J Fourier Anal Appl DOI 10.1007/s00041-015-9454-6

On the Relationship Between Two Kinds of Besov Spaces with Smoothness Near Zero and Some Other Applications of Limiting Interpolation Fernando Cobos1 · Óscar Domínguez1

Received: 9 April 2015 © Springer Science+Business Media New York 2015

Abstract Using limiting interpolation techniques we study the relationship between 0,−1/q Besov spaces B p,q with zero classical smoothness and logarithmic smoothness −1/q defined by means of differences with similar spaces B 0,b,d p,q defined by means 0,−1/2 0,0,1/2 = B2,2 . We of the Fourier transform. Among other things, we prove that B2,2 0,−1/q

also derive several results on periodic spaces B p,q (T), including embeddings in generalized Lorentz–Zygmund spaces and the distribution of Fourier coefficients of 0,−1/q functions of B p,q (T). Keywords Besov spaces · Logarithmic smoothness · Limiting interpolation spaces · Fourier coefficients · Lorentz–Zygmund spaces Mathematics Subject Classification 42A16

Primary 46E35 · 46M35; Secondary 46E30 ·

To the memory of Professor Manuel Antonio Fugarolas. Communicated by Winfried Sickel.

B

Fernando Cobos [email protected] Óscar Domínguez [email protected]

1

Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad Complutense de Madrid, Plaza de Ciencias 3, 28040 Madrid, Spain

J Fourier Anal Appl

1 Introduction Besov spaces with classical smoothness zero and logarithmic smoothness with exponent b are attracting a lot of attention in recent years. See, for example, the papers by Caetano, Gogatishvili and Opic [9,10], Besov [7], Triebel [35] and by the present authors [12–14]. They can be introduced by using differences B0,b p,q or by means of a smooth dyadic resolution of unity and the Fourier transform B 0,b p,q (precise definitions are given in Sect. 2). In contrast to the case when the classical smoothness is positive 0,b (see [24, Theorem 2.5] and [34, 2.5.12]), now B0,b p,q  = B p,q . In [14, Theorem 3.3] we established that 0,b+1/ min{2, p,q}

B p,q

0,b+1/ max{2, p,q}

→ B0,b p,q → B p,q

(1.1)

provided that 1 < p < ∞, 0 < q ≤ ∞ and b > −1/q. In particular, for b > −1/2 0,b+1/2 we have that B0,b . 2,2 = B2,2 It is natural to investigate the connection between these spaces when b = −1/q. Accordingly, we study in this paper such a question. We show that if b = −1/q then 0,−1/q are related to there is another jump in the scale, with the result that spaces B p,q 0,b,d spaces of Fourier type B p,q which in addition to the logarithmic smoothness with exponent b, they also have smoothness of the type of an iterated logarithm to the power 0,−1/2 0,0,1/2 d. In particular, B2,2 = B2,2 . Our approach is based on limiting interpolation (see [13,16,17]). The key result is given in Lemma 3.1. It is a reiteration formula that relates the limiting real interpolation space ((A0 , A1 )θ, p , A1 )(0,1/q),q with interpolation spaces obtained by making logarithmic perturbations of the real method. Using this lemma we derive, among 0,0,1/q 0,−1/q other things, that if 1 < p < ∞ and 0 < q ≤ min{2, p} then B p,q → B p,q , 0,−1/q 0,0,1/q → B p,q . Then we extend this result while if max{2, p} ≤ q < ∞ then B p,q to the case q  min{2, p} (respectively, max{2, p}  q) by using the fact that B 0,b,d p,q is a retract of q ((1 + j)b (1 + log(1 + j))d L p ) and some interpolation formulae for vector-valued sequence spaces. We also study the limit cases p = 1 and ∞. All these results are given in Sect. 3. In the previous Sect. 2 we review the interpolation methods that we use in the paper and we introduce the function spaces that we consider here. Lemma 3.1 has also some other interesting applications that we describe in Sect. 4. 0,−1/q We consider there Besov spaces B p,q (T) of periodic functions on the unit circle T. 0,−1/q First we prove that B p,q (T) is embedded in a certain generalized Lorentz–Zygmund space which closes a limit case left open in [13,23]. Afterwards we study the relationship between the smoothness of the derivatives of f and the smoothness of f . We show the optimality of the result of [13, Theorem 4.7] on the required smoothness for f to have that D k f belongs to B0,b p,q (T) with b > −1/q. Then we consider the case b = −1/q. Finally we investigate the distribution of Fourier coefficients of functions in B0,b p,q (T). We start by showing the optimality of the result given in [13, Theorem 5.1] for b > −1/q and then we describe the behaviour of the Fourier coefficients of func0,−1/q tions in B p,q (T). We close the paper with a result on Lorentz–Zygmund spaces which are near to L 1 (T).

J Fourier Anal Appl

2 Preliminaries If A = (A,  ·  A ) is a quasi-Banach space and λ > 0, we denote by λA the space A with the quasi-norm λ ·  A . Given two quasi-Banach spaces A0 , A1 , we put A1 → A0 to mean that the embedding from A1 into A0 is continuous. Let N0 = N ∪ {0}. If (λ j ) j∈N0 is a sequence of positive numbers, (A j ) j∈N0 is a sequence of quasi-Banach spaces and 0 < q ≤ ∞, we write q (λ j A j ) for the collection of all vector-valued sequences a = (a j ) such that a j ∈ A j for any j ∈ N0 and ⎛ ⎞1/q ∞  aq (λ j A j ) = ⎝ (λ j a j  A j )q ⎠ < ∞ j=0

(as usual, the sum should be replaced by the supremum if q = ∞). Let A0 , A1 be quasi-Banach spaces with A1 → A0 . The Peetre’s K -functional is defined by K (t, a) = K (t, a; A0 , A1 ) = inf{a0  A0 +ta1  A1 :a = a0 +a1 , a j ∈A j }, a∈A0 , t > 0. For 0 < θ < 1 and 0 < q ≤ ∞, the real interpolation space (A0 , A1 )θ,q consists of all those a ∈ A0 having a finite quasi-norm  a(A0 ,A1 )θ,q =

∞

(t −θ K (t, a))q

0

dt t

1/q .

See [5,6,8,33]. The following extension of the real interpolation method is useful. Let g be a slowly varying function on (0, ∞) (see, for example, [19]). We put (A0 , A1 )θ,q,g



= a ∈ A0 : a(A0 ,A1 )θ,q,g =

∞

(t

−θ

q dt

g(t)K (t, a))

0

t

1/q

 <∞

(see [1,20,22]). If g(t) = (1+| log t|)γ , γ ∈ R (respectively, g(t) = (1+| log t|)γ (1+ log(1 + | log t|))δ , γ , δ ∈ R) we write (A0 , A1 )θ,q,γ (respectively, (A0 , A1 )θ,q,γ ,δ ) instead of (A0 , A1 )θ,q,g. We shall also need the limit interpolation spaces (A0 , A1 )(0,η),q formed by all those a ∈ A0 having a finite quasi-norm  a(A0 ,A1 )(0,η),q =

0

1

K (t, a) (1 − log t)η

q

dt t

1/q

(see [13,16]). Here −∞ < η < ∞. Any of these three interpolation methods, say F, has the interpolation property for bounded linear operator: If A0 , A1 , B0 , B1 are quasi-Banach spaces with A1 →

J Fourier Anal Appl

A0 , B1 → B0 and T is a linear operator such that the restrictions T : A0 −→ B0 and T : A1 −→ B1 are bounded, then the restriction T : F(A0 , A1 ) −→ F(B0 , B1 ) is also bounded. Let (U, μ) be a σ -finite measure space. Suppose that g is a slowly varying function on (0, ∞) and let 0 < p, q ≤ ∞. The Lorentz-Karamata space L p,q;g = L p,q;g(U ) consists of all (equivalence classes of) μ-measurable functions f on U such that   f  L p,q;g =

∞

(t

1/ p

∗

q dt

g(t) f (t))

0

1/q

t

<∞

(see [19, Sect. 3.4.3]). Here f ∗ is the non-increasing rearrangement of f . If g(t) = (1 + | log t|)γ , γ ∈ R (respectively, g(t) = (1 + | log t|)γ (1 + log(1 + | log t|))η , γ , η ∈ R) we get the Lorentz–Zygmund space L p,q (log L)γ (respectively, the generalized Lorentz–Zygmund space L p,q (log L)γ (log log L)η ). In particular, the special case γ = 0 produces the Lorentz space L p,q , which coincides with the Lebesgue space L p if p = q. See [5,6,19,33]. When U = N or Z and μ is the counting measure, we denote these spaces by  p,q (log )γ ,  p,q (log )γ (log log )η ,  p,q and  p , respectively. Next we recall the definition of a certain kind of Besov spaces with generalized smoothness. First we do it using differences and then using the Fourier analytic approach. Let f ∈ L p = L p (Rn ). For x, h ∈ Rn and k ∈ N, we put f )(x) = 1h ( kh f )(x). ( 1h f )(x) = f (x + h) − f (x) and ( k+1 h The k-th order modulus of smoothness is given by ωk ( f, t) p = sup  kh f  L p , t > 0, |h|≤t

where |h| stands for the Euclidean norm of h. If k = 1, we simply write ω( f, t) p . Let α ≥ 0, k ∈ N with k > α and b, d ∈ R. The Besov space Bα,b,d p,q is formed by all those f ∈ L p such that 1/q  1 −α b d q dt  f Bα,b,d =  f  L p+ [t (1− log t) (1+ log(1− log t)) ωk ( f, t) p ] < ∞. p,q t 0 0,b If α = d = 0, we simply write B0,b p,q . See [14,18]. The space B p,q has zero classical smoothness and logarithmic smoothness with exponent b. Note that B0,b p,q = L p if b < −1/q. Let ϕ0 be a C ∞ function on Rn such that

supp ϕ0 ⊂ {x ∈ Rn : |x| ≤ 2} and ϕ0 (x) = 1 if |x| ≤ 1.

J Fourier Anal Appl

For j ∈ N and x ∈ Rn put ϕ j (x) = ϕ0 (2− j x) − ϕ0 (2− j+1 x).

(2.1)

Then (ϕ j ) j∈N0 is a smooth dyadic resolution of unity in Rn . For 0 < p, q ≤ ∞ and s, b, d ∈ R, the Besov space B s,b,d p,q is formed by all tempered distributions f on Rn having a finite quasi-norm ⎛

 f  B s,b,d p,q

⎞1/q ∞  = ⎝ (2 js (1 + j)b (1 + log(1 + j))d F −1 (ϕ j F f ) L p )q ⎠ . j=0

Here F stands for the Fourier transform and F −1 for the inverse Fourier transform. We refer to [3,15,24,26,27] for properties of these spaces. If b = d = 0 then B s,0,0 p,q s,b,0 is the usual Besov space B sp,q . We also write B s,b p,q instead of B p,q . In addition to Besov spaces, we shall also work with some Triebel-Lizorkin spaces s (see [33,34]). Recall that if 1 < p < ∞ then F s coincides with the fractional F p,q p,2 k being the Sobolev space W k for k ∈ N and F 0 = L . Sobolev space H ps , with F p,2 p p p,2 s → B s Moreover B sp,min{ p,q} → F p,q p,max{ p,q} (see [34, Proposition 2.3.2/2(iii)]). As usual, if X, Y are non-negative quantities depending on certain parameters, we write X  Y if there is a constant c > 0 independent of the parameters in X and Y such that X ≤ cY . If X  Y and Y  X , we write X ∼ Y .

3 Relationships Between Besov Spaces with Logarithmic Smoothness 0,−1/q

With the aim of describing the connections between B p,q and the corresponding Besov spaces defined by using the Fourier transform, we first establish a result which relates the three interpolation methods introduced in the previous section. It corresponds to the extreme case η = −1/q in [14, Lemma 2.1/(b)]. Subsequently, we write K (t, a) for the K -functional of the couple (A0 , A1 ). The K -functional of a different couple will be pointed out explicitly in the notation. Lemma 3.1 Let A0 , A1 be quasi-Banach spaces with A1 → A0 . Assume that 0 < θ < 1, 0 < p ≤ ∞ and 0 < q < ∞. The following continuous embeddings hold (a) (A0 , A1 )θ,q,max{0,1/ p−1/q},max{1/ p,1/q} → ((A0 , A1 )θ, p , A1 )(0,1/q),q . (b) ((A0 , A1 )θ, p , A1 )(0,1/q),q → (A0 , A1 )θ,q,min{0,1/ p−1/q},min{1/ p,1/q} . Proof Let B = ((A0 , A1 )θ, p , A1 )(0,1/q),q . Suppose p < ∞. The case p = ∞ can be carried out similarly. Holmstedt’s formula [25, Remark 2.1] yields that  K (t, a; (A0 , A1 )θ, p , A1 ) ∼

0

t 1/(1−θ)



K (s, a) sθ

p

ds s

1/ p .

J Fourier Anal Appl

Hence



1  t

a B ∼ 0



0

K (s, a) sθ

p

ds s

q/ p

dt t (1 − log t)

1/q .

(3.1)

If q ≤ p, using the embedding L 1/(1−θ),q ([0, 1]) → L 1/(1−θ), p ([0, 1]) and that K (s, a)/s is a decreasing function, we derive 

1  1 

K (s, a) s χ(0,t) (s) a B ∼ s 0 0   q 1  K (s, a)    χ = (s) (0,t)  s  

p

1−θ

ds s

L 1/(1−θ), p ([0,1])

0

q/ p

dt t (1 − log t)

1/q

1/q

dt t (1 − log t)

1/q  q 1  K (s, a)  dt     s χ(0,t) (s) 0 L 1/(1−θ),q ([0,1]) t (1 − log t)  1  t  q 1/q K (s, a) dt ds = sθ s t (1 − log t) 0 0 q  1 1/q  1  K (s, a) ds dt = sθ 0 s t (1 − log t) s 1/q  1 ds = (s −θ (log(1 − log s))1/q K (s, a))q s 0  a(A0 ,A1 )θ,q,0,1/q . Suppose now q > p. Applying the weighted Hardy’s inequality of [21, Lemma 3.3] to (3.1) we obtain  a B 

1

(t

−θ

(1 − log t)

1/ p

(log(1 − log t))

1/ p

0

dt K (t, a)) t (1 − log t)

1/q

q

 a(A0 ,A1 )θ,q,1/ p−1/q,1/ p . This establishes (a). In order to prove (b) assume first that q ≥ p. Using (3.1) and the embedding between Lorentz spaces we obtain  a B ∼  

  

1  K (s, a) 0

s

q  χ(0,t) (s) 

L 1/(1−θ), p ([0,1])

dt t (1 − log t)

1/q

1/q  q 1  K (s, a)  dt    s χ(0,t) (s) 0 L 1/(1−θ),q ([0,1]) t (1 − log t)

J Fourier Anal Appl



1

=

(s

−θ

(log(1 − log s))

1/q

q ds

K (s, a))

0

1/q

s

∼ a(A0 ,A1 )θ,q,0,1/q where in the last equivalence we have used that K (t, a) ∼ a A0 for t ≥ 1. Now suppose that q < p. Put ρ = q/ p < 1, δ = 1/ρ > 1 and let β > p/q − 1 = 1/ρδ > 0. Hölder’s inequality yields 

t

[(log(1 − log s))−β (s −θ K (s, a)(1 − log s)1/ p ) p s −1/ρ (1 − log s)−1/ρ ]ρ ds

0



t

≤

[(s

−θ

1/ p p −1

K (s, a)(1 − log s)

) s

(1 − log s)

] ds

0



t

×

[s 1/ρ−1 (1 − log s)1/ρ−1 (log(1 − log s))β ]−ρδ ds

0



t

=

(s −θ K (s, a)) p

0

ds s

1/δ 

−βρ+1/δ



∼ (log(1 − log t))

t

(log(1 − log s))−βρδ

0 t

(s

−θ

K (s, a))

0

p ds

s

1/δ

−1 ρδ

q/ p

1/δ

ds s(1 − log s)

1/δ

.

Inserting this estimate in (3.1), we derive 

q

a B 

1

(log(1 − log t))βρ−1/δ 0  t × [(log(1 − log s))−β (s −θ K (s, a)(1 − log s)1/ p ) p ]ρ 0

dt ds s(1 − log s) t (1 − log t)

×  =

1

0

[(log(1 − log s))−β (s −θ K (s, a)(1 − log s)1/ p ) p ]ρ 

×  ∼

1

(log(1 − log t))βρ−1/δ

s 1

ds dt t (1 − log t) s(1 − log s)

[s −θ K (s, a)(1 − log s)1/ p ]q (log(1 − log s))q/ p

0

ds s(1 − log s)

q

∼ a(A0 ,A1 )θ,q,1/ p−1/q,1/ p .  

This completes the proof. 0,−1/q

Now we are ready to compare B p,q logarithmic smoothness.

with the other kind of Besov spaces with

J Fourier Anal Appl

Theorem 3.2 Let 1 < p < ∞ and 0 < q < ∞. Then 0,1/ min{2, p,q}−1/q,1/ min{2, p,q}

B p,q

0,−1/q

→ B p,q

0,1/ max{2, p,q}−1/q,1/ max{2, p,q}

→ B p,q

.

0,−1/q

= (L p , H p1 )(0,1/q),q . Moreover, Proof By [14, Theorem 3.1/(b)], we have that B p,q 0 −1 1 L p → B p,max{2, p} = (H p , H p )1/2,max{2, p} . Therefore, using Lemma 3.1/(b) we derive 0,−1/q

B p,q

→ ((H p−1 , H p1 )1/2,max{2, p} , H p1 )(0,1/q),q → (H p−1 , H p1 )1/2,q,1/ max{2, p,q}−1/q,1/ max{2, p,q} 0,1/ max{2, p,q}−1/q,1/ max{2, p,q}

= B p,q

.

The proof of the left-hand side embedding of the statement is similar but using now   that B 0p,min{2, p} → L p and Lemma 3.1/(a). Remark 3.3 In the previous result we have excluded the case q = ∞. Note that 0 B0,0 p,∞ = L p . On the other hand, by [34, Proposition 2.5.7], we have that L p → B p,∞ for 1 ≤ p ≤ ∞, and this embedding is known to be strict. Corollary 3.4 Let 1 < p < ∞ and 0 < q < ∞. 0,0,1/q

0,−1/q

(a) If q ≤ min{2, p}, then B p,q → B p,q . 0,−1/q 0,0,1/q → B p,q . (b) If max{2, p} ≤ q, then B p,q In particular, we have with equivalence of norms 0,−1/2

B2,2

0,0,1/2

= B2,2

.

The structure of the space B s,b,d p,q has an important peculiarity. To describe it, let ϕ j as in (2.1), put ϕ−1 ≡ 0 and let  ϕ j = ϕ j−1 + ϕ j + ϕ j+1 for j ∈ N0 . We put λ j = js b d 2 (1 + j) (1 + log(1 + j)) , and we consider the mapping J : B s,b,d p,q −→ q (λ j L p ) −1 (ϕ F f )) and the mapping assigning to any f ∈ B s,b,d the sequence J f = (F j p,q ∞ −1 ( R : q (λ j L p ) −→ B s,b,d F ϕ j F f j ) (convergence in p,q defined by R( f j ) = j=0 S ). Using that  ϕ j ϕ j = ϕ j one can check that RJ f = f for any f ∈ B s,b,d p,q . So, in the sense of [33, Definition1.2.4], the space B s,b,d is a retract of  (λ L ), q j p being R a p,q s,b,d retraction from q (λ j L p ) to B p,q and J the corresponding co-retraction from B s,b,d p,q to q (λ j L p ) (see [2,15]). Next we use this property to extend Corollary 3.4/(a) to the case q  min{2, p} [respectively, max{2, p}  q in the case (b)]. We need some auxiliary results. Lemma 3.5 Let A be a Banach space, λ > 1 and 0 < q < ∞. Then (A, λA)(0,1/q),q = (log(1 + log λ))1/q A with equivalence of quasi-norms where the constants are independent of λ.

J Fourier Anal Appl

Proof Clearly K (t, a; A, λA) = min{1, tλ}a A . Whence  a(A,λA)(0,1/q),q = 

1/λ

−1/q q dt

(λt (1 − log t)

)

t

0 1/λ

=

−1/q q dt

(λt (1 − log t)

)

t

0

Since



1/λ

(λt (1 − log t)−1/q )q

0

 +

1

(1 − log t)

1/λ

−1 dt

1/q a A

t

1/q + log(1 + log λ) a A .

dt  (1 + log λ)−1 t  

the result follows. Subsequently, we put p

p

K p (t, a; A0 , A1 ) = inf{(a0  A0 + t p a1  A1 )1/ p : a = a0 + a1 , a j ∈ A j }. Lemma 3.6 Let 0 < p ≤ ∞, 0 < q < ∞ and let (A j ), (B j ) be sequences of Banach spaces with B j → A j and sup j∈N0 I  B j ,A j < ∞, so  p (B j ) →  p (A j ). Then   (a) min{ p,q} (A j , B j )(0,1/q),q → ( p (A j),  p (B j ))(0,1/q),q. (b) ( p (A j ),  p (B j ))(0,1/q),q → max{ p,q} (A j , B j )(0,1/q),q . Proof Let Z = ( p (A j ),  p (B j ))(0,1/q),q . Since ⎛ ⎞1/ p ∞  K p (t, a;  p (A j ),  p (B j )) = ⎝ K p (t, a j ; A j , B j ) p ⎠ , j=0

we have that ⎛ ⎜ a Z ∼ ⎝

⎡



1

⎣

0

∞ 

⎤q/ p

((1 − log t)−1/q K p (t, a j ; A j , B j )) p ⎦

j=0

⎞1/q dt ⎟ ⎠ t

If q ≥ p, triangle inequality yields that ⎛ ⎞ p/q 1/ p ∞  1  dt ⎠ a Z  ⎝ [(1 − log t)−1/q K p (t, a j ; A j , B j )]q t 0 j=0 ⎛ ⎞1/ p ∞  p ∼⎝ a j (A j ,B j )(0,1/q),q ⎠ . j=0

.

J Fourier Anal Appl

If q < p, using that q →  p we obtain ⎛  a Z  ⎝

⎞1/q dt [(1 − log t)−1/q K p (t, a j ; A j , B j )]q ⎠ t j=0 ⎞1/q

∞ 1 0

⎛ ∼⎝

∞ 

a j (A j ,B j )(0,1/q),q ⎠ q

.

j=0

This establishes (a). The proof of (b) is similar but using now Minkowski’s inequality for integrals when p ≥ q.   Theorem 3.7 Let 1 < p < ∞ and 0 < q < ∞. Then 0,0,1/q

0,−1/q

B p,min{2, p,q} → B p,q

0,0,1/q

→ B p,max{2, p,q} .

0,−1/q

Proof By [14, Theorem 3.1/(b)], B p,q = (L p , H p1 )(0,1/q),q . On the other hand, 0 1 1 L p → B p,max{2, p} and H p → B p,max{2, p} . Hence 0,−1/q

B p,q

→ (B 0p,max{2, p} , B 1p,max{2, p} )(0,1/q),q .

The space B kp,max{2, p} is a retract of max{2, p} (2 jk L p ) and for the couple of vectorvalued sequence spaces, according to Lemmas 3.5 and 3.6, we get (max{2, p} (L p ), max{2, p} (2 j L p ))(0,1/q),q   → max{2, p,q} (L p , 2 j L p )(0,1/q),q = max{2, p,q} ((log(1 + j))1/q L p ). 0,−1/q

0,0,1/q

→ B p,max{2, p,q} . Consequently, we conclude that B p,q To establish the left-hand side embedding of the statement, we use now that B 0p,min{2, p} → L p and B 1p,min{2, p} → H p1 . Hence 0,−1/q

(B 0p,min{2, p} , B 1p,min{2, p} )(0,1/q),q → B p,q

.

Since min{2, p,q} ((log(1 + j))1/q L p ) → (min{2, p} (L p ), min{2, p} (2 j L p ))(0,1/q),q , 0,0,1/q

0,−1/q

we conclude that B p,min{2, p,q} → B p,q

.

 

J Fourier Anal Appl

Remark 3.8 Corollary 3.4 can be also derived from Theorem 3.7. However, if 0 < q < max{2, p}, then Theorem 3.2 yields that 0,−1/q

B p,q

0,1/ max{2, p}−1/q,1/ max{2, p}

→ B p,q

(3.2)

while Theorem 3.7 shows that 0,−1/q

B p,q

0,0,1/q

→ B p,max{2, p}

(3.3)

and spaces to the right side in (3.2) and (3.3) are not comparable. Indeed, since 

(1 + j)1/ max{2, p}−1/q (1 + log(1 + j))1/ max{2, p} (1 + log(1 + j))1/q

∈ / (1/q−1/ max{2, p})−1 ,

it follows from [11, Proposition 5.3/(i)] that 0,0,1/q

0,1/ max{2, p}−1/q,1/ max{2, p}

B p,max{2, p} → B p,q

.

On the other hand, we also have 

Therefore

(1 + log(1 + j))1/q (1 + j)1/ max{2, p}−1/q (1 + log(1 + j))1/ max{2, p} 0,1/ max{2, p}−1/q,1/ max{2, p}

B p,q

∈ / ∞ .

0,0,1/q

→ B p,max{2, p} .

Theorems 3.2 and 3.7 as well as (1.1) require the assumption 1 < p < ∞. We close this section with the corresponding results in the extreme cases p = 1 and p = ∞. Theorem 3.9 Let 0 < q ≤ ∞, b > −1/q and p = 1 or ∞. Then 0,b+1/ min{1,q}

B p,q

0,b → B0,b p,q → B p,q .

Proof Suppose p = 1. The case p = ∞ can be treated analogously. Take any 0 < θ < 1. By [14, Theorem 3.1/(b)] and [13, Lemma 2.2/(b)], we have that 1 1 B0,b 1,q = (L 1 , W1 )(0,−b),q = (L 1 , (L 1 , W1 )θ,∞ )(0,−b),q θ = (L 1 , B1,∞ )(0,−b),q . 0 → L → B 0 Moreover, B1,1 1 1,∞ (see [34, Proposition 2.5.7]). Consequently, using [13, Lemma 2.5/(b)] and [15, Theorem 5.3], we derive 0 θ B0,b 1,q → (B1,∞ , B1,∞ )(0,−b),q −θ θ θ = ((B1,∞ , B1,∞ )1/2,∞ , B1,∞ )(0,−b),q

J Fourier Anal Appl −θ θ → (B1,∞ , B1,∞ )1/2,q,b 0,b = B1,q .

Proceeding similarly, we get 0 θ B0,b 1,q ← (B1,1 , B1,∞ )(0,−b),q −θ θ θ = ((B1,∞ , B1,∞ )1/2,1 , B1,∞ )(0,−b),q −θ θ ← (B1,∞ , B1,∞ )1/2,q,b+1/ min{1,q} 0,b+1/ min{1,q}

= B1,q

.  

The following result can be derived proceeding as in Theorem 3.9 but using now Lemma 3.1. Theorem 3.10 Let 0 < q < ∞ and p = 1 or ∞. Then 0,1/ min{1,q}−1/q,1/ min{1,q}

B p,q

0,−1/q

→ B p,q

0,−1/q

→ B p,q

.

The corresponding result to Theorem 3.7 reads as follows. Theorem 3.11 Let 0 < q < ∞ and p = 1 or ∞. Then 0,0,1/q

0,−1/q

B p,min{1,q} → B p,q

0,0,1/q

→ B p,∞

.

4 Other Applications of Lemma 3.1 We start with an auxiliary result. Lemma 4.1 Let (U, μ) be a finite measure space, let 0 < p < ∞ and 0 < q < ∞. Then L p,q (log L)max{0,1/ p−1/q} (log log L)max{1/ p,1/q} → (L p , L ∞ )(0,1/q),q → L p,q (log L)min{0,1/ p−1/q} (log log L)min{1/ p,1/q} . Proof Take 0 < r < p and 0 < θ < 1 with 1/ p = (1 − θ )/r . Then L p =   (L r , L ∞ )θ, p . Now the result follows from Lemma 3.1 and [28, Lemma 6.1]. Subsequently we work with Besov spaces Bα,b,d p,q (T) of periodic functions on the unit circle T (see [12,13,18,31]). They are defined as Bα,b,d p,q (R) but replacing L p (R) by L p (T), the Lebesgue space of periodic functions on the unit circle. Embeddings of B0,b p,q (T) into Lorentz–Zygmund spaces have been studied by the present authors in [12,13] and by Gogatishvili, Opic, Tikhonov and Trebels [23] (see

J Fourier Anal Appl

also [9] and [35]). The following result refers to the limit case b = −1/q which was left open in [13, Theorem 4.4] and [23, Lemma 3.5]. Theorem 4.2 Let 0 < p < ∞ and 0 < q < ∞. Then 0,−1/q

B p,q

(T) → L p,q (log L)min{0,1/ p−1/q} (log log L)min{1/ p,1/q} (T). 0,−1/q

Proof By [13, Corollary3.5], B p,q

1/ p

(T) = (L p (T), B p,1 (T))(0,1/q),q and, according

1/ p

to [31, Theorem 3.5.5/(i)], B p,1 (T) → L ∞ (T). Therefore, using Lemma 4.1 we obtain 0,−1/q

B p,q

(T) → (L p (T), L ∞ (T))(0,1/q),q → L p,q (log L)min{0,1/ p−1/q} (log log L)min{1/ p,1/q} (T).  

Now we turn our attention to the relationship between the smoothness of derivatives of f and the smoothness of f . As one can see in [34, Theorem 2.3.8], given any f ∈ B sp,q we have that ∂∂xf j ∈ B s−1 p,q . However there is a loss of smoothness in order to have the derivative in B0,b p,q (T). We proved in [13, Theorem 4.7] (a previous result k,b+1/ min{2, p,q} can be found in [18, p. 70]) that if f ∈ B p,q (T) then D k f ∈ B0,b p,q (T) provided that k ∈ N, 1 < p < ∞, 0 < q ≤ ∞ and b > −1/q. This result is best possible in general as we show next. Proposition 4.3 Let 2 ≤ q < ∞ and b > −1/q. For any  > 0 there is a function 1,b+1/2− f ∈ B2,q (T) such that f ∈ / B0,b 2,q (T). 1,η

Proof Since B2,q (T) → B1,δ 2,q (T) if δ ≤ η, we may assume without loss of generality that 0 <  < b+1/q. Take β ∈ R such that −b−1/q −1/2 ≤ β < −b−1/q −1/2+ and let ∞  k −3/2 (1 + log k)β cos(kx), x ∈ [0, 2π ]. f (x) = k=1

This function belongs to L 2 (T) because  f 2L 2 (T) = π

∞ 

k −3 (1 + log k)2β < ∞.

k=1

We proceed to estimate the modulus of smoothness of order 2 of f with the help of [32, Lemma 3]. Let Sn f be the n-th partial sum of the Fourier series of f and let

Sn f, Sn f be its first and second derivatives, respectively. We have for n ∈ N that

ω2 ( f, 1/n)2 ∼  f − Sn f  L 2 (T) + n −2 Sn f  L 2 (T) ∞

1/2 n

1/2   −3 2β −2 2β ∼ k (1 + log k) +n k(1 + log k) k=n+1

k=1

J Fourier Anal Appl

∼ n −1 (1 + log n)β . So ω2 ( f, t)2 ∼ t (1 − log t)β for 0 < t < 1 and therefore 

1

[t −1 (1 − log t)b+1/2− ω2 ( f, t)2 ]q

0

dt ∼ t



1

(1 − log t)(b+1/2−+β)q

0

dt t

1,b+1/2−

(T). which is finite by our choice of β. Hence f ∈ B2,q ∞ −1/2

β (1 + log k) sin(kx) and Nevertheless, f (x) = − k=1 k ω1 ( f , 1/n)2 ∼  f − Sn f  L 2 (T) + n −1 Sn f  L 2 (T) ∞

1/2 n

1/2   ∼ k −1 (1 + log k)2β + n −1 k(1 + log k)2β k=n+1

k=1

∼ (1 + log n)β+1/2 because β + 1/2 < 0. This yields that ω1 ( f , t)2 ∼ (1 − log t)β+1/2 for 0 < t < 1. Using now that b + β + 1/2 + 1/q ≥ 0, we derive that 

1

[(1 − log t)b ω1 ( f , t)2 ]q

0

dt ∼ t



1

(1 − log t)(b+β+1/2)q

0

dt = ∞. t

Consequently, f ∈ / B0,b 2,q .

 

The following result refers to the limit case b = −1/q. Theorem 4.4 Let 1 < p < ∞, 0 < q < ∞ and k ∈ N. If

k,−1/q+1/ min{2, p,q},1/ min{2, p,q}

f ∈ B p,q

0,−1/q

(T) then D k f ∈ B p,q

(T).

Proof Take any α ∈ R with α > k. Then D k : H pk (T) −→ L p (T) is bounded and D k : Bαp,q (T) −→ Bα−k p,q (T) is also bounded (see [18, p. 70]). Interpolating we derive that D k : (H pk (T), Bαp,q (T))(0,1/q),q −→ (L p (T), Bα−k p,q (T))(0,1/q),q is bounded as well. By [13, Corollary 3.5] and [12, Lemma 2.1], we have that 0,−1/q (L p (T), Bα−k (T). To work with the source space, take α0 < k p,q (T))(0,1/q),q = B p,q and let 0 < θ < 1 such that k = (1 − θ )α0 + θ α. Using [30, Proposition 5.5] and Lemma 3.1/(a), we derive k,−1/q+1/ min{2, p,q},1/ min{2, p,q}

B p,q (T) α0 α = (B p,q (T), B p,q (T))θ,q,−1/q+1/ min{2, p,q},1/ min{2, p,q}

J Fourier Anal Appl α 0 (T), Bα (T)) → ((Bαp,q θ,min{2, p} , B p,q (T))(0,1/q),q p,q

= (Bkp,min{2, p} (T), Bαp,q (T))(0,1/q),q → (H pk (T), Bαp,q (T))(0,1/q),q .  

This completes the proof. Finally we deal with Fourier coefficients 1 fˆ(m) = 2π



2π

f (x)e−imx d x, m ∈ Z,

0

of functions in Besov spaces of logarithmic smoothness. Improving a result of DeVore, Riemenschneider and Sharpley [18, Corollary 7.3/(i)], we have shown in [13, Theorem 5.1] that if 1 ≤ p ≤ 2, 1/ p + 1/ p = 1, 0 < q ≤ ∞ and b > −1/q, then ˆ f ∈ B0,b p,q (T) implies that ( f (m)) ∈  p ,q (log )b+1/ max{ p ,q} . This result is the best possible in general as we show next. Proposition 4.5 Let 0 < q ≤ 2 and b > −1/q. Given any  > 0, there exists ˆ / 2,q (log )b+1/2+ . f ∈ B0,b 2,q (T) such that ( f (m)) ∈ Proof Take β ∈ R such that b + 1/2 + 1/q < β < b + 1/2 + 1/q +  and consider the function f (x) =

∞ 

k −1/2 (1 + log k)−β cos(kx), x ∈ [0, 2π ].

k=1

Since β > 1/2, we have that  f 2L 2 (T) = π

∞ 

k −1 (1 + log k)−2β < ∞.

k=1

Using [32, Lemma 3], we obtain that ω( f, 1/n)2 ∼  f − Sn f  L 2 (T) + n −1 Sn f  L 2 (T) ∞

1/2  ∼ k −1 (1 + log k)−2β k=n+1

+n

−1

n 

1/2 k(1 + log k)

k=1 −β+1/2

∼ (1 + log n)

−2β

J Fourier Anal Appl

where we have used again that β > 1/2 in the last equivalence. So ω( f, t)2 ∼ (1 − log t)−β+1/2 , 0 < t < 1. Therefore 

1

q dt

[(1 − log t) ω( f, t)2 ] b

0

t

 ∼

1

(1 − log t)(b−β+1/2)q

0

dt t

and the last integral is finite because b − β + 1/2 + 1/q < 0. Consequently, f ∈ B0,b 2,q (T). However, ( fˆ(k)) ∈ / 2,q (log )b+1/2+ . Indeed, since b + 1/2 +  − β + 1/q > 0, we get ∞  1 [k 1/2 (1 + log k)b+1/2+ k −1/2 (1 + log k)−β ]q k k=1

=

∞  k=1

(1 + log k)(b+1/2+−β)q

1 = ∞. k  

The next result refers to the limit case b = −1/q. Theorem 4.6 Let 1 ≤ p ≤ 2, 1/ p = 1 − 1/ p and 0 < q < ∞. If f ∈ B p,q then ( fˆ(m)) belongs to  p ,q (log )−1/q+1/ max{ p ,q} (log log )1/ max{ p ,q} .

0,−1/q

(T)

Proof Let F( f ) = ( fˆ(m)). By the Hausdorff–Young inequality, F : L p (T) −→  p is bounded. Take any α > 0 and let 1/r = α + 1/ p . It follows from [29, Theorem 3] that F : Bαp,r (T) −→ r is also bounded. Therefore F : (L p (T), Bαp,r (T))(0,1/q),q −→ ( p , r )(0,1/q),q

(4.1)

is bounded as well. By [13, Corollary 3.5], we know that (L p (T), Bαp,r (T))(0,1/q),q = 0,−1/q

B p,q (T). So it is enough to check that the target space in (4.1) is contained in the generalized Lorentz–Zygmund sequence space of the statement. Take 0 < θ < 1 with 1/ p = θ/r . Using Lemma 3.1 we derive ( p , r )(0,1/q),q = ((∞ , r )θ, p , r )(0,1/q),q → (∞ , r )θ,q,min{0,1/ p −1/q},min{1/ p ,1/q} =  p ,q (log )−1/q+1/ max{ p ,q} (log log )1/ max{ p ,q} where the last equality follows from [28, Lemma 6.1].

 

J Fourier Anal Appl

We finish the paper with a result on Fourier coefficients of functions in Lorentz– Zygmund spaces which are close to L 1 (T). Extending results of Hardy and Littlewood and of Bennett (see [4, Theorem 1.6/(a)]), we showed in [13, Theorem 5.2] that if b > −1/q and f belongs to b ˆ ∗ q −1 < ∞. Here ( fˆ(n)∗ ) L 1,q (log L)b+1/ min{1,q} (T) then ∞ n=1 ((1 + log n) f (n) ) n is the non-increasing rearrangement of the sequence (| fˆ(m)|). Now we cover the limit case b = −1/q. Theorem 4.7 Let 0 < q < ∞. If f ∈ L 1,q (log L)−1/q+1/ min{1,q} (log log L)1/ min{1,q} (T) then

∞  ( fˆ(n)∗ )q < ∞. n(1 + log n) n=1

Proof The operator F( f ) = ( fˆ(m)) is bounded from L 1 (T) into ∞ and from L 2 (T) into 2 . By the interpolation property, F : (L 1 (T), L 2 (T))(0,1/q),q −→ (∞ , 2 )(0,1/q),q is also bounded. According to [13, Lemma 2.2/(b)], (L 1 (T), L 2 (T))(0,1/q),q = (L 1 (T), (L 1 (T), L ∞ (T))1/2,2 )(0,1/q),q = (L 1 (T), L ∞ (T))(0,1/q),q . Hence, Lemma 4.1 yields that L 1,q (log L)−1/q+1/ min{1,q} (log log L)1/ min{1,q} (T) → (L 1 (T), L 2 (T))(0,1/q),q . As for the sequence space, it follows from [13, Corollary 3.5] that (∞ , 2 )(0,1/q),q = ∞,q (log )−1/q . This completes the proof.   Acknowledgments The authors would like to thank the referees for their comments. The authors have been supported in part by the Spanish Ministerio de Economía y Competitividad (MTM2013-42220-P). O. Domínguez has also been supported by the FPU Grant AP2012-0779 of the Ministerio de Economía y Competitividad, and F. Cobos by UCM-BS (GR3/14-910348).

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