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Dichotomies for Lp spaces
J. Math. Anal. Appl. 368 (2010) 382–390
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Dichotomies for L p spaces Szymon Głab ¸ a,∗ , Filip Strobin a,b a b
Institute of Mathematics, Technical University of Łód´z, Wólcza´ nska 215, 93-005 Łód´z, Poland Institute of Mathematics, Polish Academy of Sciences, S´niadeckich 8, 00-956 Warszawa, Poland
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 28 July 2009 Available online 10 February 2010 Submitted by T.D. Benavides
Assume that ( X , Σ, μ) is a measure space and p 1 , . . . , pn , r > 0. We prove that {( f 1 , . . . , f n ) ∈ L p1 × · · · × L pn : f 1 · · · f n ∈ L r } is either L p1 × · · · × L pn or a σ -porous subset of L p 1 × · · · × L pn . This dichotomy depends on properties of μ and the sign of the number 1 − p11 − · · · − p1n . r © 2010 Elsevier Inc. All rights reserved.
Keywords: Measure space L p space Porous sets Porosity
1. Introduction Among linear topological spaces there are spaces X consisting of sequences or functions such that a natural multiplication is defined on pairs (x1 , x2 ) ∈ X 2 , however, its result need not necessarily belong to X . It is an interesting question about the size of the set of such “bad” pairs in a various sense. Such a kind of studies was initiated in [1,4]. Balcerzak and Wachowicz proved in [1] that {( f , g ) ∈ L 1 [0, 1] × L 1 [0, 1]: f · g ∈ L 1 [0, 1]} is a meager subset of L 1 [0, 1] × L 1 [0, 1]. They also proved that
(x, y ) ∈ c 0 × c 0 :
n
∞ x(i ) y (i )
i =1
is bounded
n =1
is a meager subset of c 0 × c 0 . These meagerness results were generalized by Jachymski in the following extension of the classical Banach–Steinhaus theorem. Recall that a function ϕ : X → R+ is L-subadditive for some L 1, if ϕ (x + y ) L (ϕ (x) + ϕ ( y )) for any x, y ∈ X . Theorem 1. (See Jachymski [4].) Given k ∈ N, let X 1 , . . . , X k be Banach spaces, X = X 1 if k = 1, and X = X 1 × · · · × X k if k > 1. Assume that L 1, F n : X → R+ (n ∈ N) are lower semicontinuous and such that all functions xi → F n (x1 , . . . , xk ) (i = 1, . . . , k) are L-subadditive and even. Let E = {x ∈ X: ( F n (x))n∞=1 is bounded}. Then the following statements are equivalent: (i) E is meager; (ii) E = X; (iii) sup{ F n (x): n ∈ N, x 1} = ∞.
*
Corresponding author. E-mail addresses: [email protected] (S. Głab), ¸ [email protected] (F. Strobin).
0022-247X/$ – see front matter doi:10.1016/j.jmaa.2010.02.011
©
2010 Elsevier Inc. All rights reserved.
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
At first, we were interested in a further generalization of this theorem changing meagerness by that this is not possible. To see it, consider the following set:
E=
x ∈ R:
n |sin(k!π x)| k =1
k
∞
383
σ -porosity. It turns out
is bounded .
n =1
n
Using Theorem 1 for F n (x) = k=1 |sin(k!π x)|/k (clearly, each F n is subadditive) we obtain that this set is meager (E = R since it is of measure zero) and is not σ -upper porous [5, p. 341]. Hence we could not generalize Jachymski’s theorem in this manner. Assume that ( X , Σ, μ) is a measure space. In our paper we answer the question about a size of the set (in the following we will write L p instead of L p ( X , Σ, μ)):
( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn : f 1 · · · f n ∈ L r .
We do not restrict our attention only to Banach L p spaces for p ∈ [1, ∞], but we consider all linear metric L p spaces for p ∈ (0, ∞]. It appears that this set is either L p 1 × · · · × L pn or a σ –c-lower porous (for some c > 0) subset of L p 1 × · · · × L pn . So, it is either the whole space or a very small set. We determine this dichotomy for every type of a measure space 1 ( X , Σ, μ). Surprisingly it depends on the following parameters (in the sequel the symbol ∞ means 0):
• the sign of the number 1r − p11 − · · · − p1n ; • inf{μ( A ): μ( A ) > 0} (it is important whether it is equal or greater than zero); • sup{μ( A ): μ( A ) < ∞} (it is important whether it is finite or infinite). The dichotomy is stated in Proposition 2 and Theorems 9, 10. Let X be a metric space. B (x, R ) stands for the ball with a radius R centered at a point x. Let c ∈ (0, 1]. We say that M ⊂ X is c-lower porous [6], if
∀x ∈ M ,
γ (x, M , R )
lim inf R →0+
R
c
, 2
where
γ (x, M , R ) = sup r 0: ∃z ∈ X , B (z, r ) ⊂ B (x, R )\ M . Clearly, M is c-lower porous iff
∀x ∈ M ,
∀β ∈ (0, c /2),
∃ R 0 > 0,
∀ R ∈ (0, R 0 ),
∃z ∈ X ,
B ( z, β R ) ⊂ B (x, R )\ M .
The set is σ –c-lower porous if it is a countable union of c-lower porous sets. Note that a σ –c-lower porous set is meager, and the notion of σ -porosity is essentially stronger than that of meagerness. Note that the sets investigated in this paper will be c-porous in some stronger sense, namely,
∀x ∈ X ,
∀β ∈ (0, c /2),
∀ R > 0,
∃z ∈ X ,
B ( z, β R ) ⊂ B (x, R )\ M .
However, we do not want to define any new notion of porosity, so in the formulations of theorems we will deal only with c-lower porosity. 2. Algebraic product of functions from L p 1 × ··· × L pn Throughout the paper, ( X , Σ, μ) is a measure space. If p ∈ (0, 1), then we consider L p as a metric linear space with the metric
| f − g | p dμ.
d( f , g ) = X
Additionally we put
| f | p dμ.
f p = d ( f , 0) = X
If p ∈ [1, ∞), then we consider L p as a normed linear space with the norm
f p = X
1/ p | f | p dμ .
384
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
Finally, if p = ∞, then we consider L p as a normed linear space with the norm f ∞ = supess | f |. Note that in all cases L p is a complete space. For every n ∈ N and any p 1 , . . . , pn , r ∈ (0, ∞], we define the set (we allow n to be 1): ( p 1 ,..., pn )
Er
= ( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn : f 1 · · · f n ∈ L r .
In this paper we consider L p 1 × · · · × L pn as a space with the metric defined as the maximum of distances on all coordinates in L p 1 , . . . , L pn . Using the general Hölder inequality [3, p. 10] we obtain that: Proposition 2. Let p 1 , . . . , pn , r ∈ (0, ∞] be such that
1 r
1
=
p1
( p 1 ,..., pn )
Then E r
+ ··· +
1 pn
.
= L p 1 × · · · × L pn .
Now we will give some helpful lemmas. Lemma 3. Let h 0, h ∈ L 1 , ε > 0. Then
(i) if inf{μ( A ): A ∈ Σ, μ( A ) > 0} = 0, there is A ∈ Σ with 0 < μ( A ) ε and A h dμ ε ; (ii) if sup{μ( A ): A ∈ Σ, μ( A ) < ∞} = ∞, there is A ∈ Σ with 1/ε μ( A ) < ∞ and A h dμ ε .
Proof. (i) Follows immediately from the absolute continuity of the function B → B h dμ (B ∈ Σ ) with respect to μ. ∈ N, An be such that n < μ( An ) < ∞. Set F n = nk=1 A k . Then ( F n ) is increasing, μ( F n ) < ∞ and (ii) Let, for any n μ( F n ) → ∞. Put F = n∞=1 F n . We have
h dμ =
lim
n→∞ Fn
h dμ < ∞. F
Then there is n0 ∈ N with
h dμ − ε .
h dμ > F n0
Hence
F
h dμ < ε .
F \ F n0
On the other hand, limn→∞ μ( F n \ F n0 ) = ∞, so there is N ∈ N such that
μ( F N \ F n0 ) > 1/ε . Put A = F N \ F n0 . 2
Lemma 4. Let p 1 , . . . , pn , r ∈ (0, ∞), ( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn and let A be a measurable subset of X . Suppose that for some numbers a1 , . . . , an and for each i = 1, . . . , n, the following holds
| f i − 1| p i d μ a i . A
Then for any numbers c 1 , . . . , cn ∈ (0, 1), we have
r
| f 1 · · · f n | dμ
cr1 · · · cnr
A
μ( A ) −
a1
(1 − c 1 ) p 1
− ··· −
Proof. Observe that the above assumptions imply that any i, we have
A
Hence
| f i − 1| p i d μ
ai
(1 − cn ) pn
.
μ( A ) < ∞. Let A i = {x ∈ A: f i (x) < c i } for i = 1, . . . , n. Then for
| f i − 1| p i d μ Ai
an
|1 − c i | p i dμ = (1 − c i ) p i μ( A i ). Ai
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
| f 1 · · · f n |r dμ
| f 1 · · · f n |r dμ n
A\
A
385
i =1
n
Ai
cr1 · · · cnr μ( A ) − μ cr1 · · · cnr μ( A ) −
A\ n
i =1
c r1 · · · cnr dμ Ai
Ai
i =1
a1
(1 − c 1 ) p 1
− ··· −
an
(1 − cn ) pn
.
2
Lemma 5. Let A, A 1 , . . . , A n be measurable with A i ⊂ A and μ( A i ) > (1 − n1 )μ( A ) for any i = 1, . . . , n. Then
μ
n
Ai
> 0.
i =1
Proof. Using the induction principle, it is easy to show that
μ
k
> (1 − k/n)μ( A ) for any k = 1, . . . , n.
Ai
i =1
In particular, for k = n, we get that
n
μ(
i =1
A i ) > 0.
2 ( p ,..., p )
n The next theorem is a main result of the paper. It is rather technical, but it shows when E r 1 can be σ -porous and how good are porosity estimations in each of the considered cases. For any n ∈ N and any p 1 , . . . , pn , put c ( p 1 , . . . , pn ) = 2/(1 + m) if there is at least one finite p i , where m is the number of finite p i ’s, and put c ( p 1 , . . . , pn ) = 1 if p i = ∞ for every i = 1, . . . , n.
Theorem 6. Let n ∈ N and let p 1 , . . . , pn , r ∈ (0, ∞]. Assume that one of the following conditions holds: (i) (ii)
1 p1 1 p1
+ ··· + + ··· +
1 pn 1 pn
> <
1 r 1 r
and inf{μ( A ):
μ( A ) > 0} = 0; and sup{μ( A ): μ( A ) < ∞} = ∞.
Then for any u > 0, the set
E u = ( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn : f 1 · · · f n r u
( p 1 ,..., pn )
is c-lower porous, where c = c ( p 1 , . . . , pn ). In particular, the set E r
is σ –c-lower porous.
Proof. We will consider two cases. Case 1. p 1 = · · · = pn = ∞. Then our assumptions imply that r < ∞ and sup{μ( A ): μ( A ) < ∞} = ∞. Let ( f 1 , . . . , f n ) ∈ L ∞ × · · · × L ∞ , R > 0, α ∈ (0, 12 ) (note that in this case c ( p 1 , . . . , pn ) = 1). Fix a measurable set A of finite measure such that
μ( A ) >
ur
(( 12
− α ) R )rn
.
For any i = 1, . . . , n, we define
˜f i (x) =
f i (x) + f i (x) −
1 R, 2 1 R, 2
f i (x) 0; f i (x) < 0.
Clearly, for any i = 1, . . . , n, ˜f i − f i ∞ = R /2 and B (( ˜f 1 , . . . , ˜f n ), α R ) ⊂ B (( f 1 , . . . , f n ), R ). Now if (h1 , . . . , hn ) ∈ B (( ˜f 1 , . . . , ˜f n ), α R ), then for any i = 1, . . . , n and for μ-almost every x ∈ A, we have
h i (x) Hence
1 2
− α R.
386
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
|h1 · · · hn |r
1 2
rn −α R · μ( A ) > ur ,
A
and
h1 · · · hn r > u . This ends the proof in Case 1. Case 2. For some i = 1, . . . , n, p i < ∞. Without loss of generality, we assume that p i ∈ (0, 1) for i = 1, . . . , m, 1 p i < ∞ for i = m + 1, . . . , m + k and p i = ∞ for i = m + k + 1, . . . , m + k + j, where j is such that m + k + j = n (clearly, m, k or j can be equal to zero, but m + k = 0). Additionally define q i = pm+i for i = 1, . . . , k. Then the product space L p 1 × · · · × L pn can be written in the following way:
L p 1 × · · · × L p m × L q 1 × · · · × L qk × L ∞ × · · · × L ∞ . Let ( f 1 , . . . , f m , g 1 , . . . , gk , l1 , . . . , l j ) be a member of that space, and let R > 0, δ ∈ (0, m+1k+1 ) (note that in this case c ( p 1 , . . . , pn ) = 2/(m + k + 1)). Then, clearly, 1 − δ > (m + k)δ and hence we can take η ∈ ((m + k)δ, 1 − δ). Since δ/η < 1/(m + k) and hence (δ/η)qi < 1/(m + k) for i = 1, . . . , k, there exist c ∈ (0, 1) and ε > 0 such that
δ
η and
(1 − c ) p i m+k+ε
qi δ
η
for every i = 1, . . . , m
(1 − c )qi m+k+ε
(1)
for every i = 1, . . . , k.
(2)
Now we will define a positive number β . To define β consider three cases. If r < ∞, p1 + · · · + p1 + q1 + · · · + q1 > 1r , then r ( 1r − p1 − · · · − p1 − q1 − · · · − 1
that for any β
ur
1
∈ (0, β], we have
r j
R (1 − 2δ)
kr +r ( p1 +···+ p1 )
(η R )
m
1
r
r j kr +r ( p1 +···+ p1 ) m · c (m+k)r 1 R (1 − 2δ) (η R )
If r = ∞, then our assumptions imply we have
u
· c (m+k)r
j
1 p1
m+k+ε
1
1 p1
−···−
1 pm
< 0, so we can find β > 0 be such
(3)
− q11 − · · · − q1 ) > 0, so we can find β > 0 be such that k
<
m+k+ε
1 ) qk
r ( 1 − 1 −···− p1 − q1 −···− q1 ) m 1 k < ∞. < β r p1
−1
ε
1
1
r ( 1r − p1
1
−···− p1m − q1 −···− q1 )
β
1
k
< ∞.
(4)
+ · · · + p1m + q11 + · · · + q1 > 0 = 1r , so we can find β > 0 such that for any β ∈ (0, β], k
k+ p1 +···+ p1
R (1 − 2δ) cm+k · (η R )
−1
ε
+ · · · + p1m + q11 + · · · + q1 < 1r , then r ( 1r − k ∈ (0, β], we have
for any β
m
1
k
1 p1
If r < ∞,
u
m
m
− 1
−( p1 +···+ p1 + q1 +···+ q1 ) m 1 1 k < ∞. < β
(5)
Using Lemma 3 with h = max{| f 1 | p 1 , . . . , | f m | pm , | g 1 |q1 , . . . , | gk |qk } (note that h ∈ L 1 ) and
ε = min β, (1 − δ − η) R , (1 − δ − η) R
q1
q , . . . , (1 − δ − η) R k ,
we infer that there is A ∈ Σ with 0 < μ( A ) ε if inf{μ( A ): ∞} = ∞, such that the following conditions hold
μ( A ) > 0} = 0, or with 1/ε μ( A ) < ∞ if sup{μ( A ): μ( A ) <
| f i | p i dμ (1 − δ − η) R for every i = 1, . . . , m; A
1/qi | g i |qi dμ (1 − δ − η) R for every i = 1, . . . , k.
(6)
(7)
A
Next, let M 1 , . . . , M m , N 1 , . . . , N k be such that pi
Mi
μ( A ) = η R for every i = 1, . . . , m;
1/qi = η R for every i = 1, . . . , k. N i μ( A )
(8) (9)
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
387
Now, let us define ˜f 1 , . . . , ˜f m , g˜ 1 , . . . , g˜ k , ˜l1 , . . . , ˜l j by formulas:
˜f i (x) = ˜li (x) =
Mi ,
x ∈ A;
f i (x),
x∈ / A,
g˜ i (x) =
Ni ,
x ∈ A;
g i (x),
x∈ / A,
li (x) + (1 − δ) R ,
if li (x) 0;
li (x) − (1 − δ) R ,
if li (x) < 0.
Using (6)–(9) we obtain
d( ˜f i , f i ) =
| M i − f i | p i dμ A
p
| f i | p i dμ
M i i dμ + A
A
η R + (1 − δ − η) R = R − δ R , 1/qi
1/qi
1/qi
q g˜ i − g i qi = | N i − g i |qi dμ N i i dμ + | g i |qi dμ A
A
A
η R + (1 − δ − η) R = R − δ R , and
˜li − li ∞ = (1 − δ) R . B (( ˜f 1 , . . . , ˜f m , g˜ 1 , . . . , g˜ k , ˜l1 , . . . , ˜l j ), δ R ) ⊂ B (( f 1 , . . . , f m , g 1 , . . . , gk , l1 , . . . , l j ), R ). It is enough to show that ˜ B (( f 1 , . . . , ˜f m , g˜ 1 , . . . , g˜ k , ˜l1 , . . . , ˜l j ), δ R ) ∩ E u = ∅. Let
Hence
(h1 , . . . , hm , s1 , . . . , sk , w 1 , . . . , w j ) ∈ B ( ˜f 1 , . . . , ˜f m , g˜ 1 , . . . , g˜ k , ˜l1 , . . . , ˜l j ), δ R . Clearly, since ˜li ∞ (1 − δ) R, for
μ-almost every x ∈ A, we have
w i (x) R (1 − 2δ).
(10)
Assume now that r < ∞. For any i = 1, . . . , m, we have
δR
|h i − ˜f i | p i dμ =
A
p
|h i − M i | p i dμ = M i i A
pi
hi dμ. − 1 M A
i
Using (1) and (8) we obtain
pi
hi δR δ 1 pi M − 1 dμ M p i = η μ( A ) m + k + ε μ( A )(1 − c ) . i i A
Similarly for any i = 1, . . . , k,
q
|si − g˜ i |qi dμ = N i i
(δ R )qi A
qi
si − 1 dμ, N A
i
and using (2) and (9) we have
qi qi qi
si δ 1 − 1 d μ δ R = μ( A ) μ( A )(1 − c )qi . N N η m + k+ε i i A
By (3), (4), (8)–(10) and Lemma 4 used for c i = c, we obtain the following
|h1 · · · hm · s1 · · · sk · w 1 · · · w j |r dμ X
r j R (1 − 2δ)
|h1 · · · hm · s1 · · · sk |r dμ A
r j r = R (1 − 2δ) M 1r · · · Mm · N 1r · · · Nkr
r
h1 hm s 1 sk dμ · · · · · · · M Mm N 1 Nk 1 A
388
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
r j r r r r (m+k)r R (1 − 2δ) M 1 · · · Mm · N 1 · · · Nk · c μ( A ) − (m + k)
r j r = R (1 − 2δ) M 1r · · · Mm · N 1r · · · Nkr · c (m+k)r
ε m+k+ε
1 m+k+ε
μ( A )
μ( A )
1 1 r j p r p r r r = R (1 − 2δ) M 1 1 μ( A ) p1 · · · Mmm μ( A ) pm · N 1 μ( A ) q1 · · · Nk μ( A ) qk · c (m+k)r r ( 1 − 1 −···− p1 − q1 −···− q1 ) ε m 1 k · · μ( A ) r p1 m+k+ε r r r j = R (1 − 2δ) (η R ) p1 · · · (η R ) pm · (η R )r · · · (η R )r · c (m+k)r r ( 1 − 1 −···− p1 − q1 −···− q1 ) ε m 1 k · · μ( A ) r p1 m+k+ε r j r ( 1 − 1 −···− p1 − q1 −···− q1 ) ε kr +r ( p1 +···+ p1 ) m m 1 k · 1 = R (1 − 2δ) (η R ) · c (m+k)r · μ( A ) r p1 > ur . m+k+ε
For the last inequality, observe that if we may use (3) with β
1 p1
= μ( A ). If
Hence
1 p1
1 + q11 + · · · + q1 > 1r , then by pm k 1 + q1 + · · · + q1 < 1r , then μ(1A ) k
+ ··· +
+···+
1 pm
hypothesis, we infer that μ( A ) ε β , so ε β , and we may use (4) with β = μ(1A ) .
h1 · · · hm · s1 · · · sk · w 1 · · · w j r > u . Assume now that r = ∞. As was mentioned, this case is possible only if inf{μ( A ): we define
A 1i = x ∈ A: h i (x) cM i ,
μ( A ) > 0} = 0. For any i = 1, . . . , m,
A 2i = A \ A 1i ,
and for any i = 1, . . . , k, we define
B 1i = x ∈ A: si (x) cN i Then
p |h i − M i | p i dμ M i i (1 − c ) p i μ A 2i .
|h i − M i | p i dμ
δR > A
B 2i = A \ B 1i .
and
A 2i
Hence by (1) and (8), we have
μ A 2i < Then
δR p M i i (1 − c ) p i
=
δ
1
η
(1 − c ) p i
μ( A )
1 m+k
μ( A ).
μ( A 1i ) > (1 − m1+k )μ( A ) for each i = 1, . . . , m. The same estimations (by (2) and (9)) hold for si :
q (δ R )qi > |si − N i |qi dμ |si − N i |qi dμ N i i (1 − c )qi μ B 2i . A
Then
μ B 2i <
B 2i
δR N i (1 − c )
qi
δ η(1 − c )
qi
μ( A )
1 m+k
μ( A ).
1 μ( B 1i ) > (1 − m1+k )μ( A ) for each i = 1, . . . , k. Now by Lemma 5 we obtain that μ( A 11 ∩ · · · ∩ Am ∩ B 11 ∩ · · · ∩ B k1 ) > 0. 1 1 1 1 Also, for μ-almost every x ∈ A 1 ∩ · · · ∩ A m ∩ B 1 ∩ · · · ∩ B k , using (8)–(10) and (5) we have h1 (x) · · · hm (x) · s1 (x) · · · sk (x) · w 1 (x) · · · w j (x) j R (1 − 2δ) cm+k M 1 · · · Mm · N 1 · · · Nk 1 j −( 1 +···+ p1 + q1 +···+ q1 ) +···+ p1m m 1 k > u, = R (1 − 2δ) cm+k (η R ) p1 (η R )k μ( A ) p1
Hence
and hence
h1 · · · hm · s1 · · · sk · w 1 · · · w j r > u . This ends the proof.
2
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
389
Lemma 7. Assume that
inf
μ( A ): μ( A ) > 0 > 0.
Then: (i) for every r ∈ (1, ∞), L 1 ⊂ L r ; (ii) for every p > 0, L p ⊂ L ∞ . The proof of Lemma 7 is known (see, e.g. [2, 224X(e)]). Proposition 8. Let p 1 , . . . , pn , r ∈ (0, ∞]. If one of the following conditions holds:
μ( A ) < ∞} < ∞ and 0 < p11 + · · · + p1n < 1r ; (ii) inf{μ( A ): μ( A ) > 0} > 0 and p1 + · · · + p1 > 1r , n 1 (i) sup{μ( A ):
( p 1 ,..., pn )
then E r
= L p 1 × · · · × L pn .
Proof. Assume (i). Then r is finite and at least one p i < ∞. ∞ Let M = sup{μ( A ): μ( A ) < ∞}. For any k ∈ N, let D k be a measurable set with M − 1/k μ( D k ) M. Set D = k=1 D k .
k
Since μ( s=1 D s ) M for any k, then μ( D ) = M and for a measurable F ⊂ X \ D we have μ( F ) = 0 or μ( F ) = ∞. Hence if p < ∞ and f ∈ L p , then μ({x ∈ X \ D: f (x) = 0}) = 0. Assume that for some 1 m n, we have p 1 , . . . , pm < ∞ and pm+1 , . . . , pn are equal to ∞. Let M > 0 be such that | f i | M μ-a.e. on X for i = m + 1, . . . , n, and set h = max{| f 1 | p 1 , . . . , | f m | pm }. Then h ∈ L 1 . Since f 1 ∈ L p 1 and p 1 < ∞, we have that
μ x ∈ X \ D: f 1 (x) · · · f n (x) = 0 = 0. Hence
| f 1 · · · f n |r dμ = X
| f 1 · · · f n |r dμ M n−m
D
M n−m
| f 1 · · · f m |r dμ D
h
r(
1 p1
+···+
1 pm
)
dμ.
D
We only have to observe that
r
1 p1
+ ··· +
1
D
h
r ( p1 +···+ p1 ) m
1
dμ < ∞, but this follows from the fact that
μ( D ) < ∞ and
< 1.
pm
Now assume (ii). We have to consider two cases: Case 1. r < ∞. Then at least one of p 1 , . . . , pn is finite. Assume again, that for some 1 m n, we have p 1 , . . . , pm < ∞ and pm+1 = · · · = pn = ∞. Let ( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn . Set h = max{| f 1 | p 1 , . . . , | f m | pm }. Then h ∈ L 1 . Let M > 0 be such that | f i | M μ-a.e. on X for all i = m + 1, . . . , n. Then by Lemma 7, we obtain
| f 1 · · · f n |r dμ M n−m
X
h
r ( p1 +···+ p1 ) 1
n
dμ < ∞,
X
since r ( p1 + · · · + 1
1 ) pn
> 1.
Case 2. r = ∞. By Case 1, we obtain that for r < ∞ with
1 r
<
1 p1
+ ··· +
1 pn
,
if ( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn , then f 1 · · · f n ∈ L r . Hence by Lemma 7, we have f 1 · · · f n ∞ < ∞.
2
390
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
Note that Proposition 8 is not valid if each p i is infinite. Indeed, if we consider the following measure
μ( A ) = 0 if A = ∅ and μ( A ) = ∞ if A = ∅, and we set f = g = 1, then ( f , g ) ∈ L ∞ × L ∞ , but ( f , g ) ∈ / Er . Now we can summarize our results in the two following theorems. We write c instead of c ( p 1 , . . . , pn ), where c ( p 1 , . . . , pn ) was defined before the statement of Theorem 6. (∞,∞)
Theorem 9. Let ( X , Σ, μ) be a measure space. The following conditions are equivalent: (i) for any n ∈ N and p 1 , . . . , pn , r > 0 such that (ii) for any n ∈ N and p 1 , . . . , pn , r > 0 such that
1 p1 1 p1
(iii) there are n ∈ N and p 1 , . . . , pn , r > 0 such that (iv) there are n ∈ N and p 1 , . . . , pn , r > 0 such that (v) inf{μ( A ): μ( A ) > 0} = 0.
+ ··· + + ··· + 1 p1 1 p1
1 pn 1 pn
+ ··· + + ··· +
( p 1 ,..., pn )
is σ –c-lower porous;
( p 1 ,..., pn )
is not equal to L p 1 × · · · × L pn ;
> 1r , the set E r > 1r , the set E r 1 pn 1 pn
> >
1 r 1 r
( p 1 ,..., pn )
is σ –c-lower porous;
( p 1 ,..., pn )
is not equal to L p 1 × · · · × L pn ;
and the set E r and the set E r
Proof. The following implications are trivial: (i) ⇒ (ii), (i) ⇒ (iii), (ii) ⇒ (iv) and (iii) ⇒ (iv). Implication (iv) ⇒ (v) follows from Proposition 8. Finally, (v) ⇒ (i) follows from Theorem 6. 2 Theorem 10. Let ( X , Σ, μ) be a measure space. The following conditions are equivalent: (i) for any n ∈ N and p 1 , . . . , pn , r > 0 such that 0 < (ii) for any n ∈ N and p 1 , . . . , pn , r > 0 such that 0 <
1 p1 1 p1
(iii) there are n ∈ N and p 1 , . . . , pn , r > 0 such that 0 < (iv) there are n ∈ N and p 1 , . . . , pn , r > 0 such that 0 < (v) sup{μ( A ): μ( A ) < ∞} = ∞.
+ ··· + + ··· + 1 p1 1 p1
1 pn 1 pn
+ ··· + + ··· +
( p 1 ,..., pn )
is σ –c-lower porous;
( p 1 ,..., pn )
is not equal to L p 1 × · · · × L pn ;
< 1r , the set E r < 1 pn 1 pn
1 , r
< <
the set E r
1 r 1 r
( p 1 ,..., pn )
is σ –c-lower porous;
( p 1 ,..., pn )
is not equal to L p 1 × · · · × L pn ;
and the set E r and the set E r
Proof. The following implications are trivial: (i) ⇒ (ii), (i) ⇒ (iii), (ii) ⇒ (iv) and (iii) ⇒ (iv). Implication (iv) ⇒ (v) follows from Proposition 8. Finally, (v) ⇒ (i) follows from Theorem 6. 2 Acknowledgments We are grateful to an anonymous referee for a very careful reading of the text which has led to several improvements. In particular, thanks to this we clarified our argument in the proof of Theorem 6, we modified the formulation of Proposition 8, and we added the example after Proposition 8 which the referee had suggested.
References [1] [2] [3] [4] [5] [6]
M. Balcerzak, A. Wachowicz, Some examples of meager sets in Banach spaces, Real Anal. Exchange 26 (2) (2000/2001) 877–884. D.H. Fremlin, Measure Theory, vol. 2, Broad Foundations, Torres Fremlin, Colchester, 2001. L. Grafakos, Classical Fourier Analysis, second ed., Grad. Texts in Math., vol. 249, Springer-Verlag, Berlin/New York, 2008. J. Jachymski, A nonlinear Banach–Steinhaus theorem and some meager sets in Banach spaces, Studia Math. 170 (3) (2005) 303–320. L. Zajíˇcek, Porosity and σ -porosity, Real Anal. Exchange 13 (1987/1988) 314–350. L. Zajíˇcek, On σ -porous sets in abstract spaces, Abstr. Appl. Anal. 5 (2005) 509–534.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Dichotomies for L p spaces Szymon Głab ¸ a,∗ , Filip Strobin a,b a b
Institute of Mathematics, Technical University of Łód´z, Wólcza´ nska 215, 93-005 Łód´z, Poland Institute of Mathematics, Polish Academy of Sciences, S´niadeckich 8, 00-956 Warszawa, Poland
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 28 July 2009 Available online 10 February 2010 Submitted by T.D. Benavides
Assume that ( X , Σ, μ) is a measure space and p 1 , . . . , pn , r > 0. We prove that {( f 1 , . . . , f n ) ∈ L p1 × · · · × L pn : f 1 · · · f n ∈ L r } is either L p1 × · · · × L pn or a σ -porous subset of L p 1 × · · · × L pn . This dichotomy depends on properties of μ and the sign of the number 1 − p11 − · · · − p1n . r © 2010 Elsevier Inc. All rights reserved.
Keywords: Measure space L p space Porous sets Porosity
1. Introduction Among linear topological spaces there are spaces X consisting of sequences or functions such that a natural multiplication is defined on pairs (x1 , x2 ) ∈ X 2 , however, its result need not necessarily belong to X . It is an interesting question about the size of the set of such “bad” pairs in a various sense. Such a kind of studies was initiated in [1,4]. Balcerzak and Wachowicz proved in [1] that {( f , g ) ∈ L 1 [0, 1] × L 1 [0, 1]: f · g ∈ L 1 [0, 1]} is a meager subset of L 1 [0, 1] × L 1 [0, 1]. They also proved that
(x, y ) ∈ c 0 × c 0 :
n
∞ x(i ) y (i )
i =1
is bounded
n =1
is a meager subset of c 0 × c 0 . These meagerness results were generalized by Jachymski in the following extension of the classical Banach–Steinhaus theorem. Recall that a function ϕ : X → R+ is L-subadditive for some L 1, if ϕ (x + y ) L (ϕ (x) + ϕ ( y )) for any x, y ∈ X . Theorem 1. (See Jachymski [4].) Given k ∈ N, let X 1 , . . . , X k be Banach spaces, X = X 1 if k = 1, and X = X 1 × · · · × X k if k > 1. Assume that L 1, F n : X → R+ (n ∈ N) are lower semicontinuous and such that all functions xi → F n (x1 , . . . , xk ) (i = 1, . . . , k) are L-subadditive and even. Let E = {x ∈ X: ( F n (x))n∞=1 is bounded}. Then the following statements are equivalent: (i) E is meager; (ii) E = X; (iii) sup{ F n (x): n ∈ N, x 1} = ∞.
*
Corresponding author. E-mail addresses: [email protected] (S. Głab), ¸ [email protected] (F. Strobin).
0022-247X/$ – see front matter doi:10.1016/j.jmaa.2010.02.011
©
2010 Elsevier Inc. All rights reserved.
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
At first, we were interested in a further generalization of this theorem changing meagerness by that this is not possible. To see it, consider the following set:
E=
x ∈ R:
n |sin(k!π x)| k =1
k
∞
383
σ -porosity. It turns out
is bounded .
n =1
n
Using Theorem 1 for F n (x) = k=1 |sin(k!π x)|/k (clearly, each F n is subadditive) we obtain that this set is meager (E = R since it is of measure zero) and is not σ -upper porous [5, p. 341]. Hence we could not generalize Jachymski’s theorem in this manner. Assume that ( X , Σ, μ) is a measure space. In our paper we answer the question about a size of the set (in the following we will write L p instead of L p ( X , Σ, μ)):
( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn : f 1 · · · f n ∈ L r .
We do not restrict our attention only to Banach L p spaces for p ∈ [1, ∞], but we consider all linear metric L p spaces for p ∈ (0, ∞]. It appears that this set is either L p 1 × · · · × L pn or a σ –c-lower porous (for some c > 0) subset of L p 1 × · · · × L pn . So, it is either the whole space or a very small set. We determine this dichotomy for every type of a measure space 1 ( X , Σ, μ). Surprisingly it depends on the following parameters (in the sequel the symbol ∞ means 0):
• the sign of the number 1r − p11 − · · · − p1n ; • inf{μ( A ): μ( A ) > 0} (it is important whether it is equal or greater than zero); • sup{μ( A ): μ( A ) < ∞} (it is important whether it is finite or infinite). The dichotomy is stated in Proposition 2 and Theorems 9, 10. Let X be a metric space. B (x, R ) stands for the ball with a radius R centered at a point x. Let c ∈ (0, 1]. We say that M ⊂ X is c-lower porous [6], if
∀x ∈ M ,
γ (x, M , R )
lim inf R →0+
R
c
, 2
where
γ (x, M , R ) = sup r 0: ∃z ∈ X , B (z, r ) ⊂ B (x, R )\ M . Clearly, M is c-lower porous iff
∀x ∈ M ,
∀β ∈ (0, c /2),
∃ R 0 > 0,
∀ R ∈ (0, R 0 ),
∃z ∈ X ,
B ( z, β R ) ⊂ B (x, R )\ M .
The set is σ –c-lower porous if it is a countable union of c-lower porous sets. Note that a σ –c-lower porous set is meager, and the notion of σ -porosity is essentially stronger than that of meagerness. Note that the sets investigated in this paper will be c-porous in some stronger sense, namely,
∀x ∈ X ,
∀β ∈ (0, c /2),
∀ R > 0,
∃z ∈ X ,
B ( z, β R ) ⊂ B (x, R )\ M .
However, we do not want to define any new notion of porosity, so in the formulations of theorems we will deal only with c-lower porosity. 2. Algebraic product of functions from L p 1 × ··· × L pn Throughout the paper, ( X , Σ, μ) is a measure space. If p ∈ (0, 1), then we consider L p as a metric linear space with the metric
| f − g | p dμ.
d( f , g ) = X
Additionally we put
| f | p dμ.
f p = d ( f , 0) = X
If p ∈ [1, ∞), then we consider L p as a normed linear space with the norm
f p = X
1/ p | f | p dμ .
384
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
Finally, if p = ∞, then we consider L p as a normed linear space with the norm f ∞ = supess | f |. Note that in all cases L p is a complete space. For every n ∈ N and any p 1 , . . . , pn , r ∈ (0, ∞], we define the set (we allow n to be 1): ( p 1 ,..., pn )
Er
= ( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn : f 1 · · · f n ∈ L r .
In this paper we consider L p 1 × · · · × L pn as a space with the metric defined as the maximum of distances on all coordinates in L p 1 , . . . , L pn . Using the general Hölder inequality [3, p. 10] we obtain that: Proposition 2. Let p 1 , . . . , pn , r ∈ (0, ∞] be such that
1 r
1
=
p1
( p 1 ,..., pn )
Then E r
+ ··· +
1 pn
.
= L p 1 × · · · × L pn .
Now we will give some helpful lemmas. Lemma 3. Let h 0, h ∈ L 1 , ε > 0. Then
(i) if inf{μ( A ): A ∈ Σ, μ( A ) > 0} = 0, there is A ∈ Σ with 0 < μ( A ) ε and A h dμ ε ; (ii) if sup{μ( A ): A ∈ Σ, μ( A ) < ∞} = ∞, there is A ∈ Σ with 1/ε μ( A ) < ∞ and A h dμ ε .
Proof. (i) Follows immediately from the absolute continuity of the function B → B h dμ (B ∈ Σ ) with respect to μ. ∈ N, An be such that n < μ( An ) < ∞. Set F n = nk=1 A k . Then ( F n ) is increasing, μ( F n ) < ∞ and (ii) Let, for any n μ( F n ) → ∞. Put F = n∞=1 F n . We have
h dμ =
lim
n→∞ Fn
h dμ < ∞. F
Then there is n0 ∈ N with
h dμ − ε .
h dμ > F n0
Hence
F
h dμ < ε .
F \ F n0
On the other hand, limn→∞ μ( F n \ F n0 ) = ∞, so there is N ∈ N such that
μ( F N \ F n0 ) > 1/ε . Put A = F N \ F n0 . 2
Lemma 4. Let p 1 , . . . , pn , r ∈ (0, ∞), ( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn and let A be a measurable subset of X . Suppose that for some numbers a1 , . . . , an and for each i = 1, . . . , n, the following holds
| f i − 1| p i d μ a i . A
Then for any numbers c 1 , . . . , cn ∈ (0, 1), we have
r
| f 1 · · · f n | dμ
cr1 · · · cnr
A
μ( A ) −
a1
(1 − c 1 ) p 1
− ··· −
Proof. Observe that the above assumptions imply that any i, we have
A
Hence
| f i − 1| p i d μ
ai
(1 − cn ) pn
.
μ( A ) < ∞. Let A i = {x ∈ A: f i (x) < c i } for i = 1, . . . , n. Then for
| f i − 1| p i d μ Ai
an
|1 − c i | p i dμ = (1 − c i ) p i μ( A i ). Ai
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
| f 1 · · · f n |r dμ
| f 1 · · · f n |r dμ n
A\
A
385
i =1
n
Ai
cr1 · · · cnr μ( A ) − μ cr1 · · · cnr μ( A ) −
A\ n
i =1
c r1 · · · cnr dμ Ai
Ai
i =1
a1
(1 − c 1 ) p 1
− ··· −
an
(1 − cn ) pn
.
2
Lemma 5. Let A, A 1 , . . . , A n be measurable with A i ⊂ A and μ( A i ) > (1 − n1 )μ( A ) for any i = 1, . . . , n. Then
μ
n
Ai
> 0.
i =1
Proof. Using the induction principle, it is easy to show that
μ
k
> (1 − k/n)μ( A ) for any k = 1, . . . , n.
Ai
i =1
In particular, for k = n, we get that
n
μ(
i =1
A i ) > 0.
2 ( p ,..., p )
n The next theorem is a main result of the paper. It is rather technical, but it shows when E r 1 can be σ -porous and how good are porosity estimations in each of the considered cases. For any n ∈ N and any p 1 , . . . , pn , put c ( p 1 , . . . , pn ) = 2/(1 + m) if there is at least one finite p i , where m is the number of finite p i ’s, and put c ( p 1 , . . . , pn ) = 1 if p i = ∞ for every i = 1, . . . , n.
Theorem 6. Let n ∈ N and let p 1 , . . . , pn , r ∈ (0, ∞]. Assume that one of the following conditions holds: (i) (ii)
1 p1 1 p1
+ ··· + + ··· +
1 pn 1 pn
> <
1 r 1 r
and inf{μ( A ):
μ( A ) > 0} = 0; and sup{μ( A ): μ( A ) < ∞} = ∞.
Then for any u > 0, the set
E u = ( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn : f 1 · · · f n r u
( p 1 ,..., pn )
is c-lower porous, where c = c ( p 1 , . . . , pn ). In particular, the set E r
is σ –c-lower porous.
Proof. We will consider two cases. Case 1. p 1 = · · · = pn = ∞. Then our assumptions imply that r < ∞ and sup{μ( A ): μ( A ) < ∞} = ∞. Let ( f 1 , . . . , f n ) ∈ L ∞ × · · · × L ∞ , R > 0, α ∈ (0, 12 ) (note that in this case c ( p 1 , . . . , pn ) = 1). Fix a measurable set A of finite measure such that
μ( A ) >
ur
(( 12
− α ) R )rn
.
For any i = 1, . . . , n, we define
˜f i (x) =
f i (x) + f i (x) −
1 R, 2 1 R, 2
f i (x) 0; f i (x) < 0.
Clearly, for any i = 1, . . . , n, ˜f i − f i ∞ = R /2 and B (( ˜f 1 , . . . , ˜f n ), α R ) ⊂ B (( f 1 , . . . , f n ), R ). Now if (h1 , . . . , hn ) ∈ B (( ˜f 1 , . . . , ˜f n ), α R ), then for any i = 1, . . . , n and for μ-almost every x ∈ A, we have
h i (x) Hence
1 2
− α R.
386
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
|h1 · · · hn |r
1 2
rn −α R · μ( A ) > ur ,
A
and
h1 · · · hn r > u . This ends the proof in Case 1. Case 2. For some i = 1, . . . , n, p i < ∞. Without loss of generality, we assume that p i ∈ (0, 1) for i = 1, . . . , m, 1 p i < ∞ for i = m + 1, . . . , m + k and p i = ∞ for i = m + k + 1, . . . , m + k + j, where j is such that m + k + j = n (clearly, m, k or j can be equal to zero, but m + k = 0). Additionally define q i = pm+i for i = 1, . . . , k. Then the product space L p 1 × · · · × L pn can be written in the following way:
L p 1 × · · · × L p m × L q 1 × · · · × L qk × L ∞ × · · · × L ∞ . Let ( f 1 , . . . , f m , g 1 , . . . , gk , l1 , . . . , l j ) be a member of that space, and let R > 0, δ ∈ (0, m+1k+1 ) (note that in this case c ( p 1 , . . . , pn ) = 2/(m + k + 1)). Then, clearly, 1 − δ > (m + k)δ and hence we can take η ∈ ((m + k)δ, 1 − δ). Since δ/η < 1/(m + k) and hence (δ/η)qi < 1/(m + k) for i = 1, . . . , k, there exist c ∈ (0, 1) and ε > 0 such that
δ
η and
(1 − c ) p i m+k+ε
qi δ
η
for every i = 1, . . . , m
(1 − c )qi m+k+ε
(1)
for every i = 1, . . . , k.
(2)
Now we will define a positive number β . To define β consider three cases. If r < ∞, p1 + · · · + p1 + q1 + · · · + q1 > 1r , then r ( 1r − p1 − · · · − p1 − q1 − · · · − 1
that for any β
ur
1
∈ (0, β], we have
r j
R (1 − 2δ)
kr +r ( p1 +···+ p1 )
(η R )
m
1
r
r j kr +r ( p1 +···+ p1 ) m · c (m+k)r 1 R (1 − 2δ) (η R )
If r = ∞, then our assumptions imply we have
u
· c (m+k)r
j
1 p1
m+k+ε
1
1 p1
−···−
1 pm
< 0, so we can find β > 0 be such
(3)
− q11 − · · · − q1 ) > 0, so we can find β > 0 be such that k
<
m+k+ε
1 ) qk
r ( 1 − 1 −···− p1 − q1 −···− q1 ) m 1 k < ∞. < β r p1
−1
ε
1
1
r ( 1r − p1
1
−···− p1m − q1 −···− q1 )
β
1
k
< ∞.
(4)
+ · · · + p1m + q11 + · · · + q1 > 0 = 1r , so we can find β > 0 such that for any β ∈ (0, β], k
k+ p1 +···+ p1
R (1 − 2δ) cm+k · (η R )
−1
ε
+ · · · + p1m + q11 + · · · + q1 < 1r , then r ( 1r − k ∈ (0, β], we have
for any β
m
1
k
1 p1
If r < ∞,
u
m
m
− 1
−( p1 +···+ p1 + q1 +···+ q1 ) m 1 1 k < ∞. < β
(5)
Using Lemma 3 with h = max{| f 1 | p 1 , . . . , | f m | pm , | g 1 |q1 , . . . , | gk |qk } (note that h ∈ L 1 ) and
ε = min β, (1 − δ − η) R , (1 − δ − η) R
q1
q , . . . , (1 − δ − η) R k ,
we infer that there is A ∈ Σ with 0 < μ( A ) ε if inf{μ( A ): ∞} = ∞, such that the following conditions hold
μ( A ) > 0} = 0, or with 1/ε μ( A ) < ∞ if sup{μ( A ): μ( A ) <
| f i | p i dμ (1 − δ − η) R for every i = 1, . . . , m; A
1/qi | g i |qi dμ (1 − δ − η) R for every i = 1, . . . , k.
(6)
(7)
A
Next, let M 1 , . . . , M m , N 1 , . . . , N k be such that pi
Mi
μ( A ) = η R for every i = 1, . . . , m;
1/qi = η R for every i = 1, . . . , k. N i μ( A )
(8) (9)
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
387
Now, let us define ˜f 1 , . . . , ˜f m , g˜ 1 , . . . , g˜ k , ˜l1 , . . . , ˜l j by formulas:
˜f i (x) = ˜li (x) =
Mi ,
x ∈ A;
f i (x),
x∈ / A,
g˜ i (x) =
Ni ,
x ∈ A;
g i (x),
x∈ / A,
li (x) + (1 − δ) R ,
if li (x) 0;
li (x) − (1 − δ) R ,
if li (x) < 0.
Using (6)–(9) we obtain
d( ˜f i , f i ) =
| M i − f i | p i dμ A
p
| f i | p i dμ
M i i dμ + A
A
η R + (1 − δ − η) R = R − δ R , 1/qi
1/qi
1/qi
q g˜ i − g i qi = | N i − g i |qi dμ N i i dμ + | g i |qi dμ A
A
A
η R + (1 − δ − η) R = R − δ R , and
˜li − li ∞ = (1 − δ) R . B (( ˜f 1 , . . . , ˜f m , g˜ 1 , . . . , g˜ k , ˜l1 , . . . , ˜l j ), δ R ) ⊂ B (( f 1 , . . . , f m , g 1 , . . . , gk , l1 , . . . , l j ), R ). It is enough to show that ˜ B (( f 1 , . . . , ˜f m , g˜ 1 , . . . , g˜ k , ˜l1 , . . . , ˜l j ), δ R ) ∩ E u = ∅. Let
Hence
(h1 , . . . , hm , s1 , . . . , sk , w 1 , . . . , w j ) ∈ B ( ˜f 1 , . . . , ˜f m , g˜ 1 , . . . , g˜ k , ˜l1 , . . . , ˜l j ), δ R . Clearly, since ˜li ∞ (1 − δ) R, for
μ-almost every x ∈ A, we have
w i (x) R (1 − 2δ).
(10)
Assume now that r < ∞. For any i = 1, . . . , m, we have
δR
|h i − ˜f i | p i dμ =
A
p
|h i − M i | p i dμ = M i i A
pi
hi dμ. − 1 M A
i
Using (1) and (8) we obtain
pi
hi δR δ 1 pi M − 1 dμ M p i = η μ( A ) m + k + ε μ( A )(1 − c ) . i i A
Similarly for any i = 1, . . . , k,
q
|si − g˜ i |qi dμ = N i i
(δ R )qi A
qi
si − 1 dμ, N A
i
and using (2) and (9) we have
qi qi qi
si δ 1 − 1 d μ δ R = μ( A ) μ( A )(1 − c )qi . N N η m + k+ε i i A
By (3), (4), (8)–(10) and Lemma 4 used for c i = c, we obtain the following
|h1 · · · hm · s1 · · · sk · w 1 · · · w j |r dμ X
r j R (1 − 2δ)
|h1 · · · hm · s1 · · · sk |r dμ A
r j r = R (1 − 2δ) M 1r · · · Mm · N 1r · · · Nkr
r
h1 hm s 1 sk dμ · · · · · · · M Mm N 1 Nk 1 A
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S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
r j r r r r (m+k)r R (1 − 2δ) M 1 · · · Mm · N 1 · · · Nk · c μ( A ) − (m + k)
r j r = R (1 − 2δ) M 1r · · · Mm · N 1r · · · Nkr · c (m+k)r
ε m+k+ε
1 m+k+ε
μ( A )
μ( A )
1 1 r j p r p r r r = R (1 − 2δ) M 1 1 μ( A ) p1 · · · Mmm μ( A ) pm · N 1 μ( A ) q1 · · · Nk μ( A ) qk · c (m+k)r r ( 1 − 1 −···− p1 − q1 −···− q1 ) ε m 1 k · · μ( A ) r p1 m+k+ε r r r j = R (1 − 2δ) (η R ) p1 · · · (η R ) pm · (η R )r · · · (η R )r · c (m+k)r r ( 1 − 1 −···− p1 − q1 −···− q1 ) ε m 1 k · · μ( A ) r p1 m+k+ε r j r ( 1 − 1 −···− p1 − q1 −···− q1 ) ε kr +r ( p1 +···+ p1 ) m m 1 k · 1 = R (1 − 2δ) (η R ) · c (m+k)r · μ( A ) r p1 > ur . m+k+ε
For the last inequality, observe that if we may use (3) with β
1 p1
= μ( A ). If
Hence
1 p1
1 + q11 + · · · + q1 > 1r , then by pm k 1 + q1 + · · · + q1 < 1r , then μ(1A ) k
+ ··· +
+···+
1 pm
hypothesis, we infer that μ( A ) ε β , so ε β , and we may use (4) with β = μ(1A ) .
h1 · · · hm · s1 · · · sk · w 1 · · · w j r > u . Assume now that r = ∞. As was mentioned, this case is possible only if inf{μ( A ): we define
A 1i = x ∈ A: h i (x) cM i ,
μ( A ) > 0} = 0. For any i = 1, . . . , m,
A 2i = A \ A 1i ,
and for any i = 1, . . . , k, we define
B 1i = x ∈ A: si (x) cN i Then
p |h i − M i | p i dμ M i i (1 − c ) p i μ A 2i .
|h i − M i | p i dμ
δR > A
B 2i = A \ B 1i .
and
A 2i
Hence by (1) and (8), we have
μ A 2i < Then
δR p M i i (1 − c ) p i
=
δ
1
η
(1 − c ) p i
μ( A )
1 m+k
μ( A ).
μ( A 1i ) > (1 − m1+k )μ( A ) for each i = 1, . . . , m. The same estimations (by (2) and (9)) hold for si :
q (δ R )qi > |si − N i |qi dμ |si − N i |qi dμ N i i (1 − c )qi μ B 2i . A
Then
μ B 2i <
B 2i
δR N i (1 − c )
qi
δ η(1 − c )
qi
μ( A )
1 m+k
μ( A ).
1 μ( B 1i ) > (1 − m1+k )μ( A ) for each i = 1, . . . , k. Now by Lemma 5 we obtain that μ( A 11 ∩ · · · ∩ Am ∩ B 11 ∩ · · · ∩ B k1 ) > 0. 1 1 1 1 Also, for μ-almost every x ∈ A 1 ∩ · · · ∩ A m ∩ B 1 ∩ · · · ∩ B k , using (8)–(10) and (5) we have h1 (x) · · · hm (x) · s1 (x) · · · sk (x) · w 1 (x) · · · w j (x) j R (1 − 2δ) cm+k M 1 · · · Mm · N 1 · · · Nk 1 j −( 1 +···+ p1 + q1 +···+ q1 ) +···+ p1m m 1 k > u, = R (1 − 2δ) cm+k (η R ) p1 (η R )k μ( A ) p1
Hence
and hence
h1 · · · hm · s1 · · · sk · w 1 · · · w j r > u . This ends the proof.
2
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
389
Lemma 7. Assume that
inf
μ( A ): μ( A ) > 0 > 0.
Then: (i) for every r ∈ (1, ∞), L 1 ⊂ L r ; (ii) for every p > 0, L p ⊂ L ∞ . The proof of Lemma 7 is known (see, e.g. [2, 224X(e)]). Proposition 8. Let p 1 , . . . , pn , r ∈ (0, ∞]. If one of the following conditions holds:
μ( A ) < ∞} < ∞ and 0 < p11 + · · · + p1n < 1r ; (ii) inf{μ( A ): μ( A ) > 0} > 0 and p1 + · · · + p1 > 1r , n 1 (i) sup{μ( A ):
( p 1 ,..., pn )
then E r
= L p 1 × · · · × L pn .
Proof. Assume (i). Then r is finite and at least one p i < ∞. ∞ Let M = sup{μ( A ): μ( A ) < ∞}. For any k ∈ N, let D k be a measurable set with M − 1/k μ( D k ) M. Set D = k=1 D k .
k
Since μ( s=1 D s ) M for any k, then μ( D ) = M and for a measurable F ⊂ X \ D we have μ( F ) = 0 or μ( F ) = ∞. Hence if p < ∞ and f ∈ L p , then μ({x ∈ X \ D: f (x) = 0}) = 0. Assume that for some 1 m n, we have p 1 , . . . , pm < ∞ and pm+1 , . . . , pn are equal to ∞. Let M > 0 be such that | f i | M μ-a.e. on X for i = m + 1, . . . , n, and set h = max{| f 1 | p 1 , . . . , | f m | pm }. Then h ∈ L 1 . Since f 1 ∈ L p 1 and p 1 < ∞, we have that
μ x ∈ X \ D: f 1 (x) · · · f n (x) = 0 = 0. Hence
| f 1 · · · f n |r dμ = X
| f 1 · · · f n |r dμ M n−m
D
M n−m
| f 1 · · · f m |r dμ D
h
r(
1 p1
+···+
1 pm
)
dμ.
D
We only have to observe that
r
1 p1
+ ··· +
1
D
h
r ( p1 +···+ p1 ) m
1
dμ < ∞, but this follows from the fact that
μ( D ) < ∞ and
< 1.
pm
Now assume (ii). We have to consider two cases: Case 1. r < ∞. Then at least one of p 1 , . . . , pn is finite. Assume again, that for some 1 m n, we have p 1 , . . . , pm < ∞ and pm+1 = · · · = pn = ∞. Let ( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn . Set h = max{| f 1 | p 1 , . . . , | f m | pm }. Then h ∈ L 1 . Let M > 0 be such that | f i | M μ-a.e. on X for all i = m + 1, . . . , n. Then by Lemma 7, we obtain
| f 1 · · · f n |r dμ M n−m
X
h
r ( p1 +···+ p1 ) 1
n
dμ < ∞,
X
since r ( p1 + · · · + 1
1 ) pn
> 1.
Case 2. r = ∞. By Case 1, we obtain that for r < ∞ with
1 r
<
1 p1
+ ··· +
1 pn
,
if ( f 1 , . . . , f n ) ∈ L p 1 × · · · × L pn , then f 1 · · · f n ∈ L r . Hence by Lemma 7, we have f 1 · · · f n ∞ < ∞.
2
390
S. Głab, ¸ F. Strobin / J. Math. Anal. Appl. 368 (2010) 382–390
Note that Proposition 8 is not valid if each p i is infinite. Indeed, if we consider the following measure
μ( A ) = 0 if A = ∅ and μ( A ) = ∞ if A = ∅, and we set f = g = 1, then ( f , g ) ∈ L ∞ × L ∞ , but ( f , g ) ∈ / Er . Now we can summarize our results in the two following theorems. We write c instead of c ( p 1 , . . . , pn ), where c ( p 1 , . . . , pn ) was defined before the statement of Theorem 6. (∞,∞)
Theorem 9. Let ( X , Σ, μ) be a measure space. The following conditions are equivalent: (i) for any n ∈ N and p 1 , . . . , pn , r > 0 such that (ii) for any n ∈ N and p 1 , . . . , pn , r > 0 such that
1 p1 1 p1
(iii) there are n ∈ N and p 1 , . . . , pn , r > 0 such that (iv) there are n ∈ N and p 1 , . . . , pn , r > 0 such that (v) inf{μ( A ): μ( A ) > 0} = 0.
+ ··· + + ··· + 1 p1 1 p1
1 pn 1 pn
+ ··· + + ··· +
( p 1 ,..., pn )
is σ –c-lower porous;
( p 1 ,..., pn )
is not equal to L p 1 × · · · × L pn ;
> 1r , the set E r > 1r , the set E r 1 pn 1 pn
> >
1 r 1 r
( p 1 ,..., pn )
is σ –c-lower porous;
( p 1 ,..., pn )
is not equal to L p 1 × · · · × L pn ;
and the set E r and the set E r
Proof. The following implications are trivial: (i) ⇒ (ii), (i) ⇒ (iii), (ii) ⇒ (iv) and (iii) ⇒ (iv). Implication (iv) ⇒ (v) follows from Proposition 8. Finally, (v) ⇒ (i) follows from Theorem 6. 2 Theorem 10. Let ( X , Σ, μ) be a measure space. The following conditions are equivalent: (i) for any n ∈ N and p 1 , . . . , pn , r > 0 such that 0 < (ii) for any n ∈ N and p 1 , . . . , pn , r > 0 such that 0 <
1 p1 1 p1
(iii) there are n ∈ N and p 1 , . . . , pn , r > 0 such that 0 < (iv) there are n ∈ N and p 1 , . . . , pn , r > 0 such that 0 < (v) sup{μ( A ): μ( A ) < ∞} = ∞.
+ ··· + + ··· + 1 p1 1 p1
1 pn 1 pn
+ ··· + + ··· +
( p 1 ,..., pn )
is σ –c-lower porous;
( p 1 ,..., pn )
is not equal to L p 1 × · · · × L pn ;
< 1r , the set E r < 1 pn 1 pn
1 , r
< <
the set E r
1 r 1 r
( p 1 ,..., pn )
is σ –c-lower porous;
( p 1 ,..., pn )
is not equal to L p 1 × · · · × L pn ;
and the set E r and the set E r
Proof. The following implications are trivial: (i) ⇒ (ii), (i) ⇒ (iii), (ii) ⇒ (iv) and (iii) ⇒ (iv). Implication (iv) ⇒ (v) follows from Proposition 8. Finally, (v) ⇒ (i) follows from Theorem 6. 2 Acknowledgments We are grateful to an anonymous referee for a very careful reading of the text which has led to several improvements. In particular, thanks to this we clarified our argument in the proof of Theorem 6, we modified the formulation of Proposition 8, and we added the example after Proposition 8 which the referee had suggested.
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