Some Inequalities for the Fourier Transform and Their Limiting Behaviour

Abstract

We identify a one-parameter family of inequalities for the Fourier transform whose limiting case is the restriction conjecture for the sphere. Using Stein’s method of complex interpolation, we prove the conjectured inequalities when the target space is $L^2$, and show that this recovers in the limit the celebrated Tomas-Stein theorem.

1 Introduction

For any $0<s<1$ and $r>0$, we consider the function

$$\begin{aligned} A^{(s)}_r(x) = {\left\{ \begin{array}{ll} 0,\ \ \ \ \ \ \ \ \ \ \text {when}\ |x|\le r,\\ \frac{c(n,s) r^{2s}}{(|x|^2 - r^2)^s |x|^n},\ \ \ \ \text {when}\ |x|> r. \end{array}\right. } \end{aligned}$$

(1.1)

where, using the standard notation $\sigma _{n-1} = \frac{2\pi ^{\frac{n}{2}}}{\Gamma (\frac{n}{2})}$ for the $(n-1)$-dimensional measure of the unit sphere ${\mathbb {S}}^{n-1}\subset {\mathbb {R}}^n$, we have let

$$\begin{aligned} c(n,s) = \frac{2}{\Gamma (s) \Gamma (1-s) \sigma _{n-1}}. \end{aligned}$$

(1.2)

Define measures on ${\mathbb {R}}^n$ by letting

$$\begin{aligned} d\mu ^{(s)}_r(x) = A^{(s)}_r(x) \textrm{d}x. \end{aligned}$$

(1.3)

From (1.1) it immediately follows that $A^{(s)}_r \in L^1({\mathbb {R}}^n)$. Moreover, from [9, Lemma 15.3]) one has for every $s\in (0,1)$ and $r>0$

$$\begin{aligned} \mu ^{(s)}_r({\mathbb {R}}^n) = ||A^{(s)}_r||_{L^1({\mathbb {R}}^n)} = \int _{{\mathbb {R}}^n} A^{(s)}_r(x) \textrm{d}x = \int _{{\mathbb {R}}^n} A^{(s)}_1(x) \textrm{d}x = 1. \end{aligned}$$

(1.4)

We next define the operator

$$\begin{aligned} {\mathscr {A}}^{(s)}_r f(x) = A^{(s)}_r \star f(x). \end{aligned}$$

(1.5)

It follows from (1.4) and Young’s convolution theorem that

$$\begin{aligned} {\mathscr {A}}^{(s)}_r : L^p({\mathbb {R}}^n)\ \longrightarrow \ L^p({\mathbb {R}}^n),\ \ \ \ \ \ \ \ 1\le p \le \infty , \end{aligned}$$

and that for any $f \in L^p({\mathbb {R}}^n)$, we have

$$\begin{aligned} ||{\mathscr {A}}^{(s)}_r f||_{L^p({\mathbb {R}}^n)} \le ||f||_{L^p({\mathbb {R}}^n)}, \end{aligned}$$

(1.6)

which shows that ${\mathscr {A}}^{(s)}_r$ is a contraction in $L^p({\mathbb {R}}^n)$ for every $0<s<1$ and $r>0$.

For any $f\in {\mathscr {S}}({\mathbb {R}}^n)$, we denote by ${{\hat{f}}}(\xi ) = {\mathscr {F}}(f)(\xi ) = \int _{{\mathbb {R}}^n} e^{-2\pi i \langle \xi ,x\rangle } f(x) \textrm{d}x$ its Fourier transform, and ask the following

Question: Let $\frac{1}{2}\le s<1$. If $1\le p < \frac{2n}{n+2s -1}$, is it true that for $1\le q \le \frac{n+1-2s}{n+1} p'$ and for every $f\in {\mathscr {S}}({\mathbb {R}}^n)$, one has for some $C^{(s)}(n,p)>0$

$$\begin{aligned} \left( \int _{{\mathbb {R}}^n} |{{\hat{f}}}(\xi )|^{q} A^{(s)}_1(\xi ) d\xi \right) ^{1/q} \le C^{(s)}(n,p)\ ||f||_{L^{p}({\mathbb {R}}^n)}. \end{aligned}$$

(1.7)

We note that the restriction $s\ge \frac{1}{2}$ serves to guarantee that $\frac{2n}{n+2s -1}\le 2$. Therefore, the hypothesis $f\in L^p({\mathbb {R}}^n)$ implies that f be in the Hausdorff-Young range [1, 2]. As a consequence, ${{\hat{f}}}$ is a function in $L^{p'}({\mathbb {R}}^n)$.

Our interest in the above conjecture stems from the following observations. Assume that (1.7) does hold for any s such that $\frac{1}{2} \le s < 1$. Since $\frac{2n}{n+2s -1} \searrow \frac{2n}{n +1}$ as $s\nearrow 1$, if we take $1\le p < \frac{2n}{n+1}$ and $q\le \frac{n-1}{n+1} p'$, then it is immediate to verify that for any $s\in [\frac{1}{2},1)$

$$\begin{aligned} 1\le p < \frac{2n}{n+2s -1}\ \ \ \ \text {and}\ \ \ \ 1\le q \le \frac{n+1-2s}{n+1} p' \end{aligned}$$

(for the second of these inequalities simply note that $q\le \frac{n-1}{n+1} p' < \frac{n-1+ 2(1-s)}{n+1} p' =\frac{n+1-2s}{n+1} p'$), therefore (1.7) holds. But we now have the following fact implicitly contained in the seminal work [14] of M. Riesz (for a proof see [9, Prop. 15.4]).

Proposition 1.1

For every function $f\in {\mathscr {S}}({\mathbb {R}}^n)$ and $x\in {\mathbb {R}}^n$, one has

$$\begin{aligned} \underset{s\rightarrow 1}{\lim }\ {\mathscr {A}}^{(s)}_r f(x) = {\mathscr {M}}_r(f,x), \end{aligned}$$

where

$$\begin{aligned} {\mathscr {M}}_r(f,x) = \frac{1}{\sigma _{n-1} r^{n-1}} \int _{S(x,r)} f(y) d\sigma (y) \end{aligned}$$

is the spherical average of f over the sphere $S(x,r) = \{y\in {\mathbb {R}}^n\mid |y-x|=r\}$.

Therefore, passing to the limit in (1.7) and using Proposition 1.1, if $C^{(s)}(n,p)$ converges to a number $C(n,p)>0$ as $s\rightarrow 1$, we would infer that for $1\le p < \frac{2n}{n+1}$ and $q\le \frac{n-1}{n+1} p'$ the following limiting inequality holds

$$\begin{aligned} \left( \frac{1}{\sigma _{n-1}}\int _{{\mathbb {S}}^{n-1}} |{{\hat{f}}}(\xi )|^q d\sigma (\xi )\right) ^{1/q} \le C(n,p)\ ||f||_{L^p({\mathbb {R}}^n)}. \end{aligned}$$

(1.8)

As it is well known, this is the famous restriction conjecture of C. Fefferman and E. Stein for the Fourier transform, see [4, 6,7,8, 15] and [19].

Eli Stein lecturing on the restriction problem, UChicago, 1985

One obvious advantage of (1.7) over (1.8) is that the support of the measures (1.3) is ${\mathbb {R}}^n\setminus B(0,r)$, instead of the lower-dimensional manifold ${\mathbb {S}}^{n-1}\subset {\mathbb {R}}^n$. Notice that if we let $d\sigma$ denote the surface measure concentrated on the sphere ${\mathbb {S}}^{n-1}$, then Proposition 1.1 can be equivalently stated as follows:

$$\begin{aligned} d\mu ^{(s)}_1\ \underset{s\rightarrow 1}{\longrightarrow }\ \frac{1}{\sigma _{n-1}} d\sigma \ \ \ \ \ \ \text {in}\ {\mathscr {S}}'({\mathbb {R}}^n). \end{aligned}$$

(1.9)

Concerning the measures (1.3), we recall that in his above quoted paper, M. Riesz developed the theory of the nonlocal operators $(-\Delta )^s$ and their inverses $I_{2s}$, the operators of fractional integration which play a pervasive role in analysis, see also [16, Ch. 5]. Among other things, he solved by inversion the Dirichlet problem

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^s u = 0\ \ \ \text {in}\ B_r, \\ u = f\ \ \ \ \ \ \ \ \text {in}\ {\mathbb {R}}^n\setminus B_r, \end{array}\right. } \end{aligned}$$

(1.10)

and proved that for every $x\in B_r$, the unique solution to (1.10) is provided by

$$\begin{aligned} u(x) = c(n,s) \int _{|y|>r} \left( \frac{r^2 - |x|^2}{|y|^2 - r^2}\right) ^s \frac{f(y)}{|y-x|^n} \textrm{d}y, \end{aligned}$$

(1.11)

see formula (3) on p. 17 in [14], but also (1.6.11’) and (1.6.2) on pages 122 and 112 in [11]. It is clear from (1.11) that $u(0) = \int _{{\mathbb {R}}^n} f(y) A^{(s)}_r(y) \textrm{d}y = {\mathscr {A}}^{(s)}_r f(0)$. The role of the operators ${\mathscr {A}}^{(s)}_r$ is further elucidated by the following result, see [9, Prop. 15.6].

Proposition 1.2

(The Blaschke-Privalov fractional Laplacian) Let $0<s<1$ and suppose that $f\in {\mathscr {L}}_s({\mathbb {R}}^n)$ be in $C^{2s+\varepsilon }$ in a neighbourhood of $x\in {\mathbb {R}}^n$, for some $0<\varepsilon < 1$. One has

$$\begin{aligned} (-\Delta )^s f(x) = - \gamma (n,s) \underset{r\rightarrow 0^+}{\lim } \frac{{\mathscr {A}}^{(s)}_r f(x) -f(x)}{r^{2s}}, \end{aligned}$$

(1.12)

where $\gamma (n,s) = \frac{s 2^{2s} \Gamma (\frac{n}{2} + s)}{\pi ^{\frac{n}{2}} \Gamma (1-s)}$.

Here, for $0<s<1$, we have denoted by ${\mathscr {L}}_s({\mathbb {R}}^n)$ the space of measurable functions $f:{\mathbb {R}}^n\rightarrow {{\overline{{\mathbb {R}}}}}$ for which the norm

$$\begin{aligned} ||f||_{{\mathscr {L}}_s({\mathbb {R}}^n)} = \int _{{\mathbb {R}}^n} \frac{|f(x)|}{1+|x|^{n+2s}} \textrm{d}x < \infty . \end{aligned}$$

Returning to the inequality (1.7), we mention that, similarly to the restriction problem (1.8), it cannot possibly hold for every exponent $p\in [1,2]$ in the Hausdorff-Young range. To understand the constraint $1\le p < \frac{2n}{n+2s -1}$, denote by $T:L^p({\mathbb {R}}^n)\rightarrow L^q({\mathbb {R}}^n,d\mu ^{(s)}_1)$ the “restriction” operator in (1.7). Then its adjoint $T^\star :L^{q'}({\mathbb {R}}^n,d\mu ^{(s)}_1)\rightarrow L^{p'}({\mathbb {R}}^n)$ is easily seen to be given by

$$\begin{aligned} T^\star f(\xi ) = \int _{{\mathbb {R}}^n} e^{-2\pi i \langle \xi ,x\rangle } f(x) A^{(s)}_1(\xi ) \textrm{d}x, \end{aligned}$$

(1.13)

so that

$$\begin{aligned} T^\star 1 = \widehat{A^{(s)}_1}. \end{aligned}$$

Assuming (1.7), we would, thus, have by duality

$$\begin{aligned} ||T^\star 1||_{L^{p'}({\mathbb {R}}^n)} \le B^{(s)}(n,p)\ ||1||_{L^{q'}({\mathbb {R}}^n,d\mu ^{(s)}_1)} = B^{(s)}(n,p) < \infty , \end{aligned}$$

where in the last equality, we have used (1.4). Therefore, the validity of (1.7) for some p implies that $\widehat{A^{(s)}_1}\in L^{p'}({\mathbb {R}}^n)$. Now, in Section 3, we prove the following.

Theorem 1.3

Let $s\in (0,1)$ and $n\ge 2$. Then the Fourier transform of the kernel defined by (1.1) is given by

$$\begin{aligned} \widehat{A^{(s)}_r}(\xi ) = \frac{2^{\frac{n}{2}-s} \Gamma (\frac{n}{2})}{\Gamma (s)}\int _{2\pi r |\xi |}^\infty t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t. \end{aligned}$$

(1.14)

Using Theorem 1.3 in (3.9) of Corollary 3.2, we obtain the following important decay at infinity

$$\begin{aligned} \widehat{A^{(s)}_1}(\xi ) \cong \frac{1}{|\xi |^{\frac{n+1}{2} - s}}, \end{aligned}$$

(1.15)

which shows that

$$\begin{aligned} \widehat{A^{(s)}_1} \in L^{p'}({\mathbb {R}}^n)\ \Longleftrightarrow \ p'> \frac{2n}{n+1-2s}\ \Longleftrightarrow \ 1\le p < \frac{2n}{n+2s -1}. \end{aligned}$$

(1.16)

We conclude that the inequality (1.7) cannot possibly hold for $p\ge \frac{2n}{n+2s -1}$.

Before proceeding, we pause to comment on an aspect of Theorem 1.3. As the reader will see its proof is somewhat more involved than its well-known counterpart in the case $s=1$. This is due to the nonlocal nature of the measure $d\mu ^{(s)}_1$, compared to the surface measure $d\sigma$ of the unit sphere ${\mathbb {S}}^{n-1}$. To explain this comment, we stress that

$$\begin{aligned} \widehat{d\sigma }(\xi ) = \frac{2\pi }{|\xi |^{\frac{n}{2}-1}} J_{\frac{n}{2}-1}(2\pi |\xi |) \end{aligned}$$

(1.17)

is just a rescaled spherically symmetric eigenfunction of the (local) differential operator $\Delta$ in ${\mathbb {R}}^n$. To see this, simply observe that for every $\lambda >0$, the function

$$\begin{aligned} f_\lambda (\xi ) = \int _{{\mathbb {S}}^{n-1}} e^{i \sqrt{\lambda }\langle \xi ,\omega \rangle } d\sigma (\omega ) \end{aligned}$$

is a spherically symmetric solution of the Helmholtz equation $\Delta f_\lambda = - \lambda f_\lambda$ in ${\mathbb {R}}^n$, which shows that (1.17) solves such PDE with $\lambda = 4\pi ^2$. However, the equations (1.10), (1.11) above, and Proposition 1.2 in particular, underscore that the Fourier transform of the measure $d\mu ^{(s)}_r$ is instead connected to the pseudodifferential operator $(-\Delta )^s$, and computations with such nonlocal operator are usually more involved. In this regard, we recall the following quote from p. 51 in [11]:...“In the theory of M. Riesz kernels, the role of the Laplace operator, which has a local character, is taken...by a non-local integral operator...This circumstance often substantially complicates the theory...”. In Lemma 3.3, we show that for $n\ge 2$, one has for every $\xi \in {\mathbb {R}}^n$

$$\begin{aligned} \underset{s\rightarrow 1}{\lim }\ \widehat{A^{(s)}_1}(\xi ) = \frac{1}{\sigma _{n-1}} \widehat{d\sigma }(\xi ). \end{aligned}$$

This result (which can be seen as a comforting a posteriori confirmation of the correctness of the computations leading to Theorem 1.3) is of course not surprising in view of (1.9) above. It is worthwhile noting at this moment that (1.1) are clearly reminiscent of the classical Bochner-Riesz kernels

$$\begin{aligned} K_z(x) = \frac{\left( 1-|x|^2\right) ^z_+}{\Gamma (z+1)},\ \ \ \ \Re z>-1, \end{aligned}$$

(1.18)

whose Fourier transform is given by

$$\begin{aligned} {{\hat{K}}}_z(\xi )=\pi ^{-z}|\xi |^{-\left( \frac{n}{2}+z\right) }J_{\frac{n}{2}+z}(2\pi |\xi |),\ \ \ \ \xi \in {\mathbb {R}}^n. \end{aligned}$$

(1.19)

If we compare this formula with (1.17), it is clear that ${{\hat{K}}}_z\rightarrow \frac{1}{2} \ \widehat{d\sigma }$ as $z\rightarrow -1$. While the computation of the Fourier transform of $A^{(s)}_1$ is more involved than (1.19), there are advantages in working with (1.1) instead of (1.18). One of them is that, as we have mentioned, $\widehat{A^{(s)}_1}$ is directly connected to the nonlocal operator $(-\Delta )^s$, while this is not the case for (1.19).

We have seen that (1.7) is only possible when p satisfies (1.16). But, given a p in such range what is the optimal range of q’s? To answer this question, we use the well-known argument of Knapp, except that because of the presence of the measure $d\mu _1^{(s)}$ in (1.7), we need to work a bit more. In Proposition 2.1 below, we show that, given p within the range (1.16), a necessary condition for (1.7) to hold is that

$$\begin{aligned} 1\le q \le \frac{n+1-2s}{n+1} p'. \end{aligned}$$

(1.20)

Notice that when the target space is $L^2$, then in view of (1.20), the conjecture asks whether it is true that (1.7) holds with $q=2$ for any $f\in {\mathscr {S}}({\mathbb {R}}^n)$ and $2 \le \frac{n+1-2s}{n+1} p'$. This is equivalent to asking that $\frac{1}{p'} \le \frac{n+1-2s}{2(n+1)} = \frac{1}{2} - \frac{s}{n+1}$, or equivalently $\frac{1}{p} \ge \frac{1}{2} + \frac{s}{n+1} = \frac{n+1+2s}{2(n+1)}$, and therefore, for any

$$\begin{aligned} 1\le p \le \frac{2(n+1)}{n+1+2s}. \end{aligned}$$

(1.21)

In the next result, we prove this conjecture.

Theorem 1.4

For a given $s\in (0,1)$ let $p = \frac{2(n+1)}{n+1+2s}$. Then there exists a constant $C^{(s)}(n)>0$ such that for every $f\in {\mathscr {S}}({\mathbb {R}}^n)$ one has

$$\begin{aligned} \left( \int _{{\mathbb {R}}^n} |{{\hat{f}}}(\xi )|^2 A^{(s)}_1(\xi ) d\xi \right) ^{1/2} \le C^{(s)}(n)\ ||f||_{L^p({\mathbb {R}}^n)}. \end{aligned}$$

(1.22)

In order to establish Theorem 1.4, we exploit Plancherel as in [20] and reduce matters to proving that the operator $R^{(s)} f = \widehat{A^{(s)}_1} \star f$ maps $L^p({\mathbb {R}}^n)\rightarrow L^{p'}({\mathbb {R}}^n)$. This is ultimately achieved using Stein’s theorem of complex interpolation by embedding $R^{(s)}$ into an analytic family of operators $\{T_z\}_{z\in S}$ in the strip $S = \{z\in {\mathbb {C}}\mid - \frac{n-1}{2}\le \Re z\le 1\}$. Specifically, we show that

$$\begin{aligned} {\left\{ \begin{array}{ll} T_{-\frac{n-1}{2} + i y}: L^1({\mathbb {R}}^n)\ \longrightarrow \ L^\infty ({\mathbb {R}}^n),\\ T_{1 + i y}: L^2({\mathbb {R}}^n)\ \longrightarrow \ L^2({\mathbb {R}}^n), \end{array}\right. } \end{aligned}$$

with appropriate bounds on the operator norms. Since the constant $C^{(s)}(n)$ in (1.22) is bounded uniformly in $s\in (0,1)$, see (4.21) below, passing to the limit as $s\rightarrow 1$ we recover the Tomas-Stein restriction theorem, see [20].

Remark 1.5

One should observe that the threshold exponent $p = \frac{2(n+1)}{n+1+2s}$ in (1.21) is strictly less than 2 for any $0<s<1$. Therefore, the limitation $\frac{1}{2} \le s <1$ is not necessary in such case.

The plan of the paper is as follows. In Sect. 2, we adapt the well-known argument of Knapp to prove that, if for a given p in the range (1.16), the restriction inequality (1.7) does hold, then we must have $1\le q \le \frac{n+1-2s}{n+1} p'$. Sect. 3 is devoted to proving Theorem 1.3, from which we obtain Corollary 3.2. The representation formula (3.8) contained in it will be quite important in the remainder of the paper. Finally in Sect. 4, we prove the nonlocal restriction Theorem 1.4. As a corollary of this result, we obtain the celebrated Tomas-Stein theorem. The paper closes with an appendix in Sect. 1 in which we gather some well-known facts and collect some results needed in the rest of the paper.

2 Necessary Condition for Restriction

In (1.16) above we have seen that the inequality (1.7) can possibly hold only when $1\le p < \frac{2n}{n+2s -1}$. In this section, we adapt the well-known argument of Knapp to prove that, if for a given p in such range (1.7) does hold, then we must have $1\le q \le \frac{n+1-2s}{n+1} p'$.

Proposition 2.1

A necessary condition for (1.7) to hold for $1\le p < \frac{2n}{n+2s -1}$ is that $1\le q \le \frac{n+1-2s}{n+1} p'$.

Proof

For $\varepsilon >0$ small consider the parallelepiped $K_\varepsilon = K'_\varepsilon \times [1-\varepsilon ,1]$ with sides parallel to the coordinate axis whose projection onto ${\mathbb {R}}^{n-1}\times \{0\}$ is the $(n-1)$-dimensional cube $K'_\varepsilon$ circumscribing the intersection of the hyperplane $x_n = 1-\varepsilon$ with the unit sphere ${\mathbb {S}}^{n-1}$. If $\theta _\varepsilon$ is the angle of aperture of the right-circular cone obtained by projecting to the origin the points of the $(n-2)$-dimensional sphere obtained intersecting ${\mathbb {S}}^{n-1}\cap \{x_n=1-\varepsilon \}$, from elementary trigonometry, we have $\cos \theta _\varepsilon = 1-\varepsilon$, $\sin \theta _\varepsilon = R(\varepsilon ) = \sqrt{\varepsilon (2-\varepsilon )}$, and therefore, $K'_\varepsilon = [-R(\varepsilon ),R(\varepsilon )]^{n-1}$.

Denoting now with $B'(0,r) = \{x'\in {\mathbb {R}}^{n-1}\mid |x'|<r\}$, consider now the right-circular cylinders $C_\varepsilon = B'(0,R(\varepsilon ))\times [1-\varepsilon ,1]$ and $C^\star _\varepsilon = B'(0,\sqrt{n-1}\ R(\varepsilon ))\times [1-\varepsilon ,1]$. A moment’s thought reveals that

$$\begin{aligned} C_\varepsilon \cap ({\mathbb {R}}^n\setminus B(0,1)) \subset K_\varepsilon \cap ({\mathbb {R}}^n\setminus B(0,1)) \subset C^\star _\varepsilon \cap ({\mathbb {R}}^n\setminus B(0,1)). \end{aligned}$$

(2.1)

We note that

$$\begin{aligned} C_\varepsilon \cap ({\mathbb {R}}^n\setminus B(0,1)) = \{(x',x_n)\in {\mathbb {R}}^n\mid 1-\varepsilon \le x_n \le 1,\ \sqrt{1-x_n^2}\le |x'|\le R(\varepsilon )\}. \end{aligned}$$

As in Knapp’s argument, if ${\textbf{1}}_E$ is the indicator function of a set $E\subset {\mathbb {R}}^n$, we now consider the function $f_\varepsilon = {\mathscr {F}}^{-1}({\textbf{1}}_{K_\varepsilon })$, so that ${{\hat{f}}}_\varepsilon = {\textbf{1}}_{K_\varepsilon }$. As it is well known

$$\begin{aligned} f_\varepsilon (\xi ) = \hat{{\textbf{1}}}_{K_\varepsilon }(\xi ) = e^{i\pi (2-\varepsilon )\xi _n}\ \frac{\sin (\pi \varepsilon \xi _n)}{\pi \xi _n} \prod _{j=1}^{n-1} \frac{\sin (2\pi R(\varepsilon )\xi _j)}{\pi \xi _j}, \end{aligned}$$

and therefore, $f_\varepsilon \in L^p({\mathbb {R}}^n)$ for any $p>1$ and moreover,

$$\begin{aligned} ||f_\varepsilon ||_{L^p({\mathbb {R}}^n)} \cong \varepsilon ^{\frac{n+1}{2p'}}. \end{aligned}$$

(2.2)

Next, we want to understand the asymptotic behaviour as $\varepsilon \rightarrow 0^+$ of the quantity

$$\begin{aligned} \left( \int _{{\mathbb {R}}^n} |{{\hat{f}}}_\varepsilon (x)|^q A^{(s)}_1(x) \textrm{d}x \right) ^{1/q} = \left( \int _{K_\varepsilon \cap ({\mathbb {R}}^n\setminus B(0,1))} A^{(s)}_1(x) \textrm{d}x\right) ^{1/q}. \end{aligned}$$

(2.3)

In view of the inclusions (2.1), it suffices to understand the asymptotic behaviour of the right-hand side of (2.3) when the integral is performed on the set $C_\varepsilon \cap ({\mathbb {R}}^n\setminus B(0,1))$. With this objective in mind, we obtain from Cavalieri’s principle

$$\begin{aligned}&\left( \int _{C_\varepsilon \cap ({\mathbb {R}}^n\setminus B(0,1))} A^{(s)}_1(x) \textrm{d}x\right) ^{1/q} = \left( \int _{1-\varepsilon }^1 \int _{\sqrt{1-t^2}\le |x'|\le R(\varepsilon )} A^{(s)}_1(x',t) \textrm{d}x' \textrm{d}t\right) ^{1/q}\\&\quad = \left( \int _{1-\varepsilon }^1 \int _{\sqrt{1-t^2}\le |x'|\le R(\varepsilon )} \frac{\textrm{d}x'}{(|x'|^2 - (1-t^2))^{s} (|x'|^2 + t^2)^{\frac{n}{2}}} \textrm{d}t\right) ^{1/q}\\&\quad \cong \left( \int _{1-\varepsilon }^1 \int _{\sqrt{1-t^2}\le |x'|\le R(\varepsilon )} \frac{\textrm{d}x'}{(|x'|^2 - (1-t^2))^{s}} \textrm{d}t\right) ^{1/q}\\&\quad \cong \left( \int _{1-\varepsilon }^1 \int _{\sqrt{1-t^2}}^{R(\varepsilon )} \frac{\rho ^{n-2} d\rho }{(\rho ^2 - (1-t^2))^{s}} \textrm{d}t\right) ^{1/q}. \end{aligned}$$

We now want to show that as $\varepsilon \rightarrow 0^+$, we have

$$\begin{aligned} G(\varepsilon ) = \int _{1-\varepsilon }^1 \int _{\sqrt{1-t^2}}^{R(\varepsilon )} \frac{\rho ^{n-2} d\rho }{(\rho ^2 - (1-t^2))^{s}} \textrm{d}t \cong \varepsilon ^{\frac{n+1}{2}-s}. \end{aligned}$$

(2.4)

To see this we write $G(\varepsilon ) = \int _{1-\varepsilon }^1 F(\varepsilon ,t) \textrm{d}t$, where

$$\begin{aligned} F(\varepsilon ,t) = \int _{\sqrt{1-t^2}}^{R(\varepsilon )} \frac{\rho ^{n-2} d\rho }{(\rho ^2 - (1-t^2))^{s}}. \end{aligned}$$

The chain rule gives

$$\begin{aligned} G'(\varepsilon ) = F(\varepsilon ,1-\varepsilon ) + \int _{1-\varepsilon }^1 \frac{\partial F}{\partial \varepsilon }(\varepsilon ,t) \textrm{d}t = \int _{1-\varepsilon }^1 \frac{\partial F}{\partial \varepsilon }(\varepsilon ,t) \textrm{d}t, \end{aligned}$$

since $F(\varepsilon ,1-\varepsilon ) = 0$. A simple computation gives

$$\begin{aligned} \frac{\partial F}{\partial \varepsilon }(\varepsilon ,t) = \frac{R'(\varepsilon ) R(\varepsilon )^{n-2}}{(R(\varepsilon )^2 - (1-t^2))^{s}}, \end{aligned}$$

therefore,

$$\begin{aligned} G'(\varepsilon )&= R'(\varepsilon ) R(\varepsilon )^{n-2} \int _{1-\varepsilon }^1 \frac{\textrm{d}t}{(R(\varepsilon )^2 - (1-t^2))^{s}} \\&= R'(\varepsilon ) R(\varepsilon )^{n-2} \int _{1-\varepsilon }^1 \frac{1}{(t^2 - (1-\varepsilon )^2)^{s}} \textrm{d}t\\&\cong R'(\varepsilon ) R(\varepsilon )^{n-2} \int _{1-\varepsilon }^1 \frac{\textrm{d}t}{(t - (1-\varepsilon ))^{s}} \cong \varepsilon ^{1-s} R'(\varepsilon ) R(\varepsilon )^{n-2}. \end{aligned}$$

Since $R(\varepsilon )^{n-2} = (\varepsilon (2-\varepsilon ))^{\frac{n-2}{2}} \cong \varepsilon ^{\frac{n}{2} -1}$, and $R'(\varepsilon ) \cong \varepsilon ^{-\frac{1}{2}}$, we infer that $G(\varepsilon ) \cong \varepsilon ^{\frac{n+1}{2}-s}$, which gives the desired conclusion (2.4). In conclusion, we have shown that

$$\begin{aligned} \left( \int _{{\mathbb {R}}^n} |{{\hat{f}}}_\varepsilon (\xi )|^q A^{(s)}_1(\xi ) d\xi \right) ^{1/q} \cong \varepsilon ^{(\frac{n+1}{2}-s)\frac{1}{q}}. \end{aligned}$$

(2.5)

Combining (2.2) with (2.5), we finally infer that a necessary condition for (1.7) to hold is

$$\begin{aligned} \left( \frac{n+1}{2}-s\right) \frac{1}{q} \ge \frac{n+1}{2p'}\ \Longleftrightarrow \ 1\le q \le \frac{n+1-2s}{n+1} p'. \end{aligned}$$

$\square$

3 The Fourier Transform of the Kernel $A^{(s)}_1$

In this section, we prove Theorem 1.3. Using Lemmas 5.3 and 5.4 in the Appendix, we establish a result which provides a key step in the proof of Theorem 1.3.

Lemma 3.1

For every $0<s<1$, $r>0$ and $\xi \in {\mathbb {R}}^n\setminus \{0\}$, one has

$$\begin{aligned}&\int _1^\infty \rho ^{-\frac{n}{2}} (\rho ^2 - 1)^{-s} J_{\frac{n}{2} -1}(2\pi r |\xi |\rho )\ d\rho \\&\quad = \Gamma (1-s)(2\pi r|\xi |)^{\frac{n}{2} - 1} \left\{ \frac{ \Gamma (s)}{2^{\frac{n}{2}} \Gamma (\frac{n}{2})}- \frac{1}{2^s} \int _0^{2\pi r |\xi |} t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t\right\} . \end{aligned}$$

Proof

To prove Lemma 3.1, we use Lemma 5.3 in which we take

$$\begin{aligned} \nu = \frac{n}{2} - 1,\ \ \ \ \ \alpha = -\frac{n}{2} + 1 = - \nu ,\ \ \ \ \ \beta = 1-s,\ \ \ \ c = 2 \pi r |\xi |. \end{aligned}$$

(3.1)

Since $s<1$, the condition $\Re \beta >0$ is guaranteed. Also, $\alpha +2\beta <\frac{7}{2}$ is equivalent to $\frac{n}{2} + 2s > - \frac{1}{2}$, which is trivially satisfied. Furthermore, we have

$$\begin{aligned} \frac{\alpha + \nu }{2} = 0,\ \ \ \ \ \nu + 1 = \frac{n}{2},\ \ \ \ \ \ \frac{\alpha +\nu }{2} + \beta = 1-s,\ \ \ \ 1-\beta -\frac{\alpha +\nu }{2} = s. \end{aligned}$$

We, thus, have

$$\begin{aligned}&\frac{c^\nu \ \Gamma (\beta ) \Gamma \left( 1-\beta -\frac{\alpha +\nu }{2}\right) }{2^{\nu +1} \ \Gamma (\nu +1) \Gamma \left( 1-\frac{\alpha +\nu }{2}\right) } \ _1F_2\left( \frac{\alpha +\nu }{2}; \nu +1,\frac{\alpha +\nu }{2} + \beta ; -\left( \frac{c}{2}\right) ^2\right) \\&\quad = (2\pi r|\xi |)^{\frac{n}{2} - 1} \frac{\Gamma (1-s) \Gamma (s)}{2^{\frac{n}{2}} \Gamma \left( \frac{n}{2}\right) } \ _1F_2\left( 0; \frac{n}{2},1-s; - \left( \frac{2\pi r |\xi |}{2}\right) ^2\right) \\&\quad = (2\pi r|\xi |)^{\frac{n}{2} - 1} \frac{\Gamma (1-s) \Gamma (s)}{2^{\frac{n}{2}} \Gamma \left( \frac{n}{2}\right) }, \end{aligned}$$

since from (5.12), we have $_1F_2(0; \frac{n}{2},1-s; - (\frac{2\pi r |\xi |}{2})^2) \equiv 1$. We next want to write in a more convenient form the second hypergeometric function in the right-hand side of (5.14) below. With the above choices (3.1), we now have

$$\begin{aligned}{} & {} 1-\beta = s,\ \ \ \ \ 2-\beta - \frac{\alpha +\nu }{2} = s+1,\\{} & {} 2-\beta + \frac{\nu -\alpha }{2} = s+1 + \frac{n}{2} - 1 = \frac{n}{2} + s, \end{aligned}$$

and also

$$\begin{aligned} \alpha +2\beta -3 = - \frac{n}{2} - 2s,\ \ \ \ \ 2-\alpha -2\beta = \frac{n}{2} - 1 + 2s,\ \ \ \ \ \beta + \frac{\alpha +\nu }{2} - 1 = - s. \end{aligned}$$

This gives

$$\begin{aligned}&\frac{2^{\alpha +2\beta -3} c^{2-\alpha -2\beta } \Gamma (\beta + \frac{\alpha +\nu }{2} -1)}{\Gamma (2-\beta +\frac{\nu -\alpha }{2})} \\&\quad _1F_2\left( 1-\beta ; 2-\beta - \frac{\alpha +\nu }{2},2-\beta + \frac{\nu -\alpha }{2}; - \left( \frac{c}{2}\right) ^2\right) \\&\quad = \frac{2^{- \frac{n}{2} - 2s} (2\pi r|\xi |)^{\frac{n}{2} - 1 + 2s} \Gamma (-s)}{\Gamma \left( \frac{n}{2} + s\right) } \ _1F_2(s; s+1,\frac{n}{2} + s; - \left( \frac{2\pi r|\xi |}{2})^2\right) \\&\quad = - \frac{2^{- \frac{n}{2} - 2s}(2\pi r|\xi |)^{\frac{n}{2} - 1 + 2s} \Gamma (1-s)}{s \Gamma \left( \frac{n}{2} + s\right) } \ _1F_2\left( s; s+1,\frac{n}{2} + s; - (\frac{2\pi r|\xi |}{2})^2\right) \end{aligned}$$

If we now apply Lemma 5.4 with

$$\begin{aligned} a = 2\pi r |\xi |,\ \ \ \ c = 1,\ \ \ \ \nu = \frac{n}{2} - 1 + s,\ \ \ \ \frac{\alpha +\nu }{2} = s\ \Longrightarrow \ \alpha = s - \frac{n}{2} + 1, \end{aligned}$$

we obtain

$$\begin{aligned}&_1F_2\left( s; s+1,\frac{n}{2} + s; - \left( \frac{2\pi r|\xi |}{2}\right) ^2\right) \\&\quad = \frac{2s 2^{\frac{n}{2} - 1 + s}\Gamma \left( \frac{n}{2} + s\right) }{(2\pi r |\xi |)^{2 s}} \int _0^{2\pi r |\xi |} t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t. \end{aligned}$$

Substituting in the above, and putting everything together, we find

$$\begin{aligned}&\int _1^\infty \rho ^{-\frac{n}{2}} (\rho ^2 - 1)^{-s} J_{\frac{n}{2} -1}(2\pi r |\xi |\rho )\ d\rho \\&\quad =\Gamma (1-s)(2\pi r|\xi |)^{\frac{n}{2} - 1} \left\{ \frac{ \Gamma (s)}{2^{\frac{n}{2}} \Gamma \left( \frac{n}{2}\right) }- \frac{1}{2^s} \int _0^{2\pi r |\xi |} t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t\right\} , \end{aligned}$$

which finally gives the desired conclusion. $\square$

Proof of Theorem 1.3

We begin by observing that the left-hand side of (1.14) coincides with the right-hand side when $\xi = 0$. For this, note that on one hand (1.4) gives

$$\begin{aligned} \widehat{A^{(s)}_r}(0) = ||A^{(s)}_r||_{L^1({\mathbb {R}}^n)} = 1. \end{aligned}$$

On the other, we apply Lemma 5.1 with

$$\begin{aligned} a=1,\ \ \ \ \ \mu = s - \frac{n}{2},\ \ \ \ \ \nu = \frac{n}{2} + s - 1. \end{aligned}$$

In such case, we have $\mu < \frac{1}{2}$ and also $\mu + \nu = s - \frac{n}{2} + \frac{n}{2} + s - 1 = 2s - 1 > -1$, since $s>0$. Since $\frac{\nu -\mu +1}{2} = \frac{n}{2}$, we thus obtain

$$\begin{aligned} \int _0^{\infty } t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t = 2^{s-\frac{n}{2}} \frac{\Gamma (s)}{\Gamma \left( \frac{n}{2}\right) }. \end{aligned}$$

(3.2)

This shows that when $\xi = 0$ the right-hand side of (1.14) becomes

$$\begin{aligned} \frac{2^{\frac{n}{2}-s} \Gamma \left( \frac{n}{2}\right) }{\Gamma (s)}\int _{0}^\infty t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t = 1, \end{aligned}$$

(3.3)

and therefore (1.14) does hold in $\xi = 0$.

Let now $\xi \not = 0$ and recall the well-known formula of Bochner for the Fourier transform of a spherically symmetric function $f(x) = f^\star (|x|)$,

$$\begin{aligned} {{\hat{f}}}(\xi ) = \frac{2\pi }{|\xi |^{\frac{n}{2}-1}} \int _0^\infty r^{\frac{n}{2}} f^\star (r) J_{\frac{n}{2} -1}(2\pi r|\xi |) dr, \end{aligned}$$

(3.4)

see [3, Theor. 40 on p. 69]. Applying (3.4) to (1.1), after a simple change of variable, we find

$$\begin{aligned} \widehat{A^{(s)}_r}(\xi )&= c(n,s) \frac{2\pi }{r^{\frac{n}{2} - 1} |\xi |^{\frac{n}{2} -1}} \int _1^\infty \rho ^{-\frac{n}{2}} (\rho ^2 - 1)^{-s} J_{\frac{n}{2} -1}(2\pi r |\xi |\rho )\ d\rho . \end{aligned}$$

(3.5)

At this point, we substitute Lemma 3.1 in (3.5), obtaining

$$\begin{aligned} \widehat{A^{(s)}_r}(\xi )&= (2\pi )^{\frac{n}{2}}\ c(n,s)\Gamma (1-s) \left\{ \frac{ \Gamma (s)}{2^{\frac{n}{2}} \Gamma \left( \frac{n}{2}\right) }- \frac{1}{2^s} \int _0^{2\pi r |\xi |} t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t\right\} \nonumber \\&= \frac{\sigma _{n-1}}{2} c(n,s)\ \Gamma (1-s)\Gamma (s) \left\{ 1 - \frac{2^{\frac{n}{2}-s} \Gamma \left( \frac{n}{2}\right) }{\Gamma (s)} \int _0^{2\pi r |\xi |} t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t\right\} \nonumber \\&= \frac{\sigma _{n-1}}{2} c(n,s)\ \Gamma (1-s)\Gamma (s) \frac{2^{\frac{n}{2}-s} \Gamma \left( \frac{n}{2}\right) }{\Gamma (s)} \nonumber \\&\quad \left\{ \int _0^{\infty } t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t - \int _0^{2\pi r |\xi |} t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t\right\} , \end{aligned}$$

(3.6)

where in the last equality we have used (3.3). Using (1.2) in (3.6), we finally obtain for every $0<s<1$ and $r>0$

$$\begin{aligned} \widehat{A^{(s)}_r}(\xi ) = \frac{2^{\frac{n}{2}-s} \Gamma \left( \frac{n}{2}\right) }{\Gamma (s)}\int _{2\pi r |\xi |}^\infty t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t. \end{aligned}$$

(3.7)

This completes the proof of Theorem 1.3.$\square$

For subsequent purposes, it will be important to have the following alternative representation of $\widehat{A^{(s)}_r}$.

Corollary 3.2

Let $0<s<1$. For every $\xi \in {\mathbb {R}}^n$, we have

$$\begin{aligned} \widehat{A^{(s)}_r}(\xi )&= \frac{2^{\frac{n}{2}-s} \Gamma \left( \frac{n}{2}\right) }{\Gamma (s)}\left\{ n \int _{2\pi r |\xi |}^\infty t^{s - \frac{n}{2}-1} J_{\frac{n}{2} + s}(t) \textrm{d}t -\frac{1}{(2\pi r|\xi |)^{\frac{n}{2} -s}} J_{\frac{n}{2} + s}(2\pi r|\xi |)\right\} . \end{aligned}$$

(3.8)

The identity (3.8) implies, in particular, the existence of a universal $C(n,s)>0$, such that as $|\xi |\rightarrow \infty$

$$\begin{aligned} \left| \widehat{A^{(s)}_1}(\xi )\right| \le \frac{C(n,s)}{|\xi |^{\frac{n+1}{2} -s}}. \end{aligned}$$

(3.9)

Proof

The proof of (3.8) follows from (3.7) by applying the recursive formula (5.3) with $\nu = \frac{n}{2} + s$ and integrating by parts. One has in fact

$$\begin{aligned}&\int _{2\pi r |\xi |}^\infty t^{s - \frac{n}{2}} J_{\frac{n}{2} - 1 + s}(t) \textrm{d}t = \int _{2\pi r |\xi |}^\infty t^{- n} \frac{d}{\textrm{d}t}\left( t^{\frac{n}{2}+s} J_{\frac{n}{2}+ s}(t)\right) \textrm{d}t. \end{aligned}$$

Details are left to the interested reader, who should notice that, since $s>0$, we now have $s - \frac{n}{2}-1+ \frac{n}{2} + s = 2s-1>-1$, and therefore, the oscillatory integral $\int _{0}^\infty t^{s - \frac{n}{2}-1} J_{\frac{n}{2} + s}(t) \textrm{d}t$ is convergent near $t=0$ (and can in fact be explicitly computed via Lemma 5.1). To prove (3.9) it is enough to observe that when $|\xi |$ is sufficiently large, then by (5.10) we have for some universal constant $C>0$ and all $t \ge 2\pi |\xi |$,

$$\begin{aligned} |J_{\frac{n}{2} + s}(t)|\le \frac{C}{t^{1/2}}. \end{aligned}$$

We, thus, find for the first term in the right-hand side of (3.8)

$$\begin{aligned} \left| \int _{2\pi |\xi |}^\infty t^{s - \frac{n}{2}-1} J_{\frac{n}{2} + s}(t) \textrm{d}t\right| \le C \int _{2\pi |\xi |}^\infty t^{s - \frac{n}{2}-\frac{3}{2}} \textrm{d}t = \frac{2C}{(n+1-2s)(2\pi |\xi |)^{\frac{n+1}{2}-s}}. \end{aligned}$$

Since the second term can obviously be estimated in the same way, we are finished.

$\square$

The next result provides the limiting value of $\widehat{A^{(s)}_1}(\xi )$ as $s\rightarrow 1$. It represents the counterpart on the Fourier transform side of Proposition 1.1.

Lemma 3.3

Let $n\ge 2$. Then for every $\xi \in {\mathbb {R}}^n$, one has

$$\begin{aligned} \underset{s\rightarrow 1}{\lim }\ \widehat{A^{(s)}_1}(\xi ) = \frac{1}{\sigma _{n-1}} \widehat{d\sigma }(\xi ) \end{aligned}$$

Proof

Notice that for any $\xi \not = 0$ and $t>2\pi |\xi |$, we have from (5.9)

$$\begin{aligned} |t^{s - \frac{n}{2}-1} J_{\frac{n}{2} + s}(t)| \le C(n,\xi )\ t^{- \frac{n+1}{2}} \in L^1(2\pi |\xi |,\infty ), \end{aligned}$$

for some constant $C(n,\xi )>0$ independent of $s\in (0,1)$. By Lebesgue dominated convergence and (5.4), we, thus, have

$$\begin{aligned} \underset{s\rightarrow 1}{\lim }\ \int _{2\pi |\xi |}^\infty t^{s - \frac{n}{2}-1} J_{\frac{n}{2} + s}(t) \textrm{d}t = \int _{2\pi |\xi |}^\infty t^{- \frac{n}{2}} J_{\frac{n}{2} + 1}(t) \textrm{d}t = - \int _{2\pi |\xi |}^\infty \frac{d}{\textrm{d}t}\left( t^{- \frac{n}{2}} J_{\frac{n}{2}}(t)\right) \textrm{d}t. \end{aligned}$$

This observation and (3.8) give

$$\begin{aligned} \underset{s\rightarrow 1}{\lim }\ \widehat{A^{(s)}_1}(\xi ) = 2^{\frac{n}{2}-1} \Gamma \left( \frac{n}{2}\right) \frac{1}{(2\pi |\xi |)^{\frac{n}{2} -1}} J_{\frac{n}{2}-1}(2\pi |\xi |). \end{aligned}$$

In view of (1.17) above, we have reached the desired conclusion when $\xi \not = 0$. When instead $\xi = 0$ for the left-hand side of (1.14) we obtain from (1.4)

$$\begin{aligned} \underset{s\rightarrow 1}{\lim }\ \widehat{A^{(s)}_1}(0) = \underset{s\rightarrow 1}{\lim }\ ||A^{(s)}_1||_{L^1({\mathbb {R}}^n)}= 1. \end{aligned}$$

Convergence to the same limit of the right-hand side follows from (3.3).$\square$

4 Proof of Theorem 1.4

In this section, we prove Theorem 1.4. The proof will be based on two central ideas: 1) To exploit the $L^2$ nature of the inequality (1.22) via the Plancherel theorem. This reduces considerations to proving that the nonlocal Tomas-Stein operator $R^{(s)}$ in (4.1) below maps $L^p$ into $L^{p'}$; 2) To accomplish this step, we embed $R^{(s)}$ into an analytic family of operators $T_z$. For the latter, we show that

$$\begin{aligned} {\left\{ \begin{array}{ll} T_{-\frac{n-1}{2} + i y}: L^1({\mathbb {R}}^n)\ \longrightarrow \ L^\infty ({\mathbb {R}}^n)\\ T_{1 + i y}: L^2({\mathbb {R}}^n)\ \longrightarrow \ L^2({\mathbb {R}}^n), \end{array}\right. } \end{aligned}$$

with appropriate bounds on the operator norms.

Proof of Theorem 1.4

Similarly to [20] we write for $f\in {\mathscr {S}}({\mathbb {R}}^n)$

$$\begin{aligned} \int _{{\mathbb {R}}^n} |{{\hat{f}}}(\xi )|^2 A^{(s)}_1(\xi ) d\xi = \int _{{\mathbb {R}}^n} \overline{{{\hat{f}}}(\xi )} {{\hat{f}}}(\xi ) A^{(s)}_1(\xi ) d\xi = \int _{{\mathbb {R}}^n} \overline{{{\hat{f}}}(\xi )}\ \widehat{R^{(s)} f}(\xi ) d\xi , \end{aligned}$$

where we have defined

$$\begin{aligned} R^{(s)} f(\xi ) = \widehat{A^{(s)}_1} \star f(\xi ), \end{aligned}$$

(4.1)

so that $\widehat{R^{(s)} f}(\xi ) = {{\hat{f}}}(\xi ) A^{(s)}_1(\xi )$. By Plancherel, we, thus, obtain

$$\begin{aligned} \int _{{\mathbb {R}}^n} |{{\hat{f}}}(\xi )|^2 A^{(s)}_1(\xi ) d\xi&= \int _{{\mathbb {R}}^n} \overline{f(x)}\ R^{(s)} f(x) \textrm{d}x\\&\le ||f||_{L^p({\mathbb {R}}^n)} ||R^{(s)} f||_{L^{p'}({\mathbb {R}}^n)}, \end{aligned}$$

where in the last inequality, we have used Hölder. The proof will be finished if we can show that for every $0<s<1$, there exists $M_s = M_s(n)>0$ such that for $f\in {\mathscr {S}}({\mathbb {R}}^n)$, one has

$$\begin{aligned} ||R^{(s)} f||_{L^{p'}({\mathbb {R}}^n)} \le M_s\ ||f||_{L^p({\mathbb {R}}^n)}, \end{aligned}$$

(4.2)

for $p = \frac{2(n+1)}{n+1+2s}$. We want to accomplish (4.2) by interpolating between the two endpoints $L^1\rightarrow L^\infty$ and $L^2\rightarrow L^2$. This means we have to choose $\theta \in [0,1]$ such that

$$\begin{aligned} \frac{n+1+2s}{2(n+1)} = \frac{1}{p} = \frac{1-\theta }{1} + \frac{\theta }{2} = 1 - \frac{\theta }{2}, \end{aligned}$$

which gives

$$\begin{aligned} \frac{\theta }{2} = \frac{1}{p'} = \frac{n+1-2s}{2(n+1)}, \end{aligned}$$

and, therefore,

$$\begin{aligned} \theta = \theta (s) = \frac{n+1-2s}{n+1} = 1 - \frac{2s}{n+1},\ \ \ \ \ \ 1-\theta = \frac{2s}{n+1}. \end{aligned}$$

(4.3)

For $z\in {\mathbb {C}}$ such that $\Re z\le 1$ we define a linear operator $T_z:{\mathscr {S}}({\mathbb {R}}^n)\rightarrow {\mathbb {C}}$ by letting

$$\begin{aligned} \widehat{T_z f}(\xi ) = A^{(1-z)}_1(\xi ) {{\hat{f}}}(\xi ), \end{aligned}$$

(4.4)

where for $\Re z \le 1$, we have let

$$\begin{aligned} A^{(z)}_1(x) = \frac{c(n,z)}{(|x|^2 - 1)_+^z |x|^n}, \end{aligned}$$

(4.5)

with $c(n,z) = \frac{2}{\sigma _{n-1}\Gamma (z)\Gamma (1-z)}$. Notice that $c(n,z) = c(n,1-z)$ and that for $y\in {\mathbb {R}}$, we have

$$\begin{aligned} ||A^{(i y)}_1||_{L^\infty ({\mathbb {R}}^n)} \le \frac{2}{\sigma _{n-1}|\Gamma (i y)||\Gamma (1-i y)|}. \end{aligned}$$

(4.6)

From Plancherel theorem, (4.4), (4.6), and (5.1) we conclude for some universal $C(n)>0$ and for every $y\in {\mathbb {R}}$

$$\begin{aligned} ||T_{1+i y} f||_{L^2({\mathbb {R}}^n)} \le M_1(y)\ ||f||_{L^2({\mathbb {R}}^n)}, \end{aligned}$$

(4.7)

with $M_1(y) = C(n)\ e^{\pi |y|}$.

We now introduce the kernels

$$\begin{aligned} K_z(\xi )= & {} \frac{2^{\frac{n}{2}-1+z} \Gamma \left( \frac{n}{2}\right) }{\Gamma (1-z)}\bigg \{n \int _{2\pi |\xi |}^\infty t^{-z - \frac{n}{2}} J_{\frac{n}{2} + 1-z}(t) \textrm{d}t\nonumber \\{} & {} \quad -\frac{1}{(2\pi |\xi |)^{\frac{n}{2} -1+z}} J_{\frac{n}{2} + 1-z}(2\pi |\xi |)\bigg \}. \end{aligned}$$

(4.8)

Notice that, according to (3.8) in Corollary 3.2, when $z=1-s$ we have $K_{1-s} = \widehat{A^{(s)}_1}$, and therefore, (4.4) gives

$$\begin{aligned} T_{1-s} f(\xi ) = \widehat{A^{(s)}_1} \star f(\xi ) = R^{(s)} f(\xi ), \end{aligned}$$

(4.9)

where in the last equality, we have used (4.1). Also notice that, since by analytic continuation, (3.8) continues to be valid for any $z\in {\mathbb {C}}$ in the strip $0<\Re z < 1$, for any such z, we have from (4.4)

$$\begin{aligned} T_z f = K_z \star f. \end{aligned}$$

(4.10)

Since by (4.8), the kernel $K_z$ defines an analytic function of z for $-\frac{n+1}{2}< \Re z<1$ (see Lemma 5.1), we can use (4.10) to analytically extend the operator $T_z$ to the whole strip $S = \{z\in {\mathbb {C}}\mid -\frac{n-1}{2}<\Re z<1\}$. If we let $z = x + i y$, then we define

$$\begin{aligned} S_{x+iy} = T_{-\frac{n-1}{2} + \frac{n+1}{2} x+i y}. \end{aligned}$$

Note that $S_z$ is now defined on the strip $\Sigma = \{z\in {\mathbb {C}}\mid 0<\Re z < 1\}$, and that (4.7) now reads

$$\begin{aligned} ||S_{1+i y} f||_{L^2({\mathbb {R}}^n)} \le C(n)\ \sinh (\pi |y|)\ ||f||_{L^2({\mathbb {R}}^n)}. \end{aligned}$$

(4.11)

Since $S_{i y} = T_{-\frac{n-1}{2} +i y}$, we next analyse the behaviour of $T_z$ on the line $L_0 = \{z\in {\mathbb {C}}\mid \Re z = - \frac{n-1}{2}\}$. Notice that (4.3) gives

$$\begin{aligned} (1-\theta )\left( -\frac{n-1}{2}\right) + \theta \cdot 1 = 1-s. \end{aligned}$$

Note that the equation $-\frac{n-1}{2} + \frac{n+1}{2} x = (1-s)$ gives

$$\begin{aligned} x = \frac{(1-s) + \frac{n-1}{2}}{\frac{n+1}{2}} = \frac{n+1-2s}{n+1} = 1 - \frac{2s}{n+1} = \theta , \end{aligned}$$

see (4.3). This shows that $S_\theta = T_{1-s}$. In view of (4.7), we conclude that, if we can show that $T_{-\frac{n-1}{2} + i y} :L^1({\mathbb {R}}^n) \rightarrow L^\infty ({\mathbb {R}}^n)$, with appropriate bounds on the operator norms, then by Stein’s theorem of complex interpolation for an analytic family of operators, see [17] or [18, Theor. 4.1, p. 205], it will follow that $T_{1-s} : L^p({\mathbb {R}}^n) \rightarrow L^{p'}({\mathbb {R}}^n)$, as desired.

In order to show that $T_{-\frac{n-1}{2} + i y} :L^1({\mathbb {R}}^n) \rightarrow L^\infty ({\mathbb {R}}^n)$, we will prove that $K_{-\frac{n-1}{2} + i y}\in L^\infty ({\mathbb {R}}^n)$ and that moreover, for some universal constant $C>0$ depending only on n, one has

$$\begin{aligned} ||K_{-\frac{n-1}{2} + i y}||_{L^\infty ({\mathbb {R}}^n)} \le C e^{\frac{3\pi }{2} |y|},\ \ \ \ \ \ \ \ y\in {\mathbb {R}}. \end{aligned}$$

(4.12)

From (4.8), we obtain

$$\begin{aligned} K_{-\frac{n-1}{2} + i y}(\xi )= & {} \frac{2^{-\frac{1}{2} + i y} \Gamma (\frac{n}{2})}{\Gamma \left( \frac{n+1}{2} -i y\right) }\bigg \{n \int _{2\pi |\xi |}^\infty t^{-\frac{1}{2} - i y} J_{n+\frac{1}{2} - i y}(t) \textrm{d}t \nonumber \\{} & {} \quad -\frac{1}{(2\pi |\xi |)^{-\frac{1}{2} + i y}} J_{n +\frac{1}{2} - i y}(2\pi |\xi |)\bigg \}. \end{aligned}$$

(4.13)

Note that from (5.10), there exists a universal $R = R(n)>0$ such that when $|\xi | \ge R$, one has

$$\begin{aligned} \left| \frac{1}{(2\pi |\xi |)^{-\frac{1}{2} + i y}} J_{n +\frac{1}{2} - i y}(2\pi |\xi |)\right| \le 1. \end{aligned}$$

(4.14)

On the other hand, for $|\xi |\le R$, we have from (5.8)

$$\begin{aligned} \left| \frac{1}{(2\pi |\xi |)^{-\frac{1}{2} + i y}} J_{n +\frac{1}{2} - i y}(2\pi |\xi |)\right|&\le \frac{\Gamma (n+1)}{|\Gamma (n+1-i y)|\ \Gamma \left( n + \frac{1}{2}+1\right) } \frac{\left( 2\pi |\xi |\right) ^{n+1}}{2^{n+\frac{1}{2}}} \nonumber \\&\le \frac{\sqrt{2} \pi ^{n+\frac{1}{2}} \Gamma (n) R^{n+1}}{\Gamma \left( n+\frac{1}{2}\right) \ |\Gamma (n+1-i y)|}. \end{aligned}$$

(4.15)

We now have for $n\ge 2$

$$\begin{aligned} |\Gamma (n+1-i y)|&= |(n-i y)||\Gamma (n-i y)|\\&= |(n-i y)||(n-1-i y)|\ldots |(1-i y)||\Gamma (1-i y)|\\&\ge (1+y^2)^{\frac{n}{2}} |\Gamma (1-i y)| = (1+y^2)^{\frac{n}{2}} \sqrt{\frac{\pi |y|}{\sinh \pi |y|}}, \end{aligned}$$

where in the last equality, we have used (5.1). This gives

$$\begin{aligned} \frac{1}{|\Gamma (n+1-i y)|}\le (1+y^2)^{-\frac{n}{2}} \sqrt{\frac{\sinh \pi |y|}{\pi |y|}}\le \sqrt{\frac{\sinh \pi |y|}{\pi |y|}} \le \frac{3}{2} e^{\frac{\pi }{2} |y|}. \end{aligned}$$

(4.16)

Inserting this information in (4.15), and combining the resulting estimate with (4.14), we conclude that there exists $C(n)>0$ such that for every $\xi \in {\mathbb {R}}^n$ and any $y\in {\mathbb {R}}$, one has

$$\begin{aligned} \left| \frac{1}{(2\pi |\xi |)^{-\frac{1}{2} + i y}} J_{n +\frac{1}{2} - i y}(2\pi |\xi |)\right| \le C e^{\frac{\pi }{2} |y|}. \end{aligned}$$

(4.17)

Next, we show that for some $C = C(n)>0$ one has for every $\gamma \in {\mathbb {R}}$

$$\begin{aligned} \left| \frac{1}{\Gamma \left( \frac{n+1}{2} -i y\right) }\right| \le C e^{\frac{\pi }{2} |y|}. \end{aligned}$$

(4.18)

To see this, we apply Legendre duplication formula, see e.g. (1.2.3) in [12], to write

$$\begin{aligned} 2^{n-2i y}\Gamma \left( \frac{n+1}{2} -i y\right) \Gamma \left( \frac{n}{2}+1 -i y\right) = \sqrt{\pi }\Gamma (n+1 - 2i y). \end{aligned}$$

Using the estimate $|\Gamma (z)|\le |\Gamma (\Re z)|$, this gives

$$\begin{aligned}&\left| \frac{1}{\Gamma \left( \frac{n+1}{2} -i y\right) }\right| \le \frac{2^n |\Gamma \left( \frac{n}{2}+1 -i y\right) |}{\sqrt{\pi }|\Gamma (n+1 - 2i y)|} \le \frac{2^n \Gamma \left( \frac{n}{2}+1\right) }{\sqrt{\pi }|\Gamma (n+1 - 2i y)|} \le C(n) e^{\frac{\pi }{2} |y|}, \end{aligned}$$

where in the last inequality, we have used (4.16). This proves (4.18). Next, we show that for every $\xi \in {\mathbb {R}}^n$ and every $y\in {\mathbb {R}}$

$$\begin{aligned} \left| \int _{2\pi |\xi |}^\infty t^{-\frac{1}{2} - i y} J_{n+\frac{1}{2} - i y}(t) \textrm{d}t\right| \le C (1+|y|) e^{\frac{\pi }{2} |y|}. \end{aligned}$$

(4.19)

To prove (4.19) observe that, in view of (5.9), there exists $R>0$ depending on n such that for $|\xi |\ge R$, we have for $t\in [2\pi |\xi |,\infty )$

$$\begin{aligned} \left| J_{n+\frac{1}{2} - i y}(t) - \sqrt{\frac{2}{\pi t}}\cos \left( t-\frac{n+1}{2} \pi + i \frac{\pi }{2} y\right) \right| \le t^{-3/2}. \end{aligned}$$

Since

$$\begin{aligned} \cos \left( t-\frac{n+1}{2} \pi + i \frac{\pi }{2} y\right)&= \cos \left( t+i \frac{\pi }{2} y\right) \cos \left( \frac{n+1}{2} \pi \right) \\&\quad + \sin \left( t+i \frac{\pi }{2}y\right) \sin \left( \frac{n+1}{2} \pi \right) \\&= {\left\{ \begin{array}{ll} (-1)^{n+1}\sin \left( t+i \frac{\pi }{2}\gamma \right) ,\ \ \ \ n\ \text {even},\\ (-1)^{n}\cos \left( t+i \frac{\pi }{2}\gamma \right) ,\ \ \ \ n\ \text {odd}, \end{array}\right. } \end{aligned}$$

with obvious meaning of the notation, we infer

$$\begin{aligned}&\left| \int _{2\pi |\xi |}^\infty t^{-\frac{1}{2} - i\gamma } J_{n+\frac{1}{2} - i\gamma }(t) \textrm{d}t\right| \le \left| \sqrt{\frac{2}{\pi }}\int _{2\pi |\xi |}^\infty \frac{\begin{pmatrix}\sin (t+i \frac{\pi }{2}\gamma )\\ \cos \left( t+i \frac{\pi }{2}\gamma \right) \end{pmatrix}}{t^{1 + i\gamma }} \textrm{d}t\right| + \int _{2\pi |\xi |}^\infty t^{-2} \textrm{d}t. \end{aligned}$$

Keeping in mind that for $z = x+iy$ we have

$$\begin{aligned} \cos z = \cos x \cosh y - i \sin x \sinh y,\ \ \ \ \sin z = \sin x \cosh y + i \cos x \sinh y, \end{aligned}$$

and that integrating by parts, we obtain

$$\begin{aligned} \left| \int _{2\pi |\xi |}^\infty \frac{\begin{pmatrix}\sin t\\ \cos t\end{pmatrix}}{t^{1 + i y}} \textrm{d}t\right| \le C \frac{1+|y|}{|\xi |}\le C (1+|y|), \end{aligned}$$

we conclude that (4.19) holds when $|\xi |\ge R$. If instead $|\xi |\le R$, then we write

$$\begin{aligned} \int _{2\pi |\xi |}^\infty t^{-\frac{1}{2} - i y} J_{n+\frac{1}{2} - i y}(t) \textrm{d}t&= \int _{0}^\infty t^{-\frac{1}{2} - i y} J_{n+\frac{1}{2} - i y}(t) \textrm{d}t \\&\quad - \int _0^{2\pi |\xi |} t^{-\frac{1}{2} - i y} J_{n+\frac{1}{2} - i y}(t) \textrm{d}t, \end{aligned}$$

and then use Lemma 5.1 and (5.8) to estimate

$$\begin{aligned} \left| \int _{2\pi |\xi |}^\infty t^{-\frac{1}{2} - i y} J_{n+\frac{1}{2} - i y}(t) \textrm{d}t\right|&\le \frac{|\Gamma \left( \frac{n}{2} - i y\right) |}{\sqrt{2} \Gamma \left( \frac{n}{2} + 1\right) } + \frac{C(n) R^{n+2}}{|\Gamma \left( n-i y + \frac{1}{2}\right) |} \\&\le C(n) \left( 1 + \frac{1}{|\Gamma \left( n-i y + \frac{1}{2}\right) |}\right) . \end{aligned}$$

Using again Legendre duplication formula, similarly to the proof of (4.18) we recognise

$$\begin{aligned} \frac{1}{|\Gamma \left( n-i y + \frac{1}{2}\right) |} \le C(n) e^{\frac{\pi }{2} |y|}. \end{aligned}$$

If we now use (4.17), (4.18) and (4.19) in (4.13), we conclude that (4.12) does hold. It follows that for every $f\in {\mathscr {S}}({\mathbb {R}}^n)$ and any $y\in {\mathbb {R}}$ one has

$$\begin{aligned} ||T_{-\frac{n-1}{2} + i y} f||_{L^\infty ({\mathbb {R}}^n)} \le M_0(y) ||f||_{L^1({\mathbb {R}}^n)}, \end{aligned}$$

(4.20)

with $M_0(y) = C e^{\frac{3\pi }{2}|y|}$. By [18, Theor. 4.1 on p. 205] we infer that there exists $M_s = M_{\theta (s)}>0$ such that (4.2) holds. This proves Theorem 1.4.$\square$

From p. 209 in [18] we see that the constant $M_s$ is given by

$$\begin{aligned} M_s = \exp \left[ \frac{\sin \pi \theta (s)}{2} \int _{{\mathbb {R}}} \left\{ \frac{\log M_0(y)}{\cosh \pi y - \cos \pi \theta (s)} + \frac{\log M_1(y)}{\cosh \pi y + \cos \pi \theta (s)}\right\} \textrm{d}y\right] , \end{aligned}$$

From (4.3) we have

$$\begin{aligned} \sin \pi \theta (s) = \sin \left( \pi - \frac{2\pi s}{n+1}\right) = \sin \left( \frac{2\pi s}{n+1}\right) ,\ \ \ \ \ \ \cos \pi \theta (s) = - \cos \left( \frac{2\pi s}{n+1}\right) . \end{aligned}$$

We thus find

$$\begin{aligned} M_s= & {} \exp \left[ \sin \left( \frac{2\pi s}{n+1}\right) \int _0^\infty \left\{ \frac{\log M_0(y)}{\cosh \pi y + \cos \left( \frac{2\pi s}{n+1}\right) } \right. \right. \nonumber \\{} & {} \left. \left. + \frac{\log M_1(y)}{\cosh \pi y - \cos \left( \frac{2\pi s}{n+1}\right) }\right\} \textrm{d}y\right] . \end{aligned}$$

(4.21)

Since $\cos (\frac{2\pi }{n+1}) \not = \pm 1$ for any $n\ge 2$, by Lebesgue dominated convergence we find

$$\begin{aligned} \underset{s\rightarrow 1}{\lim }\ M_s&= \exp \left[ \sin \left( \frac{2\pi }{n+1}\right) \int _0^\infty \left\{ \frac{\log M_0(y)}{\cosh \pi y + \cos \left( \frac{2\pi }{n+1}\right) } \right. \right. \nonumber \\&\quad \left. \left. + \frac{\log M_1(y)}{\cosh \pi y - \cos \left( \frac{2\pi }{n+1}\right) }\right\} \textrm{d}y\right] \nonumber \\&= M(n)\ <\ \infty . \end{aligned}$$

(4.22)

Since the function $s\rightarrow \frac{2(n+1)}{n+1+2s}$ is decreasing on (0, 1), with range $(\frac{2(n+1)}{n+3},2)$, if now $1\le p \le \frac{2(n+1)}{n+3}$, then for any $s\in (0,1)$ we also have $1\le p \le \frac{2(n+1)}{n+1+2s}$. From the proof of Theorem 1.4 we infer for every $f\in {\mathscr {S}}({\mathbb {R}}^n)$ one has

$$\begin{aligned} \left( \int _{{\mathbb {R}}^n} |{{\hat{f}}}(\xi )|^2 A^{(s)}_1(\xi ) d\xi \right) ^{1/2} \le \sqrt{M_s}\ ||f||_{L^p({\mathbb {R}}^n)}, \end{aligned}$$

(4.23)

with $M_s$ give by (4.21) above. Passing to the limit for $s\rightarrow 1$ in (4.23), and using Proposition 1.1 and (4.22), we conclude the celebrated Tomas-Stein theorem for the sphere

$$\begin{aligned} \left( \frac{1}{\sigma _{n-1}}\int _{{\mathbb {S}}^{n-1}} |{{\hat{f}}}(\xi )|^2 d\sigma (\xi )\right) ^{1/2} \le \sqrt{M(n)}\ ||f||_{L^p({\mathbb {R}}^n)}. \end{aligned}$$

5. Appendix: Some Classical Material

In this section, we collect some results needed in this paper. We recall the gamma function

$$\begin{aligned} \Gamma (z) = \int _0^\infty t^{z-1} e^{-t} \textrm{d}t,\ \ \ \ \ \ \ \ \ \Re z>0. \end{aligned}$$

Using the well-known formula $\Gamma (z+1) = z \Gamma (z)$, this function can be extended as a meromorphic function to the whole ${\mathbb {C}}$ having simple poles in $z = -k, k=0,1,\ldots$, with residues ${\text {Res}}(\Gamma ;-k) = \frac{(-1)^k}{k!}$. The reciprocal gamma function $1/\Gamma (z)$ is an entire function with zeros at the negative integers. The following special values of such function will be useful subsequently, see e.g. formulas 6.1.29 and 6.1.31 on p. 256 in [1],

$$\begin{aligned} \frac{1}{|\Gamma (\pm i y)|} = |y| \sqrt{\frac{\sinh (\pi |y|)}{\pi |y|}},\ \ \ \ \ \ \frac{1}{|\Gamma (1\pm i y)|} = \sqrt{\frac{\sinh (\pi |y|)}{\pi |y|}}. \end{aligned}$$

(5.1)

Since this note is about the Fourier transform, Bessel functions (introduced and developed in the seminal work [2]) play a pervasive role in it. We recall the series expansion of the Bessel function $J_\nu (z)$ for $|\arg z|<\pi$,

$$\begin{aligned} J_\nu (z) = \left( \frac{z}{2}\right) ^{\nu }\sum _{k=0}^\infty \frac{(-1)^k}{\Gamma (\nu +k+1) k!} \left( \frac{z}{2}\right) ^{2k}, \end{aligned}$$

(5.2)

see [21]. From (5.2), it is easy to obtain the following well-known formulas

$$\begin{aligned} \frac{d}{dz}\left[ z^\nu J_\nu (z)\right]= & {} z^\nu J_{\nu -1}(z), \end{aligned}$$

(5.3)

$$\begin{aligned} - \frac{d}{dz}\left[ z^{-\nu } J_\nu (z)\right]= & {} z^{-\nu } J_{\nu +1}(z), \end{aligned}$$

(5.4)

and

$$\begin{aligned} \frac{2\nu }{z} J_\nu (z) - J_{\nu +1}(z) = J_{\nu -1}(z). \end{aligned}$$

(5.5)

For every $\nu \in {\mathbb {C}}$ such that $\Re \nu >-\dfrac{1}{2}$, the function $J_\nu$ admits the following Poisson representation

$$\begin{aligned} J_\nu (z) =\frac{1}{\Gamma \left( \frac{1}{2}\right) \Gamma \left( \nu +\frac{1}{2}\right) } \left( \frac{z}{2}\right) ^\nu \int ^1_{-1} e^{izs}(1-s^2)^{\frac{2\nu -1}{2}}\textrm{d}s, \end{aligned}$$

(5.6)

where z ranges in the complex plane cut along the negative real axis $(-\infty ,0]$. When $\nu \in {\mathbb {N}}_0$ then we can take $z\in {\mathbb {C}}$. From (5.2) one easily recognises that

$$\begin{aligned} J_\nu (z)\cong \frac{1}{\Gamma (\nu +1)}\left( \frac{z}{2}\right) ^\nu ,\quad \text {as }z\rightarrow 0. \end{aligned}$$

(5.7)

Again from (5.6) one sees that when $z\in {\mathbb {R}}$ and $z>0$, then for $\Re \nu >-\dfrac{1}{2}$ one has

$$\begin{aligned} |J_\nu (z)| \le \frac{\Gamma (\Re \nu +\frac{1}{2})}{|\Gamma (\nu +\frac{1}{2})|\ \Gamma (\Re \nu +1)} \left( \frac{z}{2}\right) ^{\Re \nu }. \end{aligned}$$

(5.8)

This estimate is not useful for large $z>0$ since $J_\nu (z)$ decays at infinity with an oscillatory behaviour. When $\Re \nu >-\dfrac{1}{2}$ the following result due to Hankel holds, see for instance [18, Lemma 3.11], or also (5.11.6) on p. 122 in [12]. One has for $0<\delta <\pi$

$$\begin{aligned} J_\nu (z)&=\sqrt{\frac{2}{\pi z}}\cos \left( z-\frac{\pi \nu }{2}-\frac{\pi }{4}\right) + O(z^{-\frac{3}{2}})\nonumber \\&\quad \text {as }|z|\rightarrow \infty ,\quad -\pi +\delta<\arg z<\pi -\delta . \end{aligned}$$

(5.9)

In particular,

$$\begin{aligned} J_\nu (z) = O(z^{-1/2}),\ \ \ \ \ \text {as}\ z\rightarrow +\infty . \end{aligned}$$

(5.10)

The following beautiful formula can be found in 6.561.14 on p. 684 of [10], or also (19) on p. 49 in vol.2 of [5].

Lemma 5.1

Let $a>0$, $-\Re \nu -1<\Re \mu <\frac{1}{2}$. Then

$$\begin{aligned} \int _0^\infty t^\mu J_\nu (at) \textrm{d}t = 2^\mu a^{-\mu -1} \frac{\Gamma \left( \frac{\nu +\mu +1}{2}\right) }{\Gamma \left( \frac{\nu -\mu +1}{2}\right) }. \end{aligned}$$

We next recall the definition of the hypergeometric functions. The Pochammer’s symbols are defined by

$$\begin{aligned} \alpha _0 = 1,\ \ \ \ \alpha _k \overset{def}{=}\ \frac{\Gamma (\alpha + k)}{\Gamma (\alpha )} = \alpha (\alpha +1)...(\alpha + k -1),\ \ \ \ \ \ \ \ \ \ k\in {\mathbb {N}}.\nonumber \\ \end{aligned}$$

(5.11)

Notice that since the gamma function has a pole in $z=0$, we have

$$\begin{aligned} 0_k = {\left\{ \begin{array}{ll} 1\ \ \ \ \ \ \text {if}\ k = 0\\ 0\ \ \ \ \ \ \text {for}\ k\ge 1. \end{array}\right. } \end{aligned}$$

Definition 5.2

Let $p, q\in {\mathbb {N}}\cup \{0\}$ be such that $p\le q+1$, and let $\alpha _1,\ldots ,\alpha _p$ and $\beta _1,\ldots ,\beta _q$ be given parameters such that $-\beta _j\not \in {\mathbb {N}}_0$ for $j=1,\ldots ,q$. Given a number $z\in {\mathbb {C}}$, the power series

$$\begin{aligned} _p F_q(\alpha _1,\ldots ,\alpha _p;\beta _1,\ldots ,\beta _q;z) = \sum _{k=0}^\infty \frac{(\alpha _1)_k \ldots (\alpha _p)_k}{(\beta _1)_k . . . (\beta _q)_k} \frac{z^k}{k!} \end{aligned}$$

is called the generalised hypergeometric function. When $p = 2$ and $q=1$, then the function $_2 F_1(\alpha _1,\alpha _2;\beta _1;z)$ is the Gauss’ hypergeometric function, and it is usually denoted by $F(\alpha _1,\alpha _2;\beta _1;z)$.

Using the ratio test one easily verifies that the radius of convergence of the above hypergeometric series is $\infty$ when $p\le q$, whereas it equals 1 when $p = q+1$. Thus for instance it is 1 for Gauss’ hypergeometric function $F(\alpha _1,\alpha _2;\beta _1;z)$. For later reference we record the following facts that follow easily from Definition 5.2:

$$\begin{aligned} _p F_q(0,\alpha _2,\ldots ,\alpha _p;\beta _1,\ldots ,\beta _q;z) = 1, \end{aligned}$$

(5.12)

and (see also p. 275 in [12])

$$\begin{aligned} F(\alpha ,\beta ;\beta ;-z) =\ _1F_0(\alpha ;-z) = (1+z)^{-\alpha }. \end{aligned}$$

(5.13)

The following result plays a key role in this work. The interested reader can find it in formula 2.12.4.16 on p. 178 in [13].

Lemma 5.3

Let $c>0$, $\Re \beta >0$, $\Re (\alpha +2\beta )<\frac{7}{2}$. Then

$$\begin{aligned}&\int _1^\infty \rho ^{\alpha -1} (\rho ^2 - 1)^{\beta -1} J_\nu (c \rho ) d\rho \nonumber \\&\quad = \frac{c^\nu \ \Gamma (\beta ) \Gamma \left( 1-\beta -\frac{\alpha +\nu }{2}\right) }{2^{\nu +1} \ \Gamma (\nu +1) \Gamma \left( 1-\frac{\alpha +\nu }{2}\right) } \ _1F_2\left( \frac{\alpha +\nu }{2}; \nu +1,\frac{\alpha +\nu }{2} + \beta ; - \left( \frac{c}{2}\right) ^2\right) \nonumber \\&\qquad + \frac{2^{\alpha +2\beta -3} c^{2-\alpha -2\beta } \Gamma \left( \beta + \frac{\alpha +\nu }{2} -1\right) }{\Gamma \left( 2-\beta +\frac{\nu -\alpha }{2}\right) } \nonumber \\&\quad _1F_2\left( 1-\beta ; 2-\beta - \frac{\alpha +\nu }{2},2-\beta + \frac{\nu -\alpha }{2}; - \left( \frac{c}{2}\right) ^2\right) . \end{aligned}$$

(5.14)

In order to fully exploit Lemma 5.3, we next derive a useful representation formula of a certain hypergeometric function $_1 F_2$ in terms of an integral involving a Bessel function. We stress that, although for convenience we have used the same letters, the parameters $\alpha , \nu$ in the statement of the next result are not the same as those in Lemma 5.3.

Lemma 5.4

Let $a, c>0$ and $\Re (\alpha +\nu )>0$. Then

$$\begin{aligned}{} & {} \ _1F_2\left( \frac{\alpha +\nu }{2};\frac{\alpha +\nu }{2}+1,\nu +1; - \left( \frac{ac}{2}\right) ^2\right) \\{} & {} \quad = \frac{(\alpha +\nu ) 2^{\nu }\Gamma (\nu +1)}{a^{\alpha + \nu } c^\nu }\int _0^a t^{\alpha -1} J_\nu (ct) \textrm{d}t. \end{aligned}$$

Proof

Performing the change of variable $\tau = ct$ and using (5.2) we find

$$\begin{aligned} \int _0^a t^{\alpha -1} J_\nu (ct) \textrm{d}t&= c^{-\alpha }\int _0^{ac} \tau ^{\alpha -1} J_\nu (\tau ) d\tau \\&= 2^\alpha c^{-\alpha } \sum _{k=0}^\infty \frac{(-1)^k}{\Gamma (\nu +k+1) k!} \int _0^{ac} \left( \frac{\tau }{2}\right) ^{2(k+\frac{\alpha + \nu }{2})} \frac{d\tau }{\tau }. \end{aligned}$$

The change of variable $z = \left( \frac{\tau }{2}\right) ^2$, for which $\frac{dz}{z} = 2 \frac{d\tau }{\tau }$, gives

$$\begin{aligned}&\int _0^a t^{\alpha -1} J_\nu (ct) \textrm{d}t = 2^{\alpha -1} c^{-\alpha } \sum _{k=0}^\infty \frac{(-1)^k}{\Gamma (\nu +k+1) k!} \int _0^{\left( \frac{ac}{2}\right) ^2} z^{k+\frac{\alpha + \nu }{2}}\frac{dz}{z}\\&\quad = 2^{\alpha -1} c^{-\alpha } \left( \frac{ac}{2}\right) ^{\alpha + \nu } \sum _{k=0}^\infty \frac{(-1)^k}{\left( \frac{\alpha + \nu }{2} + k\right) \Gamma (\nu +k+1) k!} \left( \frac{ac}{2}\right) ^{2k}\\&\quad = \frac{a^{\alpha + \nu }c^\nu }{2^{\nu +1}} \sum _{k=0}^\infty \frac{\left( \frac{\alpha +\nu }{2}\right) _k}{\left( \frac{\alpha +\nu }{2} + 1\right) _k (\nu +1)_k k!} \left( \frac{ac}{2}\right) ^{2k}\\&\qquad \frac{\Gamma \left( \frac{\alpha +\nu }{2}\right) }{\left( \frac{\alpha + \nu }{2} + k\right) \Gamma \left( \frac{\alpha +\nu }{2}+k\right) }\frac{ \Gamma \left( \frac{\alpha +\nu }{2}+k+1\right) }{\Gamma \left( \frac{\alpha +\nu }{2}+1\right) } \frac{\Gamma (\nu +k+1)}{\Gamma (\nu +1)\Gamma (\nu +k+1)}\\&\quad = \frac{a^{\alpha + \nu } c^\nu }{(\alpha +\nu ) 2^{\nu }\Gamma (\nu +1)}\ _1F_2\left( \frac{\alpha +\nu }{2};\frac{\alpha +\nu }{2}+1,\nu +1; - \left( \frac{ac}{2}\right) ^2\right) , \end{aligned}$$

where the details of the last equality are left to the reader. $\square$