J. Math. Anal. Appl. 436 (2016) 1082–1101
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Integrability spaces for the Fourier transform of a function of bounded variation E. Liflyand Department of Mathematics, Bar-Ilan University, 52900 Ramat-Gan, Israel
a r t i c l e
i n f o
Article history: Received 24 April 2015 Available online 29 December 2015 Submitted by L. Fialkow Keywords: Bounded variation Fourier transform Integrability Hilbert transform Hardy space
a b s t r a c t New relations between the Fourier transform of a function of bounded variation and the Hilbert transform of its derivative are revealed. After various preceding works of the last 25 years where the behavior of the Fourier transform has been considered on specific subspaces of the space of functions of bounded variation, in this paper such problems are considered on the whole space of functions of bounded variation. The widest subspaces of the space of functions of bounded variation are studied for which the cosine and sine Fourier transforms are integrable. The main result of the paper is an asymptotic formula for the sine Fourier transform of an arbitrary locally absolutely continuous function of bounded variation. Interrelations of various function spaces are studied, in particular, the sharpness of Hardy’s inequality is established and the inequality itself is strengthened in certain cases. A way to extend the obtained results to the radial case is shown. © 2015 Elsevier Inc. All rights reserved.
1. Introduction It apparently was Bochner who first paid attention to the fact that the Fourier transform for functions of bounded variation deserves a unified approach. In fact, many properties of the Fourier transform of a monotone function are given in [8], but the same machinery amounts to functions with bounded variation. We cite a 1959 edition but the book was published already in the 30s. However, this approach has not been taken up, though various results within this scope continued to appear in literature. The paper [24] is also from that period and contains certain contributions to the topic. However, most of them concern either Fourier series of a function of bounded variation or trigonometric series with the sequence of coefficients of bounded variation (see, e.g., [45,7,44,25], and various journal publications). In the last two monographs [44] and [25] many applications of such results are indicated. Let us also mention [10]. We are going to study an important line of this topic in which the Fourier transform of a function of bounded variation is integrable. This subject is relatively recent, essential results have been obtained within E-mail address: lifl
[email protected]. http://dx.doi.org/10.1016/j.jmaa.2015.12.042 0022-247X/© 2015 Elsevier Inc. All rights reserved.
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the last 25 years. There are various arguments to justify our interest in this piece. First of all, and this goes back to the mentioned [8], this gives us an opportunity to understand the Fourier transform in a relatively convenient, improper sense. Second, many of the results obtained in the past and in this paper, have their analogs, or prototypes, in the developed theory of the integrability of trigonometric series, a topic that was born about a hundred years ago. Among numerous results from this theory (see, e.g., [25, Ch. 3, 3.3]) those due to Telyakovskii (the so-called Boas–Telyakovskii condition, see, e.g., [42]) are best matched to the problems under consideration. Further, the spaces of integrability considered till recently are of interest by themselves and have applications in other areas of analysis (see, e.g., [9,17,28]). In a recent survey paper [33] integrability of the Fourier transform is considered in the context of belonging to Wiener’s algebra and its relations to the multipliers theory and comparison of operators. Many of integrability conditions originated from the noteworthy result of Trigub [43] on the asymptotic behavior of the Fourier transform of a convex function. However, the most advanced onward integrability results are related to the real Hardy space (see [29,16,25]). This gives rise to an additional argument. We will see in what follows that the well-known extension of Hardy’s inequality (see, e.g., [18, (7.24)]) R
| g (x)| dx gH 1 (R) |x|
(1)
plays crucial role in these considerations. Proposition 4 below asserts that if the integral on the left-hand side of (1) is finite, then g is the derivative of a function of bounded variation f which is locally absolutely continuous (for short, LAC), lim f (t) = 0 and its Fourier transform is integrable. This shows that the |t|→∞
study of analytic properties of the real Hardy space is unavoidably concerned with integrability properties for the Fourier transform of functions of bounded variation. Here and in what follows we use the notations “” and “” as abbreviations for “≤ C” and “≥ C”, with C being an absolute positive constant. Let us summarize what is done in this paper. We establish new conditions for the integrability of the Fourier transform in terms of belonging of the derivative of the considered function to certain function space. For the cosine Fourier transform, belonging to one of the considered subspaces of the space of functions of bounded variation ensures only integrability. For the sine Fourier transform, in most cases an asymptotic formula of the form fs (x) =
∞ f (t) sin xt dt =
1 π f + F (x) x 2x
0
is obtained, with F integrable. We study in detail asymptotic behavior of the sine Fourier transform of an arbitrary function of bounded variation. The relation is of the asymptotic form 1 π fs (x) = f + one more term + F (x). x 2x This result can be considered as the main result of the paper; that additional leading term is a key to the full generality within the scope of all functions of bounded variation. By these considerations, we add new spaces to the known scales of integrability spaces. We thoroughly study the relations between them. For instance, proving that the embedding between some of them is proper, we establish the sharpness of Hardy’s inequality (1), apparently for the first time in the literature. Further, we find a new scale of nested spaces, intermediate between the known ones. More precisely, we prove that an analog of the known Aq scale forms a class of intermediate spaces between two Hardy type spaces. This is the first step in finding other such scales, more convenient in applications. Returning to analysis of the newly introduced spaces, we realize
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that, in fact, various refinements of the mentioned Hardy’s inequality are obtained. Last but not least, the described approach should be extended to the multivariate case. Again, as a first step, we generalize the obtained results to appropriate classes of radial functions. There, a special function depending on dimension is subject to familiar assumptions posed on it, rather than on the derivative in dimension one. The paper is organized as follows. In the next section we present known results on the integrability of the Fourier transform of a function of bounded variation relevant to our analysis. Most of them are used in the sequel either for comparison or for constructing counterexamples. In Section 3 we study in detail asymptotic behavior of the sine Fourier transform of a function of bounded variation. In Section 4 we first establish certain embeddings, then we show that they are proper and find intermediate spaces between two Hardy type spaces. In Section 5 we show that some of the obtained results improve the well-known Hardy’s inequality. In the last Section 6 a multidimensional extension is proved for a special class of radial functions for which a convenient formula for the Fourier transform is obtained earlier in [36]. 2. Integrability of the Fourier transform In this section we present results on the Fourier transform of a function of bounded variation that will be either compared with the new results or used in the sequel, for constructing counterexamples say. We start with some preliminaries on the Hilbert transform. 2.1. Hilbert transform and Hardy spaces The Hilbert transform of an integrable function g is defined by Hg(x) =
1 π
R
g(t) dt, x−t
(2)
where the integral is also understood in the improper (principal value) sense, now as lim
δ→0+ |t−x|>δ 1
. It is not
necessarily integrable, and when it is, we say that g is in the (real) Hardy space H 1 := H (R). If g ∈ H 1 (R), then g(t) dt = 0. (3) R
It was apparently first mentioned in [27]. An odd function always satisfies (3). However, not every odd integrable function belongs to H 1(R), for counterexamples, see [35] and [30]. When in the definition of the Hilbert transform (2) the function g is odd, we will denote this transform by Ho , and it is equal to Ho g(x) =
2 π
∞ 0
tg(t) dt. x2 − t 2
If it is integrable, we will denote the corresponding Hardy space by H01 (R+ ), or sometimes simply H01 . Correspondingly, when in the definition of the Hilbert transform (2) the function g is even, we will denote this transform by He , and it is equal to He g(x) =
2 π
∞ 0
xg(t) dt. x2 − t 2
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There exist clear relations between the even and odd Hilbert transforms. Proposition 1. Let g ∈ L1 (R+ ). Then 2 He g(x) = Ho g(x) + πx
x g(t) dt + G(x), 0
where ∞
2 ln 2e |G(x)| dx ≤ π
0
∞ |g(t)| dt. 0
Proof. The first step is as follows: 2 He g(x) = Ho g(x) + π
∞
g(t) dt. x+t
0
We then split the last integral in two, the one over (0, x) and the other over (x, ∞). To integrate the latter, we apply ∞ ∞
|g(t)| dt dx = ln 2 x+t
x
0
∞ |g(t)| dt.
(4)
0
Then, integrating by parts, we obtain x
g(t) 1 dt = x+t x
0
x
x g(t) dt −
0
1 (x + t)2
x g(u) du dt. t
0
The integral of the last value on the right is estimated by ∞ x 0
1 (x + t)2
x
∞ |g(u)| du dt dx ≤
t
0
∞ u|g(u)|
dx du = x2
u
0
x
0
∞ =
1 x2
u|g(u)| du dx 0
∞ |g(u)| du, 0
which, with ∞ G(x) = x
completes the proof. 2
g(t) dt − x+t
x 0
1 (x + t)2
x g(u) du dt, t
(5)
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2.2. Known integrability spaces We are going to tie up the Fourier transform of a function of bounded variation with the Hilbert transform of a related function. The latter will give rise to a scale of integrability spaces for the Fourier transform. As a reasonable starting point, let us present certain known results. The first one is given in [29, Thm. 2] (see also [16]). Theorem 1. Let f : R+ → C be locally absolutely continuous on (0, ∞), of bounded variation and lim f (t) = t→∞
0. Let also f ∈ H01 (R+ ). Then the cosine Fourier transform of f fc (x) =
∞ f (t) cos xt dt
(6)
0
is Lebesgue integrable on R+ , with fc L1 (R+ ) f H01 (R+ ) ;
(7)
and for the sine Fourier transform, we have, with x > 0, fs (x) =
∞ f (t) sin xt dt =
1 π f + F (x), x 2x
(8)
0
where F L1 (R+ ) f H01 (R+ ) .
(9)
There is a scale of subspaces of H01 (R+ ). For 1 < q < ∞, set ∞ gAq =
⎛
⎝1 x
0
⎞1/q |g(t)|q dt⎠
dx.
x≤t≤2x
These spaces and their sequence analogs first appeared in the paper by D. Borwein [9], but became – for sequences – widely known after the paper by G.A. Fomin [14]; see also [19,20]. On the other hand, these spaces are a partial case of the so-called Herz spaces (see first of all the initial paper by Herz [23], and also a relevant paper of Flett [13]). Further, for q = ∞ let ∞ gA∞ =
ess sup |g(t)|dx. 0
x≤t≤2x
The role of integrable monotone majorant for problems of almost everywhere convergence of singular integrals is known from the work of D.K. Faddeev (see, e.g., [1, Ch. IV, §4]; also [40, Ch. I]); for spectral synthesis problems it was used by Beurling [6], for more details see [5]. In fact, in the previous works integrability properties of the Fourier transform have been considered for functions of bounded variation with the derivative in such spaces. Many of these results are summarized in [25]. Let us mention also [34] where similar problems are considered in a specific context. In this paper we study the outlined range of questions for the space of functions of bounded variation in whole.
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The space H01 (R+ ) is one of the widest spaces the belonging of the derivative f to which ensures the integrability of the cosine Fourier transform of f . However, the possibility of existence of a wider space of such type is of considerable interest. In addition to the conditions mentioned in Theorem 1 and further, the following one ensures the integrability of the cosine transform of a function of bounded variation f (see [31]): f ∈ A1,2 , where A1,2 is an amalgam type space endowed with the norm
gA1,2
⎧ ⎡ ⎤2 ⎫ 12 m (j+1)2 ⎪ ⎪ ∞ ∞ ⎨ ⎬ ⎢ ⎥ = |g(t)| dt⎦ dx < ∞. ⎣ ⎪ ⎪ ⎭ m=−∞ ⎩ j=1 m
(10)
j2
Remark 2. As for the sine transform, we have (8) but with the integral of F dominated by f A1,2 . The spaces H01 (R+ ) and A1,2 are incomparable, while all the Aq spaces are embedded in A1,2 like they are embedded in H01 (R+ ). There is a possibility to indicate the maximal space of integrability. In fact, it has in essence been introduced (for different purposes) in [26] as Q = {g : g ∈ L (R), 1
R
| g (x)| dx < ∞}. |x|
With the obvious norm gL1 (R) + R
| g (x)| dx |x|
it is a Banach space and ideal in L1 (R). The following result is a simple effect of the structure of this space (see [32]). Theorem 3. Let f : R+ → C be locally absolutely continuous on (0, ∞), of bounded variation and lim f (t) = t→∞ 0. a) The cosine Fourier transform of f given by (6) is Lebesgue integrable on R+ if and only if f ∈ Q. b) The sine Fourier transform of f given by (8) is Lebesgue integrable on R+ if and only if f ∈ Q. Discussion and comments are in order. First a) appeared in [30], it gave rise to the whole subject of this paper. On the other hand, Theorem 3 does not seem to be a result at all, at most a technical reformulation of the definitions (6) and (8). This could be so but not after appearance of the analysis of Q in [26]. Indeed, (1) implies H 1 (R) ⊆ Q ⊆ L10 (R),
(11)
where the latter is the subspace of all the functions g in L1 (R) which satisfy the cancellation property (3). Such functions are sometimes called wavelet functions, see, e.g., [38]. In fact, in a) and b) the conditions are given in terms of different subspaces of Q. We shall use the space Qo of the odd functions from Q rather than Q: ∞ Qo = {g : g ∈ L1 (R), g(−t) = −g(t), 0
|gs (x)| dx < ∞}; x
E. Liflyand / J. Math. Anal. Appl. 436 (2016) 1082–1101
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such functions naturally satisfy (3). This is exactly the space that works in a). As for b), an even counterpart of Qo is used, more precisely, ∞ Qe = {g : g ∈ L (R), g(−t) = g(t), 1
|gc (x)| dx < ∞}. x
0
This makes sense only if ∞ g(t) dt = 0. 0
In conclusion, let us give a simple consequence of Theorem 3 for the Fourier transform on the whole axis (cf. [21]). Corollary 4. Let f : R → C be locally absolutely continuous on R \{0}, of bounded variation and lim f (t) = |t|→∞
0. Then the Fourier transform of f f(x) =
∞
f (t)e−ixt dt
−∞
is Lebesgue integrable on R if and only if f (0) = 0 and f ∈ Q. 2.3. Near-proof of Theorem 1 Both proofs of Theorem 1, in [29, Thm. 2] and in [16], are quite technical and somewhat tedious. We are now in a position to give a simple proof not of Theorem 1 but of an almost equivalent statement. It will employ only a few routine calculations. First, (11) immediately proves (7). Here, we use (1) applied to Qo . Similarly, we can prove a necessary and sufficient condition for the integrability instead of (8) and (9) with just a shoe-string effort. This condition is ∞
|f (t)| dt < ∞ t
0
and follows from Theorem 1 by integrating the leading term on the right-hand side of (8). We will prove its sufficiency directly. Indeed, using (1) for f ∈ Qe from b) in Theorem 3, we have to estimate the L1 norm π of He f (x). It is the same as to consider 2xπ2 He f ( 2x ). We have π 2x2
∞ 0
π 2x f (t) π2 2 4x2 − t
π dt = 2 2x
∞ 0
π + 2 2x
∞ 0
Since I1 is exactly by parts, we get
π π 2x2 Ho f ( 2x ),
tf (t) dt π2 2 4x2 − t f (t) dt = I1 + I2 . π 2x + t
it will give the H01 bound. It remains to consider I2 . Using (4) and integrating
E. Liflyand / J. Math. Anal. Appl. 436 (2016) 1082–1101 π
π 2x2
2x 0
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π 2x π f (t) 2x π f (t) f (t) dt = 2 π + 2 dt π π 2x 2x + t 0 2x ( 2x + t)2 2x + t π
0
=
π 2x
π π 1 f( ) + 2 2x 2x 2x
0
f (t) dt. π ( 2x + t)2
Since π
π 2x2
2x 0
1 1 , dt = π ( 2x + t)2 2x
we obtain π
π 2x2
2x 0
π
1 π π f (t) dt = f ( ) + 2 π + t x 2x 2x 2x
2x 0
π f (t) − f ( 2x ) dt. π ( 2x + t)2
The last integral can be represented as π
2x − 0
π
1 π ( 2x + t)2
2x
f (u) du dt,
t
and we have to estimate ∞ 0
π
π 2x2
π
2x 0
π ( 2x
1 + t)2
2x
∞ x
|f (u)| du dt dx = t
0
0
1 (x + t)2
x
|f (u)| du dt dx.
t
It is just (5), which completes the proof. 2 Remark 5. The above proof and Proposition 1 show how and why the odd Hilbert transform plays a special role in the considered problems. In the next sections we will see even more on this issue. 3. The sine Fourier transform For the cosine Fourier transform, we cannot suggest more than a) in Theorem 3. The situation is more delicate with the sine Fourier transform, where a sort of asymptotic relation can be obtained. On the one hand, (8) and (9) do not follow from Theorem 3. On the other hand, Theorem 1 is already a hint of non-stoppage after getting b) in Theorem 3 and of search for something more developed. Somewhat surprisingly, in a situation more general than that in Theorem 1, the form of the answer is different from that in (8). In what follows we shall denote g(x) =
gs (x) . x
Theorem 6. Let f : R+ → C be locally absolutely continuous on (0, ∞), of bounded variation and lim f (t) = t→∞
0. Then for the sine Fourier transform of f given in (8) there holds for any x > 0
E. Liflyand / J. Math. Anal. Appl. 436 (2016) 1082–1101
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1 π fs (x) = f − Ho f (x) + G(x), x 2x
(12)
where GL1 (R+ ) f L1 (R+ ) . To compare with Theorem 3, it should be mentioned that the necessary and sufficient condition in a) is, in fact, f ∈ Qo . Theorem 6 also deals with Qo but in a more sophisticated way and makes it natural to 1 consider a Hardy type space HQ (R+ ) which consists of Qo functions g with integrable Ho g. 1 Corollary 7. If a function f satisfies the assumptions of Theorem 6 and is such that f ∈ HQ (R+ ), then
1 π fs (x) = f + G(x), x 2x where GL1 (R+ ) f HQ1 (R+ ) . Technically, this is an obvious corollary of Theorem 6. We shall discuss it later on. Proof of Theorem 6. Let us start with integration by parts in π
2x 0
π 2x 2x 1 − cos xt 1 f (t) + f (t) sin xt dt = f (t)[cos xt − 1] dt x x 0 π
0
=
1 π 1 f + x 2x x
π 2x
f (t)[cos xt − 1] dt.
0
The last value is bounded by
π 2x
0
t|f (t)| dt, and π
∞ 2x 0
π t|f (t)| dt dx = 2
0
∞
|f (t)| dt.
(13)
0
Going over to ∞ I = I(x) =
f (t) sin xt dt, π 2x
we start with the following statement. Lemma 1. If f satisfies the assumptions of Theorem 6, then −Ho
f (x)
2 = xπ
∞
f (t) sin xt ∞
xt
f (t) cos xt 0
cos v dv dt v
xt
0
2 + xπ
∞
0
sin v dv dt. v
(14)
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Proof. Denoting, as usual, ∞ Ci(u) = −
cos t dt t
u
and u Si(u) =
π sin t dt = − t 2
∞
sin t dt, t
u
0
we will make use of the formula (see [4, Ch. II, §2.2 (18)]) ∞ 0
1 sin ut du = lim 2 δ→0+ x − u2
x2
|x−u|>δ
1 1 sin ut du = [sin xt Ci(xt) − cos xt Si(xt)], 2 −u x
(15)
where the integrals are understood in the principal value sense and x, t > 0. Since f ∈ L1 (R+ ) and the limit in (15) is uniform in t, we have
Ho f (x) =
2 π
∞
fs (u)
0
2 = π
∞
1 du x 2 − u2 ∞
f (t) 0
x2 0
1 sin ut du dt. − u2
(16)
Applying (15) to the inner integral on the right-hand side of (16) and using the expressions for Ci and Si completes the proof of the lemma. 2 With this in hand, we are going to prove that I(x) = −Ho f (x) + G(x),
(17)
where GL1 (R+ ) f L1 (R+ ) . We denote the two summands on the right-hand side of (14) by I1 and I2 . For both, we make use of the fact that ∞
1 cos v dv = O( ). v xt
xt
The same is true when cos v is replaced by sin v. We begin with the integral of I1 . For t ≥ x1 , we have ∞ 0
1 x
∞ 1 x
|f (t)|
1 dt dx = xt
∞ 0
|f (t)| t
∞ 1 t
1 dx dt = x2
∞ 0
|f (t)| dt.
(18)
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For t ≤ x1 , we split the inner integral in I1 into two. First, ∞
cos v dv = O(1), v
1
and using sinxxt ≤ t, we arrive at the relation similar to (13). Further, we have 1 cos v dv = O ln 1 . v xt xt
By this, integrating in x over (0, ∞), we end up with ∞
1/t ∞ 1 dx dt = |f (t)| dt. |f (t)|t ln xt
0
0
0
Here we use that 1
t
ln
1 1 dx = . xt t
0
In conclusion, I1 can be treated as G. Let us proceed to the integral of I2 . Using that xt 1 sin v dv = O(t), x v 0
we arrive, for t ≤
π 2x ,
at (13). Let now t ≤ xt
π 2x .
We have
π sin v dv = − v 2
∞
sin v dv. v
xt
0
Now 2 xπ ∞ For the integral
∞ π 2x
f (t) cos xt dt
π = I. 2
sin v dv, the estimates are exactly like those in (18). v
xt
Combining (17) and the estimates before Lemma 1, we complete the proof of the theorem. 2 Remark 8. Formally, there exists certain symmetry between the two Fourier transforms. Indeed, using the formula (see [4, Ch. II, §1.2 (15)])
E. Liflyand / J. Math. Anal. Appl. 436 (2016) 1082–1101
∞ a2 0
1093
π 1 sin ay, cos yx dx = 2 −x 2a
we obtain Ho
fc (u) u
2 (x) = π
∞ 0
=
2 π
∞
u 1 2 2 x −u u f (t)
0
=
1 x
∞
∞ 0
∞
f (t) cos ut dt du
0
cos ut du dt x 2 − u2
f (t) sin xt dt.
(19)
0
Comparing this with the proof of a) in Theorem 3, on the one hand, and with (12) on the other hand, we see that (19) represents the cosine Fourier transform in the form of a counterpart of (12) in the sense that it gives an analog of the second term on the right-hand side of (12), while the “remainder term” is identical zero. Of course, the first term cannot be reproduced, since it is an ultimate feature of the sine transform. 4. Intermediate spaces It is highly improbable that Q (or Qo ) can be defined in terms of g itself rather than its Fourier transform. Hence it is of interest to find certain proper subspaces of Qo wider than H01 belonging to which can be verified in a reasonable way. For example, in the paper [41] a family of nested subspaces between H 1 and L1 is introduced and duality properties of that family are studied. However, it is not clear how to compare that family with Qo . Before searching for other subspaces, the relations between those already considered should be made more exact. 4.1. Embeddings Back to Theorem 6, let us analyze (17). On the one hand, we have ∞
∞ |I(x)| dx =
0
|Ho f (x)| dx + O(f H01 (R+ ) ).
0
On the other hand, it is proved in [29] that ∞
|I(x)| dx = O(f H01 (R+ ) ).
0
This leads to Proposition 2. If g is an integrable odd function, then Ho gL1 (R+ ) gH01 (R+ ) .
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The above proof of Proposition 2 looks “artificial”. Let us give a direct proof. Proof of Proposition 2. The proof will be molecular. The idea of molecular decomposition is due to Coifman and Weiss [11]; good description can be found in [18, Ch. III, §7]. Recall that a function M is called a molecule (centered at x0 ∈ R) if it satisfies (3) and ⎛ ⎝
⎞ 14 ⎛ |M (x)|2 dx⎠ ⎝
R
⎞ 14 |x − x0 |2 |M (x)|2 dx⎠ < ∞.
R
The left-hand side is called a molecular norm of M and is denoted by N (M ). A function g belongs to H 1 (R) ∞ ∞ Mj (x) for almost all x, with the Mj -s being molecules which satisfy N (Mj ) < ∞. if and only if g(x) = j=1
j=1
We now represent g as a molecular sum. To prove the proposition, it suffices to show that, for every s (x) M is also a molecule. Indeed, since we consider the odd Hilbert transform, molecule M , the function x s (x) M . We have (3) is satisfied automatically for the odd extension of x s (x)| ≤ |M
x
∞ |M (t)| dt + x
|M (t)| t dt x
0
and ⎛ ⎞ 12 ⎞2 ⎞ 12 ⎛∞ ⎛ x ∞ 2 ⎟ −2 ⎝ |Ms (x)| dx⎠ ≤ ⎜ ⎝ x ⎝ |M (t)| dt⎠ dx⎠ x2 0
0
0
⎛ ⎛ ⎞2 ⎞ 12 ∞ ∞ ⎜ ⎟ + ⎝ ⎝ |M (t)| t dt⎠ dx⎠ = I1 + I2 . 0
x
We will estimate both I1 and I2 by means of Hardy’s inequalities (see, e.g., [22, (330)]): ⎡ ⎣
∞
⎤ 12 ⎡∞ ⎤ 12 2 2b x R(x) dx⎦ ⎣ x2b+2 ψ(x)2 dx⎦ ,
0
0
x where, for ψ(x) ≥ 0, either R(x) =
∞ ψ(t) dt and 2b < −1, or R(x) =
ψ(t) dt and 2b > −1. For I1 , we x
0
have the first case with ψ(t) := M (t) and 2b = −2 < −1, correspondingly I1 M L2 (R+ ) . For I2 , we have the second case with ψ(t) := tM (t), 2b = 0 > −1, and ⎡ I2 ⎣
∞
⎤ 12 x2 M (x)2 dx⎦ .
0
s immediately follows from Parseval’s identity and the square integrability of M , Square integrability of M which completes the proof. 2
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This implies a broadened chain of embeddings: 1 H01 (R+ ) ⊆ HQ (R+ ) ⊆ Qo ⊆ L10 (R+ ).
It is very interesting to figure out which of these embeddings are proper. Correspondingly, intermediate spaces are of interest, both theoretical and practical. 4.2. A counterexample We will prove that H01 (R+ ) = Qo . The above considerations reduce this problem to constructing a function of bounded variation such that its cosine Fourier transform is integrable, while its derivative is not in H01 (R+ ). Counterexamples for sequences can be found in [2] and [15], we will build an example for functions in a similar manner, more precisely, an example of a function g from A1,2 (see (10)) but not from H01 (R+ ) will be constructed. First, we may think that g is good enough on (0, 1) and concentrate on its behavior on (1, ∞). Therefore, we compare the values t ∞ 2 g(t − s) − g(t + s) ds dt N1 = s 1 0 and ⎧ ⎡ ⎤2 ⎫ 12 m (j+1)2 ⎪ ⎪ ∞ ⎨ ⎬ ⎢ ⎥ N2 = |g(t)| dt⎦ dx < ∞. ⎣ ⎪ ⎪ ⎭ m=0 ⎩ j=1 m ∞
j2
We can restrict ourselves to the upper limit 2t in the inner integral for N1 , since that integral, named the Telyakovskii transform in [16], differs from the Hilbert transform of the odd function g by an integrable function (see, e.g., [25, Ch. 3, Lemma 3.40]). Let the function g be non-negative, vanishing everywhere except the intervals (2m , 2m + 1) and satisfying m 2 +1
g(t) dt =
1 (m + 1)2
2m
for m = 0, 1, 2, . . . . Obviously, the integral does not vanish only when j is a power of 2, hence
N2 ≤
∞
⎛ ∞ ⎝
m=0
j=0
⎞ 12
∞ 3 1 ⎠ ≤ (m + 1)− 2 < ∞. (m + 1 + j)4 m=0
On the other hand, t m+1 t ∞ 2 2 ∞ 2 g(t − s) − g(t + s) g(t − s) − g(t + s) ds dt = ds dt s s m=0 2m 0 1 0 m m−1 t ∞ 2 +2 2 g(t − s) − g(t + s) ≥ ds dt s m=2 2m +1 0
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E. Liflyand / J. Math. Anal. Appl. 436 (2016) 1082–1101
=
k+1 2 m−1 ∞ 2m +2 −1 m=2
Since 0 < k − 2m − 1 < k − 2m + 1 <
k 2
k=2m +2
0
k
g(t − s) ds dt. s
≥ 2t , we have t
2
g(t − s) ≥ s
0
1 ≥ k − 2m + 1
t
m k−2 +1
k−2m −1
g(t − s) ds s
m k−2 +1
1 g(t − s) ds = k − 2m + 1
k−2m −1
m 2 +1
g(s) ds. 2m
Consequently, ∞
1 N1 ≥ (m + 1)2 m=2
m−1 2m +2 −1
k=2m +2
1 . k − 2m + 1
The right-hand side is equivalent to ∞
1 = ∞, m +1 m=2 which gives the desired counterexample. Taking into account Remark 2 and Theorem 6 and supplementing the above counterexample with pro1 to be integrable, we can construct f with the derivative in A1,2 but not from HQ . viding |f (x)| x 1 4.3. Intermediate spaces between H01 and HQ 1 We are now going to find a nested sequence of intermediate spaces between H01 and HQ . They will be applicable to both sine and cosine Fourier transforms. Look back at the Aq spaces from Subsection 2.2, which are the subspaces of H01 . We define a symmetric scale but with respect to g rather than for all g. Let for 1 < q < ∞,
∞ gΣq = 0
⎛ ⎝1 x
⎞ q1 |g(t)|q dt⎠ dx.
x≤t≤2x
The case q = ∞ is treated in an obvious way as above. Analogously to Aq , these spaces are subspaces of 1 HQ . Proposition 3. For 1 < q < 2, we have 1 H01 (R+ ) ⊂ Σq ⊂ HQ (R+ ).
Proof. It has been proved above that for each molecule in the molecular representation of g ∈ H01 (R+ ) the function g is also a molecule with the norm bounded by a constant, which is the same for all such molecules. 1 Therefore such g also belongs to HQ . To prove the present proposition, it suffices to show that g also belongs to corresponding Σq . In turn, it will be proven if we show that the function
E. Liflyand / J. Math. Anal. Appl. 436 (2016) 1082–1101
⎛
⎞ 12
⎝1 x
1097
|g(t)|2 dt⎠ dx
x≤t≤2x
and this function times x are both square integrable provided that g is a molecule. Indeed, all Σq with 1 1 < q < 2 will be between Σ2 and HQ then. So, let ⎛ LM = ⎝
⎞ 12
1 x
|M (t)|2 dt⎠ dx,
x≤t≤2x
where M is a molecule. This is enough since, as is proven above, g is a molecule if g is. We have ∞
∞ |LM (x)|2 dx =
0
1 x
0
∞
|M (t)|2 dt dx = ln 2
|M (t)|2 dt, 0
x≤t≤2x
which is ln 2M 22 . Further, ∞
∞ x |LM (x)| dx ≤ 2
2
0
0
1 x
∞ t |M (t)| dt dx = ln 2 2
t2 |M (t)|2 dt,
2
0
x≤t≤2x
which is ln 2tM (t)22 . The proof is complete. 2 By this, we have arrived to the following extended chain of embeddings A∞ ⊂ . . . ⊂
Aq 1
⊂ . . . H01 (R+ ) ⊂ Σ2 . . . ⊂
Σq 1
1 ⊂ . . . HQ (R+ ) ⊆ Qo ⊂ L10 (R+ ).
1 It is clear that though the spaces Σq deliver a scale of intermediate spaces between H01 and HQ , the search for other such spaces, more convenient in applications, is an open problem of considerable importance.
5. Hardy type inequalities Back to (1), this inequality shows that for the Fourier transform of a function from the Hardy space we have more than just the Riemann–Lebesgue lemma for an integrable function and its Fourier transform. There is a possibility to derive a statement complement, in a sense, to Corollary 4. Proposition 4. Let the integral on the left-hand side of (1) be finite, or, equivalently, let g ∈ Q. Then g is the derivative of a function of bounded variation f which is locally absolutely continuous, lim f (t) = 0 |t|→∞
and its Fourier transform is integrable. t Proof. Denoting f (t) =
g(u) du, we integrate by parts as follows −∞
∞
−ixt
g(t)e −∞
" #∞ # t ∞ " t ∞ −ixt −ixt dt = e g(u) du + ix g(u) du e dt = ix f (t)e−ixt dt. −∞
−∞
−∞ −∞
−∞
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E. Liflyand / J. Math. Anal. Appl. 436 (2016) 1082–1101
The integrated terms vanish because of (3). It follows now from g ∈ Q that the L1 norm of f is controlled by the Q norm of its derivative. 2 The obtained estimates lead to refinement of (1) in certain particular cases. More precisely, for odd functions (1) reduces to gQo gH01 . It is proved in Subsection 4.2 that this inequality is meaningful, that is, the left-hand side can be finite, with the infinite right-hand side for the same function. Surprisingly, it seems that no such example has been given in literature before. A refinement of (1) may be an inequality with either a) a smaller right-hand side in place of gH01 , in other words, for a space wider than H01 but still smaller than Qo ; or b) a larger left-hand side in place of gQo , that is, for a space smaller than Qo but still wider than H01 ; or c) both a left-hand side larger than gQo and a right-hand side smaller than gH01 . In fact, we have obtained inequalities of all three types. For a), we have gQo gHQ1 gΣq . For b), we have gHQ1 gΣq gH01 . It is easy to see that in both cases we have double inequalities, since an inequality of type c) holds true: gHQ1 gΣq . Recall that in all the cases 1 < q < 2. 6. Multidimensional case: radial functions Of course, this approach should be and already is, in part, extended to the multivariate case. For example, a multidimensional version of Theorem 1 is given in the same paper [29] where the theorem itself is proved for the first time. This is a meticulous work, partially because of a variety of multidimensional variations, each with its specific properties and of its special significance. However, one particular setting can be treated immediately. Using a formula for the Fourier transform of a radial function from [36], we can generalize the obtained above results to the radial case. We first need to dwell upon a notion of fractional derivative. For 0 < δ < 1 and a locally integrable function g on (0, ∞), define the fractional (Weyl type) integral of order δ by $ Wωδ g(t)
=
1 Γ(δ)
0,
ω t
g(r)(r − t)δ−1 dr,
0 < t < ω, t ≥ ω,
and, following Cossar [12], a fractional Weyl derivative of order α by
E. Liflyand / J. Math. Anal. Appl. 436 (2016) 1082–1101
g (α) (t) = lim − ω→∞
1099
d 1−α W g(t) dt ω
when 0 < α < 1 and dp (δ) g (t) dtp
g (α) (t) =
when α = p + δ with p = 1, 2, . . . , and 0 < δ < 1. One of the reasons that just this type of fractional integral (and derivative) is chosen is that the Weyl integral of a function with compact support has, in turn, compact support, unlike the better known Riemann–Liouville integral 1 Rα (f0 ; t) = Γ(α)
t f0 (r) (t − r)α−1 dr. 0
All these notions may be found, for example, in [3, Ch. 13] (see also [39]). Let α∗ be the greatest integer less than α. If α is fractional α∗ = [α], while for α integer α∗ = α − 1. b Consider the class M Vα+1 , with α > 0 and b ≥ 0, of C(0, ∞)-functions satisfying the following conditions g, g , . . . , g (α
∗
are LAC on (0, ∞);
)
lim tα+b g (α) (t) = 0;
lim g(t),
t→∞
t→∞
and ∞ gM Vα+1 := sup |t g(t)| + b b
t>0
d[(tα+b g (α) (t)] < ∞.
0
Let us first describe for the introduced class, the passage from the multidimensional Fourier transform of a radial function to the one-dimensional Fourier transform of a related function. The most important partial case will give us a possibility to extend the obtained one-dimensional results to the radial case in a n−1 (α) natural way. Let us denote Fα (t) = t 2 f0 (t) and F := F n+1 (t). 2
b Theorem 9. Let f0 ∈ M Vα+1 with 0 < α ≤ f (x) = f0 (|x|) of f0 ,
n+1 n−1 ∗ f(x) = 2 2 π 2 (−1)α +1
+
Cα |x|α+
Fα
n+1 2
π 2|x|
n−1 2
and b = ∞
1 n−1 |x|α+ 2
% π(n + 2α − 1) & dt Fα (t) cos |x|t − 4
0
+O
− α. Then there holds, for the radial extension
n−1 2
∞
1 |x|α+
n+1 2
' 2|x|t π ( , |dFα (t)| . min π 2|x|t
(20)
0
0 Here Cα is an absolute constant depending only on α. In particular, if f0 ∈ M V n+1 , then 2
f(x) = 2
n+1 2
π
n−1 2
C n−1 + where
∞ 0
2
|x|n
0 |G(r)| dr ≤ f0 M V n+1 . 2
F
[ n+3 2 ]
(−1) π 2|x|
1 |x|n−1
+O
∞ 0
G(|x|) |x|n−1
% πn & dt F (t) cos |x|t − 2 ,
(21)
E. Liflyand / J. Math. Anal. Appl. 436 (2016) 1082–1101
1100
We limit ourselves to an extension of Theorem 3. Theorem 10. Let f0 ∈ M V n+1 ; assume additionally F to be locally absolutely continuous on (0, ∞). Then 2 for |x| > 0, Cn F f(x) = |x|n where
∞ 0
π 2|x|
+O
G(|x|) |x|n−1
,
|G(r)| dr ≤ F Q .
We can be more precise by observing that the last norm is in Qe for n even and in Qo for n odd. The theorem follows readily by applying Theorem 3 to the one-dimensional Fourier transform in (21). It is completely clear how to generalize to the radial case many other above results. Except [36], there are two more approaches for the Fourier transform of a radial functions to be represented via a one-dimensional Fourier transform (without Bessel functions as in the classical well-known formula, see, e.g., [40, Ch. IV]): Leray’s formula (see, e.g., [39, Ch. 5, Lemma 25.1 ]) and the one from [37]. However, just (20) seems to be perfectly suitable for generalizations to the radial case of known one-dimensional results for the Fourier transform of a function of bounded variation. Acknowledgment The author is grateful to the referee for extremely thorough reading and valuable remarks and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26]
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