Fourier-Type Kernels and Transforms on the Real Line

FOURIER-TYPE KERNELS AND TRANSFORMS ON THE REAL LINE

GOONG CHEN AND DAOWEI MA

Abstract. Kernels of integral transforms of the form k(xy) are called Fourier kernels. Hardy and Titchmarsh [6] and Watson [15] studied self-reciprocal transforms with Fourier kernels on the positive half-real line R+ and obtained nice characterizing conditions and other properties. Nevertheless, the Fourier transform has an inherent complex, unitary structure and by treating it on the whole real line more interesting properties can be explored. In this paper, we characterize all unitary integral transforms on L2(R) with Fourier kernels. Characterizing conditions are obtained through a natural splitting of the kernel on R+ and R−, and the conversion to convolution integrals. A collection of concrete examples are obtained, which also include ”discrete” transforms (that are not integral transforms). Furthermore, we explore algebraic structures for operator-composing integral kernels and identify a certain Abelian group associated with one of such operations. Fractional powers can also be assigned a new interpretation and obtained.

1. Introduction

The Fourier transform is one of the most powerful methods and tools in mathematics (see, e.g., [3]). Its applications are especially prominent in signal processing and diﬀerential equations, but many other applications also make the Fourier transform and its variants universal elsewhere in almost all branches of science and engineering. Due to its utter signiﬁcance, it is of interest to investigate the Fourier transform from a fundamental point of view. In Hardy and Titchmarsh [6] and Watson [15], those authors look at some fundamental properties of the Fourier transform and ask the following question: [Q] What types of integral transforms behave in a way similar to the Fourier transform? There, by “similar-to-Fourier” they used the following criteria: (1) the integral transform is self-reciprocal on R = (0, ∞): ∫ ∫ + π f(x) = k(xy)k(yz)f(z) dz dy 2 R+ R+ just like the Fourier cosine transform: ∫ ∫ π f(x) = cos(xy) cos(yz)f(z) dz dy; 2 R+ R+ (2) the kernel of the integral transform is a Fourier kernel, i.e., of the form k(xy).

2010 Mathematics Subject Classiﬁcation. Primary: 42A38. Key words and phrases. Unitary transformation, convolution operator, analytic functions. 1 2 GOONG CHEN AND DAOWEI MA

A characterizing condition on such k(xy) in terms of the Mellin transform has been obtained in [6] and [15]: k˜(s)k˜(1 − s) = 1, where ∫ k˜(s) = xs−1k(x) dx, Re s > 0, R+ is the Mellin transform of k. This is an outstanding result, among others such as concrete examples of k, summability results, etc., in [6], [15] and [13]. They have stimulated many other papers, see [2], [4], [8], [9], [10], [16], for example. It is noteworthy that the original papers of Hardy and Titchmarsh and Watson have also greatly influenced (almost) all follow-up studies in the 2 2 sense that the space of all [2], [4], [8], [9], [10], [16] dwells on L (R+) rather than on L (R). Important and interesting as those cited references are, we know that the Fourier transform has an inherent complex structure and is (more) naturally defined on R (rather than R+). Why don’t we revisit the same problem studied by those analysts on the entire R as a complex transform? This motivates the study in this paper, i.e., we wish to consider [Q], but use the following criteria: (A) the integral transform is unitary; (B) the integral kernel is Fourier, i.e., of the form k(xy). (Same as (2) above.) Items (A) and (B) above are motivated by the well known Fourier-transform properties ∫ 1 ∞ fˆ(x) = F(f)(x) = √ e−ixyf(y) dy, 2π ∫−∞ 1 ∞ f(x) = F(fˆ)(x) = √ eixyfˆ(y) dy, 2π −∞ so that√ we have FF = id, where id denotes the identity operator, and the kernel is k(xy) = − e ixy/ 2π. Thus, the Fourier transform itself is unitary, with a kernel k√(xy), satisfying both (A) and (B) above. In other words, the Fourier kernel k(x) = e−ix/ 2π satisfies ∫ ∫ (1) f(x) = k(xu) du k(uy)f(y) dy, f ∈ L2(R). R R

To answer the posed question [Q], we adopt a basic approach diﬀerent from the earlier references by using the Fourier transform itself instead of the Mellin transform, as we feel that this seems to be a more natural thing to do here. A key technique converts a Fourier kernel into a convolution kernel of the form k(x − y); cf. (8)-(11). The integral transform is split into a sum of two with convolution kernels. Since much is well understood for convolution integrals, we are able to ﬁnd characterizing conditions, given in Theorem 2.3, for such kernels. Comparison with the Mellin transform result is also made in Theorem 2.4. This is done in Section 2 of the paper. In Section 3, a set of concrete examples are given and tabulated from the characterizing conditions. FOURIER-TYPE TRANSFORMS 3

In Section 4, we explore the algebraic structures of the associated operators and kernels. An Abelian group is identiﬁed. In Section 5, we further introduce the notion of “Fourier-like” kernels, oﬀering an even closer resemblance to the Fourier transform kernel. In Section 6, we investigate the possibility of taking “fractional powers” and assign a new meaning in order for an analogue to be possible. Concluding remarks are given in Section 7.

2. Fourier-type transforms, splitting and conversion to convolution

We begin by considering a function (or a distribution in a certain class) k and associate to it an operator K by ∫

(2) Kf(x) = Kkf(x) = k(xy)f(y) dy. R Its complex-conjugate operator is then ∫

Kf(x) = Kkf(x) = k(xy)f(y) dy. R

The main task of this section is to characterize all k such that KK = id. Such an operator K satisfying KK = id is known to be a unitary operator on the Hilbert space L2(R√). Let K be the collection of all such k. It is clear that the kernel function τ(x) := e−ix/ 2π of the Fourier transform belongs to K .

We initially consider K a linear operator on S (R−) ⊕ S (R+), where S (R±) is the Schwartz class of C∞ functions that decrease rapidly at 0 and ±∞; see [5], for example. Let ∫ (f, g) = fg dx.

Deﬁnition 2.1. Let f be any function or distribution on R. We deﬁne f+ and f− by { { f(x) if x ≥ 0, f(−x) if x > 0, f (x) = f−(x) = + 0 if x < 0, 0 if x ≤ 0. Proposition 2.2. If k ∈ K , then (Kf, Kg) = (f, g).

Proof. (Kf, Kg) = (KKf, g) = (f, g).

Thus K can be extended to L2(R). For the amenability of the problem to subsequent analysis, at this point we perform a natural splitting of the kernel k: (3) k(x) = p(x) + q(−x), 4 GOONG CHEN AND DAOWEI MA

where p(x) = k+(x) and q(x) = k−(x). From the functions p and q we deﬁne operators P and Q by ∫ ∫ P f(x) = p(xy)f(y) dy, Qf(x) = q(xy)f(y) dy. Then the induced operator splitting for K is (4) Kf(x) = P f(x) + Qf(−x). Thus f(x) = KKf(x) = (PP + QQ)f(x) + (PQ + QP )f(−x), or (5) f(x) = Lf(x) + ιSf(x), where L = PP + QQ, S = PQ + QP, ιg(x) = g(−x).

2 2 2 Note that each of the four operators P, Q, L, S maps L (R+) to L (R+), and L (R−) to 2 2 2 2 L (R−). For f ∈ L (R+), we have Lf ∈ L (R+) and ιSf ∈ L (R−). It follows from (5) 2 that, for f ∈ L (R+), the equalities (6) Lf = f, Sf = 0 2 2 hold. Similarly, (6) holds for f ∈ L (R−). It follows that (6) holds for f ∈ L (R) and therefore (7) PP + QQ = id, and PQ + QP = 0.

Consider an operator T defined by T f(x) = f(ex)ex/2. The operator T is an isometry 2 2 from L (R+) to L (R), and it maps a kernel defined on R+ to a kernel defined on R. Let f ∈ L2(0, ∞). Then ∫ ∫ (8) T P f(x) = ex/2p(exy)f(y) dy = ex/2p(exeu)f(eu)eu du, R+ R so ∫ ∫ (9) T P f(x) = e(x+u)/2p(ex+u)f(eu)eu/2 du = T p(x + u)T f(u) du, R R or ∫ Pˆ g(x) := TPT −1g(x) = T p(x + u)g(u) du, g ∈ L2(R), R or ∫ Pˆ g(x) = pˆ(x + y)g(y) dy, g ∈ L2(R), R where (10) pˆ(x) = T p(x) = ex/2p(ex), qˆ(x) = T q(x) = ex/2q(ex). It follows that √ FPˆ g(t) = 2πFpˆ(t)Fg(−t). Let √ √ (11) u(t) = 2πFpˆ(t), v(t) = 2πFqˆ(t). FOURIER-TYPE TRANSFORMS 5

Then √ √ u(−t) = 2πFpˆ(t), v(−t) = 2πFˆq(t). Hence, (12) FPˆ g(t) = u(t)Fg(−t), FPˆ g(t) = u(−t)Fg(−t).

Combining these two equations, we obtain FPˆ Pˆ g(t) = u(−t)u(−t)Fg(t). Combining the above equality and a similar equality for Qˆ Qˆ, we obtain F(Pˆ Pˆ + Qˆ Qˆ)g(t) = (u(−t)u(−t) + v(−t)v(−t))Fg(t). Since Pˆ Pˆ + Qˆ Qˆ = T (PP + QQ)T −1 = id, we see that u(−t)u(−t) + v(−t)v(−t) = 1, or u(t)u(t) + v(t)v(t) = 1. Similarly, from PQ + QP = 0, we derive u(t)v(t) + v(t)u(t) = 0. In summary, we have proved that if k ∈ K , then u(t)u(t) + v(t)v(t) = 1 and u(t)v(t) + v(t)u(t) = 0. It is clear that all reasoning and steps are reversible. Thus, we obtain the following. Theorem 2.3. (Characterizing conditions) Let k be a kernel, and let k’s splitting be (3). Deﬁne u, v in terms of p, q as in (10) and (11). Then k ∈ K if and only if (13) uu + vv = 1 and uv + vu = 0, or, equivalently, (14) |u|2 + |v|2 = 1, ℜ(u/v) = 0. Corollary 2.4. For u and v given in Theorem 2.3, there are real functions α and φ for which (15) u = eiα cos φ, v = ieiα sin φ.

By (10) and (11), we have ∫ ∞ ∫ ∞ u(t) = T p(x)e−itx dx = p(x)x−1/2−it dx = Mp(1/2 − it), −∞ 0 (16) ∫ ∞ ∫ ∞ v(t) = T q(x)e−itx dx = q(x)x−1/2−it dx = Mq(1/2 − it), −∞ 0 where ∫ Mh(s) = xs−1h(x) dx R+ is the Mellin transform. A crucial property of the Mellin transform is ∫ (17) Mh(s) = Mp(s)Mf(1 − s), where h(x) = p(xy)f(y) dy. R+ 6 GOONG CHEN AND DAOWEI MA

Even though our treatment here does not use the Mellin transform, it is still useful to see what the characterizing conditions may look like if that transform is actually used. It also serve as a connection and comparison with the earlier works. Thus we present the following. Theorem 2.5. (Characterizing conditions in terms of the Mellin transform) Let k(xy) be a Fourier kernel (on R) such that (A) and (B) are satisﬁed. This holds if and only if ( )( ) κ+(1 − s) κ−(1 − s) κ+(s) κ−(s) (18) = I2, for all s ∈ C : 0 < Re s < 1, κ−(1 − s) κ+(1 − s) κ−(s) κ+(s)

where κ+ and κ− are, respectively, the Mellin transforms of k+ and k− of the Fourier kernel k, κ±(s) = κ±(s), I2 is the 2 × 2 identity matrix, and s can be analytically continued on C.

∞ Proof. For any C function f with compact support on R, we have f(x) = f+(x) + f−(x). Let k(xy) be a Fourier kernel satisfying (A) and (B), and let ∫ g(x) = k(xy)f(y) dy. R Then ∫ ∫

(19) g+(x) = k+(xy)f+(y) dy + k−(xy)f−(y) dy; R+ R+ and ∫ ∫

(20) g−(x) = k−(xy)f+(y) dy + k+(xy)f−(y) dy. R+ R+

Here (19) and (20) are exactly (4) in long hand by noting that k+ = p and k− = q. Taking the Mellin transforms of (19) and (20) and using (17), We obtain

γ+(s) = κ+(s)ϕ+(1 − s) + κ−(s)ϕ−(1 − s),

γ−(s) = κ−(s)ϕ+(1 − s) + κ+(s)ϕ−(1 − s), where γ±, κ± and ϕ± are, respectively, the Mellin transforms of g±, k± and f±. We thus have ( ) ( )( ) γ (s) κ (s) κ−(s) ϕ (1 − s) (21) + = + + . γ−(s) κ−(s) κ+(s) ϕ−(1 − s) Similarly, from the reciprocal relation ∫ f(x) = k(xy)g(y) dy, R we obtain ( ) ( )( ) ϕ (s) κ (s) κ−(s) γ (1 − s) (22) + = + + . ϕ−(s) κ−(s) κ+(s) γ−(1 − s) From (21) and (22), we now conclude (18).

The equation (18) is actually a generalization of the characterizing condition K(s)K(1− s) = 1 obtained by Hardy and Titchmarsh [6] and Watson [15]. FOURIER-TYPE TRANSFORMS 7

Corollary 2.6. Let κ+ and κ− satisfy (18). Then

2 2 (23) |κ+(1/2+it)| +|κ−(1/2+it)| = 1, Re(κ+(1/2+it)κ−(1/2 + it)) = 0, t ∈ R.

Note that (23) is equivalent to (14). Remark 2.1 In comparison between Theorem 2.3, Corollary 2.4 and Theorem 2.5, we feel that Corollary 2.4 oﬀers perhaps the simplest form of the characterizing conditions, in terms of circular complex functions. Its group structure is well understood. This fact, and other advantages, will be exploited in the subsequent sections.

3. Examples of discrete and integral Fourier-akin transforms

In this section, we apply the results of Section 2 to calculate a number of linear transformations whose kernels k are in K . A mixture of of discrete and integral transforms are obtained, based on the special forms of u and v.

For k ∈ K , let uk, vk, pk, qk,... denote the corresponding functions. √ Example 3.1. (The Fourier transform itself) Let τ = e−ix/ 2π be the Fourier kernel. Then Kτ is the√ Fourier transform.√ We now work out the functions u = uτ and v = vτ . We have p = e−ix/ 2π, q = eix/ 2π, so v(t) = u(−t), and ∫ ∞ u(t) = p(x)x−1/2−it dx 0 ∫ 1 ∞ =√ e−ixx−1/2−it dx 2π 0 (24) ∫ e−πi(1/2−it)/2 ∞ = √ e−xx−1/2−it dx 2π 0 e−πi/4e−πt/2 = √ Γ(1/2 − it), 2π where Γ is the usual gamma function [1, Chap. 6], and

eπi/4eπt/2 eπi/4eπt/2 (25) v(t) = √ Γ(1/2 + it) = √ Γ(1/2 − it), 2π 2π

eπt + e−πt |u|2 + |v|2 = |Γ(1/2 − it)|2 2π cosh πt π = = 1, π sin(π(1/2 − it)) (cf. [1, (6.1.29)–(6.1.31), p. 256]) (v/u)(t) = ieπt. Example 3.2. Let u(t) = cos(t2), v = i sin(t2). We now work out k. 8 GOONG CHEN AND DAOWEI MA ∫ ∞ it2 7→ πi/4 Let I = 0 e dt. By a substitution t e t (this is formal calculation; to be rigorous, one needs to consider path integrals on the complex plane), we obtain ∫ ∞ √ I = eπi/4 e−t2 dt = eπi/4 π/2. 0 Thus ∫ ∫ ∞ √ ∞ √ eit2 dt = eπi/4 π, e−it2 dt = e−πi/4 π. −∞ −∞ It follows (cf. also [7, p. 116]) that ∫ ∫ ∞ ∞ √ eit2+itx dt = e−ix2/4 ei(t+x/2)2 dt = ei(π−x2)/4 π. −∞ −∞ Thus ∫ ∫ ∞ √ ∞ √ eit2±itx dt = ei(π−x2)/4 π, e−it2±itx dt = e−i(π−x2)/4 π. −∞ −∞ Therefore ∫ ∫ ∞ √ π − x2 ∞ √ π − x2 cos t2eitx dt = π cos , sin t2eitx dt = π sin . −∞ 4 −∞ 4 Since ∫ ∫ 1 ∞ i ∞ pˆ(x) = cos t2eitx dt, qˆ(x) = sin t2eitx dt, 2π −∞ 2π −∞ we obtain  2 | |−  √1 cos log x π , x > 0, 2 π|x| 4 k(x) = 2 | |  √i cos log x +π , x < 0. 2 π|x| 4

−1 Example 3.3. u = 1, v = 0,p ˆ = δ, p = δ1, k = δ1, Kf(x) = |x| f(1/x). −1 Example 3.4. u = 0, v = 1,q ˆ = δ, q = δ1, k = δ−1, Kf(x) = |x| f(−1/x).

Remark 3.1. Note that on R+, the map 1 1 Kf(x) = f( ), x ∈ R , x x + was discovered earlier by Hardy and Titchmarsh [6, p. 123], which could be viewed as the restriction of our Example 3.3 on R+. Nevertheless, an analogue of our Example 3.4, i.e., 1 1 Kf(x) = f(− ), x ∈ R , x x + was not mentioned in [6]. Example 3.5. u = cos θ, v = i sin θ, Kf(x) = |x|−1(f(1/x) cos θ + if(−1/x) sin θ).

−iat ′ Example 3.6. Suppose u = e , v = 0. Using the formula δφ(a)(φ(x))|φ (a)| = δa, where a/2 δa(x) = δ(x − a), we obtain that ifp ˆ(x) = δa(x), then p(x) = e δea (x). It follows that a/2 k = e δea , and ea/2 Kf(x) = f(ea/x). |x| FOURIER-TYPE TRANSFORMS 9

Example 3.7. u(t) = cos t, v(t) = i sin t, e−1/2 e1/2 e−1/2 e1/2 k = δ −1 + δ + δ− −1 − δ− . 2 e 2 e 2 e 2 e

Note that Examples 3.3–3.7 above furnish linear transformations that are “discrete” (in the sense that they are not really integral transforms). On the other hand, Examples 3.8– 3.12 to follow furnish “hybrid-type” integral transformations in the sense that they are Fredholm integral operators of the second kind.

Example 3.8. Let b > 0, u = 1 − χ[−b,b], v = χ[−b,b],q ˆ(x) = sin bx/(πx), x sin(b log |x|) k(x) = δ (x) − . 1 π|x|3/2 log |x|

Example 3.9. Let u = 1 + (cos(2π/n) − 1)χ[−b,b], v = i sin(2π/n)χ[−b,b]. Then pˆ(x) = δ(x) + (cos(2π/n) − 1) sin bx/(πx),

qˆ(x) = i sin(2π/n) sin bx/(πx).

2πi/n Example 3.10. Let u = 1 + (e − 1)χ[−b,b], v = 0. Then pˆ(x) = δ(x) + (e2πi/n − 1) sin bx/(πx), qˆ(x) = 0. Example 3.11. Let h(t) be a polynomial of complex coeﬃcients such that h is non-vanishing on R. Then the functions u(t), v(t) deﬁned by h(t) h(t) u(t) = Re , v(t) = iIm h(t) h(t) satisfy (14). Putting h(t) = t + i yields u = (t2 − 1)/(t2 + 1), v = 2it/(t2 + 1). Since u = 1 − 2/(t2 + 1), we have ∫ 1 ∞ 2eixt − − −|x| pˆ(x) = δ(x) 2 dt = δ(x) e , 2π −∞ t + 1 and ∫ 1 ∞ 2iteixt − −|x| qˆ(x) = 2 dt = sgn(x)e . 2π −∞ t + 1 Hence

−1/2 −1 p(x) = δ1(x) − x · min(x, x ), q(x) = − sgn(ln x)x−1/2 · min(x, x−1). Example 3.12. Let φ(t) be the continuous, odd function on R such that 1 − t2 2t cos φ(t) = , sin φ(t) = . 1 + t2 1 + t2 10 GOONG CHEN AND DAOWEI MA

Let e−πi/4 1 − 2t − t2 u(t) = e−iπ/4 cos(φ(t) + π/4) = √ , 2 t2 + 1 eπi/4 1 + 2t − t2 v(t) = eiπ/4 sin(φ(t) + π/4) = √ . 2 t2 + 1 Then u, v satisfy (37); we will see later, in Section 5, that the corresponding function k(x) is called “Fourier-like”. Now we have 1 − i pˆ(x) = [−δ(x) + (1 − i sgn(x)) · e−|x|], 2 1 + i qˆ(x) = [−δ(x) + (1 + i sgn(x)) · e−|x|], 2 1 − i p(x) = (−δ (x) + (1 − i sgn(ln x))x−1/2 · min(x, x−1)), 2 1 1 + i q(x) = (−δ (x) + (1 + i sgn(ln x))x−1/2 · min(x, x−1)). 2 1 The following Table 1 gives a somewhat organized summary of the above examples. To conclude this section, we mention that there remain a large number of interesting cases of u and v not included above, but, unfortunately, the evaluation and determination of explicit expressions of their kernels are so far intractable. FOURIER-TYPE TRANSFORMS 11 √ k(x) e−ix/ 2π u(t) (2π)−1/2e−π(i+2t)/4Γ(1/2 − it) (i) v(t) (2π)−1/2eπ(i+2t)/4Γ(1/2 − it) pˆ(x) (2π)−1/2ex/2 exp(−iex) qˆ(x) (2π)−1/2ex/2 exp(iex) − 2 | |− k(x) (1+i)+(1√ i) sgn x cos ln x π sgn(x) 4 π|x| 4 u(t) cos(t2) (ii) v(t) i sin(t2) √1 x2−π pˆ(x) 2 π cos 4 √i x2+π qˆ(x) 2 π cos 4 k(x) δ(x − 1) u(t) 1 (iii) v(t) 0 pˆ(x) δ(x) qˆ(x) 0 k(x) δ(x + 1) u(t) 0 (iv) v(t) 1 pˆ(x) 0 qˆ(x) δ(x) k(x) δ(x − 1) cos θ + iδ(x + 1) sin θ u(t) cos θ (v) v(t) i sin θ pˆ(x) δ(x) cos θ qˆ(x) iδ(x) sin θ k(x) ea/2δ(x − ea) u(t) e−iat (vi) v(t) 0 pˆ(x) δ(x − a) qˆ(x) 0 1 −1/2 − −1 1/2 − −1/2 −1 − 1/2 k(x) 2 [e δ(x e ) + e δ(x e) + e δ(x + e ) e δ(x + e)] u(t) cos t (vii) v(t) i sin t pˆ(x) (δ(x − 1) + δ(x + 1))/2 qˆ(x) (δ(x + 1) − δ(x − 1))/2 k(x) δ(x − 1) − √sgn x sin(b ln |x|) π |x| ln |x| u(t) 1 − χ − (t) (viii) [ b,b] v(t) χ[−b,b](t) pˆ(x) δ(x) − sin bx/(πx) qˆ(x) sin bx/(πx) 12 GOONG CHEN AND DAOWEI MA

sin(√b ln |x|) 1+sgn x 2π 1−sgn x 2π k(x) δ1(x) + [ (cos − 1) + i sin ] π |x| ln |x| 2 n 2 n u(t) 1 + (cos(2π/n) − 1)χ − (t) (ix) [ b,b] v(t) i sin(2π/n)χ[−b,b](t) pˆ(x) δ(x) + (cos(2π/n) − 1) sin bx/(πx) qˆ(x) i sin(2π/n) sin bx/(πx) 1+sgn x 2πi/n sin(√b ln |x|) k(x) δ1(x) + (e − 1) 2 π |x| ln |x| 2πi/n u(t) 1 + (e − 1)χ − (t) (x) [ b,b] v(t) 0 pˆ(x) δ(x) + (e2πi/n − 1) sin bx/(πx) qˆ(x) 0 1+sgn√ x −1 1−√sgn x −1 k(x) δ1(x) − min(x, x ) − sgn(ln |x|) min(|x|, |x| ) 2 |x| 2 |x| u(t) (t2 − 1)/(t2 + 1) (xi) v(t) 2it/(t2 + 1) pˆ(x) δ(x) − e−|x| qˆ(x) − sgn(x)e−|x| 1−i 1+sgn x 1−i sgn(ln√ |x|) −1 k(x) (−δ1(x) + · min(|x|, |x| )) 2 2 |x| 1+i 1−sgn x 1+i sgn(ln√ |x|) −1 + (−δ−1(x) + · min(|x|, |x| )) 2 2 |x| (xii) − u(t) e √πi/4 1−2t−t2 2 t2+1 v(t) e√πi/4 1+2t−t2 2 t2+1 1−i − − · −|x| pˆ(x) 2 [ δ(x) + (1 i sgn(x)) e ] 1+i − · −|x| qˆ(x) 2 [ δ(x) + (1 + i sgn(x)) e ] Table 1: This table summarizes all the results in Exam- ples 3.1–3.12, in corresponding sequential order.

4. Operations on K and certain algebraic group structures

For a function k, let Kk be the associated operator as in (2), and denote k ⋆ f = Kkf.

Theorem 4.1. If k, ℓ ∈ K , then k ⋆ ℓ ∈ K , i.e., the set K is closed under the operation ⋆.

Proof. Denote the splitting of k as k(x) = p(x) + q(−x) = pk(x) + qk(−x). Then we have

pk⋆ℓ = pk ⋆ pℓ + qk ⋆ qℓ, qk⋆ℓ = pk ⋆ qℓ + qk ⋆ pℓ.

It follows from (12) that

uk⋆ℓ = uk(t)uℓ(−t) + vk(t)vℓ(−t), (26) vk⋆ℓ = uk(t)vℓ(−t) + vk(t)uℓ(−t). FOURIER-TYPE TRANSFORMS 13

From these equations and (15), we obtain

uk⋆ℓ(t) = exp(iαk(t) + iαℓ(−t)) cos(φk(t) + φℓ(−t)), (27) vk⋆ℓ(t) = i exp(iαk(t) + iαℓ(−t)) sin(φk(t) + φℓ(−t)).

It is then clear that uk⋆ℓ, vk⋆ℓ satisfy the characterizing conditions (13) in Theorem 2.3. Therefore k ⋆ ℓ ∈ K .

From (24), (25) and (26), we obtain − Γ(1/2 it) −π(i/4+t/2) π(i/4+t/2) uFk(t) = √ [uk(−t)e + vk(−t)e ], 2π − Γ(1/2 it) π(i/4+t/2) −π(i/4+t/2) vFk(t) = √ [uk(−t)e + vk(−t)e ]. 2π We have | |−1 − − (28) δ1 ⋆ ℓ(x) = x ℓ(1/x), uδ1⋆ℓ(t) = uℓ( t), vδ1⋆ℓ(t) = vℓ( t). Deﬁnition 4.2. For k, ℓ ∈ K , we deﬁne

k ∗ ℓ = k ⋆ (δ1 ⋆ ℓ). (Note the distinction between “∗” and “⋆”.)

By (27) and (28), we obtain

uk∗ℓ(t) = exp(iαk(t) + iαℓ(t)) cos(φk(t) + φℓ(t)), (29) vk∗ℓ(t) = i exp(iαk(t) + iαℓ(t)) sin(φk(t) + φℓ(t)). It follows that the operation ∗ is closed, commutative and associative in K .

Since uδ1 = 1 and vδ1 = 0, it follows from (26) that − − (30) uδ1⋆k(t) = uk( t), vδ1⋆k(t) = vk( t).

Thus δ1 ⋆ (δ1 ⋆ k) = k. By (26) and (30), we obtain − − − − u(δ1⋆k)⋆ℓ = uk( t)uℓ( t) + vk( t)vℓ( t) = uδ1⋆(k⋆(δ1⋆ℓ))(t), − − − − v(δ1⋆k)⋆ℓ = uk( t)vℓ( t) + vk( t)uℓ( t) = vδ1⋆(k⋆(δ1⋆ℓ))(t), and hence (δ1 ⋆ k) ⋆ ℓ = δ1 ⋆ (k ⋆ (δ1 ⋆ ℓ)). Therefore,

Kδ1⋆k = Kδ1 KkKδ1 . Similarly, we obtain

(31) Kk⋆ℓ = KkKℓKδ1 ,Kk∗ℓ = KkKδ1 Kℓ, and

(32) KkKℓ = Kk⋆ℓKδ1 = Kδ1 Kℓ⋆k. ∗ Since id = KkKk = Kk⋆kKδ1 , we see that k ⋆ k = k ⋆ k = δ1. Thus k (δ1 ⋆ k) = δ1. We therefore conclude the following. 14 GOONG CHEN AND DAOWEI MA

Theorem 4.3. (Group structure) The set K with the operation ∗ constitutes an abelian group, with unit element δ1. For −1 any k ∈ K , its inverse is given by k = δ1 ⋆ k.

Let T = {eiα : α ∈ R} be the unit circle, and let G be the group of functions (eiα, eiφ): R → T × T , where α, φ are real measurable functions on R. The map L : G → K given by iα iα (33) uL(α,φ) = e cos φ, vL(α,φ) = ie sin φ is a group homomorphism. The kernel of the homomorphism is E := {(eiα, eiφ) ∈ G :(eiα(t), eiφ(t)) ∈ N ∀t}, ∼ where N is the subgroup of T × T consisting of (1, 1) and (−1, −1). Therefore K = G /E. Thus each element of K has uncountably many square roots. Each k in Example 3.8 is a square root of δ1 and satisfies δ1 ⋆ k = k (since uk, vk are even functions). Each k in Examples 3.9 or 3.10 is an n-th root of δ and satisfies δ ⋆ k = k. √ 1 1 Let τ = e−ix/ 2π. We consider the “square root” k such that k ∗ k = τ. As noted in the previous paragraph, there exist infinitely many such k’s. But exactly four of them have analytic u and v. One of such k’s has the following u and v: √ √ u(t) = e−πi/8(8π)−1/4Γ1/2(1/2 − it) eπt + e−πt + e−πt/2, √ (34) √ v(t) = ie−πi/8(8π)−1/4Γ1/2(1/2 − it) eπt + e−πt − e−πt/2.

We emphasize that when we say k is a square root of some ℓ, we mean k ∗ k = ℓ. This does not imply KkKk = Kℓ. In fact, we have the following. ∈ K ̸ Proposition 4.4. For any k1, k2, ℓ , Kk1 Kk2 = Kℓ.

2 Proof. We ﬁrst consider the case that uℓ ̸≡ 0. Suppose that f ∈ L (0, ∞). Recall that T f(t) = fˆ(t) = f(et)et/2. Let Π be the projection from L2(−∞, 0)⊕L2(0, ∞) onto L2(0, ∞). By (12) and (32), we have F F ˆ T ΠKk1 Kk2 f(t) = uk1⋆k2 (t) f(t), ˆ FT ΠKℓf(t) = uℓ(t)Ff(−t). Since the two right-side expressions cannot be equal for arbitrary f, regardless of what ̸ uk1⋆k2 (t) and uℓ(t) are, we see that Kk1 Kk2 = Kℓ.

In case uℓ ≡ 0, we have vℓ ≠ 0. The statement can be proved in a similar way by 2 considering the projection Π− onto L (−∞, 0) in place of Π.

Note that k ∗ k = δ1 does not imply KkKk = id. A member k ∈ K satisﬁes KkKk = id if and only if k is real. This is because KkKk = id. The following proposition shows that a k ∈ K satisﬁes both KkKk = id and k ∗ k = δ1 if and only if k, uk, vk are real. Proposition 4.5. For k ∈ K , we have

(i) k ⋆ k = δ1; FOURIER-TYPE TRANSFORMS 15

(ii) k ∗ k = δ1 ⇔ k = δ1 ⋆ k ⇔ uk(−t) = uk(t), vk(−t) = vk(t);

(iii) k ∗ k = δ−1 ⇔ k = δ−1 ⋆ k ⇔ uk(−t) = vk(t), vk(−t) = uk(t);

(iv) k ⋆ k = δ1 ⇔ k = k ⇔ uk(−t) = uk(t), vk(−t) = vk(t);

(v) k ⋆ k = δ−1 ⇔ k = δ−1 ∗ k ⇔ uk(−t) = vk(t), vk(−t) = uk(t);

(vi) k ∗ k = δ1 ⇔ k = δ1 ⋆ k ⇔ uk = uk, vk = vk;

(vii) k ∗ k = δ−1 ⇔ k = δ−1 ⋆ k ⇔ uk = vk, vk = uk.

Proof. Since Kk⋆kKδ1 = KkKk = id = Kδ1 Kδ1 , it follows that (i) holds. The proofs of (ii-iiv) are all similar. Here we will prove only (v).

Since k ⋆ k = δ1 for every k ∈ K , it follows that

k = δ−1 ∗ k ⇔ (δ−1 ∗ k) ⋆ k = δ1

⇔ δ−1 ∗ (k ⋆ k) = δ1

⇔ k ⋆ k = δ−1. − The second part of (v) is true because uk(t) = uk( t), uδ−1∗k(t) = vk(t).

5. Fourier-like kernels √ In addition to belonging to K , the Fourier kernel τ : τ(x) = e−ix/ 2π also enjoys the properties that

(35) k(−x) = k(x),KkKkf(x) = f(−x), when k = τ. The two conditions in (35) may be further used as selective criteria of how closely an integral transform resembles the Fourier transform (in addition to the two criteria (A) and (B) cited in Section 1). ∗ The ﬁrst equation in (35) is equivalent to δ−1 k = k. Since KkKk = Kk⋆kKδ1 , and since − Kδ−1 Kδ1 f(x) = f( x), we see that the second equation in (35) is equivalent to k ⋆ k = δ−1. By Proposition 4.5 (v), the two equations are equivalent, and they are both equivalent to

(36) k ⋆ k = δ−1. Definition 5.1. A member k ∈ K is said to be Fourier-like if it satisfies (35) or, equivalently, if it satisfies (36). Theorem 5.2. (i) A member k ∈ K is Fourier-like if and only if

(37) uk(−t) = vk(t), vk(−t) = uk(t). (ii) A member ℓ ∈ K is Fourier-like if and only if ℓ = k ∗ τ for some k ∈ K with k real. Hence there exist inﬁnitely many Fourier-like kernels.

Proof. (i) By Proposition 4.5 (v), (36) and (37) are equivalent. (ii) Since (K , ∗) is a group, every member ℓ has the form ℓ = k ∗ τ, where k ∈ K . Since

(k ∗ τ) ⋆ (k ∗ τ) = (k ⋆ k) ∗ (τ ⋆ τ) = (k ⋆ k) ∗ δ−1, 16 GOONG CHEN AND DAOWEI MA

it follows that (36) holds with k ∗ τ in place of k if and only if k ⋆ k = δ1, which holds if and only if k is real, by Proposition 4.5 (iv). The proof is complete. Example 5.3. Using Theorem 5.2, we see that e−πi/4e−πt/2 eπi/4eπt/2 u(t) = √ , v(t) = √ 2 cosh πt 2 cosh πt correspond to a Fourier-like kernel k ∈ K .

6. A new meaning of fractional powers √ We now work out “fractional powers” of the Fourier transform. For τ(x) = e−ix/ 2π, the Fourier transform operator is Kτ . By Proposition 4.4, no k ∈ K satisﬁes KkKk = Kτ . Literally speaking, fractional powers of the Fourier transform therefore do not exist, in K . We shall give a diﬀerent meaning to the term of fractional powers. ∈ K L { ∈ K } For k , put Lk = KkKδ1 . Let = Lk : k . By (31),

Lk∗ℓ = Kk∗ℓKδ1 = KkKδ1 KℓKδ1 = LkLℓ.

It follows that L is a group with the operator multiplication, and k 7→ Lk is a group isomorphism from K to L . We say Kℓ is a fractional power of Kk if Lℓ is a fractional power of Lk. Thus Kℓ is a fractional power of Kk if and only if ℓ is a fractional power of k with respect to the multiplication ∗. Define real functions φ(t) on R by −π/4 < φ(t) < π/4 and e−πt/2 eπt/2 (38) cos(φ(t) + π/4) = √ , sin(φ(t) + π/4) = √ . 2 cosh πt 2 cosh πt Then φ is analytic and odd. √ The function ψ(t) := (cosh πt)/πΓ(1/2 − it) is analytic and satisfies |ψ(t)| = 1 and ψ(−t) = ψ(t). Thus, there is an analytic, odd function α(t) such that √ cosh πt (39) eiα(t) = Γ(1/2 − it). π By (24), (38) and (39), we have i(α−π/4) i(α−π/4) uτ = e cos(φ + π/4), vτ = ie sin(φ + π/4). For −∞ < s < ∞, define τ ∗s by is(α−π/4) is(α−π/4) uτ ∗s = e cos(s(φ + π/4)), vτ ∗s = ie sin(s(φ + π/4)). Then ∗0 ∗1 ∗s ∗p ∗(s+p) τ = δ1, τ = τ, τ ∗ τ = τ . Thus {τ ∗s} is a family of fractional powers of τ. Let λ : R → K be defined by λ(s) = τ ∗s. Then λ or the graph Hλ of λ is a one-parameter subgroup of K , with λ(1) = τ. Note that such λ is not unique. For instance, µ :(−∞, ∞) → K defined by is(α+7π/4) is(α+7π/4) uµ(s) = e cos(s(φ + π/4)), vµ(s) = ie sin(s(φ + π/4)) FOURIER-TYPE TRANSFORMS 17 is also a one-parameter subgroup of K with µ(1) = τ. Fractional powers of a Fourier-like kernel are not necessarily Fourier-like. In fact, a square root of a Fourier-like kernel can never be Fourier-like, which is equivalent to the statement that if k is Fourier-like, then k ∗k is not Fourier-like. This can be seen as follows. Assuming that k is Fourier-like, we have

(k ∗ k) ⋆ (k ∗ k) = (k ⋆ k) ∗ (k ⋆ k) = δ−1 ∗ δ−1 = δ1; hence k ∗ k is not Fourier-like. Proposition 6.1. A kernel k ∈ K is Fourier-like if and only if there are odd, real functions α, φ, an odd integer m, and an ε = ±1, such that i(α−mπ/4) i(α−mπ/4) (40) uk = εe cos(φ + mπ/4), vk = iεe sin(φ + mπ/4).

Furthermore, if uk, vk are continuous, the functions α, φ can be chosen to be continuous.

Proof. Suppose that uk, vk are of the form (40). Without loss of generality, we assume that ε = 1. First, we see that uk, vk are indeed induced by some k ∈ K , because uk, vk are of the form (15) and therefore satisfy (14). To see that the corresponding k is Fourier-like, it suﬃces to verify that

(41) uk(−t) = vk(t). Now, we have i(α(t)+mπ/4) uk(−t) = e cos(φ(t) − mπ/4) = ei(α(t)+mπ/4) sin(φ(t) − (m − 2)π/4) = iei(α(t)−mπ/4)ei(m−1)π/2 sin(φ(t) + mπ/4 − (m − 1)π/2) = iei(α(t)−mπ/4)(−1)(m−1)/2 sin(φ(t) + mπ/4)(−1)(m−1)/2

= vk(t). Therefore, k is Fourier-like. Conversely, suppose that k is Fourier-like. Then 2 2 2 uk(0) = vk(0), |uk(0) + vk(0)| = |uk(0)| + |vk(0)| = 1.

It is straightforward to verify that there are exactly four solutions for uk(0): √ −nπi/4 uk(0) = e / 2, n = 1, 3, 5, 7. It follows that i(−mπ/4) i(−mπ/4) (42) uk(0) = εe cos(mπ/4), vk(0) = iεe sin(mπ/4), where m = 1, 3, ε = ±1.

Fix the ε and m. There are measurable functions α, φ deﬁned on {t ≥ 0} such that α(0) = φ(0) = 0 and such that (40) holds for t ≥ 0. Furthermore, if uk, vk are continuous, the functions α, φ can be chosen to be continuous on {t ≥ 0}. 18 GOONG CHEN AND DAOWEI MA

Extend α, φ to R so that the resulting functions are odd. Deﬁne U, V by (43) U = εei(α−mπ/4) cos(φ + mπ/4),V = iεei(α−mπ/4) sin(φ + mπ/4). Then U, V correspond to a Fourier-like kernel in K , and

(44) U(t) = uk(t),V (t) = vk(t)

for t ≥ 0. Since both pairs U, V and uk, vk satisfy (37), it follows that (44) actually holds for all t ∈ R. Therefore, (40) holds. The proof is complete. Proposition 6.2. Let n > 1 be an odd integer. Then there is a one-parameter subgroup of K with µ(1) = τ such that µ(1/n) is Fourier-like.

Proof. Since the greatest common divisor of n and 8 is 1, there are integers m, j such that nm + 8j = 1. Note that m must be odd. Deﬁne µ :(−∞, ∞) → K by is(α−(nmπ/4)) uµ(s) = e cos(s(φ + (nmπ/4))), is(α−(nmπ/4)) vµ(s) = ie sin(s(φ + (nmπ/4))), where α, φ are analytic functions deﬁned in (39) and (38). It is straightforward to verify that µ(1) = τ. By Proposition 6.1, µ(1/n) is Fourier-like.

7. Concluding remarks

We have explored some fundamental properties of the important Fourier transform through the unitarity and Fourier kernel properties by generalizing earlier works done by Hardy and Titchmarsh [6] and Watson [15]. A collection of interesting new transformations and group structures have been discovered through our derived characterizing conditions. These serve as evidence to support our point of view of the advantages of using unitarity (rather than self-reciprocity) to characterize the Fourier transform. Obviously, there remains to be an even larger set than that in Table 1 of concrete examples to be worked out. They could oﬀer some more insights for Fourier analysis or even new applications to computational science. We hope to continue the investigation in the future. Acknowledgment. The authors gratefully acknowledge partial ﬁnancial support by the Texas Norman Hackman Advanced Research Program Grant #010366-0149-2009 from the Texas Higher Education Coordinating Board, and Qatar National Research Fund (QNRF) National Priority Research Program (NPRP) Grant #NPRP5-674-1-114.

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[email protected], Department of Mathematics, Texas A&M University, College Station, TX 77843, USA, and Science Program, Texas A&M University at Qatar, Education City, Doha, Qatar

[email protected], Department of Mathematics, Wichita State University, Wichita, KS 67260-0033, USA