Hardy Space Decompositions of $ L^ P (\Mathbb {R}^ N) $ for $0< P< 1

$Hardy Space Decompositions of $ L^ P (\Mathbb {R}^ N) $ for $0< P< 1$

arXiv:1711.04918v1 [math.CV] 14 Nov 2017 mi:egtbueuc.Ti okwsprilyspotdb SC( NSFC by supported partially 20100003110004) was (Grant SRFDP work This Email:[email protected]. est,Gaghu hn.Eal c21@cueuc o,lihaicho (or, [email protected] Email: China. Guangzhou, versity, [email protected] a atal upre yRsac Gra Research by supported partially 098/2012 FDCT was Government work Macao MYRG115(Y1-L4)-FST13-QT, The [email protected]. ebgnwt uvyo h aeo n iesoa ad spaces Hardy dimensional one of case the on survey a with begin We Introduction 1 0 ht o an for that, H p < 2 ∗ † ‡ ( colo ahmtclSine,BiigNra nvriy Beijin University, Normal Beijing Sciences, Mathematical of School orsodn uhr olg fMteaisadifrais So informatics, and Mathematics of College author. Corresponding eateto ahmtc,Uiest fMcu aa VaHong (Via Macao Macau, of University Mathematics, of Department C + ∞ ≤ fadol fsupp if only and if ) ad pc eopstosof Decompositions Space Hardy 5 eetyotie uhHrysaedcmoiinrsl:for result: decomposition space index Hardy function the such any for obtained recently tubes [5] on functions 0 space range Hardy corresponding the hsppram ooti eopstoso ihrdimensional higher of decompositions obtain to L aims paper This f a approximation nal such of words issue weKey uniqueness the paper, discuss present result and decomposition. dimensions, decomposition the higher space In the Hardy to one-dimensional in planes. the functions lower-half space generalize and Hardy upper-half some of the limits boundary non-tangential 0 2 p o h lsia case classical the For . ( L uhthat such R < p < 2 n ( R ucin nosm fnntneta onaylmt of limits boundary non-tangential of sums into functions ) function ) unTeDeng Guan-Tie < p < ad pcso uedmi,dcmoiin ratio- decomposition, domain, tube on spaces Hardy : f f ihRtoa Approximation Rational with 1 ∈ = .I h n-iesoa ae egadQian and Deng case, one-dimensional the In 1. f ˆ L f, f ⊂ p 1 ( ti h o-agnilbudr ii fafnto in function a of limit boundary non-tangential the is it + R [0 ) , f , ∞ 2 f ˆ where , 0 p ) = a-huLi Hai-Chou , ∗ , < p < Abstract ,tefmu ae-inrTermstates Theorem Paley-Wiener famous the 2, = h atrbigeuvln with equivalent being latter the χ [0 , 1 ∞ ) f f, ,teeeitfunctions exist there 1, 1 ˆ and f 2 a Qian Tao , † r,rsetvl,the respectively, are, /A3 [email protected]). L t hn giutrlUni- Agricultural China uth to nvriyo Macau, of University of nt p rn 1705 n by and 11271045) Grant ( R ‡ ,107,China. 100875, g, n f 1 for ) og.Email: Kong). and H p ( C ± for ) (1) 2 − where χ[0,∞) is the characteristic function of [0, ∞). Similarly, f ∈ H (C ) is equivalent with

fˆ = χ(−∞,0]f.ˆ (2)

From the relations (1) and (2), a canonical decomposition result is obtained as

f = f + + f −,

2 + ˆ ∨ − ˆ ∨ for f ∈ L (R), where f =(χ[0,∞)f) and f =(χ(−∞,0]f) belong to, respectively, the Hardy spaces H2(C+) and H2(C−). In the present paper we call such decomposition as Hardy space decomposition. It can be alternatively obtained through the Hilbert transformation in view of the Plemelj formula in relation to the Cauchy integral transform. For the relevant knowledge we refer the reader to, for instance, [10] and [15]. The significance of the L2(R) space decomposition into the Hardy spaces H2(C±) lays on the fact that the H2 functions have much better properties than the L2- functions. The functions f ± are non-tangential boundary limits of H2(C±) functions, the latter being analytically defined in their respective domains. We note that a function in Lp is only a.e. determined, that is, if f = g, a.e., then f and g are considered as the same function in Lp. In this regard one cannot assume a general Lp-function to have any smoothness. On the other hand, any two a.e. identical Lp functions correspond to the same Hardy space decomposition, the latter, being unique, having all kinds of smoothness in their domains of definition. The Hardy H2 spaces are reproducing kernel Hilbert spaces. In particular, Cauchy’s theory and techniques are available to Hardy space functions. Furthermore, there is an isometric correspondence between the Hardy space functions and their non-tangential boundary limits. Based on what just mentioned, analysis on the L2-functions can be reduced to that of their Hardy space components. Such treatment of L2-functions has found significant applications in both the pure and applied mathematics. In particular, some demonstrative results in signal analysis have recently been obtained ( [3], [13], [14],). There have been studies on Hardy space decomposition for extended index range of p. Both the above mentioned Paley-Wiener Theorem related Fourier spectrum characterization and the Plemelj formula can be extended to Hardy Hp(R) spaces with p =26 . Systematic studies on the spectrum properties as well as the Lp decomposition are carried out in the works of Qian et al. [15] among others. Their work is summarized as f ∈ Hp(C+) if and only if f ∈ Lp(R) and in the distributional sense suppfˆ ⊂ [0, ∞) for all 1 ≤ p ≤ ∞. Comparatively the Plemelj formula approach is more applicable than Fourier transformation, for on the Lp(R), p> 2, spaces Fourier transforms are distributions. As a consequence, the Hardy space decomposition results also hold for f ∈ Lp(R), 1 Fourier series other than Fourier transforms ([10], [2]). Turning to the case of 0

2 not distributions. One, however, can still have the Hardy space decompositions. The reference [2] uses the real analysis methods of harmonic analysis, making use of a dense subclass of the Lp(R)-functions with vanishing moment conditions and the Hilbert transforms. In contrast, [5] uses the complex analysis methods, and, in particular, rational functions approximation to achieve Hardy space decompositions. The methods in [5] are direct and constructive. The following Theorem A is obtained by Deng and Qian in [5]. Theorem A ([5]) Suppose that 0 sequences of rational functions {Pk} and {Qk} such that p p Pk ∈ H (C+), Qk ∈ H (C−) and

∞ p p p kPkk p + kQkk p ≤ Apkfkp, H+ H− Xk=1 n

lim ||f − (Pk + Qk)||p =0. n→∞ Xk=1 Moreover,

∞ ∞ p p g(z)= Pk(z) ∈ H (C+), h(z)= Qk(z) ∈ H (C−), Xk=1 Xk=1 p and g(x) and h(x) are the non-tangential boundary limits of functions for g ∈ H (C+) p and h ∈ H (C−), respectively, f(x)= g(x)+ h(x) almost everywhere, and

p p p p kfkp ≤kgkp + khkp ≤ Apkfkp. That is, in the sense of Lp(R),

p R p R p R L ( )= H+( )+ H−( ),

where H+(R) and H−(R) denote the set of non-tangential boundary limits of functions p p in the Hardy spaces H (C+) and H (C−), respectively. Recently we, in [7], obtain the Hardy space decomposition of Lp for 0 < p ≤ 1 on the unite circle ∂D by using polynomial approximation other than general rational p D p D approximation. The result is stated as follows. Denoting by LI (∂ ) and LO(∂ ) the closed subspaces of Lp(∂D), consisting of respectively, the non-tangential boundary p p limits of the functions of H (D) and H (DO), then we have p D p D p D L (∂ )= LI (∂ )+ LO(∂ ), where the right-hand-side is not a direct sum. As a matter of fact, the intersection p D p D LI (∂ ) and LO(∂ ) contains non zero functions. The work on the unit circle exposes the particular features adaptable to higher dimensions. All the Hardy space results mentioned above for dimension one can be generalized to dimension n in the setting of Hardy spaces on tubes with correspondingly the right notions. In fact, [17] already contain some basic results, mostly for p = 2, while [8]

3 gives a systematic treatment for general indices p ∈ [1, ∞], including Fourier spectrum characterization of Hardy spaces on tubes, the Cauchy integral and Poisson integral representation of the Hardy space functions, the Plemelj formulas in relation to Hilbert transforms, and the Hardy space decompositions of functions in the Lp(Rn) spaces. The purpose of this article is to prove the Hardy space decomposition of the Lp(Rn) space functions for p ∈ (0, 1). In doing so neither the Cauchy integral formula nor the Fourier transformation can be directly used, for the functions deﬁned on Rn are lack of integrability. They are even not distributions. In the present paper, inspired by the idea of [5], with the rational approximation method, we obtain decompositions of functions in Lp(R) for 0

P (x1, ..., xn) f(x1, x2, ..., xn)= n m , (xk − akj) k=1 j=1 Q Q l1 ln s1 sn where P (x1, ..., xn) = ··· as1...sn x1 ··· xn , l = max{l1, ..., ln}, (m − l)p > 1, s1=0 sn=0 R and akj =6 akm ∈ as jP=6 m forPk =1, 2, ..., n. The writing plan of this paper is as follows: In § 2, some basic deﬁnitions and notations are given. In § 3 we devote to establishing the higher dimensional Hardy space decomposition of Lp(Rn), 0

2 Preliminary Knowledge

In this section, we introduce some useful basic deﬁnitions and notions. For more p information, see e.g. [10] and [17]. The classical Hardy spaces H (Ck), 0 0}. They are Banach spaces for 1 ≤ p< ∞ under the norms

1 ∞ p p kfk p = sup |f(x + iy)| dx ; Hk ky>0 Z−∞ and complete metric spaces for 0 0 Z−∞ 4 n n Let B be an open subset of R . Then the tube TB with base B ⊂ R is the set

n n TB = {z = x + iy ∈ C : x ∈ R , y ∈ B}.

For example, when n = 1, the classical upper-half complex plane C+ and lower-half − complex plane C are the tubes in C with the base B+ = {y ∈ R : y > 0} and the base R C+ R B− = {y ∈ : y< 0}, respectively. That is, = TB+ = {z = x+iy : x ∈ , y> 0} C− R and = TB− = {z = x + iy : x ∈ , y< 0}. Obviously, the tube TB are generalizations of C+ and C−. It is known that n-dimensional real Euclidean space Rn has 2n octants. To denote the octants we adopt the following notations. Let σk =(σk(1), σk(2), ..., σk(n)), where n n Rn σk(j)=+1 or − 1 for k =1, 2, ..., 2 . Then 2 octants of are denoted by Γσk , k = 1, 2, ..., 2n, where

Rn Γσk = {y =(y1,y2, ..., yn) ∈ : σk(j)yj > 0, j =1, 2, ..., n}.

Rn In this paper, we denote Γσ1 the ﬁrst octant of , that is

Rn Γσ1 = {y =(y1,y2, ..., yn) ∈ : yj > 0, j =1, 2, ..., n}.

Correspondingly, Cn can be decomposed into 2n tubes, denoted by T , k =1, 2, ..., 2n. Γσk That is T = {z = x + iy ∈ Cn : x ∈ Rn, y ∈ Γ }. Γσk σk p A function F (z) is said to belong to the space H (TB), 0

1 p p kF kHp = sup |F (x + iy)| dx : y ∈ B < ∞. Rn (Z ) Hence, p H (TB)= {F : F holomorphic on TB and kF kHp < ∞}. The spaces Hp(T ) are deﬁned through replacing B by Γ , k =1, ..., 2n. Γσk σk A function, f, deﬁned in tube TΓ, is said to have non-tangential boundary limit n (NTBL) l in each component of the variable at x0 ∈ R if f(z)= f(x + iy)= f(x1 + iy1, ..., xn + iyn) tends to l as the point z = (x1,y1; x2,y2; ...; xn,yn) tends to x0 = (1) (2) (n) (x0 , 0; x0 , 0; ...; x0 , 0) within the Cartesian product

(1) (2) (n) γα(x0)=Γα1 (x0 ) × Γα2 (x0 ) ×···× Γαn (x0 ) ⊂ TΓ,

for all n-tuples α =(α1,α2, ..., αn) of positive real numbers, where

(j) C+ (j) Γαj (x0 )= (xj,yj) ∈ : |xj − x0 | <αjyj , j =1, 2, ..., n. n o As an important property of the Hardy spaces, it is shown that if f is a function in a p Hardy space H , 0

5 Since the mapping that maps the functions in the Hardy spaces to their NTBLs p Rn is an isometric isomorphism, we denote by Hσk ( ) for 0

Fσ (x) = lim F (x + iy) = lim F (x1 + iy1, ..., xn + iyn). (3) k + + y∈Γσk ,y→0 σk(1)y1→0 ,...,σk(n)yn→0

3 Hardy Space Decomposition of Lp(Rn)

As previously mentioned in the introduction, for one dimension, in [5], the authors use the rational approximation method to obtain the Hardy space decomposition for the range 0 sequence of rational p functions whose poles are in the lower-half plane, and f− ∈ H−, through a sequence of rational functions whose poles are in the upper-half plane. We note that the Hardy spaces decompositions for functions in Lp(R), 0

Theorem 3.1 Suppose that f ∈ Lp(Rn), 0

6 where Anp is a constant only depending on (n, p); (ii) m 2n

lim ||f − Rkσ ||p =0; (5) m→∞ j j=1 Xk=1 X (iii) m

lim ||fσ − Rkσ ||Hp =0, (6) m→∞ j j Xk=1 for all j =1, 2, ..., 2n;

(iv) fσj (x) are the non-tangential boundary limits of functions fσj (z), and

2n Rn f(x)= fσj (x), a.e. x ∈ ; j=1 X (v) 2n p p p kfkp ≤ kfσj kp ≤ Anpkfkp. j=1 X (vi) In summary, 2n p Rn p Rn L ( )= Hσj ( ). j=1 X Remark 3.1 For the non-uniqueness issue of the Hardy space decomposition,we will 2n p Rn show that Hσj ( ) is a non-empty set. We will prove it in § 4. j=1 T In order to prove Theorem 3.1, we need the following lemmas. We note that the proof of Theorem 3.1 is at the end of this section.

Rn Lemma 3.1 Let Γσ1 be the ﬁrst octant of . Suppose that 0

where P (z) is a polynomial of z, and Q(z) = Q1(z1)Q2(z2) ····Qn(zn), Q(zk) is a polynomial of z , k =1, 2, ..., n. If R ∈ Lp(Rn) and R(z) is holomorphic in T , then k Γσ1 R(z) ∈ Hp(T ). Γσ1 Deng-Qian [5] proved the special case n = 1 of Lemma 3.1, and obtained the following Lemma 3.1-1, which is needed for the proof of Lemma 3.1. Lemma 3.1-1 ([5]) Suppose that 0

7 Lemma 3.1-2 p n ([9]) If f(z) ∈ H (TΓσj ), 0

boundary limit of f(z). Then ϕ(y) is continuous convex and bounded in Γσj , moreover, p p kfk p = sup ϕ(y)= ϕ(0, ..., 0) = kfkp, Hσj y∈Γσj

p n where ϕ(y)= Rn |f(x + iy)| dx, y ∈ Γσj , j =1, 2, ..., 2 . Proof of Lemma 3.1 We are to prove Lemma 3.1 by mathematical induction. R When n = 1, Lemma 3.1 is just Lemma 3.1-1. Next, when n > 1, we assume that Lemma 3.1 holds for n − 1. Take n1 = n − 1, and ﬁx tn ∈ R\Z0(Qn), where Z0(Qn)= {zn : Qn(zn)=0}. We consider the function r of n − 1 real variables deﬁned by

r(x1, x2..., xn−1)= R(x1, ..., xn−1, tn), n−1 where (x1, x2..., xn−1) ∈ R . The Fubini Theorem ensures that r(x1, x2..., xn−1) be- p n−1 longs to L (R ) for almost all tn ∈ R\Z0(Qn). Moreover, it is easy to see that the rational function r(z1, ..., zn−1) satisﬁes the assumptions of Lemma 3.1. p Therefore, by the induction hypothesis, we obtain that r(z1, ..., zn−1) ∈ H (T n−1 ), Γσ1 n−1 where T n−1 denotes the tube with the ﬁrst octant of R as base. Γσ1 By Lemma 3.1-2,

p p |r(x1 + iy1, ..., xn−1 + iyn−1)| dx1 ··· dxn−1 ≤ |r(x1, ..., xn−1)| dx1 ··· dxn−1. Rn−1 Rn−1 Z Z That is

p |R(x1 + iy1, ..., xn−1 + iyn−1, xn)| dx1 ··· dxn−1 Rn−1 Z (7) p ≤ |R(x1, ..., xn−1, xn)| dx1 ··· dxn−1. Rn−1 Z Integrating both sides of the last inequality with respect to xn, we have

p p |R(x1 + iy1, ..., xn−1 + iyn−1, xn)| dx1 ··· dxn ≤ |R(x1, ..., xn)| dx1 ··· dxn. (8) Rn Rn Z Z By Fubini Theorem,

p p |R(x1 + iy1, ..., xn−1 + iyn−1, xn)| dxn dx1 ···dxn−1 ≤ |R(x)| dx. (9) Rn−1 R Rn Z Z Z n−1 So, ﬁx (y1, ..., yn−1), for almost all (x1, ..., xn−1) ∈ R , we have

p |R(x1 + iy1, ..., xn−1 + iyn−1, xn)| dxn < ∞. R Z That is, rational function R(x1 + iy1, ..., xn−1 + iyn−1, xn) as a function of xn belongs to Lp(R). Moreover, since R(z , ..., z ) is holomorphic in T , R(z , ..., z ) as a func- 1 n Γσ1 1 n + tion of zn is also holomorphic in upper-half plane C . Due to the result for n = 1, p + R(z1, ..., zn−1, zn) as a function of zn is a member of H (C ). By Lemma 3.1-2 again,

p p R(z1, ..., zn−1, xn + iyn)| dxn ≤ |R(z1, ..., zn−1, xn)| dxn. R R Z Z 8 Integrating both sides of the last inequality with respect to x1, ..., xn−1, we have

|R(x + iy)|p dx ≤ |R(x)|p dx. Rn Rn Z Z Due to the analyticity of the rational function R(z) in T , we get R(z) ∈ Hp(T ). Γσ1 Γσ1 The proofs for the other octants are similar. So the proof of Lemma 3.1 is complete.

Similarly to prove Lemma 3.1, we also can get the analogous results about the other octants as follows.

n Rn Corollary 3.1 Let Γσj (j = 1, 2, ..., 2 ) be all the octants of . Suppose that 0

where P (z) is a polynomial of z, and Q(z)= Q1(z1)Q2(z2) ····Qn(zn), where for each p n k, Qk(zk) is a polynomial of zk, k =1, 2, ..., n. If R ∈ L (R ) and R(z) is holomorphic p n in TΓσj , then R(z) ∈ H (TΓσj ), j =1, 2, ..., 2 . Lemma 3.2 If 0 0, there exist a sequence of rational functions {Rk(x)}, Rk ∈A, such that

∞ p p ||Rk||p ≤ (1 + ε)kfkp, (11) Xk=1 and n

lim ||f − Rk||p =0, (12) n→∞ Xk=1 where P (x) A = R(x)= : 2lj > degjP, j =1, 2, ..., n , (1 + x2)l1 ··· (1 + x2 )ln 1 n k1 kn Rn P (x)= (k) αkx1 ···xn is a polynomial of x =(x1, x2, ..., xn) ∈ , αk are constants, sn s1 k = (k1,kP2, ..., kn), and the notation (k) means that (k) = ··· , degjP = kn=0 k1=0 sj, j =1, 2, ..., n. P P P P

9 Proof We assume that kfkp > 0, and let

n n C0(R )= f : f is continuous in R , lim f(x)=0 . |x|→∞ n n It is obviously that A is a subalgebra of C0(R ) and A separates points. Since R is a n local compact Hausdorﬀ space, C0(R ) is a Banach algebra with the supremum norm kfk = sup{f(x): x ∈ Rn}. The Stone-Weierstrass theorem assures that A is dense in n C0(R ). Suppose that 0

fN (x)= f(x)X{x:|x|≤N;|f(x)|≤N}(x).

It is clear to see that, |fN (x)| ≤ N, and

suppfN ⊂ B(0, N)= {x : |x| ≤ N}.

p n p Moreover, |fN − f| ≤ 2|f| ∈ L (R ), and, lim |fN − f| = 0. Thus, by the Lebesgue N→∞ dominated convergence theorem, we have

p p lim |fN − f| dx = lim |fN − f| dx =0. N→∞ Rn Rn N→∞ Z Z

Therefore, for ε0 > 0, there exists an integer N > 1, such that

p ||fN − f||p <ε0/3. (13)

Because fN (x) is a measurable function, according Lusin Theorem, there exists a func- n tion g0 ∈ C0(R ), such that

p suppg0 ⊂ B(0, N), |g0(x)| ≤ N, m(EN ) <ε0/3(2N) ,

where EN = {x : g0(x) =6 fN (x)}. Therefore,

p p |g0(x) − fN | dx = |g0(x) − fN | dx Rn Z ZB(0,N) (14) p ≤ (2N) dx<ε0/3. ZEN Thus we obtain

p p p p kg0 −fkp = kg0 −fN +fN −fkp ≤kg0 −fN kp +kfN −fkp <ε0/3+ε0/3=2ε0/3. (15)

Taking integers lk such that plk > 1, k =1, 2, ..., n, the fact suppg0 ⊂ B(0, N) implies

n e e 2 lk Rn g0(x) (1 + xk)e ∈ C0( ). kY=1

10 n Since A is dense in C0(R ), there exists rational functions r(x) ∈A, such that

n 2 lk 1/p r(x) − g0(x) (1 + xk)e < (ε0/3) . (16)

k=1 Y

In fact, r(x) can be written as P (x) r(x)= n , 2 lk (1 + xk) k=1 Q k1 kn where P (x)= (k) αkx1 ··· xn , degjP < 2lj, j =1, 2, ..., n. Then, by (16),

P n n 2 lk 2 lk n P (x)/ (1 + xk) − g0(x) (1 + xk)e <ε0/3π ,

k=1 k=1 Y Y or n 1/p (ε0/3) 2 lk+elk P (x)/ (1 + xk) − g0(x) < n . n 2 lk k=1 π (1 + xk)e Y k=1

n Q 2 lk+lk Obviously, Q(x)= P (x)/ (1 + xk) e ∈A. Thus, k=1 Q p 1 (ε0/3) |Q(x) − g0| dx < n n dx<ε0/3. Rn Rn π 2 plk Z Z (1 + xk) e k=1 Q Therefore, for any ε0 > 0, there exists Q(x) ∈A such that

p p p p kQ − fkp = kQ − g0 + g0 − fkp ≤kQ − g0kp + kg0 − fkp <ε0.

p kfkpε0 Thus, for any ε> 0, taking εk = 4k+3 , k =1, 2, ..., there exist Qk(x) ∈A, such that

p kQk − fkp <εk.

Moreover, the function Qk(z) is a rational function satisfying ε kQ kp = kQ − f + fkp ≤kQ − fkp + kfkp <ε + kfkp =(1+ )kfkp. k p k p k p p k p 4k+3 p

Thus the sequence of rational functions Qk(z) can be chosen such that

p p p p kQk −Qk−1kp = kQk −f +f −Qk−1kp ≤kQk −fkp +kf −Qk−1kp < 2εk, (k =2, 3, ···). Let R1(z)= Q1(z), Rk(z)= Qk(z) − Qk−1(z), (k =2, 3, ···).

Then {Rk(z)} is a sequence of rational functions satisfying (11) and (12). This com- pletes the proof of Lemma 3.2.

11 Lemma 3.3 Suppose that 0

R(z)= Rσj (z), j=1 X and 2n p p kRσj kp ≤ CnpkRkp, (17) j=1 X − n n 21 pπ where Cnp =2 1−p . Proof Let R(x) ∈ Lp(Rn) A be a rational function. R(z) can be written as

T k1 kn P (z) (k) αkz1 ··· zn R(z)= n = n , 2 lj 2 lj (1 + zj ) P (1 + zj ) j=1 j=1 Q Q where p(2lj − degjP ) > 1 for all j =1, 2, ..., n. Let i − ξ β(ξ)= , ξ ∈ C. i + ξ It is easy to know that β(x) = eiθ(x), where θ(x) = arg(i − x) − arg(i + x) ∈ (−π, π) for x ∈ R. Rn For each ϕ = (ϕ1,ϕ2, ..., ϕn) ∈ , z ∈ TΓσj . Rational functions Rσj (z,ϕ) (j = 1, 2, ..., 2n) are deﬁned as follows.

n mkσj (k) (β(zk)) R (z,ϕ)= k=1 R(z), σj n Q m σj (k) iϕ σj (k) ((β(zk)) k − e k ) k=1 Q where mk >lk + n (k =1, 2, ..., n) are positive integers .

12 Then Rσj (z,ϕ) can be written as,

mk −mk (β(zk)) (β(zk)) Rσj (z,ϕ)= m iϕ −m −iϕ R(z) (β(zk)) k − e k (β(zk)) k − e k σj (Yk)=+1 σj (Yk)=−1 mk iϕk mk (β(zk)) e (β(zk)) = m iϕ iϕ m m R(z) (β(zk)) k − e k (e k − (β(zk)) k )(β(zk)) k σj (Yk)=+1 σj (Yk)=−1 mk mk iϕk (β(zk)) (β(zk)) −e = m iϕ m iϕ m R(z) (β(zk)) k − e k (β(zk)) k − e k (β(zk)) k σj (Yk)=+ σj (Yk)=−1 σj (Yk)=−1 n mk iϕk (β(zk)) −e = m iϕ m R(z) (β(zk)) k − e k (β(zk)) k Yk=1 σj (Yk)=−1 −eiϕk m (β(zk)) k σj (k)=−1 = n Q R(z). eiϕk 1 − m (β(zk)) k k=1 Q Therefore, 2n −eiϕk n m 2 (β(zk)) k j=1 σj (k)=−1 Rσj (z,ϕ)= Pn Q R(z)= R(z). eiϕk j=1 1 − m (β(zk)) k X k=1 Next we are to prove that R (z,ϕ)Q∈ Hp(T ), for all j =1, 2, ..., 2n. σj Γσj Now we only consider the case that the base is the ﬁrst octant of Rn, because proofs of the other octants are similar. For any z ∈ T , Γσ1

n mk (β(zk)) k=1 Rσ1 (z,ϕ)= n R(z) Q m iϕ ((β(zk)) k − e k ) k=1 Q n ( i−zk )mk i+zk (18) k=1 P (z) = n n Q m iϕ l l ((β(zk)) k − e k ) (zk − i) k (zk + i) k k=1 k=1 n Q (−1)mj (zQ− i)mj −lj = P (z) j . (z + i)lj +mj ((β(z ))mj − eiϕj ) j=1 j j Y

iϕk Since mk >lk, and |β(zk)| < 1, |e | =1, for all k =1, 2, ..., n, the function Rσ1 (z) is a rational function which is holomorphic in the tube T . Γσ1 Moreover, set p Iσ1 = |Rσ1 (x, ϕ)| dϕdx. Rn n Z Z(−π,π)

n n 2 2 1−p n p n 2 π p Iσk = |Rσk (x, ϕ)| dxdϕ ≤ 2 kRkp. (−π,π)n Rn 1 − p Xk=1 Z Xk=1 Z n Thus, there exists a ϕ =(ϕ1, ..., ϕn) ∈ (−π, π) , such that

n 2 21−pπ n kR kp ≤ 2n kRkp, σk p 1 − p p Xk=1 n where Rσk (x) = Rσk (x, ϕ), k = 1, 2, ..., 2 . This shows that the inequality (17) holds. It is easy to know that 21−pπ n kR kp ≤ 2n kRkp, σk p 1 − p p n p Rn for all k =1, 2, ..., 2 . So rational functions Rσk (x) ∈ L ( ). By Lemma 3.1, and that R (z) is holomorphic in T , we have σk Γσk R (z) ∈ Hp(T ) σk Γσk for all k =1, 2, ..., 2n. Thus, the proof of Lemma 3.3 is complete. We still need the following lemmas.

14 n p Lemma 3.4 ([17]) Let B be an open cone in R . Suppose F ∈ H (TB), p> 0, and c c B0 ⊂ B satisﬁes d(B0, B ) = inf{|y1 − y2|; y1 ∈ B0, y2 ∈ B } ≥ ε > 0, then there exists a constant Cp(ε), depending on ε and p but not on F , such that

sup |F (z)| ≤ Cp(ε)kF kHp . z∈TB0

Given below oﬀers a more precise estimation than that obtained in above Lemma 3.4.

p Lemma 3.5 Suppose that f ∈ H (TΓ), p> 0, and TΓ is the tube with its base Γ n as the ﬁrst octant of R . If let fδ(z) = f(z + iδ), for any z = x + iy ∈ TΓ and δ =(δ1, δ2, ..., δn) ∈ Γ, then there holds

1 − p sup |fδ(z)| ≤ CpkfkHp (δ1 ··· δn) , z∈TΓ

n 2 p where Cp =( π ) . We note that the proof of Lemma 3.5 is obtained in our recently work [9].

3.1 Proof of Theorem 3.1 Based on the above lemmas, we are now to prove Theorem 3.1. Proof of Theorem 3.1 When 0 0, there exists a sequence of rational functions {Rk(x)}, such that

∞ p p ||Rk||p ≤ (1 + ε)kfkp, (21) Xk=1 and m

lim ||f − Rk||p =0. (22) m→∞ Xk=1 n For each k = 1, 2, ..., by Lemma 3.3, there exist 2 rational functions Rkσj (z) ∈ p n H (TΓσj ) (j =1, 2, ..., 2 ), such that

Rk(z)= Rkσj (z), (23) j=1 X and 2n p p kRkσj kp ≤ CnpkRkkp. (24) j=1 X Therefore, ∞ 2n ∞ p p p ||Rkσj ||p ≤ CnpkRkkp ≤ (1 + ε)Cnpkfkp, j=1 Xk=1 X Xk=1

15 m 2n

lim ||f − Rkσ ||p =0. m→∞ j j=1 Xk=1 X By Lemma 3.1-2, there are

p p n kRkσj k p = kRkσj kp, j =1, 2, ..., 2 , Hσj which imply that the properties (4) and (5) hold. n Moreover, for any δ = (δ1, δ2, ..., δn) ∈ Γσj , and j = 1, 2, ..., 2 , by Lemma 3.4 and 3.5, there exists a positive constant Mδ < ∞ such that

Rkσj (z + iδ) ≤ MδkRkσj kp.

Hence, m p m m p p Rkσj (z + iδ) ≤ |Rkσj (z + iδ)| ≤ Mδ kRσj kp,

k=1 k=1 k=1 X X X ∞ for z ∈ TΓσj . This implies that the series Rkσj (z) uniformly converges to a function k=1 n n δ C R fσj (z) in the tube domain TΓ = {z = x + iy ∈ : x ∈ , σj(l)yl > σj(l)δl, l = σj P

1, 2, ..., n} for any δ =(δ1, δ2, ..., δn) ∈ Γσj . As a consequence, the function f (z) is holomorphic in T . Property (4) implies σj Γσj that property (6) holds. By properties of Hardy spaces on tubes, the non-tangential p n boundary limit fσj (x) of function fσj (z) ∈ H (TΓσj ) exists for every j =1, 2, ..., 2 . Therefore, property (5) implies that

f(x)= fσj (x) j=1 X holds almost everywhere, and moreover,

2n p p p kfkp ≤ kfσj kp ≤ Apkfkp. j=1 X Thus, the proof of Theorem 3.1 is complete.

4 Non-uniqueness of Hardy Space Decomposition

In this section, we are to answer the questions asked in the Remark 3.1 of Theorem 3.1. For the one dimension, A.B. Aleksandrov ([1], [2]) obtained the following theorem. Theorem B ([1] and [2]) Let 0

N c f(x)= j , a ∈ R, c ∈ C. x − a j j j=1 j X 16 Then p p R p R X = H+( ) ∩ H−( ). We note that, A.B. Aleksandrov’s proof of Theorem M ([1], [2]) is rather long in- volving vanishing moments and the Hilbert transformation. Deng and Qian [5] present a more straightforward proof for Theorem M. In this section, our aim is to extend The- orem M to higher dimensions. In order to do this, we need first to extend the following Theorem C obtained by J.B. Garnett ([10]) and Theorem D obtained by Deng-Qian ([5]) to higher dimensions. Theorem C ([10]) Let N be a positive integer. For 0 1, the class p + p + wN is dense in H (C ), where wN is the family of H (C ) functions satisfying (i) f(z) is infinitely differentiable in C+, (ii) |z|N f(z) → 0 as z →∞. Theorem D ([5]) Let N be a positive integer. For 0 1, the class p + p − RN (i) is dense in H (C ), and the class RN (−i) is dense in H (C ). Where α ∈ C R −N−1 1 and N (α) is the family of rational functions f(z)=(z + α) P ( z+α ), P (w) are polynomials. We obtain the following three theorems for higher dimensions.

Theorem 4.1 Let N be a positive integer. For 0 1, the class wN is p p dense in H (TΓ), where wN is the family of H (TΓ) functions satisfying (i) f(z) is inﬁnitely diﬀerentiable in T Γ, N (ii) |z| f(z) → 0 as |z|→∞, where |z|→∞ means that zj →∞, 1 ≤ j ≤ n, z = (z1, z2, ..., zn) ∈ T Γ.

p Proof We can approximate f(z) ∈ H (TΓ) by the smooth function f(z + i/m) = f(z1 + i/m, ..., zn + i/m). In fact, the property that the existence of the boundary limits functions of Hardy space functions assures that

kfm − fkHp → 0, m →∞.

We will construct the special functions gk(z) such that (a) gk(z) ∈ wN , (b) |gk(z)| ≤ 1, z ∈ TΓ, (c) |gk(z)|→ 1, z ∈ TΓ, as k →∞. Before we construct the above functions gk(z), we note that the functions

fm(z)= gm(z)f(z + i/m)

in wN and then obtain the desired approximation. That is to say, if there exist the special functions gk(z), we can complete the proof of Theorem 4.1. As the heart of the proof, we are to construct the functions gk(z) in the following. n Let (w1,w2, ..., wn) ∈ TΓ, (αk,αk, ..., αk) ∈ R , 0 <αk < 1, and αk → 1 as k →∞. Consider the function n

hk(w)= ϕj(wj), j=1 Y 17 N+1 wj +αk where ϕj(wj)= , j =1, 2, ..., n has inﬁnite (N + 1)−fold zero at −αk. 1+αkwj Then, n n N+1 wj + αk |hk(w)| = ϕj(wj) = < 1. (25) 1+ αkwj j=1 j=1 Y Y n Fixing N, hk(w) converges to 1 uniformly on the compact set D \ Ek, where k=1 S D = D1 × D2 ×···× Dn,Dk = |z| < 1, k =1, 2, ..., n, and Ek = {(w1, ..., wj−1, −αk,wj, ..., wn): wj ∈ Dk, k =6 j}.

Then, for w =(w ,w , ..., w )= i−z1 , ..., i−zn and α =(α ,α , ..., α ), we deﬁne the 1 2 n i+z1 i+zn k k k functions gk(z)= hk(αkw).

Below we verify that the function gk(z) satisﬁes the three conditions (a) (b) and (c). In fact, for condition (a), there is

n α w + α N+1 g (z)= h (α w)= k j k k k k 1+ α2w j=1 k j Y N+1 n N+1 n i−zj w +1 i+z +1 = αN+1 j = αN+1 j k 1+ α2w k 2 i−zj j=1 k j j=1 1+ αk ! Y Y i+zj n 2i N+1 = αN+1 . k (1 + α2)i + (1 − α2)z j=1 k k j Y

It is clearly that gk(z) satisfies the first condition (i) of the class wN . Moreover, there N holds |z| gk(z) → 0 as |z|→∞. This implies that gk(z) satisfies the second condition (ii) of the class wN . Therefore, gk(z) ∈ wN , which shows that gk(z) satisfies the condition (a). For condition (b), from (25), we can get that

|gk(w)| = |hk(αkw)| < 1.

For (c), it is clear that

gk(w)= hk(αkw) → 1, k →∞, z ∈ T Γ. Thus, the proof is complete. We shall notice that the condition Np > 1 implies that the above class wN is p n contained in H (TΓ). Let α =(α1,α2, ..., αn) ∈ C and let RN (α) be the family of the rational functions

−N−1 −N−1 −N−1 1 1 f(z)=(z1 + α1) (z2 + α2) ··· (zn + αn) P ( , ..., ), z1 + α1 zn + αn

18 n where z = (z1, z2, ..., zn) ∈ C and P (w) are polynomials. We notice that the class RN (α) is contained in the class wN for Imαj > 0, j =1, 2, ..., n. Thus, we obtain the following results.

Theorem 4.2 Let N be a positive integer. For 0 1, the class p RN (i, i, ..., i) is dense in H (TΓ).

Corollary 4.1 Let N be a positive integer. For 0 1, the class R p n N (σj(1)i, ..., σj(n)i) is dense in H (TΓσj ), j =1, 2, ..., 2 .

The proof of Corollary 4.1 is similar to the proof of Theorem 4.2, so we only prove Theorem 4.2. p Proof of Theorem 4.2 If f(z) ∈ H (TΓ), Np> 1, then, for any ε> 0, by Theorem p ∞ 4.1, there exists function fN in H (TΓ) C (T Γ) such that

lim T|z|N+1f(z)=0, |z|→∞,z∈TΓ

and kfN − fkHp < ε. The fractional linear mapping

1 − wj zj = α(wj)= i , j =1, 2, ..., n, 1+ wj is a conformal mapping from the n-tuple unit disc

n D = D1 × D2 ×···Dn = {(w1,w2, ..., wn) ∈ C : |wj| < 1, j =1, 2, ..., n}

n n to the ﬁrst octant of C , TΓ = {(z1, z2, ..., zn) ∈ C : Imzj > 0, j =1, 2, ..., n}. Its inverse mapping is

i − zj wj = β(zj)= , j =1, 2, ..., n. i + zj Let hN (w)= fN (α(w1), ..., α(wn))

and hN (w1, ..., wj, −1,wj+1, ..., wn)=0, j =1, 2, ..., n. Then hN (w) is continues in the closed disc D, and

1 − w N+1 1 − w N+1 h (w) i 1 ··· i n → 0, N 1+ w 1+ w 1 n

D as wj → w0 ∈ F, |w − w0 | → 0, for j = 1, 2, ..., n and w ∈ \F , where F = n Fj, Fj = {(w1, ..., wj, −1,wj+1, ..., wn): |wj| ≤ 1, j =1, 2, ..., n}. j=1 S

19 Therefore, for w0 ∈ F , hN (w) n → 0, N+1 (1 + wj) j=1 Q as wj → w0, |wj| ≤ 1, j =1, 2, ..., n. Let

hN (w) n w ∈ D\F N+1 Q (1+wj ) hN (w)=  j=1 (26) 0 w ∈ F e Then hN (w) is holomorphic in then-tuple D and continues in the closed set D. By the Stone-Weierstrass Theorem, there exists a polynomial PN (w)= PN (w1, ..., wn) such thate

hN (w) − PN (w + 1) < ε, |wj| ≤ 1, wj =6 −1, j =1, 2, ..., n, that is e n N+1 hN (w)/ (1 + wj) − PN (w + 1) < ε, |wj| ≤ 1, wj =6 −1, j =1, 2, ..., n.

j=1 Y Thus,

n n N+1 N+1 fN (α(w1), ..., α(wn)) − (1 + wj) PN (w + 1) ≤ ε |1+ wj| ,

j=1 j=1 Y Y where |w | ≤ 1, w =16 , j =1, 2, ..., n. j j Since z = α(w )= i 1−wj and w = β(z )= i−zj for all j =1, 2, ..., n, the last inequality j j 1+wj j j i+zj becomes n 2i N+1 2i 2i n 2i N+1 fN (z) − PN , ..., ≤ ε , i + zj i + zn i + zn i + zj j=1 j=1 Y Y for z ∈ T , Imz > 0, j =1, 2, ..., n. Therefore, we can obtain that Γ j n +∞ p p (N+1)p 2 −(N+1)p |fN (x + iy) − R(x + iy)| dx<ε 2 (1 + xj ) dxj < ∞, Rn Z j=1 Z−∞ Y where z =(z1, z2, ..., zn) ∈ TΓ and n 2i N+1 2i 2i R(x + iy)= P , ..., ∈ R (i, i, ..., i). i + z N i + z i + z N j=1 j n n Y p This concludes that the class RN (i, i, ..., i) is dense in H (TΓ). Therefore, the proof of Theorem 4.2 is complete. The following result shows that the Hardy space decomposition of Lp(Rn) for 0 < 2n p Rn p< 1 is not unique and the intersection space Hσj ( ) is a non-empty set. j=1 T 20 Theorem 4.3 Let 0

P (x1, ..., xn) f(x1, x2, ..., xn)= n m , (xk − akj) k=1 j=1 Q Q l1 ln s1 sn where P (x1, ..., xn) = ··· as1...sn x1 ··· xn , l = max{l1, ..., ln}, (m − l)p > 1, s1=0 sn=0 R and akj =6 akm ∈ for Pj =6 m andP k =1, 2, ..., n. Then 2n p p Rn X = Hσj ( ) =6 ∅. j=1 \ 2n Proof p p Rn Firstly, we will show that X ⊆ Hσj ( ). Let j=1 T l1 ln s1 sn ··· as1...sn z1 ··· zn f(z , z , ..., z )= s1=0 sn=0 , 1 2 n P n Pm (zk − akj) k=1 j=1 Q Q n ak =(ak1, ..., akm) ∈ R , k =1, 2, ..., n. n Then f(z) is a rational function with singular part being contained in Gl, where l=1 C Gl = {(z1, ..., zn) : Imzl = 0, zk ∈ , k =6 l}. It is clear that the set Gj Sis not in the n 2n n octant TΓσj and Gl ⊆ ∂TΓσj . So f(z) is holomorphic in TΓσj , for all j =1, 2, ..., 2 . l=1 j=1 Moreover, S T

p l1 ln s1 sn ··· as1...sn z1 ··· zn p s1=0 sn=0 |f(x + iy)| dx = p dx Rn Rn P n Pm Z Z (zk − akj) k=1 j=1 (27)

QnQ +∞ sk p p |zk | ≤ |as1...sn | p dxk. m s k=1 Z−∞ X Y (zk − akj) j=1

Q Observe that, the function sk zk Rk(zk)= m , (zk − akj) j=1 Q

21 is holomorphic in the upper-half and the lower-half complex planes. Moreover, we can p prove that Rk(xk) ∈ L (R). In fact,

+∞ +∞ sk p p |xk | |Rk(xk)| dxk = p dxk m Z−∞ Z−∞ (xk − akj) j=1

Q sk p sk p |xk | |xk | = p dxk + p dxk. m m Z|xk|≤M Z|xk|>M (xk − akj) (xk − akj) j=1 j=1

Q Q

where M is suﬃciently large so that the interval (−M, M) contains all the poles of Rk(xk). Then, for the ﬁrst integral in the right hand of the last inequality,

sk p |xk | skp dxk p dxk ≤ M m , m |xk|≤M |xk|≤M p Z (x − a ) Z |xk − akj| k kj j=1 j=1 Q Q which is ﬁnite since 0

sk p sk p |xk | |xk | p dxk = p dxk |x |>M m |x |>M m Z k Z k m (xk − akj) xk (1 − akj/xk) j=1 j=1

Q Q dxk ≤ m |xk|>M p (m−sk)p Z |1 − akj/xk| |x|k j=1 m Q p |M − akj| dxk ≤ p , |M| (m−sk)p j=1 |xk|>M |x| Y Z k p which is also ﬁnite since (m − sk)p> 1. Therefore, Rk(xk) ∈ L (R). p By Lemma 3.1, together with Rk(xk) ∈ L (R) and the analyticity of Rk(xk), we get p ± that Rk(zk) ∈ H (C ). Thus, kRkHp ≤kRkLp ; and the relation (27) becomes

p p np |f(x + iy)| dx ≤ |as1...sn | kRkLp < ∞. Rn s Z X Hence, we have 2n p n f(z) ∈ H (TΓσj ), j =1, 2, ..., 2 . j=1 \ This shows that 2n p Rn f(x) ∈ Hσj ( ). j=1 \ 22 2n p p Rn Hence X ⊆ Hσj ( ) as desired. j=1 T 2n p p Rn Next, we will show X ⊇ Hσj ( ). j=1 p n Let there exist fσj (z) ∈ HT(TΓσj ), j =1, 2, ..., 2 , such that

f(x)= fσj (x)= fσl (x),

for all 1 ≤ j, l ≤ 2n, a.e. x ∈ Rn. By Theorem 4.2 and Corollary 4.1, for any ε> 1, there exist R Rσj ∈ N (σji)

such that p p ε kfσj − Rσj kHp = kf − Rσj kp < , σj 4 for j =1, 2, ..., 2n. R n The fact Rσj ∈ N (σji) implies that there exist polynomials Pσj , j =1, 2, ..., 2 , such i−zk that for β(zk)= , k =1, 2, ..., n, i+zk

n N+1 σj (1) σj (n) σj (k) Rσj (z)= Pσj ((β(z1)) +1, ..., (β(zn)) + 1) β (zk)+1 , k=1 Y n where (N + 1)p> 1, N +1 > max{degPσj : j =1, 2, ..., 2 }. n

It is easy to see that the singular part of Rσj (z) are contained in Hk which is k=1 C not in TΓσj , where Hk = {(z1, ..., zn): zk = −σj(k)i, zl ∈ , k =6 l}.S So Rσj (z) is a . holomorphic rational function in TΓσj , j =1, 2, ..., 2 Let

n − 2 n 1+σj (k) n 1 σj (k) iϕk 2 2 m (−e ) (β(zk)) (Rσj (z) − Rσ1 (z)) j=1 k=1 k=1 R(z,ϕ)= R (z)+ , σ1 P Q n Q m iϕ ((β(zk)) − e k ) k=1 Q n Rn where (m − max{degPσj : j = 1, 2, ..., 2 } − N − 1)p> 1, and ϕ = (ϕ1, ..., ϕn) ∈ . We thus are aware that the singular part of the rational function R(z,ϕ) is contained 2n 2n Rn in ∂TΓσj = . So R(z,ϕ) is holomorphic in TΓσj . j=1 j=1 iθ(xk) NoteT that β(xk) = e , where θ(xk) = arg(iSk − xk) − arg(ik + xk) ∈ [−π, π] for xk ∈ R. Moreover, set

p J = |R(x, ϕ) − Rσ1 (x)| dϕdx. Rn n Z Z[−π,π]

23 Then p n − 2 n 1+σj (k) n 1 σj (k) iϕk 2 2 m (−e ) (β(zk)) (Rσj (x) − Rσ1 (x)) j=1 k=1 k=1

|R(x, ϕ) − f(x)|p dx Rn Z p p ≤ |R(x, ϕ) − Rσj (x)| dx + |Rσj (x) − f(x)| dx (30) Rn Rn Z Z 2π n ε ≤ ε + . 1 − p 4 Hence, R(x, ϕ) ∈ Lp(Rn). This, together with the deﬁnition of R(x, ϕ), implies that R(x, ϕ) ∈ Xp. Therefore, f(x) ∈ Xp. The proof is complete.

24 References

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Hardy Space Decompositions of $ L^ P (\Mathbb {R}^ N) $ for $0&lt; P&lt; 1

Hardy Space Decompositions of $ L^ P (\Mathbb {R}^ N) $ for $0< P< 1