RECTANGULAR DIFFERENTIATION OF INTEGRALS OF ... - Math KSU

RECTANGULAR DIFFERENTIATION OF INTEGRALS OF BESOV FUNCTIONS H. AIMAR, L. FORZANI, AND V. NAIBO

Abstract. We study the differentiation of integrals of functions in the Besov spaces Bpα,1 (Rn ), α > 0, 1 ≤ p < ∞, with respect to the basis of arbitrarily oriented rectangular parallelepipeds in Rn . We show that positive results hold if α ≥ n−1 p and we give counterexamples for the case 0 < α < n − 1. Similar results hold for p Bpα,q (Rn ), q > 1. For more general bases we can also prove negative results for n − 1 ≤ α < n−1 . p p

1. definitions and main results. Following de Guzm´ an [3, 4] we shall say that a differentiation basis A is a collection of open bounded sets in Rn such that for each x ∈ Rn there is a sequence {Aj } ⊂ A with x ∈ Aj for every j and diameter of Aj tending to 0. A differentiation basis A is said to differentiate the integral of a locally integrable function f defined in Rn if Z 1 lim f (y) dy = f (x) diam(A)→0, x∈A∈A |A| A for almost every x ∈ Rn , where |A| denotes the Lebesgue measure of the set A. If A differentiates the integral of every function of a given class we say that A differentiates that class. We denote by B the basis of all arbitrarily oriented rectangular parallelepipeds in Rn with diameter smaller than 1. Each elementQof B can be regarded as a proper Pn rigid motion n of a multidimensional interval of the form j=1 (−bj , bj ), bj > 0 and j=1 (2bj )2 ≤ 1. The basis B does not differentiate the spaces Lp (Rn ). Actually, it does not even differentiate the characteristic functions of measurable sets in Rn , as was observed by Zygmund (in Nikodym [6]) (see [3, 8]). Then, no restriction on the global growth of functions is sufficient for differentiation of integrals with respect to that basis and it seems natural to impose additional restrictions on the integral smoothness of functions. In that direction, Stokolos [9, 10] considered the differentiation of the integral of functions in terms of the integrability properties of the Lp modulus of continuity. We define the Besov spaces Bpα,q (Rn ) for α > 0 and 1 ≤ p, q ≤ ∞. Let α ¯ be the smallest integer greater than α and denote by uf (x, t) = (f ∗ Pt )(x), x ∈ Rn , t > 0, the Poisson 2000 Mathematics Subject Classification. 42B35, 42B25 . Key words and phrases. Differentiation, Besov spaces, maximal functions. Supported by CONICET and Universidad Nacional del Litoral. 1

2

H. AIMAR, L. FORZANI, AND V. NAIBO

integral of f ∈ Lp (Rn ), where Pt (x) = cn t/(t2 + |x|2 )(n+1)/2 , cn = Γ((n + 1)/2)/π (n+1)/2 . Then Bpα,q (Rn ) is the set of functions for which the norm Z ∞ 1 ¯ ∂α uf q ¯ dt q (·, t)k ) (1.1) kf kBpα,q (Rn ) = kf kLp (Rn ) + ( ktα−α p n α ¯ L (R ) t ∂t 0

is finite (with the obvious changes for q = ∞). In this paper we study the differentiation of integrals of functions in Bpα,q (Rn ) with respect to B. The problem only has interest when n ≥ 2 and p < ∞ since otherwise we are dealing with intervals or continuous functions. The main results are the following: (n−1)/p,1

Theorem 1.2. a) If 1 ≤ p < ∞ then B differentiates Bp (Rn ). n b) If 1 ≤ p < n and 0 < α < p − 1 then B does not differentiate Bpα,1 (Rn ). Using the immersion theorems for Besov spaces we can extend the positive and negative results in the following way: Corollary 1.3. a) Let 1 ≤ p < ∞. If either 1 < q ≤ ∞ and α > n−1 or q = 1 and p α,q n α ≥ n−1 then B differentiates B (R ). p p b) If 1 ≤ p < n, 0 < α < np − 1 and 1 ≤ q ≤ ∞ then B does not differentiate Bpα,q (Rn ). Let us point out that we do not have a result for np − 1 ≤ α < n−1 p . Part a) of Theorem 1.2 will follow from the next local weak type inequality: Theorem 1.4. Let M be the maximal operator associated to B, Z 1 Mf (x) = sup |f (y)| dy. x∈R∈B |R| R If 1 ≤ p < ∞ and r > 0 then (1.5)

|{x ∈ Rn : Mf (x) > λ}| ≤

0 c (r + 1)n−1 r1/p kf kB (n−1)/p,1 (Rn ) p λ

(n−1)/p,1

for every f ∈ Bp (Rn ) with supp(f ) ⊂ B(0, r) = {x ∈ Rn : |x| < r}. Here c is a constant independent of f and r and p1 + p10 = 1. Part b) of Corollary 1.3 can be improved to get the following result: Theorem 1.6. If 1 ≤ p < n, 0 < α < np − 1 and 1 ≤ q ≤ ∞ then there exists f ∈ Bpα,q (Rn ) such that Z 1 lim sup f (y) dy = +∞ diam(R)→0, x∈R∈B |R| R for every x ∈ Q0 = {y = (y1 , · · · , yn ) : − 21 < yi ≤ 12 , i = 1, · · · , n}. In §2 we enumerate different equivalent norms in Bpα,q (Rn ) that we will use throughout this paper and we state immersion and regularization results for Besov spaces. In §3 we prove a trace inequality which is interesting by itself and will be useful in the proof of Theorem 1.4. In §4 we give the proofs of parts a) of Theorem 1.2 and Corollary 1.3 (positive results). In §5 we present the proofs of parts b) of Theorem 1.2 and Corollary 1.3 and Theorem 1.6 (negative results). In §6 we consider the differentiation of integrals of Besov functions with respect to a somehow more general basis. In particular we show that the gap np − 1 ≤ α < n−1 p is filled with negative results. In what follows c will denote a constant that can vary even within a single chain of inequalities.

RECTANGULAR DIFFERENTIATION OF INTEGRALS OF BESOV FUNCTIONS

3

2. norms, inmersions, regularization and localization of besov functions. Throughout this paper we will use different norms equivalent to (1.1) in Bpα,q (Rn ) : 1. For k an integer larger than α, an equivalent norm is given by Z ∞ 1 ∂k u q. (2.7) kf kLp (Rn ) + ( ktk−α ∂tkf (·, t)kqLp (Rn ) dt ) t 0

(Taibleson [11]). 2. For k ∈ N and f ∈ Lp (Rn ) we introduce the Lp modulus of continuity of f of order k, ωk (f, t)p =

sup |h|≤t,h∈Rn

k∆kh f kLp (Rn )


Pk where ∆kh f (x) = i=0 ki (−1)k−i f (x + ih) is the finite difference operator of order k and step h. Then, if k > α, Z ∞
1 q p n (2.8) kf kL (R ) + (t−α ωk (f, t)p )q dt t 0

is an equivalent norm in Bpα,q (Rn ) (Triebel [12]). 3. Let ϕ ∈ S(Rn ), the Schwartz class, be such that supp(ϕ) ˆ ⊂ {1/2 ≤ |ξ| ≤ 2} and |ϕ(ξ)| ˆ ≥ c for 3/5 ≤ |ξ| ≤ 5/3 for some constant c > 0, where ϕˆ denotes the Fourier transform of ϕ. Another equivalent norm in Bpα,q (Rn ) is given by Z ∞
q
1/q (2.9) kf kLp (Rn ) + t−α kϕt ∗ f kLp (Rn ) dt t 0

where ϕt (x) = t−n ϕ(x/t) (Triebel [12],Bui-Paluszy´ nski-Taibleson [2]). R∞ In (2.7), (2.8) and (2.9) we have the obvious changes for q = ∞ and if q < ∞, 0 may R1 be replaced by 0 . We have the following immersion results for Besov spaces (Taibleson [11]): n β,v Proposition 2.10. Bpα,q (Rn ) ,→ Bw (Rn ) if and only if either p ≤ w and α− np > β − w n n or p ≤ w, α − p = β − w and q ≤ v. Localization and regularization are two valid procedures in Besov spaces. The following lemma will be helpful in the proofs of the results in §4 (see Taibleson [11] Pn and Triebel [12]). In the following if ν = (ν1 , · · · , νn ) ∈ Nn0 is a multiindex then |ν| = i=1 νi and Dν g denotes the derivative of order ν of g. C0∞ (Rn ) is the set of indefinitely differentiable functions on Rn , with compact support. Lemma 2.11. Let α > 0 and 1 ≤ p, q ≤ ∞. a) If p, q < ∞ and f ∈ Bpα,q (Rn ) with supp(f ) ⊂ B(0, r), r > 0, then, given  > 0, there exists g ∈ C0∞ (Rn ) such that supp(g) ⊂ B(0, 2r) and kf − gkBpα,q (Rn ) < . b) If g ∈ C0∞ (Rn ) then gf ∈ Bpα,q (Rn ) for every f ∈ Bpα,q (Rn ). Moreover if m ∈ N, m > α, then X kgf kBpα,q (Rn ) ≤ c kDν gkL∞ (Rn ) kf kBpα,q (Rn ) . |ν|≤m

where c is a constant independent of f and g. c) We have C0∞ (Rn ) ⊂ Bpα,q (Rn ). Moreover, if p, q < ∞ then C0∞ (Rn ) is a dense subset in Bpα,q (Rn ).

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H. AIMAR, L. FORZANI, AND V. NAIBO

3. a trace inequality. If x ∈ Rn we set x = (x0 , xn ) where x0 ∈ Rn−1 and xn ∈ R. For a function f of variable x ∈ Rn , we write kf (·, xn )kBpα,1 (Rn−1 ) for the norm in Bpα,1 (Rn−1 ) of f (x0 , xn ) as a function of x0 . In this section we are going to prove the following theorem: Theorem 3.12. Let α > 0, 1 ≤ p ≤ ∞, r > 0 and f ∈ C0∞ (Rn ). If supp(f ) ⊂ B(0, r) then Z r 0 kf (·, xn )kBpα,1 (Rn−1 ) dxn ≤ c r1/p kf kBpα,1 (Rn ) −r

where c is a constant independent of f and r. For the proof of Theorem 3.12 we need the following lemmata: Lemma 3.13. If f is an indefinitely differentiable function on Rn , k ∈ N and h ∈ Rn then Z X k ν ν k! ∆h f (x) = ν! (D f )(x + ξh)h Mk (ξ) dξ, R |ν|=k

where M1 = χ(0,1) and Mk = M1 ∗ Mk−1 for k ≥ 2. Here hν = hν11 · · · hνnn and ν! = ν1 ! · · · νn ! if h = (h1 , · · · , hn ) and ν = (ν1 , · · · , νn ). Lemma 3.14. If 1 ≤ p ≤ ∞ and f ∈ C0∞ (Rn ) then uf (x0 , xn , y) → f (x0 , xn ) in Lp (Rn−1 ) as y → 0 for every xn ∈ R. Proof of Theorem 3.12. Fix r > 0, α > 0, 1 ≤ p ≤ ∞ and k ∈ N, k > α. Let f ∈ C0∞ (Rn ) with supp(f ) ⊂ B(0, r), y, t > 0, y < t, and consider the following integral version of Taylor’s formula uf (x0 , xn , y) =

k−1 X

m 1 ∂ uf m! ∂tm

(x0 , xn , t)(y − t)m −

1 k−1!

m=0

Z

t

(y − s)k−1

y

∂ k uf ∂sk

(x0 , xn , s) ds.

For h ∈ Rn−1 let us perform the finite difference operator of order k and step h in the variable x0 , then ∆kh uf (x0 , xn , y) =

k−1 X

k ∂ uf 1 m! ∆h ∂tm

1 k−1!

Z

m

(x0 , xn , t)(y − t)m

m=0

−

t

(y − s)k−1 ∆kh

y

∂ k uf ∂sk

(x0 , xn , s) ds.

∂mu

By Lemma 3.13 applied to ∂tmf (x0 , xn , t) as a function of x0 for m = 0, · · · , k − 1, where the order of derivation ν belongs to Nn−1 , we obtain 0 ∆kh uf (x0 , xn , y)

=

k−1 X

1 m!

1 k−1!

Z

m ν ∂ uf k! ν! (D ∂tm

)(x0 + ξh, xn , t)hν Mk (ξ) dξ (y − t)m

R |ν|=k

m=0

−

Z X

y

t

(y − s)k−1 ∆kh

∂ k uf ∂sk

(x0 , xn , s) ds.

RECTANGULAR DIFFERENTIATION OF INTEGRALS OF BESOV FUNCTIONS

Applying Minkowski’s integral inequality and noting that ∂k u k∆kh ∂skf

(·, xn , s)kLp (Rn−1 ) ≤

∂k u 2k k ∂skf

k∆kh uf (·, xn , y)kLp (Rn−1 ) ≤

R

Mk (ξ) dξ = 1 and that

(·, xn , s)kLp (Rn−1 ) it turns out that

k−1 X

1 m!

2k k−1!

Z

m ν ∂ uf k! ν! kD ∂tm

X

m=0

+

R

5

(·, xn , t)kLp (Rn−1 ) |h|k (t − y)m

|ν|=k t

(s − y)k−1 k

y

∂ k uf ∂sk

(·, xn , s)kLp (Rn−1 ) ds.

Since by Lemma 3.14 ∆kh uf (x0 , xn , y) → ∆kh f (x0 , xn ) in Lp (Rn−1 ) as y → 0 then k∆kh f (·, xn )kLp (Rn−1 ) ≤

k−1 X

1 m!

2k k−1!

Z

m=0

+

m ν ∂ uf k! ν! kD ∂tm

X

(·, xn , t)kLp (Rn−1 ) |h|k tm

|ν|=k t

sk−1 k

0

∂ k uf ∂sk

(·, xn , s)kLp (Rn−1 ) ds.

Taking supremun for |h| ≤ t, multiplying by t−α and integrating in (0, ∞) with respect to dt t we have Z ∞ Z ∞ k−1 X X ∂mu −α dt 1 k! t ωk (f (·, xn ), t)p t ≤ ktm+k−α Dν ∂tmf (·, xn , t)kLp (Rn−1 ) dt m! ν! t 0

m=0 2k k−1!

+

0

|ν|=k

Z

∞

t−α

0

Z

t

sk−1 k

0

∂ k uf ∂sk

(·, xn , s)kLp (Rn−1 ) ds dt t .

Applying Hardy’s inequality to the second term on the right hand side, integrating in (−r, r) with respect to dxn and then applying H¨older’s inequality, we get Z r Z ∞ t−α ωk (f (·, xn ), t)p dt t dxn −r

0

≤

k−1 X

X

1 m!

m=0

+

k

2 k−1!α

Z

k! ν!

|ν|=k ∞

∞

Z

0

(2r)1/p ktm+k−α Dν

0 0

(2r)1/p ktk−α

0

∂ k uf ∂tk

∂ m uf ∂tm

(·, t)kLp (Rn )

(·, t)kLp (Rn )

dt t

dt t .

Using that, for |ν| = k, we have (see Taibleson [11, Lemma 4 and Theorem 1]) Z ∞ Z ∞ m m+k−α ν ∂ uf dt kt D ∂tm (·, t)kLp (Rn ) t ≤ c ktk−α Dν uf (·, t)kLp (Rn ) dt t 0 0 Z ∞ ∂k u ≤c ktk−α ∂tkf (·, t)kLp (Rn ) dt t , 0

then Z

r

−r

Z 0

∞

t−α ωk (f (·, xn ), t)p

dt t

This last inequality and the fact that the desired result.

dxn ≤ c r1/p

0

Z 0

Rr

−r

∞

ktk−α

∂ k uf ∂tk

(·, t)kLp (Rn ) 0

dt t .

kf (·, xn )kLp (Rn−1 ) dxn ≤ (2r)1/p kf kLp (Rn ) give 

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H. AIMAR, L. FORZANI, AND V. NAIBO

Proof of Lemma 3.13. We follow Bennett-Sharpley [1]. We first prove the lemma for n = 1 by induction on k. Noting by f 0 the first derivative of f, we have for k = 1, Z

0

f (x + ξh)hM1 (ξ) dξ =

Z

1

f 0 (x + ξh)h dξ

0

R

= f (x + h) − f (x) = ∆1h f (x). Assume the formula is true for k = 1, · · · , m and let us prove it for k = m + 1. We note by f (k) the derivative of f ofRorder k. Applying ∆1h to both sides of the induction (m) hypothesis for k = m, ∆m (x + ξh)hm Mm (ξ) dξ, and using the induction h f (x) = R f hypothesis for k = 1 applied to f (m) we obtain h

−(m+1)

∆m+1 f (x) h

Z

h−1 (∆1h f (m) )(x + ξh)Mm (ξ) dξ Z Z
−1 = h f (m+1) (x + ξh + τ h)hM1 (τ ) dτ Mm (ξ) dξ R ZR Z
(m+1) = f (x + uh)M1 (u − ξ) du Mm (ξ) dξ ZR R Z
(m+1) = f (x + uh) M1 (u − ξ)Mm (ξ) dξ du R ZR (m+1) = f (x + uh)Mm+1 (u) du. =

R

R

In the next to the last equality we have used Fubini’s Theorem since the integrals converge absolutely. h Assume now n ≥ 2. Let g(t) = f (x + t |h| ), t ∈ R. Observe that ∆k|h| g(t) = (∆kh f )(x +
P h k! h ν ν t |h| ) and that g (k) (ξ|h|) = . Applying the thesis of the |ν|=k ν! (D f )(x + ξh) |h| lemma for n = 1 to g in t = 0 we obtain ∆kh f (x) = ∆k|h| g(0) =

Z

g (k) (ξ|h|)|h|k Mk (ξ) dξ

R

=

Z X

ν k! ν! (D f )(x

+ ξh)

=

Z X

ν k! ν! (D f )(x

+ ξh)hν Mk (ξ) dξ.

R |ν|=k


h ν |h|k Mk (ξ) dξ |h|

R |ν|=k

 Proof of Lemma 3.14. Let f ∈ C0∞ (Rn ). Observe that uf (x, y) → f (x) uniformly in Rn as y → 0, so in particular uf (x0 , xn , y) → f (x0 , xn ) in L∞ (Rn−1 ) as y → 0 for every xn ∈ R. So we may assume 1 ≤ p < ∞. We follow some ideas of Taibleson [11, Theorem 12].

RECTANGULAR DIFFERENTIATION OF INTEGRALS OF BESOV FUNCTIONS

7

Let j ∈ N0 . Then Z Z ∂j u ∂j u | ∂yjf (x0 ,xn , y)|p dx0 = | ∂yjf (x0 , xn , y/2) ∗ Py/2 (x0 , xn )|p dx0 n−1 Rn−1 Z Z R
p ∂j u ≤ | ∂yjf (z, y/2)| Py/2 (x0 − z 0 , xn − zn ) dz dx0 0 n−1 n Zx ∈R Z z∈R ∂j u ≤ | ∂yjf (z, y/2)|p Py/2 (x0 − z 0 , xn − zn ) dz dx0 0 n−1 n x ∈R z∈R Z ∂j u −1 ≤ cy | ∂yjf (z, y/2)|p dz. z∈Rn p n−1

In particular, uf (·, xn , y) ∈ L (R (3.15)

k

∂uf ∂y

) and

(·, xn , y)kLp (Rn−1 ) ≤ c y −1/p k

∂uf ∂y

(·, y/2)kLp (Rn ) .

0

Observe that if 0 < γ < 1/p then (3.16)

k

∂uf ∂y

0

(·, y)kLp (Rn ) ≤ c y γ−1/p . ∂u

In fact, taking into account that k ∂yf (·, y)kLp (Rn ) is a non increasing function of y for y ∈ (0, +∞)(Stein [7, page 154, Lemma 6 ]) we have Z y 0 ∂u 0 −1 1/p0 −γ ∂uf p n (1/p − γ) y k ∂y (·, y)kL (R ) = t1/p −γ k ∂yf (·, y)kLp (Rn ) dt t Z0 y 0 ∂u ≤ t1/p −γ k ∂tf (·, t)kLp (Rn ) dt t Z0 ∞ 0 ∂u ≤ t1/p −γ k ∂tf (·, t)kLp (Rn ) dt t . 0

1−1/p0 +γ,1 Bp (Rn )

Since f ∈ C0∞ (Rn ) ⊂ (Lema 2.11) the last integral is finite and then we have (3.16). ∂u From (3.15) and (3.16) we obtain k ∂yf (·, xn , y)kLp (Rn−1 ) ≤ c y γ−1 . So, if y 0 < y, Z y ∂u 0 kuf (·, xn , y) − uf (·, xn , y )kLp (Rn−1 ) ≤ k ∂tf (·, xn , t)kLp (Rn−1 ) dt y0

≤c

Z

y

tγ−1 dt.

y0

We then have that uf (x0 , xn , y) converges in Lp (Rn−1 ) to a function gxn (x0 ) ∈ Lp (Rn−1 ) as y → 0. Since uf (x, y) → f (x) uniformly in Rn as y → 0, it must be gxn (x0 ) = f (x0 , xn ) and we obtain the desired result.  4. proofs of 1.2.a and 1.3.a (positive results). Since the proof of Theorem 1.4 follows that of Stokolos [9, Lemma1] which makes use of the non-increasing rearrangement of a function, we give its definition here and we state a property (equality (4.17)) which will be useful in the proofs of both Theorem 1.4 and Lemma 4.18 below. Suppose f is a measurable function in Rn . The non-increasing rearrangement of f is the function f ∗ defined in [0, +∞) by f ∗ (t) = inf{λ ≥ 0 : mf (λ) ≤ t}

t ≥ 0,

where mf is the distribution function of f, that is, mf (λ) = |{x ∈ Rn : |f (x)| > λ}|.

8

H. AIMAR, L. FORZANI, AND V. NAIBO

Let t > 0, then (4.17)

Z 0

t

f ∗ (s) ds =

sup |E|=t, E⊂Rn

Z

|f (x)| dx.

E

See for example Bennett-Sharpley [1, page 53, Propotition 3.3 ]. For the proof of Theorem 1.4 we also need the following estimate of rearrangements in terms of Besov norms. Lemma 4.18. For 1 ≤ p < ∞ we have Z 1 t ∗ f (s) ds ≤ c kf kB n/p,1 (Rn ) p t 0 for all t > 0. Here c is a constant independent of f and t. Proof of Theorem 1.4. Fix 1 ≤ p < ∞ and r > 0. By part a) of Lemma 2.11 and using standard arguments it is enough to prove inequality (1.5) for f ∈ C0∞ (Rn ) with supp(f ) ⊂ B(0, r). We follow Stokolos [9, Lemma 1] to get the proof of the theorem. Let f ∈ C0∞ (Rn ) with supp(f ) ⊂ B(0, r) and x = (x0 , xn ) ∈ R ∈ B. It can be shown ¯ containing R, whose measure and diameter are that there exists a measurable set R, comparable to those of R, whose projection onto a coordinate axis, yn for instance, ¯ by the hyperplane yn = t, denoted by R ¯ t (i.e. is an interval I and the section of R ¯ t = {y 0 ∈ Rn−1 : (y 0 , t) ∈ R}), ¯ R has constant measure for t ∈ I. We denote by f ∗ (t, yn ) the non-increasing rearrangement of f (y 0 , yn ) for each fixed yn as a function of y 0 . Then, using (4.17) and Lemma 4.18 we have Z Z 1 1 |f (y)| dy ≤ c ¯ |f (y)| dy |R| R |R| R¯ Z Z 1 1 0 0 =c ¯ y | R¯ |f (y , yn )| dy dyn |I| I |R n yn Z Z |R¯ yn | 1 1 ≤c f ∗ (t, yn ) dt dyn ¯y | 0 |I| I |R n Z 1 ≤c kf (·, yn )kB (n−1)/p,1 (Rn−1 ) dyn . p |I| I Then, if Mhl denotes the Hardy-Littlewood maximal operator on R, it turns out that Mf (x) ≤ c Mhl (kf (·, ·)kB (n−1)/p,1 (Rn−1 ) )(xn ), p

and since the sets in B have diameter smaller than 1, supp(f ) ⊂ B(0, r) and Mhl is of weak type (1, 1) , it follows that |{x ∈ Rn : Mf (x) > λ}| = |{x ∈ B(0, r + 1) : Mf (x) > λ}| Z (r + 1)n−1 r ≤c kf (·, xn )kB (n−1)/p,1 (Rn−1 ) dxn . p λ −r Taking into account the result of Theorem 3.12 we complete the proof of the theorem.  Proof of part a) of Theorem 1.2. By part b) of Lemma 2.11 it is enough to prove the (n−1)/p,1 theorem for functions of compact support. Let f ∈ Bp (Rn ), supp(f ) ⊂ B(0, r),

RECTANGULAR DIFFERENTIATION OF INTEGRALS OF BESOV FUNCTIONS

9

r > 0 and Z 1 |f (y) − f (x)| dy R→x |R| R Z 1 = inf sup{ |f (y) − f (x)| dy, x ∈ R ∈ B, diam(R) < δ}. δ>0 |R| R

Γf (x) = lim sup

Set t > 0 and Et (f ) = {x ∈ Rn : Γf (x) > t}. We are done if we prove that |Et (f )|e = 0 for every t > 0, where |.|e denotes outer measure. By part a) of Lemma 2.11, given  > 0 there exists g ∈ C0∞ (Rn ) such that supp(g) ⊂ B(0, 2r) and kf − gkB (n−1)/p,1 (Rn ) < . We have p

Γf (x) ≤ Γ(f − g)(x) + Γg(x) ≤ M(f − g)(x) + |f (x) − g(x)| + Γg(x) = M(f − g)(x) + |f (x) − g(x)|. So, using Theorem 1.4, |Et (f )|e ≤ |{x ∈ Rn : M(f − g)(x) > 2t }| + |{x ∈ Rn : |f (x) − g(x)| > 2t }| 0

≤ (c/t)(2r + 1)n−1 (2r)1/p kf − gkB (n−1)/p,1 (Rn ) + (2/t)p kf − gkpLp (Rn ) p

0

≤ (c/t)(2r + 1)n−1 (2r)1/p  + (2/t)p p . Since  is arbitrary, this ends the proof of the theorem.



Proof of part a) of Corollary 1.3. It is a direct consequence of part a) of Theorem 1.2 and (n−1)/p,1 Proposition 2.10. In fact, we have Bpα,q (Rn ) ⊂ Bp (Rn ) if α > n−1 p , 1 ≤ q ≤ ∞ and 1 ≤ p < ∞.  Proof of Lemma 4.18. We will prove the lemma using a result which is essentially Calder´ on’s representation formula for tempered distributions. More precisely, it can be shown that for µ a finite Borel measure satisfying the standard Tauberian condition (for all ξ ∈ Rn , ξ 6= 0, there exists s > 0 such Rthat µ ˆ(sξ) 6= 0) there exists η ∈ S(Rn ) such that ∞ supp(ˆ η ) is contained in an annulus and 0 µ ˆ(sξ)ˆ η (sξ) ds s = 1 for ξ 6= 0 (Heideman [5]). n/p,1

Fix 1 ≤ p < ∞. We assume f ∈ Bp (Rn ) since otherwise there is nothing to prove. n/p,1 n Consider in Bp (R ) the norm (2.9) and let η ∈ S(Rn ) be associated to µ = ϕ(x) dx according to Heideman’s result above. We then define  R∞ ϕ(sξ)ˆ ˆ η (sξ) ds ˆ s , ξ 6= 0, 1 ψ(ξ) = 1, ξ = 0. Observe that ψˆ ∈ C0∞ (Rn ) and ψˆ = 1 in a neighborhood of the origin. In fact, if ˆ ˆ supp(ˆ η ) ⊂ {ξ : 0 < a < |ξ| < b}, then ψ(ξ) = 1 if |ξ| < a and ψ(ξ) = 0 if |ξ| > b. It is R ∞ n n now clear that ψˆ ∈ C0 (R ). Then ψ ∈ S(R ) and Rn ψ(x)dx = 1. So that ψ is a good approximation toR the identity. a 1 n Set I,a (x) =  (ϕs ∗ ηs )(x) ds s ,  < a. The function I,a ∈ L (R ) since Z Z aZ |I,a (x)| dx ≤ |(ϕs ∗ ηs )(x)| dx ds s n n R  R
≤ kϕkL1 (Rn ) kηkL1 (Rn ) log a .

10

H. AIMAR, L. FORZANI, AND V. NAIBO

By Fubini’s theorem, ˆ (ξ) = I,a =

Z

Z

a

(ϕs ∗ ηs )(x)e−ixξ

n ZRa Z 

Rn



ds s

dx

(ϕs ∗ ηs )(x)e−ixξ dx ds s

a

=

Z

=

Z a

(ϕs ∗ ηs )ˆ(ξ) ds s ϕ(sξ)ˆ ˆ η (sξ) ds s



ˆ ˆ = ψ(ξ) − ψ(aξ) = ψˆ (ξ) − ψˆa (ξ). So I,a ∈ S(Rn ) and I,a = ψ − ψa . Then, since convolution with f commutes with the integral, we arrive at Z a (4.19) (ψ ∗ f )(x) − (ψa ∗ f )(x) = (I,a ∗ f )(x) = (ϕs ∗ ηs ∗ f )(x) ds s . 

Applying Minkowski’s integral inequality and Young’s inequality we have Z a kψ ∗ f − ψa ∗ f kLp (Rn ) ≤ kηkL1 (Rn ) kϕs ∗ f kLp (Rn ) ds s 

and since ψ ∗ f → f in Lp (Rn ) as  → 0, it turns out that Z a (4.20) kf − ψa ∗ f kLp (Rn ) ≤ kηkL1 (Rn ) kϕs ∗ f kLp (Rn ) 0

ds s .

On the other hand, taking a = 1 (ψa = ψ) in (4.19) and applying H¨older’s inequality, Z 1 |(ψ ∗ f )(x)| ≤ |(ψ ∗ f )(x)| + |(ϕs ∗ ηs ∗ f )(x)| ds s 

≤ kψkLp0 (Rn ) kf kLp (Rn ) +

Z

1

kηs kLp0 (Rn ) kϕs ∗ f kLp (Rn )



= kψkLp0 (Rn ) kf kLp (Rn ) + kηkLp0 (Rn )

Z

ds s

1

s−n/p kϕs ∗ f kLp (Rn )



≤ c kf kLp (Rn ) +

(4.21)

Z 

ds s

1

s−n/p kϕs ∗ f kLp (Rn )

ds s




.

Then, using (4.17), H¨ older’s inequality, (4.20) and (4.21) we have, for a < 1, Z t Z 1 1 f ∗ (s) ds = sup |f (x)| dx t 0 t E⊂Rn , |E|=t E Z 1 ≤ sup |f (x) − (ψa ∗ f )(x)| dx + kψa ∗ f kL∞ (Rn ) t E⊂Rn , |E|=t E ≤

1

kf − ψa ∗ f kLp (Rn ) + kψa ∗ f kL∞ (Rn ) Z Z 1 kηkL1 (Rn ) a ds p n p n ≤ kϕs ∗ f kL (R ) s + c kf kL (R ) + s−n/p kϕs ∗ f kLp (Rn ) t1/p 0 a t1/p

ds s




.

RECTANGULAR DIFFERENTIATION OF INTEGRALS OF BESOV FUNCTIONS

11

If t < 1 and a = t1/n we obtain 1 t1/p

t1/n

Z

kϕs ∗

0

f kLp (Rn ) ds s

t1/n

Z

≤

s−n/p kϕs ∗ f kLp (Rn )

0

ds s

and then 1 t

Z

Z

t

t

0

f ∗ (s) ds ≤ c kf kB n/p,1 (Rn ) ,

t < 1.

p

If t ≥ 1 then 1 t

Z 0

t

f ∗ (s) ds ≤

1 t1/p

f ∗ (s)p ds

0


1

Z

p≤

∞

f ∗ (s)p ds

0


1

= kf kLp (Rn ) ≤ kf kB n/p,1 (Rn ) .

p

p



5. proofs of 1.2.b, 1.3.b and 1.6 (negative results). Proof of part b) of Theorem 1.2. Fix 1 ≤ p < n and 0 < α < np − 1. Following Stokolos [9, 10], we consider in Q0 = {(y1 , · · · , yn ) : − 12 < yi ≤ 12 , i = 1, · · · , n}, mn disjoint equal cubes Ijm of measure |Ijm | = m−n . Let Qm j be the cube concentric with (n−1)/p,∞

−nm Ijm of measure |Qm . Let ψ be a non-negative function in B∞ (Rn ) with j | = 2 1 m m m+1 supp(ψ) ⊂ B(0, 1) and ψ ≡ 1 in B(0, 2 ). We define fj (x) = 2 ψ(2 (x − xm j )), m m m m where xj is the center of Qj . Then supp(fj ) ⊂ Qj and since, from (2.8), |∆nh ψ(x)| ≤ c |h|(n−1)/p for every h ∈ Rn and x ∈ Rn , we have

|∆nh fjm (x)| = |

n   X n i=0 m

= |2 = ≤

i

(−1)n−i fjm (x + ih)|

n   X n

i

(−1)n−i ψ(2m+1 (x + ih − xm j ))|

i=0 m n |2 ∆2m+1 h ψ(2m+1 (x n−1 c 2m |2m+1 h| p

− xm j ))|

n−1 n−1 m(1+ p ) |h| p .

= c2 Set

n

f (x) =

m ∞ X X

fjm (x)

m=1 j=1

for x ∈ Rn . Clearly f ∈ Lp (Rn ) since n

kf kLp (Rn ) ≤

∞ X m X

m=1 j=1

n −p

kfjm kLp (Rn ) = 2

kψkLp (Rn )

∞ X

m=1

mn 2

n m(1− p )

< ∞.

12

H. AIMAR, L. FORZANI, AND V. NAIBO

Moreover f ∈ Bpα,1 (Rn ). In fact, if t > 0 and l ∈ N are such that 2−(l+1) ≤ t < 2−l and h ∈ Rn with |h| < t, we have n

k∆nh f kLp (Rn )

≤

m l X X

n

k∆nh fjm kLp (cn Qm j )

+2

m=1 j=1

≤

n−1 c (|h| p

n

∞ m X X

m=l+1 j=1 l X

n−1 nm m(1+ p ) − p mn 2 2

∞ X

+

m=1

≤ c (|h|

kfjm kLp (Rn ) n m(1− p )

mn 2

)

m=l+1

1 n−1 p ln+1 2l(1− p )

∞ X

+

n m(1− p )

mn 2

).

m=l+1 n n P∞ m(1− p ) (l+1)(1− p ) Since |h| < t, m=l+1 mn 2 ≤ c (l + 1)n+1 2 for 1 ≤ p < n and t ≥ −(l+1) 2 , then both terms on the right hand side of the above inequality are bounded n−1

1 l(1− p )

above by c t p ln+1 2 Z 1 t−α ωn (f, t)p 0

. We then have, ∞ Z 2−l n−1 1 X −α n+1 l(1− p ) ≤c t p l 2

dt t

l=0

=c

∞ X

2−(l+1)

1 l(1− p )

ln+1 2

n−1 −l( p −α)

(2

dt t n−1 −(l+1)( p −α) )

−2

0, 1 ≤ p, q ≤ ∞, m, n ∈ N, m ≤ n. For φ ∈ Bpα,q (Rm ) and g ∈ Bpα,q (Rn−m ) let f (x1 , · · · , xn ) = φ(x1 , · · · , xm )g(xm+1 , · · · , xn ). Then f ∈ Bpα,q (Rn ). The proof of Theorem 6.22 we present here is essentially that of Stokolos [9, Theorem 1]. Fix 0 < α < n−1 p , 1 ≤ p < ∞ and 1 ≤ q ≤ ∞, then there exists a non-negative function v ∈ Bpα,q (Rn−1 ) such that v ∈ / L∞ (Rn−1 ). Let {mi }i∈N be a dense subset in Rn−1 and define ∞ X φ(y) = 2−i v(y + mi ), y ∈ Rn−1 . i=1

Then Let g

φ ∈ Bpα,q (Rn−1 ) and φ is unbounded in every neighborhood of every point of Rn−1 . ∈ Bpα,q (R) be a non-negative function such that g = 1 in (−1, 1) and consider f (x1 , · · · , xn ) = φ(x1 , · · · , xn−1 )g(xn ).

Then, by Lemma 6.23, f ∈ Bpα,q (Rn ). We will see that P does not differentiate the integral of f by showing that for every x = (x1 , · · · , xn ) with |xn | < 1 we have Z 1 lim sup f (y) dy = +∞. diam(E)→0, x∈E∈P |E| E Fix x = (x1 , · · · , xn ) ∈ Rn with |xn | < 1. Given an n − 1 dimensional neighborhood N of R (x1 , 0· · · ,0xn−1 ) let Q be an (n − 1)−dimensional ball contained in N such that 1 |Q| Q φ(y ) dy is great enough. Let w = (w1 , · · · , wn−1 ) be the center of Q and a < b < c numbers such that b − a = c − b, (a, c) ⊂ (−1, 1) and xn ∈ (b, c). Let J be the segment through x starting at (w1 , · · · , wn−1 , b) and such that its projection onto the xn −axis is the closed interval [b, c]. We define E1 as the union of all balls congruent to Q with center in J and E2 = Q × (a, b). Let E = E1 ∪ E2 , so E ∈ P, x ∈ E, |E| = 2|Q|(b − a) and the

14

H. AIMAR, L. FORZANI, AND V. NAIBO

diameter of E can be made small enough taking N and c − a small enough; moreover we have Z Z Z 1 1 1 f (y) dy ≥ f (y) dy = φ(y 0 ) dy 0 . |E| E 2|Q|(b − a) E2 2|Q| Q  Proof of Lemma 6.23. Let φ, g, f be as in the statement of the lemma. Observe that f ∈ Lp (Rn ), since kf kLp (Rn ) = kφkLp (Rm ) kgkLp (Rn−m ) . Let t > 0, h ∈ R, |h| < t and k ∈ N, k > α. Then, if j = 1, · · · , m, we have k   X k k ∆hej f (x1 , · · · , xn ) = (−1)k−i φ(x1 , · · · , xj + ih, · · · , xm )g(xm+1 , · · · , xn ) i i=0 = g(xm+1 , · · · , xn )∆khej φ(x1 , · · · , xm ). So, k∆khej f kLp (Rn ) = kgkLp (Rn−m ) k∆khej φkLp (Rm ) and then ωkj (f, t)p :=

sup |h|