Sharp Multiplier Theorem for Multidimensional Bessel Operators

Abstract

Consider the multidimensional Bessel operator

$$\begin{aligned} B f(x) = -\sum _{j=1}^N \left( \partial _j^2 f(x) +\frac{\alpha _j}{x_j} \partial _j f(x) \right) , \quad x\in (0,\infty )^N. \end{aligned}$$

Let $d = \sum _{j=1}^N \max (1,\alpha _j+1)$ be the dimension of the space $(0,\infty )^N$ equipped with the measure $x_1^{\alpha _1}\ldots x_N^{\alpha _N} dx_1\ldots dx_N$. In the general case $\alpha _1,\ldots ,\alpha _N >-1$ we prove multiplier theorems for spectral multipliers m(B) on $L^{1,\infty }$ and the Hardy space $H^1$. We assume that m satisfies the classical Hörmander condition

$$\begin{aligned} \sup _{t>0} \left\| \eta (\cdot ) m(t\cdot ) \right\| _{W^{2,\beta }(\mathbb {R})}<\infty \end{aligned}$$

with $\beta > d/2$. Furthermore, we investigate imaginary powers $B^{ib}$, $b\in \mathbb {R}$, and prove some lower estimates on $L^{1,\infty }$ and $L^p$, $1<p<2$. As a consequence, we deduce that our multiplier theorem is sharp.

1 Introduction and Main Results

1.1 The Bessel Operator

Let $N\in \mathbb {N}$ and $\alpha = (\alpha _1, \ldots , \alpha _N)$, where $\alpha _j > -1$ for $j=1,\ldots ,N$. Consider the space $X= (0,\infty )^N$ equipped with the Euclidean metric and the measure $d\nu (x) = x^\alpha dx = x_1^{\alpha _1}\ldots x_N^{\alpha _N}\, dx_1 \ldots dx_N$. It is well-known that X satisfies the doubling property, i.e.

$$\begin{aligned} {\nu (B(x,2r)) \le C \nu (B(x,r)), \qquad x\in X, \, r>0,} \end{aligned}$$

where $B(x,r) = \left\{ y\in X \ : \ |x-y|<r \right\}$ . In other words, there exist $d, C_d>0$ such that

$$\begin{aligned} \nu (B(x,\gamma r)) \le C_d (1+\gamma )^d \nu (B(x,r)), \qquad x\in X, \, r, \gamma >0. \end{aligned}$$

(D)

We choose the constant d (dimension of X) as small as possible. In this case

$$\begin{aligned} { d = \sum _{j=1}^N \max (1,\alpha _j+1).} \end{aligned}$$

(1.1)

The multidimensional Bessel operator is given by $B = B_1 + \cdots + B_N$, where

$$\begin{aligned} B_j f(x) = - \partial _j^2 f(x) - \frac{\alpha _j}{x_j} \partial _j f(x), \qquad x\in X. \end{aligned}$$

The operator B, initially defined on, say, $(C_c^2((0,\infty )))^n$, extends to a self-adjoint operator on $L^2(X)$. Slightly abusing notation, we shall denote this extension by the same symbol B. For a precise definition of B we refer the reader to e.g.[6, Sec. 2] (see also [32]). Also, B is the infinitesimal generator of the Bessel semigroup $\mathbf {T}_t f(x) = \int _X T_t(x,y) f(y)\, d\nu (y)$, where $T_t(x,y) = T_t^{[1]}(x_1,y_1) \cdot \ldots \cdot T_t^{[N]}(x_N,y_N)$ and

$$\begin{aligned} T_t^{[j]}(x_j,y_j)= & {} \frac{1}{2t} (x_jy_j)^{-(\alpha _j-1)/2} I_{(\alpha _j-1)/2}\left( \frac{x_jy_j}{2t}\right) \nonumber \\&\quad \exp \left( -\frac{x_j^2+y_j^2}{4t}\right) , \quad x_j,y_j, t>0. \end{aligned}$$

(1.2)

Here $I_\tau (x) = \sum _{m=0}^\infty \frac{1}{m!\Gamma (m+\tau +1)} \left( \frac{x}{2}\right) ^{2m+\tau }$ is the modified Bessel function of the first kind. The kernel $T_t(x,y)$ satisfies the upper and lower gaussian bounds, i.e. there exist constants $c_1,c_2, C_1, C_2 > 0$, such that

$$\begin{aligned}&C_1 \nu (B(x,\sqrt{t}))^{-1} \exp \left( - \frac{|x-y|^2}{c_1 t}\right) \le T_t(x,y) \\&\quad \le C_2 \nu (B(x,\sqrt{t}))^{-1} \exp \left( - \frac{|x-y|^2}{c_2 t}\right) . \end{aligned}$$

(G)

This fact is well known and follows from the asymptotics for $\nu (B(x,\sqrt{t}))$ and $I_\tau$. For details see e.g. [20, Lem. 4.2].

Since B is self-adjoint and nonnegative, for a Borel function $m: (0,\infty ) \rightarrow \mathbb {C}$ the spectral theorem defines the operator

$$\begin{aligned} m(B) = \int _0^\infty m(\lambda ) \, dE_B(\lambda ), \end{aligned}$$

where $E_B$ is the spectral resolution of B.

1.2 Multiplier Theorems for B

Multiplier theorems for B and other operators are one of the main topics in harmonic analysis. Many authors investigated assumptions on m that guarantee boundedness of m(B) on various function spaces, such as $L^p(X)$, $H^p(X)$, $L^{p,q}(X)$ and others.

For example, in [23] the authors proved weak type (1,1) estimates on m(B) assuming $N=1$, $\alpha >0$ and

$$\begin{aligned} \left( \int _{R/2}^R |m^{(s)}(\lambda )|^2 d\nu (\lambda ) \right) ^{1/2} \le C R^{(\alpha +1)/2-s}, \quad R>0, \end{aligned}$$

where $s=0,\ldots ,K$ and K is the least even integer greater than $(\alpha +1)/2 = d/2$ (see also [27]). In [18], assuming still $N=1$ and $\alpha >0$, it is proved that if

$$\begin{aligned} \sup _{t>0} \left\| \eta (\cdot )m(t \cdot ) \right\| _{W^{2,\beta }(\mathbb {R})} < \infty \end{aligned}$$

(S)

with some $\beta > d/2$, then m(B) is bounded on the Hardy space $H^1(B)$ related to B. Here and thereafter $W^{2,\beta }(\mathbb {R})$ is the $L^2$-Sobolev space on $\mathbb {R}$ and $\eta$ is a fixed nonnegative smooth cut-off function such that $\mathrm {supp} \, \eta \subseteq (2^{-1},2)$.

In the multidimensional case $N\ge 1$ in [5] the authors prove weak type (1, 1) estimates for m(B), where m is of Laplace transform type, i.e. there exists $\phi \in L^\infty (0,\infty )$, such that

$$\begin{aligned} m(x) = |x|^2 \int _0^\infty e^{-t|x|^2} \phi (t) \, dt, \quad x\in (0,\infty )^N. \end{aligned}$$

Notice, that if m is of Laplace transform type, then m is radial and (as a function on $(0,\infty )$) satisfies (S) with any $\beta >0$. Another multidimensional result can be found in [20], where it is proved that m(B) is weak type (1, 1) and bounded on the Hardy space $H^1(X)$ provided that $\alpha _j>1$ for $j=1,\ldots ,N$ and m satisfies (S) with $\beta >d/2$. See also e.g. [21, 22, 38] for other multiplier results for the Bessel operator.

Our first main goal is to obtain multiplier theorem for B in the most general case $N\ge 1$ and $\alpha _j >-1$, $j=1,\ldots ,N$. Let us notice that many of the results before assumed that $\alpha _j>0$ and the case $\alpha _j<0$ is more difficult and less known. One reason for that is the singularity at zero of the measure $x^{\alpha _j} dx_j$ when $\alpha _j<0$. Also, so-called “generalized translation” operators and convolution structure for B (see, e.g. [7, Sec. 2]), does not help when $\alpha _j<0$. This is strictly related to the fact, that the generalized eigenfunctions of B are no longer bounded if $\alpha _j<0$ for some j and, therefore, the generalized translation is not even bounded on $L^2$. Let us also notice that, we are interested in multiplier results that are sharp in the sense that we assume (S) with $\beta$ as small as possible. In this case this is expected to be $\beta >d/2$ (we shall discuss this in Sect. 1.3 below).

To state the multiplier result let us recall that the weak $L^1$ space is given by the semi-norm

$$\begin{aligned} \left\| f \right\| _{L^{1,\infty }(X)} = \sup _{\lambda>0} \lambda \nu \left\{ x\in X \ : \ \left| f(x) \right| >\lambda \right\} , \end{aligned}$$

and the Hardy space $H^1(B)$ related to B can be defined by the norm

$$\begin{aligned} \left\| f \right\| _{H^1(B)} = \left\| \sup _{t>0} \left| \mathbf {T}_t f \right| \right\| _{L^1(X)}. \end{aligned}$$

In the case $N=1$ the space $H^1(B)$ was studied in [7], where $H^1(B)$ was characterized by means of atomic decompositions and the Riesz transforms. In the general case $N\ge 1$ and $\alpha _j>-1$, $j=1,\ldots ,N$ the atomic characterization of $H^1(B)$ can be found in [19] (see also [16, 20]). We shall recall this characterization in Sect. 2.3 below.

Theorem A

Let $N\ge 1$ and $\alpha _j >-1$ for $j=1,\ldots ,N$. Assume that $m:(0,\infty ) \rightarrow \mathbb {C}$ satisfies (S) with $\beta >d/2$, see (1.1). Then:

• 1. m(B) is bounded from $L^{1}(X)$ to $L^{1,\infty }(X)$,

• 2. m(B) is bounded from $H^{1}(B)$ to $H^{1}(B)$,

• 3. m(B) is bounded from $L^{p}(X)$ to $L^{p}(X)$, $1<p<\infty$.

Part 1. of Theorem A will be proved by using results of [36]. More precisely, we shall check the assumptions of [36, Theorem 3.1]. The proof of 2. will be given in Sect. 2. In fact, in the proof we shall only use general properties of B, such as e.g. (D), (G), and $P_q$ below. Thus, the multiplier result in Sect. 2 will be formulated in a more general context. This section can be read independently of the rest of the paper and we shall use different notation. As usual, 3. is a consequence of either 1. or 2. by duality and interpolation, see e.g. [4].

1.3 Imaginary Powers of B

Another goal of this paper is to study the imaginary powers $B^{ib}$, $b\in \mathbb {R}$, of the Bessel operator and establish lower bounds of these operators on some function spaces. We shall concentrate our attention on the dependence of the lower estimates on b for large b. This is related with sharpness of multiplier theorems and may be of independent interest. To state these estimates let us restrict ourselves to the one-dimensional case $N=1$ ($X=(0,\infty )$, $d\nu (x) = x^\alpha \, dx$, $\alpha >-1$). Motivated by the identity

$$\begin{aligned} {B}^{ib} = \Gamma (-ib)^{-1} \int _0^\infty t^{-ib} e^{-t{B}} \frac{dt}{t}\end{aligned}$$

(1.3)

let us define for $x\ne y$ the integral kernel

$$\begin{aligned} K_b(x,y) = \Gamma (-ib)^{-1} \int _0^\infty t^{-ib} {T}_t(x,y) \frac{dt}{t}. \end{aligned}$$

(1.4)

Notice, that the integral in (1.3) is not absolutely convergent, thus we have to explain how the kernel $K_b(x,y)$ is related to the operators $B^{ib}$. Indeed, in Sect. 3.3 we shall prove that for $f\in L^\infty (X)$ with compact support we have

$$\begin{aligned} B^{ib} f (x) = \int _X K_b(x,y) f(y) \, d\nu (y), \qquad x\notin \mathrm {supp} f \end{aligned}$$

(1.5)

One of our goals is to provide lower estimates for $B^{ib}$.

Theorem B

Assume that $\alpha >-1$. Then there exist a constant $C>0$ and a function f such that $\left\| f \right\| _{L^1(X)}=1$ and for |b| large enough we have

$$\begin{aligned} \left\| B^{ib}f \right\| _{L^{1,\infty }(X)} \ge C|b|^{d/2} . \end{aligned}$$

Theorem C

Assume that $\alpha >0$ and $p\in (1,2)$. Then there exist $C_p>0$ and f such that $\left\| f \right\| _{L^p(X)}=1$ and for |b| large enough we have

$$\begin{aligned} \left\| B^{ib} f \right\| _{ L^p(X)} \ge C_p |b|^{\frac{d}{2}\frac{(2-p)}{p}}. \end{aligned}$$

The proofs of Theorems B and C are presented in Sect. 3.3. To prove Theorem B we shall carefully analyze the kernels $K_b(x,y)$. More precisely, we prove the following lemma.

Lemma 1.1

Assume that $\alpha >-1$ and $b\in \mathbb {R}$. Then

$$\begin{aligned} \begin{aligned} K_b(x,y) =&\,c_1(b)\left( {x^2+y^2} \right) ^{-ib-(\alpha +1)/2}\\&+ c_2(b) (xy)^{-\alpha /2} |x-y|^{-2bi-1}\chi _{\{y/2<x<2y\}}(x,y)\\&+ c_3(b) R_b(x,y), \end{aligned} \end{aligned}$$

(1.6)

where

$$\begin{aligned} c_1(b)= & {} \frac{2^{2ib+1}}{\Gamma \left( (\alpha +1)/{4} \right) } \frac{\Gamma \left( ib + (\alpha +1)/2 \right) }{\Gamma (-ib)}, \quad \\ c_2(b)= & {} \frac{ 2^{2ib}}{\sqrt{\pi }} \frac{\Gamma \left( ib+1/2 \right) }{\Gamma \left( -ib \right) }, \quad c_3(b) = \Gamma (-ib)^{-1}. \end{aligned}$$

Moreover, there exists $C>0$ that does not depend on b, such that

$$\begin{aligned} |R_b(x,y)| \le C xy(x+y)^{-\alpha -3}. \end{aligned}$$

Notice that the kernel $R_b(x,y)$ is related to an operator that is bounded on every $L^p(X)$, $1\le p \le \infty$, uniformly in $b\in \mathbb {R}$. Thus we may think of $R_b(x,y)$ as of some kind of “error term”. However, for $|b| > 1$ the sizes of the constants are the following:

$$\begin{aligned} |c_1(b)| \simeq |b|^{(\alpha +1)/2}, \quad |c_2(b)| \simeq |b|^{1/2}, \quad |c_3(b)| \simeq |b|^{1/2} \exp \left( \frac{\pi |b|}{2} \right) , \end{aligned}$$

(1.7)

c.f. Lemma 4.1. Thus, $c_3(b)$ grows exponentially when $|b|\rightarrow \infty$, while the constants $c_1(b)$ and $c_2(b)$ are much smaller. It appears that the growth of the constant $c_3(b)$ will lead to a problem in deriving lower estimates for $B^{ib}$ (since our goal is to find the exact dependence on b). However, we can overcome this difficulty when analyzing weak (1, 1) norm as in Theorem B. The same trick seems not to work in other function spaces (such as $H^1(B)$, $L^p(X)$ and $L^{p,\infty }(X)$ with $p>1$), thus the proof of Theorem C is different and uses the integral representation of the Bessel function $I_\tau$ instead of Lemma 1.1.

As a corollary of Theorems B and C we obtain that Theorem A is sharp (at least for $N=1$) in the sense that d / 2 cannot be replaced by a smaller number. The argument is standard, but we shall present it now for the convenience of the reader. One can check that for $m_b(\lambda ) = \lambda ^{ib}$ we have

$$\begin{aligned} M_b :=\sup _{t>0} \left\| \eta (\cdot ) m_b(t\cdot ) \right\| _{W^{2,\beta }(\mathbb {R})} \le |b|^\beta . \end{aligned}$$

Also, Theorem A actually gives that $\left\| m_b(B)f \right\| _{L^{1,\infty }(X)} \le C M_b \left\| f \right\| _{L^1(X)},$ where C does not depend on b. Combining these estimates with Theorem B for |b| large enough we have

$$\begin{aligned} {|b|^{d/2}} \le C \left\| m_b(B) \right\| _{L^1(X) \rightarrow L^{1,\infty }(X)} \le C {|b|^{\beta }}. \end{aligned}$$

Therefore $\beta \ge d/2$. Actually, one expects that $\beta \ne d/2$, but this question is beyond the scope of this paper.

Similarly, the constant d / 2 cannot be improved for the Hardy spaces. If $\alpha <0$ then $d/2=1/2$ and (S) with $\beta <1/2$ would not even guarantee that m is bounded. On the other hand, for $\alpha >0$ if we could prove multiplier theorem on $H^1(B)$ with a constant lower than d / 2, then by interpolation we would have better upper bounds for $m_b(B)$ on $L^p(X)$ for $1<p<2$, which contradicts Theorem C by an argument similar to the one above.

1.4 Organization of the Paper and Notation

In Sect. 2 we state and prove a “sharp” multiplier theorem on Hardy spaces for self-adjoint operators on spaces of homogeneous type with certain assumptions (Theorem D). This is a slight generalization of Theorem A 2. in the spirit of [36, Theorem 3.1]. In Sect. 2 we shall use different notation, so that it can be read independently of the rest of the paper. In Sect. 3 we prove the results stated above. More precisely, first we check that B satisfies assumption $(P_2)$ (see Sect. 2 below) in the full generality $N\ge 1$, $\alpha _j >-1$ for $j=1,\ldots ,N$. Thus Theorem D can be applied for B. Then we prove Lemma 1.1 and Theorems B and C. We shall use standard notations, i.e. C and c denote positive constants that may change from line to line.

2 Sharp Multiplier Theorem on Hardy Spaces

2.1 Background and General Assumptions

In this section we consider a space Y with a metric $\rho$ and a nonnegative measure $\mu$. We shall assume that the triple $(Y,\rho , \mu )$ is a space of homogeneous type, i.e. there exists $C>0$ such that $\mu (B(x,2r)) \le C \mu (B(x,r)),$ for all $x\in Y$ and $r>0$, where $B(x,r) = \left\{ y\in Y \ : \ \rho (x,y)<r \right\}$, c.f. [14]. It is well-known that this implies the existence of $d,C_d>0$ such that

$$\begin{aligned} \mu (B(x,\gamma r)) \le C_d (1+\gamma )^d \mu (B(x,r)), \qquad x\in Y, \, \, r, \gamma > 0. \end{aligned}$$

(D)

As usual, we choose d as small as possible, even at the cost of enlarging $C_d$.

Let A denote a self-adjoint positive operator and let $E_A$ be its spectral measure, i.e. $A = \int _0^\infty \lambda \, dE_A(\lambda )$. Denote by $\mathbf {P}_t= \exp (-tA)$ the semigroup generated by A. Assume that there exists an integral kernel $P_t(x,y)$ such that $\mathbf {P}_t f(x) = \int _Y P_t(x,y) f(y) \, d\mu (y)$ and that satisfies the upper gaussian bounds, i.e. there exist $c_2, C_2 > 0$ such that

$$\begin{aligned} P_t(x,y) \le C_2 \mu (B(x,\sqrt{t}))^{-1} \exp \left( - \frac{\rho (x,y)^2}{c_2 t}\right) , \quad t>0, \, x,y\in Y. \end{aligned}$$

(UG)

2.2 Multiplier Theorems

By the spectral theorem, for a Borel function m on $(0,\infty )$, we have the operator

$$\begin{aligned} m(A) = \int _0^\infty m(\lambda ) \, dE_A(\lambda ). \end{aligned}$$

In the classical case $A=-\Delta$, $Y=\mathbb {R}^D$, the Hörmader multiplier theorem states that if m satisfies (S) with $\beta >D/2$, then $m(-\Delta )$ is weak type (1, 1) and bounded on $L^p(\mathbb {R}^D)$ for $1<p<\infty$. It is well-known that the constant D / 2 is sharp in the sense that it cannot be replaced by a smaller constant, see e.g. [34].

At this point let us recall one of many multiplier theorems on spaces of homogeneous type. Suppose Y and A are as in Sect. 2.1. Following [36] we introduce additional assumption. Suppose that there exists $C>0$ and $q\in [2,\infty ]$, such that for $R>0$ and every Borel function m on $\mathbb {R}$ satisfying $\mathrm {supp}\, m\subseteq [R/2, 2R]$ we have

Theorem 2.1

[36, Theorem 3.1] Assume that on a space of homogeneous type $(Y,\rho , \mu )$ there is a self-adjoint positive operator A that satisfies (UG). Moreover, assume that $P_q$ holds with some $q\in [2,\infty ]$ and m satisfies

with some $\beta >d/2$. Then m(A) is of weak type (1, 1) and bounded on $L^p(Y)$ for $p\in (1,\infty )$.

At this point let us make a few comments.

• 1. Assuming (UG) the operators m(A) appearing in $P_q$ always have integral kernels $K_{m(A)}(x,y)$, c.f. [36, Lemma 2.2].

• 2. For the Bessel operator we are interested in $S_q$ and $P_q$ for $q=2$ only, $(S) = (S_2)$. However, in Sect. 2 the results are stated and proved with an arbitrary $q\in [2,\infty ]$.

• 3. The assumption $P_q$ in some sense plays a role of Plancherel theorem in the proof of Theorem 2.1. It is a key to obtain the sharp range $\beta >d/2$. For example, if we would allow m to satisfy $S_q$ with $\beta >d/2+1/2$, then $P_q$ would be superfluous.

• 4. The assumption $P_q$ is written in [36] for m having support in [0, R] not in [R / 2, 2R]. However, a simple inspection of the proof shows that $P_q$ is needed only for m with $\mathrm {supp} \, m \subseteq [R/2,2R]$. This makes no difference for many operators. However, it matters e.g. when considering the Bessel operator with negative parameters $\alpha _j$.

• 5. Assumption $P_q$ in [36] is written for $m(\sqrt{A})$, but we use equivalent version with m(A) (therefore we replace $B(y,R^{-1})$ by $B(y,R^{-1/2})$).

One of the main goals of this paper is to establish a multiplier theorem on Hardy spaces. We shall use the definition of the Hardy space $H^1(A)$ associated with A by means of the maximal operator of the semigroup $\mathbf {P}_t$, namely

$$\begin{aligned} H^1(A) = \left\{ f\in L^1(Y) \ : \ \left\| f \right\| _{H^1(A)}:= \left\| \sup _{t>0} \left| \mathbf {P}_t f \right| \right\| _{L^1(Y)} < \infty \right\} . \end{aligned}$$

To state our result we shall assume additionally that $P_t(x,y)$ satisfies also the the lower Gaussian bounds, namely there exist $c_1, C_1 >0$, such that

$$\begin{aligned} P_t(x,y) \ge C_1 \mu (B(x,\sqrt{t}))^{-1} \exp \left( - \frac{\rho (x,y)^2}{c_1 t}\right) , \quad t>0, \, x,y\in Y, \end{aligned}$$

(LG)

and that the space $(Y,\rho ,\mu )$ satisfies the following assumption:

$$\begin{aligned} \text { for all } x\in Y \text { the function } r \mapsto \mu (B(x,r)) \text { is a bijection on } (0,\infty ). \end{aligned}$$

(Y)

Notice that (Y) implies that $\mu (Y)=\infty$ and that $\mu$ is non-atomic. Now we are ready to state the theorem.

Theorem D

Assume that $(Y,\rho ,\mu )$ is a space of homogeneous type, d is as in (D), and (Y) is satisfied. Suppose that there is a self-adjoint positive operator A such that (UG), (LG), and $P_q$ hold with some $q\in [2,\infty ]$. If m satisfies $S_q$ and $\beta >d/2$, then m(A) is bounded from $H^1(A)$ to $H^1(A)$, i.e. there exists $C>0$, such that

$$\begin{aligned} \left\| m(A) f \right\| _{H^1(A)} \le C \left\| f \right\| _{H^1(A)}. \end{aligned}$$

The history of multiplier theorems for spaces of homogeneous type is long and wide. The interested reader is referred to [2, 9,10,11, 13, 17, 20, 21, 24, 26, 31, 33, 35, 36] and references therein. Let us concentrate for a moment on the range of parameters $\beta$ in Theorem D. Obviously, in general, the range $\beta >d/2$ is optimal. However, it may happen that for some particular operators one may obtain multiplier results assuming that $\beta >\widetilde{d}/2$ with $\widetilde{d}<d$, see e.g. [29, 30, 33]. On the other hand, there are known families of operators for which the constant d / 2 cannot be lower. One of the methods to prove this is to derive lower estimates for $A^{ib}$ in terms of $b\in \mathbb {R}$, see [12, 30, 34, 36]. Lastly, let us mention that some multiplier results hold also in the non-doubling case, see e.g. [15].

Boundedness of operators on the Hardy space $H^1$ is a natural counterpart of weak type (1, 1) bound. For example, it is a good end point for the interpolation, see e.g. [4]. However, the Hardy spaces are strictly related to some cancellation conditions and it is usually more involving to study properties of operators on the Hardy space, than on $L^p$ or $L^{p,\infty }$ spaces. Let us also mention that boundedness from $H^1$ to $H^1$ obviously implies boundedness from $H^1$ to $L^1$, which is usually much easier to prove.

2.3 Hardy Spaces

The Hardy spaces on spaces of homogeneous type are studied extensively from the 60’s, see e.g. [14]. In particular, now we have many atomic decompositions for $H^p$ on various spaces and operators acting on this spaces. We refer the reader to e.g. [3, 7, 19, 25] and references therein.

In this subsection we recall some results on Hardy spaces related to A, assuming that (D), (UG), (LG) and (Y) are satisfied. For the proofs and more details we refer the reader to [19]. Firstly, there exists the unique (up to a multiplicative constant) A-harmonic function $\omega : Y \rightarrow \mathbb {R}$ such that

$$\begin{aligned} C^{-1} \le \omega (x) \le C {, \qquad x\in Y}. \end{aligned}$$

The function $\omega$ plays a special role in the analysis of A and $\mathbf {P}_t$. In particular we have the following Hölder-type estimate.

Theorem 2.2

Suppose that the semigroup $\mathbf {P}_t$ satisfies (UG), (LG). Then there exist positive constants $\gamma , c, C$, such that if $\rho (y,z)\le \sqrt{t}$, then

$$\begin{aligned} \left| \frac{P_t(x,y)}{\omega (y)} - \frac{P_t(x,z)}{\omega (z)} \right| \le C \mu (B(x,\sqrt{t}))^{-1}\left( \frac{\rho (y,z)}{\sqrt{t}}\right) ^\gamma \exp \left( - \frac{\rho (x,y)^2}{ct} \right) . \end{aligned}$$

Theorem 2.2 is quite well-known and follows from a general theory. For a short and independent proof see [19, Sec. 4].

Corollary 2.3

There exist $\gamma , C >0$ such that if $\rho (y,z)\le \sqrt{t}$, then

$$\begin{aligned} \int _Y \left| \frac{P_t(x,y)}{\omega (y)} - \frac{P_t(x,z)}{\omega (z)} \right| \, d\mu (x) \le C \left( \frac{\rho (y,z)}{\sqrt{t}}\right) ^\gamma . \end{aligned}$$

Using Theorem 2.2 the authors of [19] obtained the following atomic decomposition for the elements of $H^1(A)$. Let us call a function $a: Y \rightarrow \mathbb {C}$ an $(\mu ,\omega )$-atom, if there exists a ball B in Y, such that:

$$\begin{aligned} \mathrm {supp} \,a \subseteq B, \qquad \left\| a \right\| _\infty \le \mu (B)^{-1}, \qquad \int _B a(x) \omega (x) \, d\mu (x)=0. \end{aligned}$$

Theorem 2.4

[19, Theorem 1] There exists a constant $C>0$ such that for each $f\in H^1(A)$ there exist $\lambda _k\in \mathbb {C}$ and $(\mu ,\omega )$-atoms $a_k$ ($k\in \mathbb {N}$), such that

$$\begin{aligned} f(x) = \sum _{k\in \mathbb {N}} \lambda _k a_k(x), \quad \text {and} \quad C^{-1} \left\| f \right\| _{H^1(A)} \le \sum _{k\in \mathbb {N}} |\lambda _k| \le C \left\| f \right\| _{H^1(A)}. \end{aligned}$$

Let us start by recalling a few consequences of (D) and (UG).

Lemma 2.5

[36, Lemma 2.1] Suppose that (D) and (UG) hold. Then

$$\begin{aligned} \int _{ B(y,r)^c} |P_t(x,y)|^2 \, d\mu (x) \le C \mu (B(y,\sqrt{t}))^{-1} \exp \left( - \frac{r^2}{c_2t}\right) . \end{aligned}$$

In particular

$$\begin{aligned} \left\| P_t(x,\cdot ) \right\| ^2_{L^2(Y)} \le C \mu (B(x,\sqrt{t}))^{-1}. \end{aligned}$$

Lemma 2.6

[36, Lemma 4.1] For $\kappa \ge 0$ there exists a constant $C=C(\kappa )>0$ such that

$$\begin{aligned} \int _Y |P_{(1+i\tau )R^{-1}}(x,y)|^2 (1+ R^{1/2}\rho (x,y))^\kappa \, d\mu (x) \le C \mu \left( B\left( y,R^{-1/2}\right) \right) ^{-1} (1+|\tau |)^\kappa . \end{aligned}$$

Lemma 2.7

[36, Lemma 4.4] Suppose that (D) holds and $\delta > 0$. Then

$$\begin{aligned} \int _{B(y,r)^c} (1+R^{1/2}\rho (x,y))^{-d-2\delta } \, d\mu (x) \le C \mu \left( B\left( y,R^{-1/2}\right) \right) (1+rR^{1/2})^{-2\delta }. \end{aligned}$$

2.4 Key Kernel Estimates

This subsection is devoted to obtain key estimates needed for the proof ofTheorem D. We shall assume (temporarily) that m satisfies $\mathrm {supp}\, m \subseteq [R/2, 2R]$ with some $R>0$. Later we shall use a partition of unity for general m. Denote $m_R(\lambda ) = m(R\lambda )$, so that $\mathrm {supp}\, m_R \subseteq [2^{-1},2]$. Let us notice that below the letter $q\in [2,\infty ]$ is always the exponent related to $P_q$ and $S_q$. Moreover, all the spectral operators below admit related integral kernels, which can be seen by using an argument identical as in [36, Lem. 2.2]. Let us denote $\widetilde{m}_t(\lambda ) = \exp (-t\lambda ) m(\lambda )$ and let $M_t(x,y)$ be the kernel associated with $\widetilde{m}_t(A) = \mathbf {P}_t m(A)$.

Proposition 2.8

Assume that $\mathrm {supp} \, m \subseteq [R/2, 2R]$ and $m_R \in W^{q,\beta }(\mathbb {R})$ with $\beta > d/2$. Then, there exist $\delta , \gamma , C>0$ such that for $y,z\in Y$ and $r>0$ we have

$$\begin{aligned} { \int _{B(y,r)^c} \sup _{t>0} \, \left| M_t(x,y) \right| \, d\mu (x) \le C \left( 1 + rR^{1/2}\right) ^{-\delta } \left\| m_R \right\| _{W^{q,\beta }(\mathbb {R})},} \end{aligned}$$

(2.1)

and, for $\rho (y,z) < R^{-1/2}$,

$$\begin{aligned} { \int _{B(y,r)^c} \sup _{t>0}\left| \frac{M_t(x,y)}{\omega (y)} - \frac{M_t(x,z)}{\omega (z)} \right| \, d\mu (x) \le C \left( R^{1/2}\rho (y,z)\right) ^{\gamma } \left\| m_R \right\| _{W^{q,\beta }(\mathbb {R})}. }\nonumber \\ \end{aligned}$$

(2.2)

Let us start by showing the following lemma.

Lemma 2.9

For $\varepsilon >0,\kappa \ge 0$ there exists a constant $C=C(\kappa ,\varepsilon )$ such that

$$\begin{aligned}&\int _Y \sup _{t>0} |M_{t}(x,y)|^2 \left( 1+R^{1/2}\rho (x,y)\right) ^{\kappa } \, d\mu (x) \\&\quad \le C \mu \left( B\left( y,R^{-1/2}\right) \right) ^{-1} \left\| m_R \right\| ^2_{W^{q,\kappa /2+\varepsilon }(\mathbb {R})}. \end{aligned}$$

Proof

Fix a cut-off function $\psi \in C_c^{\infty }(4^{-1},4)$, such that $\psi \equiv 1$ on $[2^{-1}, 2]$. Set

$$\begin{aligned} n_{t,R}(\lambda ) = m_R(\lambda ) \underbrace{e^{-tR\lambda } e^\lambda \psi (\lambda )}_{\lambda _{t,R}(\lambda )}. \end{aligned}$$

By the Fourier inversion formula,

$$\begin{aligned} \widetilde{m}_{t}(A) = n_{t,R}(AR^{-1})e^{-AR^{-1}} = \frac{1}{2\pi } \int _\mathbb {R}\widehat{n}_{t,R}(\tau ) \exp \left( (i\tau -1)AR^{-1} \right) \, d\tau \end{aligned}$$

and

$$\begin{aligned} M_t(x,y) = \frac{1}{2\pi } \int _\mathbb {R}\widehat{n}_{t,R}(\tau ) P_{(1-i\tau )R^{-1}}(x,y) \, d\tau . \end{aligned}$$

(2.3)

Notice that $\mathrm {supp} \, \lambda _{t,R}^{(N)} \subseteq (4^{-1}, 4)$ for arbitrary $N\in \mathbb {N}$. By simple calculus we can find a constant $C_N$ such that

$$\begin{aligned} \begin{aligned} \sup _{R>0,t>0} \left| \widehat{\lambda }_{t,R}(\tau ) \right| \le C_N(1+|\tau |)^{-N}. \end{aligned} \end{aligned}$$

Since $\widehat{n}_{t,R} = \widehat{m}_R *\widehat{\lambda }_{t,R}$ and $(1+|\tau |) \le (1+|\theta |)(1+|\tau -\theta |)$, for $\kappa \ge 0$ and $\varepsilon >0$ we use the Cauchy-Schwarz inequality, getting

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}} \sup _{t>0} |\widehat{n}_{t,R}(\tau )| (1+|\tau |)^{\kappa /2} \, d\tau&\le \int _{\mathbb {R}} \int _{\mathbb {R}} \sup _{t>0} |\widehat{m}_R(\theta )||\widehat{\lambda }_{t,R}(\tau -\theta )| (1+|\tau |)^{\kappa /2} \, d\theta \,d\tau \\&\le \int _{\mathbb {R}} \int _{\mathbb {R}} \sup _{t>0} |\widehat{m}_R(\theta )||\widehat{\lambda }_{t,R}(\tau -\theta )|\\&\quad (1+|\theta |)^{\kappa /2} \, (1+|\tau -\theta |)^{\kappa /2} \,d\tau \, d\theta \\&\le C \int _{\mathbb {R}} |\widehat{m}_R(\theta )| (1+|\theta |)^{(\kappa +1)/2+\varepsilon } (1+|\theta |)^{-1/2-\varepsilon } \, d\theta \\&\le C \left\| m_R \right\| _{W^{2,(1+\kappa )/2+\varepsilon }(\mathbb {R})} \left( \int _{-\infty }^\infty (1+|\theta |)^{-1-\varepsilon } \, d\theta \right) ^{1/2} \\&\le C \left\| m_R \right\| _{W^{2,(1+\kappa )/2+\varepsilon }(\mathbb {R})}. \end{aligned} \end{aligned}$$

Hence, by (2.3), the Minkowski inequality, and Lemma 2.6 we obtain

$$\begin{aligned} \begin{aligned}&\left( \int _Y \sup _{t>0} |M_t(x,y)|^2 (1+R^{1/2}\rho (x,y))^\kappa \ d\mu (x) \right) ^{1/2} \\&\quad \le \int _\mathbb {R}\sup _{t>0} |\widehat{n}_{t,R}(\tau )| \left( \int _Y |P_{(1-i\tau )R^{-1}}(x,y)|^2 (1+R^{1/2}\rho (x,y))^{\kappa } \, d\mu (x) \right) ^{1/2} \, d\tau \\&\quad \le C \mu \left( B\left( y,R^{-1/2}\right) \right) ^{-1/2} \int _\mathbb {R}\sup _{t>0} |\widehat{n}_{t,R}(\tau )| (1+|\tau |)^{\kappa /2} \, d\tau \\&\quad \le C \mu \left( B\left( y,R^{-1/2}\right) \right) ^{-1/2} \left\| m_R \right\| _{W^{2,(1+\kappa )/2+\varepsilon }(\mathbb {R})}\\&\quad \le C \mu \left( B\left( y,R^{-1/2}\right) \right) ^{-1/2} \left\| m_R \right\| _{W^{q,(1+\kappa )/2+\varepsilon }(\mathbb {R})}. \end{aligned} \end{aligned}$$

(2.4)

In the last inequality we have used that $\mathrm {supp}\, m_R\subseteq [2^{-1},2]$ and $q\ge 2$.

Observe that (2.4) is exactly the estimate we look for, but the Sobolev parameter is higher by 1 / 2 than we want. To sharpen this estimate, we make use of known interpolation method. Notice, that $M_t(x,y) = \mathbf {P}_t(K_{m(A)}(\cdot ,y))(x)$. It is well-known that (UG) implies boundedness on $L^2(Y)$ of the maximal operator $\mathscr {M}f =\sup _{t>0} |\mathbf {P}_t f|$. A second estimate needed for an interpolation is the following

$$\begin{aligned} \begin{aligned} \left( \int _Y \sup _{t>0} |M_{t}(x,y)|^2 \, d\mu (x) \right) ^{1/2}&= \left\| \mathscr {M}K_{m(A)}(\cdot ,y) \right\| _{L^2(Y)} \\&\le C \left\| K_{m(A)}(\cdot ,y) \right\| _{L^2(Y)}\\&\le C \mu \left( B\left( y,R^{-1/2}\right) \right) ^{-1/2}\left\| m_R \right\| _{L^q(\mathbb {R})}. \end{aligned} \end{aligned}$$

(2.5)

In the last inequality we have used $P_q$. Now, Lemma 2.9 follows by interpolating (2.4) and (2.5), see e.g. proofs of [36, Lem. 4.3(a)] and [20, Lem. 2.2] for details. $\square$

Proof of (2.1)

By the Cauchy-Schwarz inequality and Lemmas 2.7 and 2.9,

$$\begin{aligned} \begin{aligned}&\int _{B(y,r)^c} \sup _{t>0} |M_{t}(x,y)| \, d\mu (x) \\&\quad \le \left( \int _Y \sup _{t>0} |M_{t}(x,y)|^2(1+R^{1/2}\rho (x,y))^{d+2\delta } \, d\mu (x) \right) ^{1/2}\\&\qquad \left( \int _{B(y,r)^c} (1+R^{1/2}\rho (x,y))^{-d-2\delta } \, d\mu (x) \right) ^{1/2}\\&\quad \le C \mu \left( B\left( y,R^{-1/2}\right) \right) ^{-1/2} \left\| m_R \right\| _{W^{q,d/2+\delta +\varepsilon }(\mathbb {R})}\mu \left( B\left( y,R^{-1/2}\right) \right) ^{1/2}(1+rR^{1/2})^{-\delta } \\&\quad \le C (1+rR^{1/2})^{-\delta } \left\| m_R \right\| _{W^{q,\beta }(\mathbb {R})}, \end{aligned} \end{aligned}$$

where $\delta , \varepsilon >0$ are such that $d/2+\delta +\varepsilon \le \beta$.

$\square$

Consider for a moment the operator $\mathbf {P}_tm(A) \exp (AR^{-1})$ and let $\widetilde{M}_{t,R}(x,y)$ be its kernel. By almost identical arguments as in the proofs of Lemma 2.9 and (2.1), we can show that for $\beta >d/2$ we also have

$$\begin{aligned} { \int _{B(y,r)^c} \sup _{t>0} |\widetilde{M}_{t,R}(x,y)| \, d\mu (x) \le C \left\| m_R \right\| _{W^{q,\beta }(\mathbb {R})}. } \end{aligned}$$

(2.6)

Proof of (2.2)

Notice, that $M_{t}(x,y) = \int _Y \widetilde{M}_{t,R}(x,u)P_{R^{-1}}(u,y) \, d\mu (u)$. For $\rho (y,z)<R^{-1/2}$, by Corollary 2.3 and (2.6),

$$\begin{aligned} \begin{aligned}&\int _{B(y,r)^c} \sup _{t>0} \left| \frac{M_t(x,y)}{\omega (y)} -\frac{M_t(x,z)}{\omega (z)} \right| \, d\mu (x) \\&\quad = \int _{B(y,r)^c} \sup _{t>0} \left| \int _Y \widetilde{M}_{t,R}(x,u)\left( \frac{P_{R^{-1}}(u,y)}{\omega (y)} - \frac{P_{R^{-1}}(u,z)}{\omega (z)} \right) \, d\mu (u) \right| \, d\mu (x) \\&\quad \le \int _Y \left| \frac{P_{R^{-1}}(u,y)}{\omega (y)} - \frac{P_{R^{-1}}(u,z)}{\omega (z)} \right| \int _{B(y,r)^c} \sup _{t>0} | \widetilde{M}_{t,R}(x,u)| \, d\mu (x) \, d\mu (u) \\&\quad \le C \left( R^{1/2}\rho (y,z)\right) ^\gamma \left\| m_R \right\| _{W^{q,\beta }(\mathbb {R})}. \end{aligned} \end{aligned}$$

$\square$

2.5 Proof of Theorem D

Theorem D follows from Proposition 2.8 by a quite standard argument. We present the details for completeness and convenience of the reader. As usual, by a continuity argument, in order to prove boundedness of the operator m(A) on $H^1(A)$ it is enough to show that there exists $C>0$ such that

$$\begin{aligned} \left\| m(A)a \right\| _{H^1(A)} = \left\| \mathscr {M}m(A)a \right\| _{L^1(Y)} \le C \end{aligned}$$

holds for every $(\mu ,\omega )$-atom a, see Theorem 2.4. Assume then that: $\mathrm {supp}\, a \subseteq B(y_0,r) = : B$, $\left\| a \right\| _\infty \le \mu (B)^{-1}$, and $\int a \, \omega d\mu = 0$. As always, the analysis on $2B = B(y_0,2r)$ follows by the Cauchy-Schwarz inequality and boundedness of $\mathscr {M}$ and m(A) on $L^2(Y)$. More precisely,

$$\begin{aligned} \begin{aligned} \left\| \mathscr {M}m(A) a \right\| _{L^1(2B)}&\le \mu (2B)^{1/2} \left\| \mathscr {M}m(A) a \right\| _{L^2(Y)} \\&\le C \mu (B)^{1/2} \left\| a (x) \right\| _{L^2(Y)} \le C. \end{aligned} \end{aligned}$$

Therefore, it is enough to prove that

$$\begin{aligned} \left\| \mathscr {M}m(A) a \right\| _{L^1((2B)^c)} \le C. \end{aligned}$$

(2.7)

Let $\eta \in C_c^\infty (2^{-1},2)$ be a fixed function such that $\sum _{j\in \mathbb {Z}}\eta (2^{-j}\lambda ) = 1$ for all $\lambda \in (0,\infty )$. By using this partition of unity, we decompose m as

$$\begin{aligned} m(\lambda ) = \sum _{j\in \mathbb {Z}} \eta (2^{-j}\lambda ) m(\lambda ) = \sum _{j\in \mathbb {Z}} m_j(\lambda ). \end{aligned}$$

Fix $N\in \mathbb {Z}$ such that $2^{-N} \le r^2 < 2^{-N+1}.$ Then

$$\begin{aligned} \begin{aligned} \left\| \mathscr {M}m(A)a \right\| _{L^1((2B)^c)}&\le \sum _{j\in \mathbb {Z}} \left\| \mathscr {M}m_j(A)a \right\| _{L^1((2B)^c)}= \sum _{j\ge N}\cdots + \sum _{j<N}\cdots = S_1 + S_2. \end{aligned} \end{aligned}$$

Denote $m_{j,t}(\lambda ) = \exp (-t\lambda ) m_j(\lambda )$ and let $M_{j,t}(x,y)$ be the kernel of $m_{j,t}(A) = \mathbf {P}_t m_j(A)$. Obviously, $\mathrm {supp}\, m_{j,t} \subseteq [2^{j-1}, 2^{j+1}]$ and applying (2.1) we obtain that

$$\begin{aligned} \begin{aligned} S_1&\le \sum _{j\ge N} \int _{(2B)^c} \int _{B} \sup _{t>0} |M_{j,t}(x,y)||a(y)| d\mu (y) d\mu (x) \\&\le \sum _{j\ge N} \int _{B} |a(y)| \int _{B^c} \sup _{t>0} |M_{j,t}(x,y)| d\mu (x) d\mu (y)\\&\le C \left\| a \right\| _{L^1(Y)} \sum _{j\ge N} (1+2^{j/2}r)^{-\delta } \left\| \eta (\cdot ) m(2^j \cdot ) \right\| _{W^{q,\beta }(\mathbb {R})} \\&\le C \sup _{t>0} \left\| \eta (\cdot ) m(t \cdot ) \right\| _{W^{q,\beta }(\mathbb {R})} \le C. \end{aligned} \end{aligned}$$

If $y\in B$ and $j < N$, then $\rho (y,y_0)< r < 2^{-j/2}$ and we can apply (2.2) for the kernel $M_{j,t}$ with $R=2^j$. Using the cancellation condition of a,

$$\begin{aligned} \begin{aligned} S_2&\le \sum _{j< N} \int _{(2B)^c} \sup _{t>0} \left| \int _{B} M_{j,t}(x,y) a(y) d\mu (y) \right| d\mu (x) \\&= \sum _{j< N} \int _{(2B)^c} \sup _{t>0} \left| \int _{B} \left( \frac{M_{j,t}(x,y)}{\omega (y)}- \frac{M_{j,t}(x,y_0)}{\omega (y_0)}\right) a(y) \omega (y) d\mu (y) \right| d\mu (x)\\&\le \sum _{j< N} \int _{B} |a(y)| \int _{B(y,r)^c} \sup _{t>0} \left| \frac{M_{j,t}(x,y)}{\omega (y)}- \frac{M_{j,t}(x,y_0)}{\omega (y_0)} \right| d\mu (x) \, \omega (y) d\mu (y)\\&\le C \sum _{j< N} 2^{\frac{j\gamma }{2}} \int _{B} |a(y)| \rho (y,y_0)^{\gamma } d\mu (y) \left\| \eta (\cdot ) m(2^j \cdot ) \right\| _{W^{q,\beta }(\mathbb {R})} \\&\le C \sup _{t>0} \left\| \eta (\cdot ) m(t \cdot ) \right\| _{W^{q,\beta }(\mathbb {R})} r^\gamma \sum _{j< N} 2^{\frac{j\gamma }{2}} \le C. \end{aligned} \end{aligned}$$

This finishes the proof of (2.7) and Theorem D.

3 The Multidimensional Bessel Operator

In this Section we turn back to the analysis related to B and prove the results stated in Sect. 1.

3.1 The Hankel Transform

Recall that $N\in \mathbb {N}$ and $\alpha _j >-1$ for $j=1,\ldots ,N$. For $x,\xi \in X = (0,\infty )^N$ denote $\varphi _{\alpha }(x\xi ) = \varphi _1(x_1\xi _1)\cdot \ldots \cdot \varphi _N(x_N\xi _N)$, where

$$\begin{aligned} \varphi _j(z) = 2^{(\alpha _j-1)/2}\Gamma \left( (\alpha _j+1)/2 \right) z^{-(\alpha _j-1)/2}J_{(\alpha _j-1)/2}(z), \quad z>0. \end{aligned}$$

Here $J_\tau$ denotes the Bessel function of the first kind. By the asymptotics of $J_\tau$ one has

$$\begin{aligned} \left| \varphi _j(z) \right| \le C (1+z)^{-\alpha _j/2}, \quad z>0. \end{aligned}$$

(3.1)

The Hankel transform is defined by

$$\begin{aligned} H_\alpha f(\xi ) = \int _X f(x)\varphi _{\alpha }(x\xi ) d\nu (x), \quad \xi \in X, \end{aligned}$$

(3.2)

As we have already mentioned, $\varphi _j \in L^\infty$ if and only if $\alpha _j \ge 0$. Nevertheless, it is known that $H_\alpha$ always extends uniquely to an isometric isomorphism on $L^2(X)$, see [5] and [8, Lem. 2.7]. The multipliers m(B) and $H_\alpha$ are related in the same way, as $m(-\Delta )$ and the Fourier transform on $\mathbb {R}^D$. In particular, if

$$\begin{aligned} n(\lambda ) = m(|\lambda |^2), \qquad \lambda \in X, \end{aligned}$$

then m is radial and

$$\begin{aligned} m(B) f = H_\alpha (n \cdot H_\alpha ). \end{aligned}$$

(3.3)

3.2 $(P_2)$ for Multidimensional Bessel Operator

Let us first recall that $\mathbf {T}_t$ satisfies (G) and, obviously, X satisfies (Y). Therefore, Theorem A follows from Theorems 2.1 and D provided that $P_q$ holds with $q=2$, which we now prove. The case $N=1$ follows by similar and simpler argument, thus we shall concentrate on $N\ge 2$. Let $k\in \left\{ 1,\ldots ,N-1 \right\}$ and $c_j<2^{-N}$ for $j=1,\ldots ,k$. Define the sets

$$\begin{aligned} S_{c_1,\ldots ,c_k} = \left\{ x\in X \ : \ 1/2<|x|^2<2 \text { and } x_{j}<c_j \text { for } j=1,\ldots ,k \right\} . \end{aligned}$$

Lemma 3.1

Suppose that $\mathrm {supp}\,m\subset [R/2,2R]$, $N\ge 2$, $k\le N-1$, and $c_j <2^{-N}$ for $j=1,2,\ldots ,k$. Then there exists $C>0$ such that

$$\begin{aligned} \int _{S_{c_1,\ldots ,c_k}} \left| m\left( R|x|^2\right) \right| ^2 x_{1}^{\alpha _{1}} \ldots x_k^{\alpha _k} \, dx_1\ldots dx_N \le C c_1^{\alpha _{1}+1} \ldots c_k^{\alpha _{k}+1} \left\| m(R\cdot ) \right\| _{L^2(\mathbb {R})}^2. \end{aligned}$$

Proof

Introduce the spherical coordinates $(r,\theta _1, \ldots , \theta _{N-1})$ on $\mathbb {R}^N$, namely

$$\begin{aligned}{ {\left\{ \begin{array}{ll} x_1 = r\sin (\theta _1), &{} \\ x_i = r \sin (\theta _i) \Pi _{j=1}^{i-1}\cos (\theta _j), \qquad \ \ \ \text { for } i=2,3,\ldots ,N-1, \\ x_N = r \Pi _{j=1}^{N-1}\cos (\theta _j)\\ dx_1\,\ldots \,dx_N = r^{N-1} \prod _{j=1}^{N-2} \cos ^{N-1-j}(\theta _j) dr\, d\theta _1 \ldots \,d\theta _{N-1}. \end{array}\right. } } \end{aligned}$$

Since $x\in X$, then $\theta _j \in (0,\pi /2)$ for $j=1,\ldots ,N-1$. We claim that if $x\in S_{c_1,\ldots ,c_k}$, then

$$\begin{aligned} { \sin (\theta _j) < 2^{-N/2} \le 2^{-1/2}} \end{aligned}$$

(3.4)

for $j=1,\ldots ,k$. Obviously, if $\sin (\beta ) < 2^{-1/2}$ for $\beta \in (0,\pi /2)$, then $(\cos \beta )^{-1} < 2^{1/2}$. Observe that $2^{-1/2}<r<2^{1/2}$, since $\mathrm {supp}\, m \subseteq [R/2, 2R]$. Therefore, (3.4) follows easily by induction, i.e. for $i=1,\ldots ,k$,

$$\begin{aligned}{ \sin \theta _i = x_i r^{-1} (\cos \theta _1)^{-1} \ldots (\cos \theta _{i-1})^{-1}\le 2^{-N} 2^{i/2} \le 2^{-N/2}. } \end{aligned}$$

Denote $S=S_{c_1,\ldots ,c_k}$. As a consequence of (3.4) we have that $\sin \theta _i \simeq \theta _i$ and $\cos \theta _i \simeq C$ for $i=1,\ldots ,k$. Using this, $x_i^{\alpha _i} \simeq r^{\alpha _i} \theta _i^{\alpha _i}$ for $i=1,\ldots ,k$ and

$$\begin{aligned} \begin{aligned}&\int _{S} \left| m(R|x|^2) \right| ^2 x_{1}^{\alpha _{1}} \ldots x_k^{\alpha _k} \, dx_1\ldots dx_N \\&\quad \le C \int _S \left| m(Rr^2) \right| ^2 r^{N-1+\alpha _1+\cdots +\alpha _k} \theta _1^{\alpha _1}\ldots \theta _k^{\alpha _k} dr d\theta _{1} \ldots d\theta _{N-1} \\&\quad \le C\int _0^{c c_1} \theta _1^{\alpha _1} \,d\theta _1 \cdot \ldots \cdot \int _0^{c c_k} \theta _k^{\alpha _k} \,d\theta _k \cdot \int _{1/2<r^2<2} \left| m\left( Rr^2\right) \right| ^2 dr \\&\quad \le C c_1^{\alpha _1+1} \cdot \ldots \cdot c_k^{\alpha _k+1} \cdot \left\| m(R\cdot ) \right\| _{L^2(\mathbb {R})}^2. \end{aligned} \end{aligned}$$

$\square$

Proposition 3.2

Assume that $N\in \mathbb {N}$ and $\alpha _j>-1$ for $j=1,..,N$. Then $P_q$ holds for B with $q=2$.

Proof

In the proof we consider only the case $N\ge 2$. Let $q=2$ and suppose that m is supported in [R / 2, 2R] for some $R>0$. Notice that by (3.3) and (3.2) we have

$$\begin{aligned} \begin{aligned} m(B)f(x)&= \int _{X} f(y) \int _{X} n(\xi ) \varphi _{\alpha }(y\xi ) \varphi _{\alpha }(x\xi ) d\nu (\xi ) \, d\nu (y)\\&= \int _{X} f(y) H_\alpha \left( n(\cdot ) \varphi _{\alpha }(y\cdot )\right) (x) \, d\nu (y) \end{aligned} \end{aligned}$$

and the kernel associated with m(B) has the form

$$\begin{aligned} {K_{m(B)}(x,y) = H_\alpha \left( n(\cdot ) \varphi _{\alpha }(y\cdot )\right) (x).} \end{aligned}$$

Therefore, by the Plancherel identity for $H_\alpha$, $(P_2)$ is equivalent to

$$\begin{aligned} { \int _{R/2<|x|^2<2R}\left| m(|x|^2) \right| ^2\left| \varphi _{\alpha }(xy) \right| ^2 d\nu (x) \le C \nu \left( B\left( y,R^{-1/2}\right) \right) ^{-1} \left\| m(R\cdot ) \right\| ^2_{L^2(\mathbb {R})}. } \end{aligned}$$

(3.5)

For each $i=1,\ldots ,N$ we consider four cases:

C1. :

$x_i < 2^{-N}$,   $\sqrt{R}y_i > 2^N$,    $\sqrt{R}y_ix_i < 1$,

C2. :

$x_i < 2^{-N}$,   $\sqrt{R}y_i \le 2^N$,

C3. :

$x_i < 2^{-N}$,   $\sqrt{R}y_{i} > 2^N$,    $\sqrt{R}y_ix_i \ge 1$,

C4. :

$2^{-N} \le x_i \le \sqrt{2}$.

Divide the set $\left\{ x\in X \ : \ 1/2<|x|^2<2 \right\}$ into several disjoint regions using the cases above. Without loss of generality we may consider the set S of points $x\in X$ such that:

• $x_i$ satisfies C1. for $i=1,\ldots , k_1$,

• $x_i$ satisfies C2. for $i=k_1+1,\ldots , k_2$,

• $x_i$ satisfies C3. for $i=k_2+1,\ldots , k_3$,

• $x_i$ satisfies C4. for $i=k_3+1,\ldots , N$,

where $0\le k_1\le k_2 \le k_3 < N$. The fact that $k_3< N$ is implied by $|x|^2>1/2$. Notice that it may happen that S is empty. Recall that $\nu (B(y,r)) \simeq \Pi _{j=1}^N \nu _j(B(y_j,r))$, where $d\nu _j(x_j) = x_j^{\alpha _j} dx_j$ is the one-dimensional measure, and

$$\begin{aligned}{ \nu _j\left( B\left( y_j, R^{-1/2}\right) \right) ^{-1} \simeq R^{(\alpha _j+1)/2} \left( 1+\sqrt{R}y_j \right) ^{-\alpha _j}. } \end{aligned}$$

Denote $d_{gl} = N+\alpha _1 +\cdots +\alpha _N$. Using (3.1) and Lemma 3.1 with $k=k_2$, we have

$$\begin{aligned} \begin{aligned}&\int _{R/2<|x|^2<2R}\left| m\left( |x|^2\right) \right| ^2\left| \varphi _{\alpha }(xy) \right| ^2 d\nu (x)\\&\quad \le C \sum _S R^{\frac{d_{gl}}{2}} \int _S \left| m\left( R|x|^2\right) \right| ^2 \prod _{j=1}^N\left( x_j^{-1}+\sqrt{R}y_j \right) ^{-\alpha _j} dx\\&\quad \le C \sum _S R^{\frac{d_{gl}}{2}} \int _S \left| m\left( R|x|^2\right) \right| ^2 x_1^{\alpha _1}\ldots x_{k_2}^{\alpha _{k_2}} \prod _{j=k_2+1}^N \left( 1+\sqrt{R}y_{j}\right) ^{-\alpha _{j}} dx_1\ldots dx_N\\&\quad \le C \sum _S R^{\frac{d_{gl}}{2}} \prod _{i=1}^{k_1}\left( \sqrt{R}y_i\right) ^{-\alpha _i-1} \prod _{j=k_2+1}^N \left( 1+\sqrt{R}y_{j}\right) ^{-\alpha _{j}} \left\| m(R\cdot ) \right\| _{L^2(\mathbb {R})}^2\\&\quad \le C \prod _{i=1}^{k_1}R^{(\alpha _i+1)/2}\left( \sqrt{R}y_i\right) ^{-\alpha _i} \prod _{k=k_1+1}^{k_2} R^{(\alpha _k+1)/2} \prod _{j=k_2+1}^N R^{(\alpha _j+1)/2} \\&\qquad \left( 1+\sqrt{R}y_{j}\right) ^{-\alpha _{j}} \left\| m(R\cdot ) \right\| _{L^2(\mathbb {R})}^2\\&\quad \le C \nu \left( B\left( y,R^{-1/2}\right) \right) ^{-1} \left\| m(R\cdot ) \right\| _{L^2(\mathbb {R})}^2. \end{aligned} \end{aligned}$$

$\square$

3.3 Imaginary Powers of B

In this subsection we prove Lemma 1.1 and Theorems B and C. From now on we consider one-dimensional Bessel operator, i.e. $N=1$, $X=(0,\infty )$, $\alpha >-1$, and $d\nu (x) = x^\alpha \, dx$.

Let us start this section by recalling well-known asymptotics of the Bessel function $I_\tau$ [28, 37], i.e.

$$\begin{aligned} I_{\tau }(x)&= \Gamma \left( \frac{\tau +1}{2} \right) ^{-1} \left( \frac{x}{2} \right) ^\tau + O\left( x^{\tau +1} \right) , x \sim 0, \end{aligned}$$

(3.6)

$$\begin{aligned} I_{\tau } (x)&= (2\pi x)^{-1/2} e^{x}\left( 1 + O(x^{-1}) \right) , x\sim \infty . \end{aligned}$$

(3.7)

Now we provide a short argument for (1.5). In [6, Sec. 4.3] it is proved that $B^{ib}$ is associated with the kernel

$$\begin{aligned} -\Gamma (-ib+1)^{-1} \int _0^\infty t^{ib} \partial _t T_t(x,y) \, dt \end{aligned}$$

in the sense as in (1.5) (let us notice that in [6] only positive values of $\alpha$’s are considered, but the proof works for $\alpha _j>-1$ as well). By integrating by parts,

$$\begin{aligned} \begin{aligned}&- \Gamma (-ib+1)^{-1} \int _0^\infty t^{-ib} \partial _t T_t(x,y)\,dt \\&\quad = - \Gamma (-ib+1)^{-1} \lim _{\varepsilon \rightarrow 0} \left( \varepsilon ^{ib} T_{\varepsilon ^{-1}}(x,y) - \varepsilon ^{-ib} T_\varepsilon (x,y) \right) \\&\qquad + \Gamma (-ib)^{-1} \int _0^{\infty } t^{-ib} T_t(x,y)\,\frac{dt}{t}= K_b(x,y) . \end{aligned} \end{aligned}$$

Proof of Lemma 1.1

Let us first notice that for $\kappa \in \mathbb {R}$ and $c,M>0$, there exists $C=C(\kappa ,c,M)$ such that

$$\begin{aligned} \begin{aligned}{ \int _{c z}^{\infty } t^\kappa \exp \left( -\frac{t}{4} \right) \frac{dt}{t}\le Cz ^{-M}, \qquad z\ge 1. } \end{aligned} \end{aligned}$$

(3.8)

Using (1.4) and (1.2) one obtains

$$\begin{aligned} \begin{aligned} 2 \Gamma (-ib) K_b(x,y)&= \int _0^\infty t^{-ib-1} (xy)^{-(\alpha -1)/2} I_{(\alpha -1)/2}\left( \frac{xy}{2t}\right) \exp \left( -\frac{x^2+y^2}{4t}\right) \frac{dt}{t}\\&= \int _0^{xy} \cdots + \int _{xy}^\infty \ldots = A_1 + A_2. \end{aligned} \end{aligned}$$

Denote $\chi _{loc}(x,y)=\chi _{\left\{ y/2<x<2y \right\} }(x,y)$ and $\chi _{glob}(x,y)= 1-\chi _{loc}(x,y)$, $x,y\in X$. In the proof below all expressions denoted by $R_k$ shall be parts of the kernel $R_b(x,y)$. Using (3.7), we write $A_1 = A_{1,1}+R_1$, where

$$\begin{aligned} \begin{aligned} A_{1,1} =&\pi ^{-1/2} \int _0^{xy} t^{-ib-1/2} (xy)^{-\alpha /2} \exp \left( -\frac{|x-y|^2}{4t}\right) \frac{dt}{t}\end{aligned} \end{aligned}$$

and

$$\begin{aligned} |R_1|&= \left| \int _0^{xy} t^{-ib-1} (xy)^{-(\alpha -1)/2} \exp \left( -\frac{x^2+y^2}{4t}\right) \right. \\&\left. \quad \left( I_{(\alpha -1)/2}\left( \frac{xy}{2t}\right) - \left( \frac{\pi xy}{t}\right) ^{-1/2}\exp \left( \frac{xy}{2t}\right) \right) \frac{dt}{t} \right| \\&\le C \int _0^{xy} t^{1/2} (xy)^{-\alpha /2-1} \exp \left( -\frac{|x-y|^2}{4t} \right) \frac{dt}{t}\\&= C |x-y| (xy)^{-\alpha /2-1} \int _{\frac{|x-y|^2}{xy}}^\infty t^{-1/2} e^{-t/4} \frac{dt}{t}\\&\le C xy (x+y)^{-\alpha -3}. \end{aligned}$$

In the last inequality we have used (3.8). Denoting

$$\begin{aligned} R_2= & {} \chi _{glob}(x,y)A_{1,1}, \quad \quad \\ R_3= & {} \pi ^{-1/2} \chi _{loc}(x,y)\int _{xy}^\infty t^{-ib-1/2} (xy)^{-\alpha /2} \exp \left( -\frac{|x-y|^2}{4t}\right) \frac{dt}{t}\end{aligned}$$

we have

$$\begin{aligned} \begin{aligned} A_{1,1} - R_2 + R_3 =&\pi ^{-1/2}\chi _{loc}(x,y)\left( \int _0^\infty t^{-ib-1/2} (xy)^{-\alpha /2} \exp \left( -\frac{|x-y|^2}{4t}\right) \frac{dt}{t} \right) \\ =&\pi ^{-1/2} {2^{2bi+1}} \Gamma \left( ib+1/2 \right) \chi _{loc}(x,y)(xy)^{-\alpha /2}|x-y|^{-2ib-1}. \end{aligned} \end{aligned}$$

Notice that $A_{1,1}$ is one of the terms from (1.6). Next, by (3.8),

$$\begin{aligned} |R_2|&\le C \chi _{glob}(x,y)\, |x-y|^{-1} (xy)^{-\alpha /2} \int _{\frac{|x-y|^2}{xy}}^\infty t^{1/2} e^{-t/4}\frac{dt}{t}\\&\le Cxy(x+y)^{-\alpha -3},\\ |R_3|&\le C \chi _{loc}(x,y)x^{-\alpha } |x-y|^{-1} \int _0^{\frac{|x-y|^2}{xy}} t^{1/2} \frac{dt}{t}\\&\simeq C \chi _{loc}(x,y)x^{-\alpha -1} . \end{aligned}$$

Now let us turn to study $A_2$. Denote $c_\alpha = 4^{-(\alpha -1)/2} \Gamma ((\alpha +1)/4)^{-1}$. Then, by using (3.6),

$$\begin{aligned} \begin{aligned} A_2 =&\, c_\alpha \int _{xy}^\infty t^{-ib-(\alpha +1)/2} \exp \left( -\frac{x^2+y^2}{4t}\right) \frac{dt}{t}+R_4 = A_{2,1}+R_4\\ \end{aligned} \end{aligned}$$

where, by (3.8),

$$\begin{aligned} |R_4|&= \left| \int _{xy}^\infty t^{-ib-1} (xy)^{-(\alpha -1)/2} \exp \left( -\frac{x^2+y^2}{4t}\right) \right. \\&\quad \left. \left( I_{(\alpha -1)/2}\left( \frac{xy}{2t}\right) - \Gamma \left( \frac{\alpha +1}{4} \right) ^{-1} \left( \frac{xy}{4t}\right) ^{(\alpha -1)/2} \right) \frac{dt}{t} \right| \\&\le C xy \int _{xy}^\infty t^{-(\alpha +3)/2} \exp \left( -\frac{x^2+y^2}{4t}\right) \frac{dt}{t}\\&\le C xy (x^2+y^2)^{-(\alpha +3)/2} \simeq C xy (x+y)^{-\alpha -3}. \end{aligned}$$

Moreover,

$$\begin{aligned} \begin{aligned} A_{2,1} +R_5&= c_\alpha \int _0^\infty t^{-ib-(\alpha +1)/2} \exp \left( -\frac{x^2+y^2}{4t} \right) \frac{dt}{t}\\&=c_\alpha 4^{ib+(\alpha +1)/2} \Gamma \left( ib+(\alpha +1)/2 \right) \left( x^2+y^2 \right) ^{-ib-(\alpha +1)/2} , \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \left| R_5 \right|&= \left| c_\alpha \int _0^{xy}t^{-ib-(\alpha +1)/2} \exp \left( -\frac{x^2+y^2}{4t}\right) \frac{dt}{t} \right| \\&\le C (x^2+y^2)^{-(\alpha +1)/2} \int _{\frac{x^2+y^2}{xy}}^\infty t^{(\alpha +1)/2} e^{-t/4} \frac{dt}{t}\\&\le C xy(x+y)^{-\alpha -3}. \end{aligned}$$

$\square$

Proof of Theorem B

  for$\varvec{\alpha <0}$. Let $|b|>1$ and $\varepsilon \in (0,10^{-1})$ (to be fixed later on). Denote $I=[1, 1+\varepsilon ]$ and $S=[1+3\varepsilon ,2]$. Put $f_\varepsilon (x) = \varepsilon ^{-1} \chi _I(x) x^{-\alpha }$, so that $\left\| f_\varepsilon \right\| _{L^1(X)}=1$. If $x\in S$, by Lemma 1.1 and the triangle inequality,

$$\begin{aligned} \begin{aligned}&\left| B^{ib} f_\varepsilon (x) \right| \le \left| c_2(b) \right| x^{-\alpha /2} |x-1|^{-1} \\&\qquad + \left| c_2(b) \right| \left| \int _I \left( (xy)^{-\alpha /2}|x-y|^{-2ib-1} - x^{-\alpha /2}|x-1|^{-2ib-1} \right) f_\varepsilon (y) \, d\nu (y) \right| \\&\qquad + \left| c_1(b) \right| \left| \int _I \left( x^2 + y^2 \right) ^{-ib-(\alpha +1)/2} f_\varepsilon (y) \, d\nu (y) \right| \\&\qquad + \left| c_3(b) \right| \left| \int _I R_b(x,y) f_\varepsilon (y) \, d\nu (y) \right| \\&\quad = \left| c_2(b) \right| x^{-\alpha /2} |x-1|^{-1} + \Lambda _1 + \Lambda _2 + \Lambda _3. \end{aligned} \end{aligned}$$

(3.9)

Observe that for $x\in S$ and $y\in I$ we have $|x-y|\simeq |x-1|$ and $x\simeq y\simeq 1$. By using the Mean Value Theorem for the function $y \mapsto y^{-\alpha /2} |x-y|^{-2ib-1}$,

$$\begin{aligned} \begin{aligned} \Lambda _1&\le C \left| c_2(b) \right| \varepsilon ^{-1} \int _{1}^{1+\varepsilon } |b||y-1| |x-1|^{-2} \, dy \le C \varepsilon \left| bc_2(b) \right| |x-1|^{-2},\\ \Lambda _2&\le C \left| c_1(b) \right| \varepsilon ^{-1} \int _{1}^{1+\varepsilon } (x^2+y^2)^{-(1+\alpha )/2} \, dy \simeq C \left| c_1(b) \right| , \\ \Lambda _3&\le C \left| c_3(b) \right| \varepsilon ^{-1} \int _{1}^{1+\varepsilon } xy(x+y)^{-\alpha -3} \, dy \simeq C \left| c_3(b) \right| . \end{aligned} \end{aligned}$$

(3.10)

Fix $|b|\ge 1$ and $\lambda$ such that $\lambda > \max (\Lambda _2,\Lambda _3, |b c_2(b)| )$. Recall that $x^{-\alpha /2} \ge 1$ for $x\in S$, so that for $\varepsilon$ small enough

$$\begin{aligned} \begin{aligned} \nu \left\{ x\in S: \left| c_2(b) \right| x^{-\alpha /2}|x-1|^{-1}> 4\lambda \right\}&\ge \nu \left\{ x\in S: \left| c_2(b) \right| |x-1|^{-1} > 4\lambda \right\} \\&= \int _{1+3\varepsilon }^{1+\left| c_2(b) \right| /(4\lambda )} x^{\alpha } \, dx\\&\ge C {\left| c_2(b) \right| /(4\lambda )} \end{aligned} \end{aligned}$$

(3.11)

and

$$\begin{aligned} \begin{aligned} \nu \left\{ x\in S: \left| \Lambda _1 \right|> \lambda \right\}&\le \nu \left\{ x\in S: C \varepsilon \left| bc_2(b) \right| |x-1|^{-2} > \lambda \right\} \\&\le \int _{1+3\varepsilon }^{1 + C(\varepsilon \lambda ^{-1} \left| bc_2(b) \right| )^{1/2}} x^\alpha \, dx \\&\le C \left( \varepsilon \lambda ^{-1} \left| bc_2(b) \right| \right) ^{1/2} \\&\le C \varepsilon ^{1/2}. \end{aligned} \end{aligned}$$

(3.12)

Hence, using (3.9)–(3.12) and (1.7) we get

$$\begin{aligned} \begin{aligned} \left\| B^{ib}f_\varepsilon \right\| _{L^{1,\infty }(X)}&\ge \lambda \nu \left\{ x\in S: \left| B^{ib}f_\varepsilon (x) \right|> \lambda \right\} \\&\ge \lambda \nu \left\{ x\in S: \left| c_2(b) \right| x^{-\alpha /2}|x-1|^{-1}> 4\lambda \right\} \\&\quad - \lambda \nu \left\{ x\in S: \left| \Lambda _1 \right|> \lambda \right\} - \underbrace{\lambda \nu \left\{ x\in S: \left| \Lambda _2 \right|> \lambda \right\} }_{=0}\\&\quad - \underbrace{\lambda \nu \left\{ x\in S: \left| \Lambda _3 \right| > \lambda \right\} }_{=0} \\&{\ge C |c_2(b)| - C\lambda \varepsilon ^{1/2} \ge C |c_2(b)|} \simeq |b|^{1/2} = |b|^{d/2}. \end{aligned} \end{aligned}$$

$\square$

Turning to the case $\alpha >0$ we could also use Lemma 1.1. In this case, the summand with $c_1(b)$ would play the first role. An alternative proof that we shall present here uses integral representation of the modified Bessel function. The same will be used in the proof of Theorem C. It is known that for $\alpha >0$

$$\begin{aligned} { I_{(\alpha -1)/2}(z) = \left( \Gamma (\alpha /2)^{-1}\sqrt{\pi } \right) ^{-1} \left( \frac{z}{2}\right) ^{(\alpha -1)/2} \int _{-1}^1 e^{-zs}(1-s^2)^{\alpha /2-1} ds, \quad z>0, }\nonumber \\ \end{aligned}$$

(3.13)

see [37, Ch. 6]. Therefore, for $\alpha >0$, using (1.4), (1.2), and (3.13) we obtain

$$\begin{aligned} \begin{aligned} K_b(x,y)&= (2\Gamma (-ib))^{-1}\int _0^\infty t^{-ib-1} (xy)^{-(\alpha -1)/2} I_{(\alpha -1)/2}\left( \frac{xy}{2t} \right) \\&\quad \exp \left( -\frac{x^2+y^2}{4t} \right) \frac{dt}{t}\\&= \left( 2^\alpha \Gamma (-ib) \Gamma (\alpha /2)\sqrt{\pi } \right) ^{-1} \int _{-1}^1 \int _0^\infty t^{-ib-(\alpha +1)/2}\\&\quad \exp \left( -\frac{x^2+y^2+2xys}{4t} \right) \frac{dt}{t}\, (1-s^2)^{\frac{\alpha }{2}-1} ds \\&= \frac{2^{2ib+1} \Gamma \left( ib+ (\alpha +1)/2\right) }{\Gamma (-ib) \Gamma (\alpha /2)\sqrt{\pi }} \int _{-1}^1 \left( x^2+y^2+2xys \right) ^{-ib-(\alpha +1)/2} \\&\quad \left( 1-s^2 \right) ^{\alpha /2-1} \, ds \\&= {C_\alpha } c_1(b) \int _{-1}^{1} \left( x^2+y^2+2xys \right) ^{-ib-(\alpha +1)/2} \left( 1-s^2 \right) ^{\alpha /2-1} ds, \end{aligned} \end{aligned}$$

(3.14)

where $C_\alpha = \pi ^{-1/2} \Gamma (\alpha /2)^{-1} \Gamma ((\alpha +1)/4)$.

Proof of Theorem B

  for$\varvec{\alpha > 0}$. Let $|b|>1$, $\varepsilon \in (0,10^{-1})$, and $f_\varepsilon (x) = x^{-\alpha } \varepsilon ^{-1} \chi _{[\varepsilon ,2\varepsilon ]}(x)$. Similarly as in (3.9), using (3.14) and the Mean Value Theorem, for $x>3\varepsilon$ we have

$$\begin{aligned} \begin{aligned} \left| B^{-ib}f_\varepsilon (x) \right|&\le \, \left| \int _\varepsilon ^{2\varepsilon } K_b(x,0) f_\varepsilon (y) \, d\nu (y) \right| \\&\quad + \left| \int _\varepsilon ^{2\varepsilon } \left( K_b(x,0) - K_b(x,y) \right) f_\varepsilon (y) \, d\nu (y) \right| \\&\le \, C|c_1(b)| \varepsilon ^{-1} \left\{ \left| \int _\varepsilon ^{2\varepsilon }\int _{-1}^{1} \left( 1-s^2 \right) ^{\alpha /2-1} x^{-2ib-(\alpha +1)}\, ds \, dy \right| \right. \\&\quad +\, \left. \int _\varepsilon ^{2\varepsilon } \int _{-1}^{1} \left( 1-s^2 \right) ^{\alpha /2-1}\right. \\&\left. \quad \left| \left( x^2+y^2+2sxy \right) ^{-ib-\frac{\alpha +1}{2}} - x^{-2ib-(\alpha +1)} \right| \, ds \, dy \right\} \\&\le \, C|c_1(b)| \left( x^{-\alpha -1} + \varepsilon ^{-1} \int _{-1}^{1} \left( 1-s^2 \right) ^{\alpha /2-1}\right. \\&\left. \quad \int _\varepsilon ^{2\varepsilon } |b|\left| y^2+2sxy \right| x^{-\alpha -3} \, dy \, ds \right) \\&\le \, C|c_1(b)| \left( x^{-\alpha -1} + \varepsilon \left| b \right| x^{-\alpha -2} \right) . \end{aligned} \end{aligned}$$

(3.15)

Let us fix $|b|>1$ and $\lambda >|b c_1(b)|$. For all $\varepsilon$ small enough, we get

$$\begin{aligned} \begin{aligned} \nu \left\{ x>3\varepsilon : C \left| c_1(b) \right| x^{-\alpha -1} > 2\lambda \right\}&= \int _{3\varepsilon }^{ C (\left| c_1(b) \right| /\lambda )^{1/(1+\alpha )}} x^{\alpha } \, dx \ge C \left| c_1(b) \right| /\lambda , \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \nu \left\{ x>3\varepsilon : C \left| bc_1(b) \right| \varepsilon x^{-\alpha -2} > \lambda \right\}&\le \, \int _0^{ C \left( \left| bc_1(b) \right| \varepsilon / \lambda \right) ^{1/(\alpha +2)}} x^{\alpha } \, dx \\&= \, C \left( \left| bc_1(b) \right| \varepsilon / \lambda \right) ^{(\alpha +1)/(\alpha +2)} \\&\le \, C \varepsilon ^{(1+\alpha )/(\alpha +2)}. \end{aligned} \end{aligned}$$

Therefore, by choosing a proper $\varepsilon$, we obtain

$$\begin{aligned} \begin{aligned} \left\| B^{ib}f_\varepsilon \right\| _{L^{1,\infty }(X)}&\ge \, \lambda \nu \left\{ x\in X: \left| B^{ib}f_\varepsilon (x) \right|> \lambda \right\} \\&\ge \, \lambda \nu \left\{ x> 3\varepsilon : C|c_1(b)| x^{-\alpha -1}> 2\lambda \right\} \\&\quad - \lambda \nu \left\{ x> 3\varepsilon : C |c_1(b)| \varepsilon x^{-\alpha -2} > \lambda \right\} \\&\ge \, C |c_1(b)| - C \lambda \varepsilon ^{(\alpha +1)/(\alpha +2)} \ge C |c_1(b)| \simeq |b|^{(\alpha +1)/2} = |b|^{d/2} . \end{aligned} \end{aligned}$$

$\square$

Proof of Theorem C

Set $\alpha >0$, $p\in (1,2)$, $|b|>1$, $\varepsilon \in (0,10^{-1})$, $\delta >1$, and a function ${f}\in {L^p(X)}$ such that $\mathrm {supp}f \subseteq (0,\varepsilon )$, and $f\ge 0$. Similarly as in (3.15), using (3.14) and Corollary 4.2,

$$\begin{aligned} \begin{aligned} \left\| B^{ib}f \right\| _{L^p(X)}^p&\ge \int _\delta ^\infty \left| \int _X \left( K_b(x,0)- (K_b(x,0) - K_b(x,y) \right) f(y)\, d\nu (y) \right| ^p\, d\nu (x) \\&\ge C \left\| f \right\| _{L^1(X)}^p \int _\delta ^\infty \left| K_b(x,0) \right| ^p \, d\nu (x) \\&\quad - C \int _\delta ^\infty \left| \int _X (K_b(x,y)-K_b(x,0)) f(y)\, d\nu (y) \right| ^p \, d\nu (x) \\&\ge C \left\| f \right\| _{L^1(X)}^p|b|^{p(\alpha +1)/2} \left( \int _\delta ^{\infty } x^{-p(\alpha +1)+\alpha } \, dx - \int _\delta ^{\infty } \varepsilon ^p |b|^p x^{-p(\alpha +2)+\alpha } \, dx \right) \\&\ge C \left\| f \right\| _{L^1(X)}^p |b|^{p(\alpha +1)/2} \delta ^{(\alpha +1)(1-p)} \left( 1 - \varepsilon ^p |b|^{p} \delta ^{-p} \right) . \end{aligned} \end{aligned}$$

Now we take $\delta =|b|$ and fix $\varepsilon$ small enough, independent of b, getting

$$\begin{aligned} \begin{aligned} \left\| B^{ib}f \right\| _{L^p(X)} \ge C_{p} {|b|^{\frac{(\alpha +1)(2-p)}{2p}}} \left\| f \right\| _{L^1(X)} \ge C_{p,\varepsilon } {|b|^{{\frac{d}{2}}\frac{2-p}{p}}} \left\| f \right\| _{L^p(X)}. \end{aligned} \end{aligned}$$

$\square$

4 Appendix: Gamma Function Estimate

Lemma 4.1

Let $a+bi\in \mathbb {C}$. For $a\ge 0$ fixed and all $|b|\ge 1$ we have

$$\begin{aligned}{ |\Gamma (a+bi)| \simeq |b|^{a-1/2} \exp \left( -\frac{\pi b}{2} \right) . } \end{aligned}$$

The result above is known. It is a consequence of the Stirling’s Formula, see [1, Ch. 6]. For the convenience of the reader we present a short proof.

Proof

Using the reflection formula

$$\begin{aligned} { \Gamma (1-z)\Gamma (z) = \pi /\sin (\pi z),} \end{aligned}$$

(4.1)

and the recursion identity

$$\begin{aligned} { z\Gamma (z)= \Gamma (z+1),} \end{aligned}$$

(4.2)

we have that $|1-ib| \left| \Gamma (ib) \right| ^2 = \left| \Gamma (ib)\Gamma (1-ib) \right| = \left| {\pi }/{\sin (\pi ib)} \right| \simeq \exp \left( -\pi |b| \right)$ for $|b|\ge 1$. Thus,

$$\begin{aligned} { \left| \Gamma (ib) \right| \simeq |b|^{-1/2}\exp \left( -\frac{\pi |b|}{2} \right) , \qquad |b|\ge 1 .} \end{aligned}$$

(4.3)

Denote $S = \left\{ z\in \mathbb {C}\ : \ 1\le \mathrm {Re}(z) \le 2, {\left| \mathrm {Im}(z) \right| } \ge 1 \right\}$ and define a holomorphic function

$$\begin{aligned}{ F(z) = \Gamma (z)z^{-z+1/2}, \quad z\in S.} \end{aligned}$$

Now, we claim that $|F(z)|\le C$ if $z\in \partial S$. This is clear for $z=a\pm i$, $a\in [1,2]$. For $z=1+ib$, $|b|\ge 1$, we use (4.2) and (4.3) getting

$$\begin{aligned} \begin{aligned} \left| F(1+ib) \right|&= \left| \Gamma \left( 1+ib\right) \right| \left| (1+ib)^{-1/2-ib} \right| = \left| b \right| \left| \Gamma \left( ib\right) \right| (1+b^2)^{-1/4} e^{b\, \mathrm {arctg} b} \\&\le C |b|^{1/2} e^{-\pi |b|/2} |b|^{-1/2} e^{b \, \mathrm {arctg}(b)} \le C. \end{aligned} \end{aligned}$$

Similarly we show boundedness of F for $z=2+bi$, $|b|\ge 1$.

Observe that $\left| F(z) \right| \le |\Gamma (z)| |z|^{|-z+1/2|} \le C e^{c|z|^2}$ for $z\in S$. Hence, applying the Phragmén-Lindelöf principle, we obtain that $|F(z)| \le C$ for $z \in S$. Therefore, for a fixed $a\in [1,2]$ and $|b|\ge 1$ we have

$$\begin{aligned} \begin{aligned} \left| \Gamma (a+bi) \right|&\le C \left| (a+ bi)^{a-1/2+bi} \right| = \left( a^2+b^2 \right) ^{(2a-1)/4} \cdot e^{-b \, \mathrm {arctg}(b/a)}\\&\simeq C|b|^{a-1/2} e^{-\pi |b|/2}. \end{aligned} \end{aligned}$$

(4.4)

This is the desired estimate from above for $a\in [1,2]$. We extend this for all $a\in [0,\infty )$ by using (4.2). Then, by (4.1), we get estimate from below for $a\in [0,1]$, and extend this for $a\in [0,\infty )$ using (4.2) once more. $\square$

Corollary 4.2

For fixed $a_1,a_2 \ge 0$ and $|b|\ge 1$ we have

$$\begin{aligned}{ \left| \frac{\Gamma (a_1+bi)}{\Gamma (a_2+bi)} \right| \simeq |b|^{a_1-a_2}. } \end{aligned}$$