  t Ip ≤ − f (s)p s1/p1 −1/p ds t1/p−1/p1 −1 dt p1 p 0 0 After applying Fubini’s Theorem we get   +∞ 1  1 1−p +∞ Ip ≤ − f (s)p s1/p1 −1/p t1/p−1/p1 −1 dt ds p1 p s 0 1 1 −p ∞ p ≤ − f (s) ds. p1 p 0

11.4 Marcinkiewicz Interpolation Theorem

135


Now we are ready to prove Marcinkiewicz’s Theorem. We say that an operator S mapping a vector space of measurable functions to measurable functions is sublinear if |S(f + g)| ≤ |Sf | + |Sg|, and for every scalar a, we have |S(af )| = |a||Sf |. Theorem 11.9 (Marcinkiewicz) Let S be a sublinear operator defined on Lp0 ,1 (µ) + Lp1 ,1 (µ) where +∞ > p0 > p1 > 1. Assume that there exist constants M0 and M1 such that
Sf
p0 ,∞ ≤ M0
f
p0 ,1 ,

and


Sf
p1 ,∞ ≤ M1
f
p1 ,1 .

Then, for every p ∈ (p1 , p0 ), S: Lp (µ) → Lp (µ) is continuous with norm
S
p ≤

p(p0 − p1 ) M 1−θ M1θ , (p0 − p)(p − p1 ) 0

where

1 θ 1−θ + . = p p0 p1

Proof. Let f be a function in Lp (µ). First we bound (Sf )∗∗ (t) using the hypotheses that S maps Lq,1 (µ) → Lq,∞ (µ) for q = p0 and q = p1 . We decompose f into two functions, f = f0 + f1 , in the following way  f (s) if |f (s)| ≤ f ∗ (at), f0 (s) = ∗ f (at) sgn(f (s)) if |f (s)| ≥ f ∗ (at);  0 if |f (s)| ≤ f ∗ (at), f1 (s) = ∗ f (s) − f (at) sgn(f (s)) if |f (s)| ≥ f ∗ (at). Here a is a parameter that we will choose later. It is easy to see that the decreasing rearrangements of these functions are given by  ∗  0 if s ≥ at f (s) if s ≥ at ∗ ∗ f1 (s) = f0 (s) = f ∗ (s) − f ∗ (at) if s ≤ at f ∗ (at) if s ≤ at Since f0 ∈ Lp0 ,1 and f1 ∈ Lp1 ,1 we see that S is defined on Lp . Since we have (Sf )∗∗ (t) ≤ (Sf0 )∗∗ (t) + (Sf1 )∗∗ (t) ≤   M0 −1/p0 1 ∗ M1 −1/p1 1 ∗ 1/p0 −1 t f0 (s)s ds + t f1 (s)s1/p1 −1 ds, p0 p1 0 0 (Sf )∗∗ (t) is bounded by  M0 −1/p0 1 ∗ (M0 a1/p0 − M1 a1/p1 )f ∗ (at) + t f (s)s1/p0 −1 ds p0 at  M1 −1/p1 at ∗ t f (s)s1/p1 −1 ds. + p1 0

136

11. Spaces of functions

We consider f ∗ (s), defined for s > 1, equal to 0. By Proposition 11.4,
Sf
p =
Sf ∗
p , and since Sf ∗ ≤ Sf ∗∗ , we obtain
Sf
p ≤
(Sf )∗∗
p ≤ (M0 a1/p0 − M1 a1/p1 )a−1/p
f
p   p 1/p M0  ∞ −p/p0  1 ∗ + t f (s)s1/p0 −1 ds dt + p0 0 at   p 1/p M1  ∞ −p/p1  at ∗ t f (s)s1/p1 −1 ds dt . p1 0 0 With a change of variables we get
Sf
p ≤ (M0 a1/p0 − M1 a1/p1 )a−1/p
f
p   p 1/p M0 1/p0 −1/p  ∞ −p/p0  +∞ ∗ + a t f (s)s1/p0 −1 ds dt p0 0 t   p 1/p M1 1/p1 −1/p  ∞ −p/p1  t ∗ + a t f (s)s1/p1 −1 ds dt . p1 0 0 Now we apply Hardy’s inequality. It follows that
Sf
p ≤ (M0 a1/p0 − M1 a1/p1 )a−1/p
f
p  pM0 1/p0 −1/p  ∞ ∗ p 1/p a + f (s) ds p0 − p 0  pM1 1/p1 −1/p  ∞ ∗ p 1/p + a f (s) ds . p − p1 0 Therefore
Sf
p ≤

p M p1 M1 1/p1 −1/p  0 0 1/p0 −1/p a
f
p . + a p0 − p p − p1

Now we select the best value for a. In this case M0 = M1 a1/p0 −1/p1 is the best choice. With this election we obtain
Sf
p ≤

p(p0 − p1 ) M 1−θ M1θ , (p0 − p)(p − p1 ) 0

where

1−θ 1 θ = + . p p0 p1


Remark. In fact we have proved that  ∞ 1/p
Sf
p,p = f ∗∗ (t)p dt 0

is bounded. This is an equivalent norm, but Hardy’s inequality gives
f
p ≤
f
p,p ≤

p
f
p . p−1

11.5 Spaces near L1 (µ)

137

In the above theorem we assume that
Sf
p,∞ ≤ M
f
p,1 . The norm
Sf
p,∞ refers to (Sf )∗∗ and it is usually easier to bound (Sf )∗ . Another problem with the theorem as we have given it, is that we have excluded the case p1 = 1, because in this case
Sf
1,∞ is not a norm, and is defined in another way. Therefore we give another version. Theorem 11.10 (Marcinkiewicz) Let S be a sublinear operator defined on Lp0 ,1 (µ) ∪ Lp1 ,1 (µ) where +∞ > p0 > p1 ≥ 1. Assume that there exist constants M0 and M1 such that sup t1/p0 (Sf )∗ (t) ≤ M0
f
p0 ,1 , and sup t1/p1 (Sf )∗ (t) ≤ M1
f
p1 ,1 .

0
0
Then, for every p ∈ (p1 , p0 ), S: L (µ) → Lp (µ) is continuous with norm p


S
p ≤ 21/p

p(p0 − p1 ) M 1−θ M1θ , (p0 − p)(p − p1 ) 0

where

1−θ 1 θ = + . p p0 p1

Proof. We follow the same procedure as in the previous theorem. Instead of a bound for (Sf )∗∗ (t) we obtain a bound for (Sf )∗ (2t). It is easy to see that (Sf )∗ (2t) ≤ (Sf0 )∗ (t) + (Sf1 )∗ (t). Observe that
Sf
p =
(Sf )∗
p , and  ∞ 1/p 2−1/p
(Sf )∗
p = (Sf )∗ (2t)p dt ≤
(Sf0 )∗ + (Sf1 )∗
p . 0

Then we follow the same reasoning as before.




11.5 Spaces near L1 (µ)  We can define many spaces between L1 (µ) and p>1 Lp (µ). They will play a role in the problem of the almost everywhere convergence of Fourier series, since every f in the last space has an a. e. convergent Fourier series, and by Kolmogorov’s example there exists a function in L1 whose Fourier series is everywhere divergent. We shall define a space that we call Lϕ(L), where ϕ: [0, +∞) → [0, +∞) will be a function such that: (1) There exists a constant C > 0 such that for every t > 0, ϕ(t2 ) ≤ Cϕ(t). (2) ϕ(t) is absolutely continuous and ϕ (t) ≥ 0 a. e. (3) ϕ(0) = 0. (4) limt→+∞ ϕ(t) = +∞.

138

11. Spaces of functions

The space Lϕ(L) will be the set of measurable functions such that  +∞ |f |ϕ(|f |) dµ < +∞. 0

Proposition 11.11 Assuming that ϕ satifies the above conditions, Lϕ(L) is a Banach space whose norm is given by  1  +∞ ∗∗ f (t)   1  ∗
f
Lϕ(L) = dt. f (t)ϕ(1/t) dt = ϕ t t 0 0 Proof. The equality of the two expressions given for the norm is a consequence of Fubini’s Theorem applied to the function 1  1  ∗ f (s) χ{0<s 1). Then since ϕ ≥ 0, the second expression and the known properties of f ∗∗ prove that
f
Lϕ(L) is a norm. Now we can prove that the norm is finite precisely in the set Lϕ(L). In fact, for a given measurable f , we define A = {f ∗ (t)2 > 1/t} and we obtain, by property (1),  1   1

1 √ ϕ(1/t) dt f ∗ (t)ϕ(1/t) dt = C f ∗ (t)ϕ f ∗ (t) dt + t 0 0 A   1 ϕ(x−2 ) dx. C |f |ϕ(|f |) dµ + 2 0

The last integral is finite. In fact, it is comparable to ∞   eα log n 1 ϕ(22n ) ≤ < +∞. n 2 2n j=1

(We have used repeatedly the condition (1)). Therefore the norm is finite for every function in Lϕ(L). On the other hand, if
f
Lϕ(L) , then
f
1 < +∞. Hence tf ∗ (t) ≤
f
1 . Therefore   1  1 |f |ϕ(|f |) dµ = f ∗ (t)ϕ(f ∗ (t)) dt ≤ f ∗ (t)ϕ(
f
1 /t) dt. 0

X

0

Now if
f
1 ≤ 1 we will have   |f |ϕ(|f |) dµ ≤ X

0

1

f ∗ (t)ϕ(1/t) dt =
f
Lϕ(L) .

11.5 Spaces near L1 (µ)

And if
f
1 > 1 we will have   |f |ϕ(|f |) dµ ≤
f
1

1

0

X

 ≤
f
1

1

0

139

f ∗ (
f
1 t)ϕ(1/t) dt f ∗ (t)ϕ(1/t) dt ≤
f
1
f
Lϕ(L) < +∞.

p The proof that Lϕ(L) is a Banach space can

be given as in the L (µ) case. Given a sequence

of functions (fn ) such that n
fn
Lϕ(L) < +∞, we prove that the series n fn is absolutely convergent a. .e. and defines a measurable function F . Then it is easy to prove that F = n fn in the space Lϕ(L).


An important information about the space Lϕ(L) is the value of the norm of a characteristic function. We have 1 µ(M )ϕ(2/µ(M )) ≤
χM
Lϕ(L) ≤ Cϕ µ(M )ϕ(2/µ(M )). 2 This follows from the following Lemma 11.12 There exists a constant Cϕ , such that  x x ϕ(2/x) ≤ ϕ(1/t) dt ≤ Cϕ x ϕ(2/x), 0 < x < 1. 2 0 Proof. Since ϕ(t2 ) ≤ Cϕ(t), we have x ϕ(2/x) ≤ 2

 0



x

ϕ(1/t) dt ≤



x

C(x − x )ϕ(1/x) +

0

0



x

ϕ(1/t) dt ≤

ϕ(1/t) dt + x2

2

On the other hand we have  x2  ϕ(1/t) dt = 2

x2

x2

ϕ(1/t) dt. 0

2

ϕ(1/u )u du ≤ 2C 0  x ≤ 2Cx ϕ(1/u) du.



x

ϕ(1/u)u du 0

0

It follows that there exists some x0 such that  x2  1 x ϕ(1/t) dt ≤ ϕ(1/t) dt, 2 0 0 Therefore in [0, x0 ] we have

0 < x < x0 .

140

11. Spaces of functions

x ϕ(2/x) ≤ 2



x

ϕ(1/t) dt ≤ 2C x ϕ(1/x).

0

x In the interval [x0 , 1] the functions xϕ(2/x) and 0 ϕ(1/t) dt are continuous and non null. Therefore our lemma is true also in this interval. Observe that the conditions ϕ(t2 ) ≤ Cϕ(t) and limt→+∞ ϕ(t) = +∞ implies that ϕ(2) = 0 is impossible.
A property that is important for us is that these spaces are atomic. Theorem 11.13 There exists a constant Cϕ such that for every function f ∈ Lϕ(L) there exist a sequence of measurable sets (Aj )∞ j=1 and a sequence

∞ such that f = a χ , and of complex numbers (aj )∞ j Aj j=1 j=1
f
Lϕ(L) ≤

∞ 

|aj |
χAj
Lϕ(L) ≤ Cϕ
f
Lϕ(L) .

j=1

Proof. Given the measurable function f , as in Theorem 11.6, we can put f =

∞ j where fj has support on a measurable set Aj , aj ≥ 0,
fj
∞ ≤ 1 j=N aj f ∞ ∗ and f ≥ j=1 aj χIj , with Ij = [e−j−1 , e−j ) and µ(Aj ) = (e − 1)e−j = mj . As in Theorem 11.6 we obtain the decomposition  f= aj βj,k χTj,k . j,k

Then we have 

|aj βjk |
χTj,k
Lϕ(L) ≤ 4

∞ 

aj mj ϕ(2/mj ) ≤ 4C(e − 1)

j=1

≤C



 aj

j=1

j,k

≤ 4C

∞ 

∞ 

∞ 

µ(Aj )

ϕ(1/t) dt 0

aj e−j ϕ 2(e − 1)−1 ej

j=1 −j

aj e

−1

(1 − e

j=1

j+1

)ϕ(e

)≤C



∞  j=1

≤C



 0

1

 aj

ϕ(1/t) dt Ij

f ∗ (t)ϕ(1/t) dt = C 
f
Lϕ(L) .


11.6 The spaces L log L(µ) and Lexp (µ)

141

11.6 The spaces L log L(µ) and Lexp (µ) The space L log L(µ) is the space of measurable functions f such that  |f | log+ |f | dµ < +∞. X By the previous section we know that  1
f
L log L = f ∗∗ (t) dt, 0

is a norm in this space. That the above expression is a norm in the vector space where it is finite is a trivial consequence of the inequality (f + g)∗∗ ≤ f ∗∗ + g ∗∗ . By Fubini’s Theorem we obtain that  1
f
L log L = f ∗ (t) log(1/t) dt. 0

∗

Since f ≤ f

∗∗

we have
f
1 ≤
f
L log L .

We shall need the expression of the norm of a characteristic function


χA
L log L = µ(A) log e/µ(A) . By Theorem 11.13 we get the following. Theorem 11.14 There is an absolute constant C such that for every function f ∈ L log L there exists a sequence of measurable sets (Aj )∞ j=1 and a sequence of complex numbers (aj ) such that f=

∞ 

aj χAj ,

and


f
L log L ≤

j=1

∞ 

|aj |
χA
L log L ≤ C
f
L log L .

j=1

The dual space of L log L can be characterized in several ways. We shall call it Lexp . It is the space of measurable functions such that µ{|f | > y} ≤ Ae−By for some constants A and B (that change with f ). It is also the space  α|f |of functions such that there is some positive real number α such that e dµ < +∞. The equivalence of these two conditions is easy to prove. But to obtain a norm that defines the space we define the space as the set of measurable functions such that

−1 ∗∗ f (t) < +∞.
f
Lexp = sup log(e/t) 0
142

11. Spaces of functions

Theorem 11.15 The dual space of L log L is Lexp . There is some constant C such that
g
Lexp ≤
g
•Lexp ≤ C
g
Lexp , Where
·
•Lexp denotes the norm on Lexp as a dual of L log L. Proof. First, assume that f ∈ L log L and g ∈ Lexp , then f g is integrable and     (11.2)  f g dµ ≤ C
f
L log L
g
Lexp . X

To see it, write f =

∞

j=1

aj χAj as in Theorem 11.14. Then

 ∞      |aj |  f g dµ ≤ X

Aj

j=1

≤ ≤

∞  j=1 ∞ 

|g| dµ ≤

∞ 

|aj |µ(Aj )g ∗∗ (µ(Aj ))

j=1

|aj |µ(Aj )
g
Lexp log(e/µ(Aj )) |aj |
χAj
L log L
g
Lexp ≤ C
f
L log L
g
Lexp .

j=1

Assume that u is a continuous linear functional on L log L. Then we define ν(A) = u(χA ). This is an additive function defined on measurable sets. Now |ν(A)| ≤
u
µ(A) log(e/µ(A)). From this inequality it follows that ν is a signed countably additive measure and that ν  µ. Then the RadonNikodym Theorem gives us a measurable function g such that for every measurable set A, u(A) = A g dµ. This function g is in Lexp (µ). Indeed, we know      g dµ ≤
u
µ(A) log(e/µ(A)). A

Then, by Proposition 11.5, we obtain g ∗∗ (t) ≤
u
log(e/t). That is, g ∈ Lexp and
g
Lexp ≤
u
.  Now for every simple function we have u(ϕ) = ϕg dµ. By the density of simple functions in the space L log L (this follows from Theorem 11.14) and (11.2), we derive that  u(f ) = f g dµ, ∀f ∈ L log L. X

Then we have proved that Lexp is the dual space to L log L. Besides, we have proved the two inequalities between the norms.
Finally note the following examples of spaces of functions Take ϕ(t) = (log+ t)2 . We obtain the space L(log L)2 , that is, the set of measurable functions f such that

11.6 The spaces L log L(µ) and Lexp (µ)



143

|f |(log+ |f |)2 dµ < +∞.

X

The norm in this case is given by  1
f
L(log L)2 = f ∗ (t)(log 1/t)2 dt < +∞. 0

The norm of a characteristic function χA with µ(A) = m is equal to m(log e/m)2 + m. By the general theory this is also an atomic space Theorem 11.16 There is an absolute constant C such that for every function f ∈ L(log L)2 there exist a sequence of measurable sets (Aj )∞ j=1 and a

∞ sequence of complex numbers (aj ) such that f = j=1 aj χAj and
f
L(log L)2 ≤

∞ 

|aj |
χA
L(log L)2 ≤ C
f
L(log L)2 .

j=1

Two more spaces have a role in the theory of pointwise convergence of Fourier series. They are L log L log log L and L log L log log log L. They are defined as the set of measurable functions such that, respectively, the integrals  |f |(log+ |f |)(log+ log+ |f |) dµ X

and



|f |(log+ |f |)(log+ log+ log+ |f |) dµ,

X

are finite. Define L(t) = L1 (t) = log(1 + t) and for every integer n > 2 put Ln (t) = L(Ln−1 (t)). These are well defined functions Ln : [0, +∞) → [0, +∞), and it is easily verified, by induction, that Ln (t2 ) ≤ 2Ln (t). Then if we put ϕ(t) = L1 (t)L3 (t) we will have ϕ(t2 ) ≤ 4ϕ(t). All the other conditions (2), (3) and (4) are easily verified. It is clear that with this choice of ϕ we obtain the space L log L log log log L. The other one is analogous.