Quantitative weighted estimates for Harmonic analysis operators in the Bessel setting by using sparse domination

Abstract

In this paper we obtain quantitative weighted $L^p$-inequalities for some operators involving Bessel convolutions. We consider maximal operators, Littlewood-Paley functions and variational operators. We obtain $L^p(w)$-operator norms in terms of the $A_p$-characteristic of the weight w. In order to do this we show that the operators under consideration are dominated by a suitable family of sparse operators in the space of homogeneous type $((0,\infty ),|\cdot |,x^{2\lambda }dx)$.

1 Introduction

In this paper we obtain quantitative weighted $L^p$-inequalities for some operators involving convolutions products in the Bessel setting. We obtain the $L^p(w)$-operator norm estimates in terms of the $A_p$-characteristic $[w]_{A_p}$ of the weight w. In order to do this we prove that the operators under consideration can be dominated pointwisely by a suitable family of sparse operators.

We now define the operators we are going to deal with. Let $\lambda >0$. We denote by $m_\lambda$ the measure on $(0,\infty )$ defined by $dm_\lambda =x^{2\lambda }dx$. $m_\lambda$ is doubling with respect to the usual metric in $(0,\infty )$ defined by the absolute value $|\cdot |$ (see for instance [73, (1.5)]). In other words the triple $((0,\infty ),|\cdot |,m_\lambda )$ is a space of homogeneous type in the sense of Coifman and Weiss ([20]).

For every $f,g\in L^1((0,\infty ),m_\lambda )$ we define the convolution product $f\#_\lambda g$ of f and g by

$$\begin{aligned} (f\#_\lambda g)(x)=\int _0^\infty f(y)\; _\lambda \tau _x(g)(y) y^{2\lambda }dy,\quad x\in (0,\infty ), \end{aligned}$$

where, for every $x\in (0,\infty )$,

$$\begin{aligned} _\lambda \tau _x(g)(y)=\frac{\Gamma (\lambda +1/2)}{\Gamma (\lambda )\sqrt{\pi }}\int _0^\pi g\left( \sqrt{x^2+y^2-2xy\cos \theta }\right) \sin ^{2\lambda -1}\theta d\theta ,\;\;\;y\in (0,\infty ). \end{aligned}$$

The main properties of $\#_\lambda$ and $_\lambda \tau _x$, $x\in (0,\infty )$, can be found in [37] and [38]. We define the Hankel transform $h_\lambda f$ of $f\in L^1((0,\infty ),m_\lambda )$ as follows

$$\begin{aligned} h_\lambda (f)(x)=\int _0^\infty (xy)^{-\lambda +1/2}J_{\lambda -1/2}(xy)f(y)y^{2\lambda }dy,\quad x\in (0,\infty ), \end{aligned}$$

where $J_\nu$ represents the Bessel function of first kind and order $\nu >-1$. $h_\lambda$ plays in the harmonic analysis in the Bessel setting the same role than the one of the Fourier transform in the Euclidean harmonic analysis. The operations $\#_\lambda$ and $_\lambda \tau _x$, $x\in (0,\infty )$, are usually named Hankel convolution and Hankel translation because the following equalities hold for every $f,g\in L^1((0,\infty ),m_\lambda )$ ([37] and [38])

$$\begin{aligned} h_\lambda (f\#_\lambda g)=h_\lambda (f)h_\lambda (g), \end{aligned}$$

and

$$\begin{aligned} h_\lambda (_\lambda \tau _x(g))(y)=2^{\lambda -1/2}\Gamma \left( \lambda +\frac{1}{2}\right) (xy)^{-\lambda +1/2}J_{\lambda -1/2}(xy)h_\lambda (g)(y),\quad x,y\in (0,\infty ). \end{aligned}$$

We now consider the Bessel operator $\Delta _\lambda =-\frac{d^2}{dx^2}-\frac{2\lambda }{x}\frac{d}{dx}$ on $(0,\infty )$. $h_\lambda$, $\#_\lambda$ and $_\lambda \tau _x$, $x\in (0,\infty )$, are connected with $\Delta _\lambda$. For every $f,g\in S(0,\infty )$, the Schwartz space on $(0,\infty )$, we have that

$$\begin{aligned} h_\lambda (\Delta _\lambda f)(x)= & {} x^2h_\lambda (f)(x),\quad x\in (0,\infty ), \\ \Delta _\lambda (f\#_\lambda g)= & {} \Delta _\lambda (f)\#_\lambda g, \end{aligned}$$

and, for every $x\in (0,\infty )$,

$$\begin{aligned} \Delta _\lambda (_\lambda \tau _x(g))=\; _\lambda \tau _x(\Delta _\lambda g). \end{aligned}$$

The heat and Poisson semigroups associated with $\Delta _\lambda$ are $\#_\lambda$-convolution semigroups. Indeed, if $\{e^{-t\Delta _\lambda }\}_{t>0}$ represents the semigroup generated by $-\Delta _\lambda$ we have that

$$\begin{aligned} e^{-t\Delta _\lambda }(f)=W_{\sqrt{2t}}^\lambda \#_\lambda f,\quad t>0, \end{aligned}$$

where

$$\begin{aligned} W^\lambda (x)=\frac{2^{1/2-\lambda }}{\Gamma (\lambda +1/2)}e^{-x^2/2},\quad x\in (0,\infty ). \end{aligned}$$

The Poisson semigroup $\{e^{-t\sqrt{\Delta _\lambda }}\}_{t>0}$ associated with $\Delta _\lambda$ is defined by

$$\begin{aligned} e^{-t\sqrt{\Delta _\lambda }}(f)=P_t^\lambda \#_\lambda f,\quad t>0, \end{aligned}$$

being

$$\begin{aligned} P^\lambda (x)=\frac{2\lambda \Gamma (\lambda )}{\Gamma (\lambda +1/2)\sqrt{\pi }}\frac{1}{(1+x^2)^{\lambda +1}},\quad x\in (0,\infty ). \end{aligned}$$

If $\phi$ is a complex function defined in $(0,\infty )$ the t-dilation $\phi _t$ of $\phi$ is defined in this setting, for every $t>0$, by

$$\begin{aligned} \phi _t(x)=\frac{1}{t^{2\lambda +1}}\phi \left( \frac{x}{t}\right) ,\quad x\in (0,\infty ). \end{aligned}$$

The study of harmonic analysis associated with Bessel operators was begun by Muckenhoupt and Stein ([62]) and continued by Andersen and Kerman ([3]) and Stempak ([69]). In the last fifteen years many problems concerning to the harmonic analysis in the Bessel context has been studied ([6,7,8,9,10, 22, 25, 27, 33, 42, 43, 45, 64, 70, 72] and [73]).

We now define the operators that we are going to study. As in [73, Definition 1.4] we consider the space $Z^\lambda$ that consists of all those $\phi \in C^2(0,\infty )$ satisfying the following properties.

• (i) $|\phi (x)|\le C(1+x^2)^{-\lambda -1},\;\;\;x\in (0,\infty )$,

• (ii) $|\phi '(x)|\le Cx(1+x^2)^{-\lambda -2}, \;\;\;x\in (0,\infty )$,

• (iii) $|\phi ''(x)|\le C(1+x^2)^{-\lambda -2}, \;\;\;x\in (0,\infty )$.

Let $\phi \in Z^\lambda$. We defined the maximal operator $\phi _*^\lambda$ by

$$\begin{aligned} \phi _*^\lambda (f)=\sup _{t>0}|f\#_\lambda \phi _t|, \end{aligned}$$

and the square function $g_\phi ^\lambda$ as follows

$$\begin{aligned} g_\phi ^\lambda (f)(x)=\left( \int _0^\infty |f\#_\lambda \phi _t(x)|^2\frac{dt}{t}\right) ^{1/2},\quad x\in (0,\infty ). \end{aligned}$$

Let $\rho >2$. Suppose that $g:(0,\infty )\longrightarrow {\mathbb {C}}$ is a measurable function. The $\rho$-variation operator $V_\rho (\{g(t)\}_{t>0})$ is defined by

$$\begin{aligned} V_\rho (\{g(t)\}_{t>0})=\sup _{\begin{array}{c} 0<t_n<t_{n-1}<\cdots <t_1 \\ n\in {\mathbb {N}} \end{array}}\left( \sum _{j=1}^{n-1}|g(t_j)-g(t_{j+1})|^\rho \right) ^{1/\rho }. \end{aligned}$$

The $\rho$-variation operator $V_\rho ^\lambda (\{\phi _t\}_{t>0})$ associated with $\{\phi _t\}_{t>0}$ is defined by

$$\begin{aligned} V_\rho ^\lambda (\{\phi _t\}_{t>0})(f)(x)=V_\rho (\{(f\# _\lambda \phi _t)(x)\}_{t>0}),\quad x\in (0,\infty ). \end{aligned}$$

Maximal, $\rho$-variation operators and square functions for families of operators have had a considerable amount of attention in probability, ergodic theory, and harmonic analysis. As it is wellknown the maximal operators $\phi _*^\lambda$ are related with the convergence of $\{f\#_\lambda \phi _t\}_{t>0}$, for instance, as $t\rightarrow 0^+$. If $\phi _*^\lambda$ is bounded in $L^p((0,\infty ),m_\lambda )$ for some $1\le p<\infty$, the almost everywhere convergence of $\{f\#_\lambda \phi _t\}_{t>0}$ as $t\rightarrow 0^+$ is obtained, for every $f\in L^p((0,\infty ),m_\lambda )$ when the almost everywhere convergence is known for every $f\in {\mathcal {A}}$ being ${\mathcal {A}}$ a dense subset of $L^p((0,\infty ),m_\lambda )$. Also, the $L^p((0,\infty ),m_\lambda )$-boundedness of the $\rho$-variation operator $V_\rho ^\lambda (\{\phi _t\}_{t>0})$ implies almost everywhere convergence of $\{f\#_\lambda \phi _t\}_{t>0}$ as $t\rightarrow 0^+$, for every $f\in L^p((0,\infty ),m_\lambda )$ but in this case it is not necessary to know the almost everywhere convergence property for f in some dense subset in $L^p((0,\infty ),m_\lambda )$. Furthermore the $\rho$-variation operator $V_\rho ^\lambda (\{\phi _t\}_{t>0})$ gives extra information about the convergence of $\{f\#_\lambda \phi _t\}_{t>0}$ as $t\rightarrow 0^+$. For instance $L^p$-boundedness properties of $V_\rho ^\lambda (\{\phi _t\}_{t>0})$ allow us to obtain estimates of the $\alpha$-jump that $\{f\#_\lambda \phi _t\}_{t>0}$ has as $t\rightarrow 0^+$. $L^p$-boundedness properties for the square function $g_\phi ^\lambda$ lead to $L^p$-estimates for spectral multipliers for Bessel operators.

Suppose that $w,b\in L^1_\mathrm{loc}((0,\infty ),m_\lambda )$ and $w_\lambda (I)=\int _I w(x)x^{2\lambda }dx>0$, for every interval $I\subset (0,\infty )$. We say thet $b\in BMO((0,\infty ),w,m_\lambda )$ when

$$\begin{aligned} \Vert b\Vert _{BMO((0,\infty ),w,\lambda )}:=\sup _{\begin{array}{c} I\subset (0,\infty ) \\ I\;\text{ interval } \end{array}}\frac{1}{w_\lambda (I)}\int _I|b(y)-b_I|y^{2\lambda }dy<\infty .\end{aligned}$$

Here, for every interval $I\subset (0,\infty )$, $b_I=\frac{1}{m_\lambda (I)}\int _I b(y)y^{2\lambda }dy$.

The Riesz transform associated with $\Delta _\lambda$ is defined by

$$\begin{aligned} {\mathcal {R}}_{\Delta _\lambda }=\partial _x\Delta _\lambda ^{-1/2}. \end{aligned}$$

In [26, Theorem 1.3] it was proved that a function $b\in \bigcup _{q>1}L^q_\mathrm{loc}((0,\infty ),m_\lambda )$ is in $BMO((0,\infty ),1,m_\lambda )$ if and only if the commutator $[{\mathcal {R}}_{\Delta _\lambda },b]$ defined by

$$\begin{aligned}{}[{\mathcal {R}}_{\Delta _\lambda },b]f={\mathcal {R}}_{\Delta _\lambda }(bf)-bR_{\Delta _\lambda }(f) \end{aligned}$$

is bounded from $L^p((0,\infty ),m_\lambda )$ into itself, for every $1<p<\infty$.

Let b be a measurable function on $(0,\infty )$. We consider the following higher order commutator operators defined, for every $m\in {\mathbb {N}}\setminus \{0\}$, as follows

$$\begin{aligned} \phi _{*,b}^{\lambda ,m}(f)(x)= & {} \sup _{t>0}\left| \left( \phi _t\#_\lambda \left[ (b(\cdot )-b(x))^m f\right] \right) (x)\right| ,\quad x\in (0,\infty ), \\ g_{\phi ,b}^{\lambda ,m}(f)(x)= & {} \left( \int _0^\infty \left| \left( \phi _t\#_\lambda \left[ (b(\cdot )-b(x))^m f\right] \right) (x)\right| ^2\frac{dt}{t}\right) ^{1/2},\quad x\in (0,\infty ), \end{aligned}$$

and

$$\begin{aligned} V_{\rho ,b}^{\lambda ,m}(\{\phi _t\}_{t>0})(f)(x) =&\sup _{\begin{array}{c} 0<t_n<t_{n-1}<...<t_1 \\ n\in {\mathbb {N}} \end{array}}\Big (\sum _{j=1}^{n-1}\left| \Bigg (\phi _{t_j}\#_\lambda \left[ (b(\cdot )-b(x))^m f\right] \Bigg )(x) \right. \\&-\left. \left( \phi _{t_{j+1}}\#_\lambda \left[ (b(\cdot )-b(x))^m f\right] \right) (x)\right| ^\rho \Big )^{1/\rho },\;\;\;\;x\in (0,\infty ). \end{aligned}$$

Commutators associated with Littlewood-Paley operators in the Euclidean context were studied in [18, 23] and [59] and with Hardy-Littlewood maximal operator on spaces of homogeneous type were investigated in [30] and [34].

We say that a measurable function w in $(0,\infty )$ is a weight when $w(x)>0$, for almost all $x\in (0,\infty )$. A weight w in $(0,\infty )$ is said to be in the class $A_p^\lambda (0,\infty )$ provided that

• (i)
  $$\begin{aligned}{}[w]_{A_p^\lambda }:=\sup _{\begin{array}{c} I\subset (0,\infty ) \\ I\;\text{ interval } \end{array}}\left( \frac{1}{m_\lambda (I)}\int _I w(x)x^{2\lambda }dx\right) \left( \frac{1}{m_\lambda (I)}\int _I w(x)^{-\frac{1}{p-1}}x^{2\lambda }dx\right) ^{p-1}<\infty ,\end{aligned}$$

  when $1<p<\infty$;

• (ii)
  $$\begin{aligned}{}[w]_{A_1^\lambda }:=\sup _{\begin{array}{c} I\subset (0,\infty ) \\ I\;\text{ interval } \end{array}}\frac{1}{m_\lambda (I)}\int _I w(x)x^{2\lambda }dx\;\mathop {\mathrm {ess\, sup \;}}_{x\in I}\frac{1}{w(x)}<\infty ,\end{aligned}$$

  when $p=1$.

By $L^p((0,\infty ),w,m_\lambda )$ we understand the $L^p$-space on $(0,\infty )$ with respect to $w(x)x^{2\lambda }dx$.

In this paper we obtain quantitative weighted $L^p$-inequalities for the operators $\phi _{*}^{\lambda }$, $\phi _{*,b}^{\lambda ,m}$, $g_{\phi }^{\lambda }$, $g_{\phi ,b}^{\lambda ,m}$, $V_{\rho }^{\lambda }$ and $V_{\rho ,b}^{\lambda ,m}$. We are going to state our results.

Theorem 1.1

Let $1<p<\infty$, $\lambda >0$, $\rho >2$ and $\phi \in Z^\lambda$. Assume that $w\in A_p^\lambda (0,\infty )$. If T represents to $\phi _{*}^{\lambda }$, $g_{\phi }^{\lambda }$ or $V_{\rho }^{\lambda }(\{\phi _t\}_{t>0})$, then T can be extended to a bounded operator from $L^p((0,\infty ),w,m_\lambda )$ into itself and there exists $C>0$ that does not depend on w such that

$$\begin{aligned}&\Vert T(f)\Vert _{L^p((0,\infty ),w,m_\lambda )} \le C[w]_{A_p^\lambda }^{\max \{\frac{1}{p-1},1\}}\Vert f\Vert _{L^p((0,\infty ),w,m_\lambda )}, \end{aligned}$$

for every $f\in L^p((0,\infty ),w,m_\lambda )$.

Stein, in his monograph [68], developed a general theory of harmonic analysis associated with semigroups of operators. The heat semigroup $\{W_t^\lambda \}_{t>0}$ and the Poisson semigroup $\{P_t^\lambda \}_{t>0}$ defined by the Bessel operator $\Delta _\lambda$ can be seen as special cases of the Stein’s theory. Then, unweighted $L^p$-inequalities for maximal operators and Littlewood-Paley functions associated with $\{W_t^\lambda \}_{t>0}$ and $\{P_t^\lambda \}_{t>0}$ follow from [68, Theorem in p. 73] and [68, Theorem 10, p. 111], respectively. By using Calderón-Zygmund theory for singular integrals in homogeneous spaces, in [6] the authors obtained weighted $L^p$-inequalities for heat and Poisson maximal and Littlewood-Paley functions related to $\Delta _\lambda$. Some weak and restricted weak type inequalities for those operators were obtained in [10].

$L^p$-inequalities without weights for maximal operators, square functions and variation operators associated with the semigroups were obtained in [46] and [47]. Wu, Yang and Zhang ([72]) studied oscillation and variation operators defined by heat and Poisson semigroups in the $\Delta _\lambda$-setting.

The results in Theorem 1.1 extend the above ones in two folds. On the one hand we consider general Hankel convolution operators that include, as special cases, heat and Poisson semigroups and, on the second hand, weighted $L^p$-inequalities are established showing as the operator norms depend on the characteristic $[w]_{A_p^\lambda }$ of the weight.

Suppose that $\mathfrak {m}$ is a bounded measurable function on $(0,\infty )$. Then, $\mathfrak {m}$ defines a $\Delta _\lambda$-spectral multiplier, $T_{\mathfrak {m}}$ that is bounded from $L^2((0,\infty ),m_\lambda )$ into itself, by

$$\begin{aligned} T_{\mathfrak {m}}(\Delta _\lambda )=\int _0^\infty \mathfrak {m}(s)dE_\lambda (s), \end{aligned}$$

where $E_\lambda$ denotes the spectral measure of $\Delta _\lambda$ in $L^2((0,\infty ),m_\lambda )$. $T_{\mathfrak {m}}$ can be written by using the Hankel transformation as follows

$$\begin{aligned} T_{\mathfrak {m}}(\Delta _\lambda )(f)=h_\lambda (\mathfrak {m} (y^2)h_\lambda (f)),\quad f\in L^2((0,\infty ),m_\lambda ). \end{aligned}$$

In [28] Dziubański, Preisner and Wróbel gave sufficient conditions on the function $\mathfrak {m}$ that guarantee that the operator $T_{\mathfrak {m}}(\Delta _\lambda )$ can be extended from $L^2((0,\infty ),m_\lambda )\cap L^p((0,\infty ),m_\lambda )$ to $L^p((0,\infty ),m_\lambda )$ as a bounded operator from $L^p((0,\infty ),m_\lambda )$ into itself when $1<p<\infty$ and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$ when $p=1$, provided that $\lambda >1/2$.

The operator $\Delta _\lambda$ satisfies the conditions (A.1) and (A.2) in [12] (also in [15]) (see [13, p. 7269]). As in [12, p. 880] and in [15, p. 5] we consider a real valued and even function $\psi$ in the Schwartz space $S({\mathbb {R}})$ such that $\int _0^\infty \psi ^2(s)\frac{ds}{s}<\infty$. We define the square function $G_{\psi ,\lambda }$ as follows

$$\begin{aligned} G_{\psi ,\lambda }(f)(x)=\left( \int _0^\infty |T_{\psi _{(t)}}(\Delta _\lambda )f(x)|^2\frac{dt}{t}\right) ^{1/2},\quad x\in (0,\infty ). \end{aligned}$$

Here $\psi _{(t)}(x)=\psi (t\sqrt{x})$, $t,x\in (0,\infty )$. We can write

$$\begin{aligned} G_{\psi ,\lambda }(f)(x)=\frac{1}{\sqrt{2}}\left( \int _0^\infty |T_{\psi _{(\sqrt{t})}}(\Delta _\lambda )f(x)|^2\frac{dt}{t}\right) ^{1/2},\quad x\in (0,\infty ). \end{aligned}$$

We have that

$$\begin{aligned} T_{\psi _{(\sqrt{t})}}(\Delta _\lambda )(f)&= h_\lambda (\psi (ty)h_\lambda (f)(y)) \\&=h_\lambda (\psi (ty))\#_\lambda f \\&=\phi _t\#_\lambda f,\quad f\in L^2((0,\infty ),m_\lambda ), \end{aligned}$$

where $\phi =h_\lambda (\psi )$. Since $\psi \in S({\mathbb {R}})$ is even, we have that $\phi ={\widetilde{\phi }}_{|(0,\infty )}$ where ${\widetilde{\phi }}$ is also even and belongs to $S({\mathbb {R}})$ ( [2, Satz 5 and p. 201]). Then, $\phi \in Z^\lambda$. Furthermore, we get

$$\begin{aligned} G_{\psi ,\lambda }=g_{\phi }^{\lambda }. \end{aligned}$$

Our result in Theorem 1.1 for the square function extends the ones in [12, Theorem 1.6] and in [15, Theorem 2.3] for Bessel operators $\Delta _\lambda$. The method we use to prove Theorem 1.1 is different than the one in the mentioned papers. Furthermore, we note that in our estimate the exponent of $[w]_{A_p}$ is $\max \{1,\frac{1}{p-1}\}$ while in [12, Theorem 1.6] and [15, Theorem 2.3] the corresponding exponent is $\max \{\frac{1}{2},\frac{1}{p-1}\}$.

In order to get this exponent $\max \{\frac{1}{2},\frac{1}{p-1}\}$, in [12] and [15] quadratic sparse domination is used. According to [60, Theorem 1.1] a quadratic sparse domination is obtained provided that our $g_\phi ^\lambda$ square function satisfies a kind of 2-sublinearity property (see the third bullet in [60, Theorem 1.1] for the definition of r-sublinearity with $0<r<\infty$). But this 2-sublinearity property does not hold for $g_\phi ^\lambda$. Note that $\phi$ has not compact support. It is also remarkable that the Hankel translation $\,_\lambda \tau _x$ is not associated to any group, that is, there is not an operation $\otimes$ in $(0,\infty )$ such that $((0,\infty ), \otimes )$ is a group and $\,_\lambda \tau _x(f)(y)=f(x\otimes y)$, $x,y\in (0,\infty )$. In [53, p. 28] the authors assert that [60, Theorem 1.1] does not apply even to localized (in time) square functions in the Euclidean case. In order to get the exponent $\max \{\frac{1}{2},\frac{1}{p-1}\}$ for our square function $g_\phi ^\lambda$, we purpose to exploit the ideas in [12, 49] and [51] by using local mean oscillations in a future work.

An important spectral multiplier is that one named Bochner-Riesz multiplier. The $\alpha$-Bochner-Riesz multiplier associated with $\Delta _\lambda$ is defined by

$$\begin{aligned} B_t^{\lambda ,\alpha }(f)=h_\lambda \left( \left( 1-\left( \frac{y}{t}\right) ^2\right) _+^\alpha h_\lambda (f)\right) ,\quad t>0,\end{aligned}$$

where $z_+=\max \{0,z\}$, $z\in {\mathbb {R}}$, and $\alpha >0$. Since ([29, (7) p. 48])

$$\begin{aligned} h_\lambda \left( 2^\alpha \Gamma (\alpha +1)y^{-\alpha -\lambda -1/2}J_{\alpha +\lambda +1/2}(y)\right) (x)=(1-x^2)_+^\alpha ,\quad x\in (0,\infty ),\end{aligned}$$

we have that

$$\begin{aligned} B_t^{\lambda ,\alpha }(f)=\phi _{1/t}^{\lambda ,\alpha }\#_\lambda f,\quad t>0, \end{aligned}$$

where $\phi ^{\lambda ,\alpha }(x)=2^\alpha \Gamma (\alpha +1)x^{-\alpha -\lambda -1/2}J_{\alpha +\lambda +1/2}(x)$, $x\in (0,\infty )$. We can see that $\phi ^{\lambda , \alpha }\in Z^\lambda$ provided that $\alpha \ge \lambda +3$ (see Sect. 2.4).

By taking into account that the Fourier transformation on ${\mathbb {R}}^n$ when it acts on radial functions reduces to the Hankel transform $h_{\frac{n-1}{2}}$, we can see that the Bochner-Riesz means $B_ t^{\frac{n-1}{2},\alpha }$ appears when the Euclidean Bochner-Riesz mean on ${\mathbb {R}}^n$ acts on radial functions.

As it is wellknown, many authors and even in the last years have studied Bochner-Riesz means in ${\mathbb {R}}^n$ and maximal and square functions associated with them ([5, 11, 16, 21, 36, 44, 48] and [71]). We recall that the critical exponent for the Bochner-Riesz means in ${\mathbb {R}}^n$ is $\alpha =\frac{n-1}{2}$. Note that, when $\lambda =\frac{n-1}{2}$, the admisible values of $\alpha$ we obtain in Corollary 2.9 are greater than $\frac{n-1}{2}$ (specifically, $\alpha \ge \frac{n+1}{2}$ for the corresponding maximal operator, and $\alpha \ge \frac{n+5}{2}$ for the associated square function and $\rho$-variation operator).

It is clear that

$$\begin{aligned}&(\phi ^{\lambda ,\alpha })_*^\lambda (f)=\sup _{t>0}|B_t^{\lambda ,\alpha }(f)|, \\&(\phi ^{\lambda ,\alpha })_{*,b}^{\lambda ,m}(f)(x)=\sup _{t>0}|B_t^{\lambda ,\alpha }((b(\cdot )-b(x))^m f)(x)|,\quad x\in (0,\infty ), \\&V_\rho ^\lambda (\{\phi _t^{\lambda ,\alpha }\}_{t>0})(f)(x)=V_\rho (\{B_t^{\lambda ,\alpha }(f)(x)\}_{t>0}),\quad x\in (0,\infty ), \end{aligned}$$

and

$$\begin{aligned}&V_{\rho ,b}^{\lambda ,m} (\{\phi _t^{\lambda ,\alpha }\}_{t>0})(f)(x)=V_\rho (\{B_t^{\lambda ,\alpha }((b(\cdot )-b(x))^m f)(x)\}_{t>0}),\quad x\in (0,\infty ). \end{aligned}$$

On the other hand, after a change of variable we get

$$\begin{aligned} g_{\psi ^{\lambda ,\alpha }}^\lambda (f)(x)=\left( \int _0^\infty |t\partial _tB_t^{\lambda ,\alpha }(f)(x)|^2\frac{dt}{t}\right) ^{1/2},\quad x\in (0,\infty ), \end{aligned}$$

and

$$\begin{aligned} g_{\psi ^{\lambda ,\alpha },b}^{\lambda , m}(f)(x)=\left( \int _0^\infty |t\partial _tB_t^{\lambda ,\alpha }((b(\cdot )-b(x))^m f)(x)|^2\frac{dt}{t}\right) ^{1/2},\quad x\in (0,\infty ), \end{aligned}$$

where $\psi ^{\lambda ,\alpha }(x)=-(2\lambda +1)\phi ^{\lambda ,\alpha }(x)-x(\phi ^{\lambda ,\alpha })'(x)$, $x\in (0,\infty )$. Note that $\psi ^{\lambda ,\alpha }\in Z^\lambda$ provided that $\alpha \ge \lambda +4$.

The square function $g_{\psi ^{\lambda ,\alpha }}^\lambda$ is a Bessel version of the Stein square function introduced in [67]. Quantitative weighted $L^p$-inequalities for the classical Stein square function were obtained in [16]. In [17] Chen, Duong and Yan established $L^p$-boundedness properties for Stein square function associated with operators in spaces of homogeneous type. The results in Theorem 1.1 suppose an extension of [17, Theorem 1.1]. By using Theorem 1.1 we can deduce for $g_{\psi ^{\lambda ,\alpha }}^\lambda$ some results that extend those ones in [17, Theorem 1.1] (see Corollary 2.9).

Theorem 1.2

Let $1<p<\infty$, $\lambda >0$, $\rho >2$, $m\in {\mathbb {N}}\setminus \{0\}$ and $\phi \in Z^\lambda$. Suppose that $w_1,w_2\in A_p^\lambda (0,\infty )$. If T represents to $\phi _{*,b}^{\lambda ,m}$, $g_{\phi ,b}^{\lambda ,m}$ or $V_{\rho ,b}^{\lambda ,m}(\{\phi _t\}_{t>0})$, then T can be extended to a bounded operator from $L^p((0,\infty ),w_1,m_\lambda )$ into $L^p((0,\infty ),w_2,m_\lambda )$ and there exists $C>0$ that does not depend on $w_1$ neither $w_2$ such that

$$\begin{aligned}&\Vert T(f)\Vert _{L^p((0,\infty ),w_2,m_\lambda )}\\&\qquad \le C\Vert b\Vert _{BMO((0,\infty ),w,m_\lambda )}^m\left( [w_1]_{A_p^\lambda }[w_2]_{A_p^\lambda }\right) ^{\frac{m+1}{2}\max \{\frac{1}{p-1},1\}}\Vert f\Vert _{L^p((0,\infty ),w_1,m_\lambda )}, \end{aligned}$$

for every $f\in L^p((0,\infty ),w_1,m_\lambda )$, provided that $b\in BMO((0,\infty ),w,m_\lambda )$ where $w=\left( \frac{w_1}{w_2}\right) ^{\frac{1}{mp}}$.

As far as we know commutators, even of the first order, related to maximal and variation operators and Littlewood-Paley functions defined by Hankel convolutions have not been yet studied.

In order to prove our results we see that the operators can be dominated by finite sums of sparse operators. This technique of controlling operators by sparse ones has been very useful in the last decade to get weighted norm inequalities. The sparse domination procedure was used at the first time by Lerner ([50]) where an alternative proof of the $A_2$-theorem was shown. The $A_2$-theorem had been established by Hytönen in [40]. Sparse arguments have bee used to study quantitative $L^p$-weighted inequalities for several class of operators (see [1, 12, 24, 32, 44, 52, 54,55,56,57,58, 60, 65] and the references therein).

One of the reasons to study sparse domination for operators is to get weighted $L^p$-inequalities that are sharp in some sense. Quantitative bounds of the norm $\Vert T\Vert _{L^p(w)}$ of an operator T in terms of the $A_p$-constant $[w]_{A_p}$ of the weight w is obtained. The last objective is to find the sharp dependence of $\Vert T\Vert _{L^p(w)}$ with respect to $[w]_{A_p}$. A typical problem is to obtain the least $\beta >0$ such that $\Vert T\Vert _{L^p(w)}\sim [w]_{A_p}^\beta$ where the equivalence constant does not depend on w. In [61] Luque, Pérez and Rela proved that the optimal lower bound of $\beta$ is closely related to the asymptotic behaviour of the unweighted $L^p$-norm $\Vert T\Vert _{L^p}$ of T as p goes to 1 and $+\infty$ ( [61, Theorem 1.2]). In order to study the sharp dependence of the weighted $L^p$-norm with respect to $[w]_{A_p}$ for the operators in Theorems 1.1 and 1.2 we purpose to employ [61, Theorem 1.2]. We have to calculate for each operator the parameters $\alpha$ and $\gamma$ introduced in [61, Definition 1.1]. This problem is related with the one that we mentioned earlier about the exponent of $[w]_{A_p}$ when the square function $g_\phi ^\lambda$ is considered and we will address in a future paper.

We recall the definition of systems of dyadic cubes. Let $\delta \in (0,1)$ and $0<a\le A<\infty$. A system of dyadic cubes with parameters $(\delta ,a,A)$ is a family $\displaystyle {\mathcal {D}}=\bigcup _{k\in {\mathbb {Z}}}{\mathcal {D}}_k$ where ${\mathcal {D}}_k=\{Q_\alpha ^k\subset (0,\infty )\;\text{ measurable },\;\alpha \in {\mathcal {A}}_k\}$ being ${\mathcal {A}}_k$ countable, for every $k\in {\mathbb {Z}}$, and the following properties are satisfied

• (i) $Q_\alpha ^k\cap Q_\beta ^k=\emptyset$, $\alpha \ne \beta$, $\alpha ,\beta \in {\mathcal {A}}_k$, $k\in {\mathbb {Z}}$ and $\displaystyle (0,\infty )=\bigcup _{\alpha \in {\mathcal {A}}_k}Q_\alpha ^k$, $k\in {\mathbb {Z}}$.

• (ii) If $\ell \ge k$, $\alpha \in {\mathcal {A}}_\ell$, $\beta \in {\mathcal {A}}_k$, then either $Q_\alpha ^\ell \subseteq Q_\beta ^k$ or $Q_\alpha ^\ell \cap Q_\beta ^k=\emptyset$.

• (iii) There exists $M\in {\mathbb {N}}$ such that for every $k\in {\mathbb {Z}}$ and $\alpha \in {\mathcal {A}}_k$

  $$\begin{aligned} \text{ card }\{\beta \in {\mathcal {A}}_{k+1}:\;Q_\beta ^{k+1}\subseteq Q_\alpha ^k\}\le M, \end{aligned}$$

  and $\displaystyle Q_\alpha ^k=\bigcup _{Q\in {\mathcal {D}}_{k+1}, Q\subseteq Q_\alpha ^k}Q$.

• (iv) For every $k\in {\mathbb {Z}}$ and $\alpha \in {\mathcal {A}}_k$ there exists $x_\alpha ^k\in (0,\infty )$ such that

  $$\begin{aligned} B(x_\alpha ^k,a\delta ^k)\subseteq Q_\alpha ^k\subseteq B(x_\alpha ^k,A\delta ^k). \end{aligned}$$

  Furthermore, if $\ell \ge k$ and $Q_\beta ^\ell \subseteq Q_\alpha ^k$, then $B(x_\beta ^\ell ,A\delta ^\ell )\subseteq B(x_\alpha ^k,A\delta ^k)$.

The set $Q_\alpha ^k$ is usually named a dyadic cube of generation k with center at $x_\alpha ^k\in Q_\alpha ^k$ and sidelength $\delta ^k$.

Main properties of dyadic cubes in quasimetric spaces with the geometric doubling property can be found in [39] (see also [19]).

Let ${\mathcal {D}}$ be a system of dyadic cubes and $0<\eta <1$. As in [24] we say that a collection $S\subset {\mathcal {D}}$ is $\eta$-sparse when there exists $c\ge 1$ such that for every $Q\in S$ there exists a measurable set $E_Q\subset Q$ such that $m_\lambda (E_Q)\ge \eta m_\lambda (Q)$ and

$$\begin{aligned} \sum _{Q\in S}{\mathcal {X}}_{E_Q}(x)\le c,\quad x\in (0,\infty ). \end{aligned}$$

(1.1)

If $S\subset {\mathcal {D}}$ is $\eta$-sparse, the sparse operator ${\mathfrak {A}}_S$ is defined by

$$\begin{aligned} {\mathfrak {A}}_S f(x)= \sum _{Q\in S} f_Q{\mathcal {X}}_{Q}(x),\quad x\in (0,\infty ). \end{aligned}$$

Here, as above, $f_Q=\frac{1}{m_\lambda (Q)}\int _Q f(x)x^{2\lambda }dx$, $Q\in S$.

The first quantitative estimates in the homogeneous setting for a “pseudo" sparse operator (not quite the sparse operator as we know it now) are due to Anderson and Vagharshakyan ([4, §6]). Recently, quantitative $L^p$-weighted estimates for sparse operators have been established in spaces of homogeneous type by Lorist ([60, Proposition 4.1]).

In order to use the sparse domination strategy we take into account that the operators that we study can be seen as Banach valued linear operators. Indeed, let $(E,\Vert \cdot \Vert _E)$ be a Banach space of measurable functions defined in $(0,\infty )$. We define

$$\begin{aligned} {\mathcal {P}}_{\phi ;E}^\lambda (f)(x)=\Vert f\#_\lambda \phi _{\centerdot }(x)\Vert _E,\quad x\in (0,\infty ), \end{aligned}$$

and

$$\begin{aligned} {\mathcal {C}}_{\phi ,b;E}^{\lambda ,m}(f)(x)=\Vert [(b(\cdot )-b(x))^m f]\#_\lambda \phi _{\centerdot }(x)\Vert _E,\quad x\in (0,\infty ). \end{aligned}$$

We have that

• (i) $\phi _*^\lambda ={\mathcal {P}}_{\phi ;L^\infty (0,\infty )}^\lambda$ and $\phi _{*,b}^{\lambda ,m}={\mathcal {C}}_{\phi ,b;L^\infty (0,\infty )}^{\lambda ,m}$.

• (ii) $g_\phi ^\lambda ={\mathcal {P}}_{\phi ;L^2((0,\infty ),\frac{dt}{t})}^\lambda$ and $g_{\phi ,b}^{\lambda ,m}={\mathcal {C}}_{\phi ,b;L^2((0,\infty ),\frac{dt}{t})}^{\lambda ,m}$.

• (iii) We define $v_\rho$ as the space of measurable functions $g:(0,\infty )\rightarrow {\mathbb {C}}$ such that

  $$\begin{aligned} \Vert g\Vert _{v_\rho }=\sup _{\begin{array}{c} 0<t_n<t_{n-1}<...<t_1 \\ n\in {\mathbb {N}} \end{array}}\left( \sum _{k=1}^{n-1}|g(t_k)-g(t_{k+1})|^\rho \right) ^{1/\rho }<\infty . \end{aligned}$$

  By identifying the constant functions the space $(v_\rho ,\Vert \cdot \Vert _{v_\rho })$ is Banach. Furthermore,

  $$\begin{aligned} V_\rho ^\lambda (\{\phi _t\}_{t>0})={\mathcal {P}}_{\phi ;v_\rho }^\lambda \;\;\text{ and }\;\;V_{\rho ,b}^{\lambda ,m}(\{\phi _t\}_{t>0})={\mathcal {C}}_{\phi ,b;v_\rho }^{\lambda ,m}. \end{aligned}$$

For every $1\le p\le \infty$ we denote by $L^p((0,\infty ),m_\lambda ,E)$ and $L^{p,\infty }((0,\infty ),m_\lambda ,E)$ the E-valued p-th and weak p-th Bochner-Lebesgue space, respectively, on the measure space $((0,\infty ),m_\lambda )$.

In the proof of our results, those ones in [24] and [60] about sparse domination on spaces of homogeneous type play an important role. Specifically, in order to prove Theorem 1.1 we apply [60, Theorem 1.1]. The results in [24] are concerned with weighted inequalities for commutators by using sparse domination. After adapting [24, Theorem 3.7] for Banach valued operators we use it in the proof of Theorem 1.2.

Throughout this paper C will always represents a positive constant that can change in each occurrence.

2 Proof of theorem 1.1

In [52] a sparse domination principle was given for an operator T by using the grand maximal truncated operator ${\mathcal {M}}_T$ of T defined by

$$\begin{aligned} {\mathcal {M}}_T(f)(x)=\sup _{Q\ni x}\Vert T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (3Q)})\Vert _{L^\infty (Q)},\quad x\in {\mathbb {R}}^n, \end{aligned}$$

where the supremum is taken over all the cubes in ${\mathbb {R}}^n$ containing x. It can be stated as follows.

Theorem 2.1

([52, Theorem 4.2]). Let $1\le q\le r<\infty$. Assume that T is a sublinear operator that is bounded from $L^q({\mathbb {R}}^n)$ into $L^{q,\infty }({\mathbb {R}}^n)$ and such that ${\mathcal {M}}_T$ is bounded from $L^r({\mathbb {R}}^n)$ into $L^{r,\infty }({\mathbb {R}}^n)$. Then, for every compactly supported $f\in L^r({\mathbb {R}}^n)$, there exists a sparse family S such that

$$\begin{aligned} |Tf(x)|\le C\sum _{Q\in S}\langle f\rangle _{r,Q}{\mathcal {X}}_Q(x),\quad \text{ a.e. } x\in {\mathbb {R}}^n, \end{aligned}$$

where $C=C_{n,q,r}\left( \Vert T\Vert _{L^q({\mathbb {R}}^n)\rightarrow L^{q,\infty }({\mathbb {R}}^n)}+\Vert {\mathcal {M}}_T\Vert _{L^r({\mathbb {R}}^n)\rightarrow L^{r,\infty }({\mathbb {R}}^n)}\right)$.

Here, $\langle f\rangle _{r,Q}=\left( \frac{1}{|Q|}\int _Q |f(x)|^r dx\right) ^{1/r}$, $Q\in S$.

In [55] Theorem 2.1 was improved by weakening some of the hypotheses. The operator ${\mathcal {M}}_T$ is replaced by the one ${\mathcal {M}}_{T,\alpha }^\#$ defined for every $\alpha >0$ by

$$\begin{aligned} {\mathcal {M}}_{T,\alpha }^\#(f)(x)=\sup _{Q\ni x}\;\mathop {\mathrm {ess\, sup \;}}_{x',x''\in Q}|T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (\alpha Q)})(x')-T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (\alpha Q)})(x'')|,\quad x\in {\mathbb {R}}^n, \end{aligned}$$

(see [55, Theorem 1.1]). Here, as above, Q represents a cube in ${\mathbb {R}}^n$.

Note that the operators ${\mathcal {M}}_T$ and ${\mathcal {M}}_{T,3}^\#$ are connected as follows. Let Q be a cube in ${\mathbb {R}}^n$ and $x\in Q$. We have that

$$\begin{aligned} |T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (3 Q)})(y)|&\le |T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (3 Q)})(y)-T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (3 Q)})(z)| \\&\quad + |Tf(z)|+|T(f{\mathcal {X}}_{3Q})(z)|, \;\;\;\;y,z\in Q. \end{aligned}$$

Then,

$$\begin{aligned} |T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (3 Q)})(y)| \le&\mathop {\mathrm {ess\, sup \;}}_{y,z\in Q}|T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (3 Q)})(y)-T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (3 Q)})(z)| \\&+ \inf _{z\in Q}\left( |Tf(z)|+|T(f{\mathcal {X}}_{3Q})(z)|\right) \\ \le&\mathop {\mathrm {ess\, sup \;}}_{y,z\in Q}|T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (3 Q)})(y)-T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (3 Q)})(z)| \\&+ C\left[ \left( \frac{1}{|Q|}\int _Q |T(f{\mathcal {X}}_{3Q})(z)|^{1/2}dz\right) ^2\right. \\&\left. +\left( \frac{1}{|Q|}\int _Q |Tf(z)|^{1/2}dz\right) ^2\right] \\ \le&\mathop {\mathrm {ess\, sup \;}}_{y,z\in Q}|T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (3 Q)})(y)-T(f{\mathcal {X}}_{{\mathbb {R}}^n\setminus (3 Q)})(z)| \\&+ C\left[ \left( {\mathcal {M}}[(T(f{\mathcal {X}}_{3Q}))^{1/2}]\right) ^2(x)+\left( {\mathcal {M}}[(Tf)]^{1/2}\right) ^2(x)\right] , \end{aligned}$$

where ${\mathcal {M}}$ denotes the Hardy-Littlewood maximal function. We get

$$\begin{aligned} {\mathcal {M}}_T(f)(x)\le C\left( {\mathcal {M}}_{T,3}^\#(f)(x)+\left( {\mathcal {M}}[(Tf)]^{1/2})(x)\right) ^2+\left( {\mathcal {M}}[(T(f{\mathcal {X}}_{3Q}))^{1/2}](x)\right) ^2\right) . \end{aligned}$$

Suppose that T is bounded from $L^1({\mathbb {R}}^n)$ into $L^{1,\infty }({\mathbb {R}}^n)$. By using Kolmogorov inequality ([35, p. 91]) we deduce that

$$\begin{aligned} {\mathcal {M}}_T(f)(x)\le C\left( {\mathcal {M}}_{T,3}^\#(f)(x)+\left( {\mathcal {M}}(|Tf|^{1/2})(x)\right) ^2+{\mathcal {M}}(|f|)(x)\right) . \end{aligned}$$

By using [63, Lemma 3.2] we conclude that ${\mathcal {M}}_T$ is bounded from $L^1({\mathbb {R}}^n)$ into $L^{1,\infty }({\mathbb {R}}^n)$ provided that ${\mathcal {M}}_{T,3}^\#$ is bounded from $L^1({\mathbb {R}}^n)$ into $L^{1,\infty }({\mathbb {R}}^n)$.

On the other hand it is clear that ${\mathcal {M}}_{T,3}^\#$ is bounded from $L^1({\mathbb {R}}^n)$ into $L^{1,\infty }({\mathbb {R}}^n)$ when ${\mathcal {M}}_T$ is bounded from $L^1({\mathbb {R}}^n)$ into $L^{1,\infty }({\mathbb {R}}^n)$.

In our Bessel contexts we consider the spaces of homogeneous type $((0,\infty ),|\cdot |,m_\lambda )$. A version of Theorem 2.1 for spaces of homogeneous type in vector valued settings was established in [60, Theorem 1.1]. We will prove that the operators under consideration in Theorem 1.1 are bounded from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$. Then the above comments concerning to ${\mathcal {M}}_T$ and ${\mathcal {M}}_{T,3}^\#$ also works in the $((0,\infty ),|\cdot |,m_\lambda )$ context.

In the sequel for each $x,r \in (0,\infty )$, we write B(x, r) to denote the interval $(x-r,x+r) \cap (0,\infty )$, and $cB(x,r)=B(x,cr)$, $c>0$.

2.1 Proof of theorem 1.1 for $\phi _*^\lambda$

Assume that $\phi$ satisfies (i) (see the definition of $Z^\lambda$). According to [73, (2.4)] we have that

$$\begin{aligned} |_\lambda \tau _x(\phi _t)(y)|\le \frac{C}{m_\lambda (B(x,t))+m_\lambda (B(x,|x-y|))}\frac{t}{t+|x-y|},\quad t,x,y\in (0,\infty ). \end{aligned}$$

Then,

$$\begin{aligned} |(f\#_\lambda \phi )(x)|&\le \sum _{k=1}^\infty \int _{B(x,2^kt)\setminus B(x,2^{k-1}t)}|_\lambda \tau _x(\phi _t)(y)||f(y)|y^{2\lambda }dy \\&\quad +\int _{B(x,t)}|_\lambda \tau _x(\phi _t)(y)||f(y)|y^{2\lambda }dy \\&\le C\sum _{k=0}^\infty \frac{1}{m_\lambda (B(x,2^kt))}\frac{1}{2^k}\int _{B(x,2^kt)}|f(y)|y^{2\lambda }dy \\&\le C{\mathcal {M}}_\lambda (f)(x),\;\;\;\;t,x\in (0,\infty ). \end{aligned}$$

Here ${\mathcal {M}}_\lambda$ represents the corresponding Hardy-Littlewood maximal operator defined by

$$\begin{aligned} {\mathcal {M}}_\lambda (f)(x)=\sup _{r>0}\frac{1}{m_\lambda (B(x,r))}\int _{B(x,r)}|f(y)|y^{2\lambda }dy,\quad f\in L^1_{loc}((0,\infty ),m_\lambda ). \end{aligned}$$

Hence,

$$\begin{aligned} \phi _*^\lambda (f)(x)\le C{\mathcal {M}}_\lambda (f)(x),\quad x\in (0,\infty ). \end{aligned}$$

Since

$$\begin{aligned} \Vert {\mathcal {M}}_\lambda (f)\Vert _{L^p((0,\infty ),w,m_\lambda )}\le C[w]_{A_p}^{\frac{1}{p-1}}\Vert f\Vert _{L^p((0,\infty ),w,m_\lambda )}, \end{aligned}$$

for every $f\in L^p((0,\infty ),w,m_\lambda )$ (see, for instance, [12, p. 890]) we conclude the proof of Theorem 1.1 for $\phi _*^\lambda$. Note that here it is sufficient to assume that $\phi$ satisfies (i).

2.2 Proof of theorem 1.1 for $g_\phi ^\lambda$

We firstly establish $L^p$-boundedness properties for the sublinear operator $g^\lambda _\phi$.

Proposition 2.2

Let $\lambda >0$. Suppose that $\phi$ is a differentiable function on $(0,\infty )$ satisfying the following three conditions:

• (a) $|\phi (x)| \le C(1+x^2)^{-\lambda -1}$, $x \in (0,\infty )$;

• (b) $\displaystyle \int _0^\infty \left| u^{2\lambda +2}\phi '(u)\right| ^2 \frac{du}{u} <\infty$;

• (c) $\displaystyle \int _0^\infty \left| h_\lambda (\phi )(u)\right| ^2\frac{du}{u} < \infty$.

Then, the operator $g_\phi ^\lambda$ is bounded from $L^p((0,\infty ), m_\lambda )$ into itself, for every $1<p<\infty$, and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$.

Proof

We have that

$$\begin{aligned} g^\lambda _\phi (f)(x) =\Vert f \#_\lambda \phi _\centerdot (x)\Vert _{L^2((0,\infty ),\frac{dt}{t})}, \quad x \in (0,\infty ). \end{aligned}$$

In order to see the $L^p$-boundedness properties for $g^\lambda _\phi$ we use vector valued singular integral theory ([66]).

We define

$$\begin{aligned} {\mathfrak {G}}^\lambda _\phi (f)(x,t)= \int _0^\infty {\mathfrak {G}}^\lambda _\phi (x,y,t) f(y)y^{2\lambda }dy,\quad x,t \in (0,\infty ), \end{aligned}$$

where ${\mathfrak {G}}^\lambda _\phi (x,y,t)=\,_\lambda \tau _x (\phi _t)(y)$, $x,y,t \in (0,\infty )$. It is clear that ${\mathfrak {G}}^\lambda _\phi (f)(x,t)=(f\#_\lambda \phi _t)(x)$, $x,t\in (0,\infty )$.

Our objective is to see that the operator ${\mathfrak {G}}_\phi ^\lambda$ is bounded from $L^p((0,\infty ),m_\lambda )$ into $L^p((0,\infty ), m_\lambda , E)$, for every $1<p<\infty$, and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda , E)$, where $E=L^2((0,\infty ),\frac{dt}{t})$.

Let $N\in {\mathbb {N}}$. We consider, for each smooth function f with compact support in $(0,\infty )$,

$$\begin{aligned} {\mathcal {G}}^\lambda _{\phi ,N}(f)(x) = \int _0^\infty {\mathfrak {G}}^\lambda _\phi (x,y,\cdot ) f(y) y^{2\lambda } dy,\quad x\in (0,\infty )\setminus \mathrm{supp } f, \end{aligned}$$

(2.1)

where the integral is understood in the $L^2((\frac{1}{N},N),\frac{dt}{t})$-Bochner sense.

We are going to justify that the integral in (2.1) is convergent in the $L^2((\frac{1}{N},N),\frac{dt}{t})$-Bochner sense. Indeed, let f be a smooth function with compact support in $(0,\infty )$ and $x\in (0,\infty )\setminus \mathrm{supp }\,f$. We consider the function $F_x$ defined by

$$\begin{aligned} F_x(y)={\mathfrak {G}}^\lambda _{\phi }(x,y,\cdot ) f(y) y^{2\lambda },\quad y\in (0,\infty ). \end{aligned}$$

In order to establish the properties of the function $F_x$ we need to prove some size and regularity estimates for the kernel ${\mathfrak {G}}^\lambda _{\phi }(x,y,\cdot )$ in the $L^2((0,\infty ),\frac{dt}{t})$-norm, for every $y\in (0,\infty )$. These properties of ${\mathfrak {G}}^\lambda _{\phi }(x,y,\cdot )$ will be also useful to apply Calderón-Zygmund theory to the vector-valued operator ${\mathcal {G}}_{\phi ,N}^\lambda$ and then we get our objective.

We have that

$$\begin{aligned} \left\| {\mathfrak {G}}^\lambda _{\phi }(z,y,\cdot )\right\| _{L^2((0,\infty ),\frac{dt}{t})} \le \frac{C}{m_\lambda (B(z,|z-y|))}, \quad z,y \in (0,\infty ),\;z \not =y. \end{aligned}$$

(2.2)

Indeed, by using Minkowski inequality we get, for every $z,y\in (0,\infty )$,

$$\begin{aligned}&\left\| {\mathfrak {G}}^\lambda _{\phi }(z,y,\cdot )\right\| _{L^2((0,\infty ),\frac{dt}{t})}\\&\quad \le C\int ^\pi _0 \sin ^{2\lambda -1}\theta \left( \int _0^\infty \Big |\frac{1}{t^{2\lambda +1}}\phi \left( \frac{\sqrt{z^2+y^2-2zy\cos \theta }}{t}\right) \Big |^{2}\frac{dt}{t}\right) ^{1/2}d\theta . \end{aligned}$$

We can write

$$\begin{aligned}&\int _0^\infty \left| \frac{1}{t^{2\lambda +1}}\phi \left( \frac{\sqrt{z^2+y^2-2zy\cos \theta }}{t}\right) \right| ^2\frac{dt}{t}\\&\quad = \frac{1}{(z^2+y^2-2zy\cos \theta )^{2\lambda +1}}\int _0^\infty |u^{2\lambda +1}\phi (u)|^2 \frac{du}{u} \\&\quad \le \frac{C}{(z^2+y^2-2zy\cos \theta )^{2\lambda +1}}, \quad z,y \in (0,\infty ) \text{ and } \theta \in (0,\pi ),\end{aligned}$$

because $\phi$ satisfies (a).

According to [6, Lemma 3.1] we obtain, for each $z,y \in (0,\infty )$, $z \not =y$,

$$\begin{aligned} \left\| {\mathfrak {G}}^\lambda _{\phi }(z,y,\cdot )\right\| _{L^2((0,\infty ),\frac{dt}{t})} \le C\int _0^\infty \frac{\sin ^{2\lambda -1}\theta }{(z^2+y^2-2zy\cos \theta )^{2\lambda +1}}d\theta \le \frac{C}{m_\lambda (B(z,|z-y|))}, \end{aligned}$$

and (2.2) is proved.

We also have that, for every $z,y\in (0,\infty )$, $z\not =y$,

$$\begin{aligned} \left\| \partial _z {\mathfrak {G}}^\lambda _{\phi }(z,y,\cdot )\right\| _{L^2((0,\infty ), \frac{dt}{t})}+ \left\| \partial _y {\mathfrak {G}}^\lambda _\phi (z,y,\cdot )\right\| _{L^2((0,\infty ), \frac{dt}{t})} \le \frac{C}{|z-y|m_\lambda (B(z,|z-y|))}.\nonumber \\ \end{aligned}$$

(2.3)

We can write, for every $z,y,t \in (0,\infty )$,

$$\begin{aligned} \partial _z {\mathfrak {G}}^\lambda _\phi (z,y,t)=\frac{\Gamma (\lambda + \frac{1}{2})}{\sqrt{\pi }\Gamma (\lambda )t^{2\lambda +1}}\int _0^\pi \phi '\left( \frac{\sqrt{z^2+y^2-2zy\cos \theta }}{t}\right) \frac{(z-y\cos \theta )\sin ^{2\lambda -1}\theta }{t\sqrt{z^2+y^2-2zy\cos \theta }}d\theta . \end{aligned}$$

Since $|a+bs| \le (a^2+b^2+2abs)^{1/2}$, $a,b \in (0,\infty )$ and $s \in (-1,1)$, it follows that

$$\begin{aligned} \left| \partial _z{\mathfrak {G}}^\lambda _\phi (x,y,t)\right| \le \frac{C}{t^{2\lambda +2}}\int _0^\pi \left| \phi '\left( \frac{\sqrt{z^2+y^2-2zy\cos \theta }}{t}\right) \right| \sin ^{2\lambda -1}\theta d\theta ,\,z,y,t\in (0,\infty ). \end{aligned}$$

By proceeding as above by using (b), we get

$$\begin{aligned}&\left\| \partial _z {\mathfrak {G}}_\phi (z,y,\cdot )\right\| _{L^2\left( (0,\infty ),\frac{dt}{t}\right) } \\&\quad \le C\int _0^\pi \sin ^{2\lambda -1}\theta \left( \int _0^\infty \left| \frac{1}{t^{2\lambda +2}}\phi '\left( \frac{\sqrt{z^2+y^2-2zy\cos \theta }}{t}\right) \right| ^2\frac{dt}{t}\right) ^{\frac{1}{2}}d\theta \\&\quad \le C\int _0^\pi \frac{\sin ^{2\lambda -1}\theta }{(z^2+y^2-2zy\cos \theta )^{\lambda +1}}d\theta \left( \int _0^\infty |u^{2\lambda +2}\phi '(u) |^2\frac{du}{u}\right) ^{\frac{1}{2}}\\&\quad \le C\int _0^\pi \frac{\sin ^{2\lambda -1}\theta }{(z^2+y^2-2zy\cos \theta )^{\lambda +1}}d\theta ,\quad z,y \in (0,\infty ). \end{aligned}$$

According to [6, Lemma 3.1] we obtain

$$\begin{aligned} \left\| \partial _z {\mathfrak {G}}^\lambda _{\phi }(x,y,\cdot )\right\| _{L^2((0,\infty ), \frac{dt}{t})}\le \frac{C}{|z-y|m_\lambda (B(z,|z-y|))},\quad z,y\in (0,\infty ),\;z\not =y. \end{aligned}$$

By symmetry we finish the proof of (2.3).

The function $F_x$ is continuous from $(0,\infty )$ into $L^2((\frac{1}{N},N),\frac{dt}{t})$. Indeed, let $y_0\in (0,\infty )$. If $y_0\not \in \mathrm{supp}\,f$, then there exists $\delta _0>0$ such that $(y_0-\delta _0,y_0+\delta _0)\cap \mathrm{supp }\,f=\emptyset$ and hence, $F_x(y)=0$, $y\in (y_0-\delta _0,y_0+\delta _0)$. Suppose that $y_0\in \mathrm{supp}\,f$. There exists $\delta _0>0$ for which $x\not \in (y_0-\delta _0,y_0+\delta _0)$. By using mean value theorem, (2.2) and (2.3) we obtain

$$\begin{aligned}&\Vert F_x(y)-F_x(y_0)\Vert _{L^2\left( \left( \frac{1}{N},N\right) ,\frac{dt}{t}\right) }\\&\qquad \le C\left( \frac{|y-y_0|}{|x-y_0|m_\lambda (B(x,|x-y_0|))}+\frac{|f(y)y^{2\lambda }-f(y_0)y_0^{2\lambda }|}{m_\lambda (B(x,|x-y_0|))}\right) , \end{aligned}$$

for every $y\in (0,\infty )$ such that $|y-y_0|<\frac{\delta _0}{2}$.

It follows that $\lim _{y\rightarrow y_0}F_x(y)=F_x(y_0)$ in $L^2((\frac{1}{N},N),\frac{dt}{t})$. Since $L^2((\frac{1}{N},N),\frac{dt}{t})$ is a separable space we obtain that $F_x$ is a $L^2((\frac{1}{N},N),\frac{dt}{t})$-strongly measurable function. Furthermore, according to (2.2) the integral in (2.1) is $L^2((\frac{1}{N},N),\frac{dt}{t})$-norm convergent.

Let f be a smooth function with compact support in $(0,\infty )$. Then ${\mathcal {G}}_{\phi ,N}^\lambda (f)(x)\in L^2((\frac{1}{N},N),\frac{dt}{t})$, for every $x\in (0,\infty )\setminus \mathrm{supp }\,f$. Furthermore, if $x\in (0,\infty )\setminus \mathrm{supp }\,f$, ${\mathcal {G}}_{\phi ,N}^\lambda (f)(x)={\mathfrak {G}}_\phi ^\lambda (f)(x,\cdot )$. Indeed, let $x\in (0,\infty )\setminus \mathrm{supp }\,f$ and $h\in L^2((\frac{1}{N},N),\frac{dt}{t})$. We have that

$$\begin{aligned} \int _{1/N}^N h(t) \left[ {\mathcal {G}}_{\phi ,N}^\lambda (f)(x)\right] (t)\frac{dt}{t} = \int _0^\infty \int _{1/N}^N h(t) {\mathfrak {G}}^\lambda _\phi (x,y,t) \frac{dt}{t}f(y) y^{2\lambda }dy. \end{aligned}$$

By using again (2.2) we get

$$\begin{aligned}&\int _0^\infty \int _{1/N}^N|h(t)||{\mathfrak {G}}^\lambda _\phi (x,y,t)|\frac{dt}{t}|f(y)|y^{2\lambda } dy \\&\quad \le \Vert h\Vert _{L^2((\frac{1}{N},N),\frac{dt}{t})}\int _0^\infty \Vert {\mathfrak {G}}^\lambda _\phi (x,y,\cdot )\Vert _{L^2((0,\infty ),\frac{dt}{t})}|f(y)|y^{2\lambda }dy \\&\quad \le \Vert h\Vert _{L^2((\frac{1}{N},N),\frac{dt}{t})}\int _0^\infty \frac{|f(y)|y^{2\lambda }}{m_\lambda (B(x,|x-y|))} dy < \infty . \end{aligned}$$

It follows that

$$\begin{aligned} \int _{1/N}^N h(t)[{\mathcal {G}}^\lambda _{\phi ,N}(f)(x)](t)\frac{dt}{t} = \int _{1/N}^N h(t){\mathfrak {G}}^\lambda _\phi (f)(x,t)\frac{dt}{t}. \end{aligned}$$

Hence, for every $x\in (0,\infty )\setminus \mathop {\mathrm {supp}}f$,

$$\begin{aligned}{}[{\mathcal {G}}^\lambda _{\phi ,N}(f)(x)](t) = {\mathfrak {G}}^\lambda _\phi (f)(x,t),\quad \text{ a.e. } t\in \big (\frac{1}{N},N\big ). \end{aligned}$$

We now observe that the function

$$\begin{aligned} G(x,y,t)={\mathfrak {G}}^\lambda _\phi (x,y,t) f(y) y^{2\lambda },\quad x,y,t\in (0,\infty ), \end{aligned}$$

is continuous. Then, G is uniformly continuous in $(x,y,t)\in K\times \mathrm{supp}\,f\times [\frac{1}{N},N]$, for every compact set $K\subset (0,\infty )$. Hence, if $x_0\in (0,\infty )$ we have that

$$\begin{aligned} \lim _{x\rightarrow x_0}\Big \Vert {\mathfrak {G}}_\phi ^\lambda (f)(x,\cdot )-{\mathfrak {G}}_\phi ^\lambda (f)(x_0,\cdot )\Big \Vert _{L^2((\frac{1}{N},N),\frac{dt}{t})}=0. \end{aligned}$$

So, as above we deduce that the function ${\mathfrak {G}}_\phi ^\lambda (f)$, where

$$\begin{aligned} \begin{array}{rccrccc} {\mathfrak {G}}^\lambda _{\phi }(f):&{}(0,\infty ) &{} \longrightarrow &{} L^2\left( \left( \frac{1}{N},N\right) ,\frac{dt}{t}\right) &{}&{}&{}\\ &{}x &{} \longrightarrow &{}{\mathfrak {G}}^\lambda _{\phi }(f)(x): &{}\left( \frac{1}{N},N\right) &{}\longrightarrow &{}{\mathbb {R}}\\ &{} &{} &{} &{}t &{}\longrightarrow &{}{\mathfrak {G}}^\lambda _\phi (f)(x,t), \end{array} \end{aligned}$$

is strongly measurable. Moreover, ${\mathfrak {G}}_\phi ^\lambda$ defines a bounded operator from $L^2((0,\infty ),m_\lambda )$ into $L^2((0,\infty ),m_\lambda ,L^2((\frac{1}{N},N),\frac{dt}{t}))$ and

$$\begin{aligned} \sup _{N\in {\mathbb {N}}}\big \Vert {\mathfrak {G}}_\phi ^\lambda \big \Vert _{L^2((0,\infty ),m_\lambda )\longrightarrow L^2((0,\infty ),m_\lambda ,L^2((\frac{1}{N},N),\frac{dt}{t}))}<\infty . \end{aligned}$$

(2.4)

Indeed, by using Plancherel formula for Hankel transformation we deduce that

$$\begin{aligned} \left\| {\mathfrak {G}}^\lambda _\phi (f)\right\| ^2_{L^2_{L^2((\frac{1}{N},N)\frac{dt}{t})}((0,\infty ),m_\lambda )}&=\left\| \Vert {\mathfrak {G}}^\lambda _\phi (f)\Vert _{L^2((\frac{1}{N},N)\frac{dt}{t})}\right\| ^2_{L^2((0,\infty ),m_\lambda )}\\&\le \int _0^\infty \int _0^\infty \left| {\mathfrak {G}}^\lambda _\phi (f)(x,t)\right| ^2\frac{dt}{t}x^{2\lambda }dx\\&= \int _0^\infty \int _0^\infty \left| (f\#_\lambda {\phi }_t)\right| ^2 x^{2\lambda }dx\frac{dt}{t}\\&= \int _0^\infty \int _0^\infty \left| h_\lambda (f)(x)\right| ^2\left| h_\lambda (\phi _t)(x)\right| ^2 x^{2\lambda }dx\frac{dt}{t}\\&= \int _0^\infty \left| h_\lambda (f)(x)\right| ^2\int _0^\infty \left| h_\lambda (\phi )(xt)\right| ^2\frac{dt}{t}x^{2\lambda }dx\\&= \int _0^\infty \left| h_\lambda (f)(x)\right| ^2\int _0^\infty \left| h_\lambda (\phi )(u)\right| ^2\frac{du}{u}x^{2\lambda }dx\\&=\Vert f\Vert ^2_{L^2((0,\infty ),m_\lambda )}\int _0^\infty \left| h_\lambda (\phi )(u)\right| ^2\frac{du}{u}. \end{aligned}$$

By taking into account property (c) we conclude that the operator ${\mathfrak {G}}^\lambda _\phi$ is bounded from $L^2((0,\infty ),m_\lambda )$ into $L^2_{L^2((\frac{1}{N},N),\frac{dt}{t})}((0,\infty ),m_\lambda )$.

By using vector valued Calderón-Zygmund theory ([66]), the properties in (2.2), (2.3) and (2.4) imply that the operator $g^\lambda _{\phi ,N}$ given by

$$\begin{aligned} g^\lambda _{\phi ,N}(f)(x)=\Vert f\#_\lambda \phi _{\centerdot }(x)\Vert _{L^2((\frac{1}{N},N),\frac{dt}{t})},\quad x\in (0,\infty ), \end{aligned}$$

can be extended from $L^2((0,\infty ),m_\lambda ) \cap L^p((0,\infty ),m_\lambda )$ to $L^p((0,\infty ),m_\lambda )$ as a bounded operator from $L^p((0,\infty ),m_\lambda )$ to

• 1. $L^p((0,\infty ),m_\lambda )$, for every $1<p<\infty$;

• 2. $L^{1,\infty }((0,\infty ),m_\lambda )$, when $p=1$,

with constants independent of N. This allows us to use monotone convergence theorem to get the same $L^p$-boundedness properties for $g_\phi ^\lambda$. $\square$

We consider the operator

$$\begin{aligned} {\mathcal {M}}^{\lambda ,\#}_\phi (f)(x)= & {} \sup _{\begin{array}{c} Q\ni x\\ Q\,\text {interval} \end{array}} \mathop {\mathrm {ess\, sup \;}}_{y_1,y_2 \in Q}\left| g^\lambda _\phi (f\chi _{(0,\infty )\setminus (3Q)})(y_1) - g^\lambda _\phi (f\chi _{(0,\infty )\setminus (3Q)})(y_2)\right| ,\\&\quad x\in (0,\infty ). \end{aligned}$$

Proposition 2.3

Let $\lambda >0$. Suppose that $\phi$ is a differentiable function on $(0,\infty )$ such that $\int _0^\infty |u^{2\lambda +2}\phi '(u)|^2\frac{du}{u}<\infty$. Then, the operator ${\mathcal {M}}_\phi ^{\lambda ,\#}$ is bounded from $L^p((0,\infty ),m_\lambda )$ into itself, for every $1<p<\infty$, and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$.

Proof

Let $x\in (0,\infty )$. We choose an interval $Q=B(x_Q,r_Q) \subset (0,\infty )$, with $x_Q,r_Q\in (0,\infty )$, and such that $x\in Q$. By using Minkowski inequality and (2.3) we obtain, for every $y_1$, $y_2\in Q$,

$$\begin{aligned}&\left| g^\lambda _\phi (f\chi _{(0,\infty )\setminus (3Q)})(y_1) - g^\lambda _\phi (f\chi _{(0,\infty )\setminus (3Q)})(y_2)\right| \\&\quad \le \left( {\int _0^\infty \left| {\mathfrak {G}}^\lambda _{\phi ,t}\left( f\chi _{(0,\infty )\setminus (3Q)}\right) (y_1) - {\mathfrak {G}}^\lambda _{\phi ,t}\left( f\chi _{(0,\infty )\setminus (3Q)}\right) (y_2)\right| ^2\frac{dt}{t}}\right) ^{1/2}\\&\quad \le \int _{(0,\infty )\setminus (3Q)} |f(y)|y^{2\lambda }\left( \int _0^\infty \left| \,_\lambda \tau _{y_1}(\phi _t)(y)-\,_\lambda \tau _{y_2}(\phi _t)(y)\right| ^2\frac{dt}{t}\right) ^{\frac{1}{2}}dy\\&\quad = \int _{(0,\infty )\setminus (3Q)} |f(y)|y^{2\lambda }\left( \int _0^\infty \left| \int _{y_1}^{y_2} \partial _z\left( \,_\lambda \tau _z(\phi _t)(y)\right) dz\right| ^2\frac{dt}{t}\right) ^{\frac{1}{2}}dy\\&\quad \le \int _{(0,\infty )\setminus (3Q)} |f(y)|y^{2\lambda }\left| \int _{y_1}^{y_2}\left( \int _0^\infty \left| \partial _z\left( \,_\lambda \tau _z(\phi _t)(y)\right) \right| ^2\frac{dt}{t}\right) ^{\frac{1}{2}}dz\right| dy\\&\quad \le C\int _{(0,\infty )\setminus (3Q)} |f(y)|y^{2\lambda }\left| \int _{y_1}^{y_2}\frac{dz}{|z-y|m_\lambda (B(z,|z-y|))}\right| dy\\&\quad \le C|y_1-y_2|\int _{(0,\infty )\setminus (3Q)} \frac{|f(y)|y^{2\lambda }}{|y_1-y|m_\lambda (B(y_1,|y-y_1|))}dy. \end{aligned}$$

Let $y_1\in Q$. We define ${{\widetilde{Q}}} = B(y_1,r_Q)$. Then, by taking into account that $m_\lambda$ is doubling and the last estimation, we obtain

$$\begin{aligned}&\left| g_\phi ^\lambda \left( f\chi _{(0,\infty )\setminus (3Q)}\right) (y_1) - g_\phi ^\lambda \left( f\chi _{(0,\infty )\setminus (3Q)}\right) (y_2)\right| \\&\quad \le Cr_Q\sum ^\infty _{k=1}\int _{2^{k+1}{{\widetilde{Q}}} \setminus 2^k{{\widetilde{Q}}}}\frac{|f(y)|}{|y_1-y|m_\lambda (B(y_1,|y-y_1|))}y^{2\lambda }dy\\&\quad \le Cr_Q\sum ^\infty _{k=1}\frac{1}{2^kr_Q} \frac{1}{m_\lambda (B(y_1,2^{k+1}r_Q))}\int _{2^{k+1}{{\widetilde{Q}}} }|f(y)|y^{2\lambda }dy \\&\quad \le C{\mathcal {M}}_\lambda (f)(x),\quad y_2 \in Q. \end{aligned}$$

We get

$$\begin{aligned} {\mathcal {M}}_\phi ^{\lambda ,\#}(f)(x) \le C{\mathcal {M}}_\lambda (f)(x). \end{aligned}$$

The $L^p$-boundedness properties of ${\mathcal {M}}^{\lambda ,\#}_\phi$ can be deduced now from the corresponding ones of ${\mathcal {M}}_\lambda$. $\square$

By using [60, Theorem 1.1], the results in Proposition 2.2 and 2.3 lead to that, for a certain $\eta \in (0,1)$ the following property holds: for every $f \in L^1((0,\infty ), m_\lambda )$ with compact support there exists an $\eta$-sparse collection of cubes S in $(0,\infty )$ such that $c=1$ in (1.1) and

$$\begin{aligned} g_\phi ^\lambda (f)(x) \le CC_{\phi ,\lambda } \sum _{Q \in S}|f|_Q\chi _Q (x),\;\;x\in (0,\infty ), \end{aligned}$$

where

$$\begin{aligned} C_{\phi ,\lambda }=\Vert g_\phi ^\lambda \Vert _{L^1((0,\infty ),m_\lambda ) \rightarrow L^{1,\infty }((0,\infty ),m_\lambda )}+\left\| {\mathcal {M}}_\phi ^{\lambda ,\#}\right\| _{L^1((0,\infty ),m_\lambda ) \rightarrow L^{1,\infty }((0,\infty ),m_\lambda )}, \end{aligned}$$

and C depends on S. Here $|f|_Q =\frac{1}{m_\lambda (Q)}\int _Q |f(y)|y^{2\lambda }dy$, $Q \in S$. The proof can be finished by using [60, Proposition 4.1].

2.3 Proof of theorem 1.1 for $V_\rho ^\lambda (\{\phi _t\}_{t>0})$

In the next proposition we establish the $L^p$-boundedness properties of the $\rho$-variation operator $V_\rho ^\lambda (\{\phi _t\}_{t>0})$. In order to prove this proposition we can not apply the vector-valued Calderón-Zygmund theory because we do not now if $V_\rho ^\lambda (\{\phi _t\}_{t>0})$ is bounded from $L^q((0,\infty ),m_\lambda )$ into itself for some $1<q<\infty$.

Proposition 2.4

Let $\lambda >0$ and $\rho >2$. Assume that $\phi \in C^2(0,\infty )$ and let

$$\begin{aligned} \Phi (z)=\int _0^\infty u^{\lambda -1}\phi (\sqrt{z^2+u})du,\quad z\in {\mathbb {R}}. \end{aligned}$$

Suppose that the following properties hold:

• (a) $\displaystyle \int _0^\infty |\phi ^{(k)}(x)| x^{2\lambda +k}dx < \infty$, $k=0,1,2$;

• (b) $\displaystyle \int _{\frac{1}{2}}^2 \frac{v^\lambda }{|1-v|}\int _0^\infty s^{k-1/2}\int _{\frac{\pi ^2vs}{4(1-v)^2}}^\infty u^{\lambda -1}\frac{|\phi ^{(k)}(\sqrt{s+u})|}{(s+u)^{k/2}}dudsdv < \infty$, $k=0,1$;

• (c) $\displaystyle \lim \nolimits _{|z|\rightarrow \infty }\Phi (z)=0$;

• (d) $\displaystyle \int _0^\infty (|\Phi (u)|+u|\Phi '(u)|)du<\infty$.

Then, the operator $V_\rho ^\lambda (\{\phi _t\}_{t>0})$ is bounded from $L^p((0,\infty ),m_\lambda )$ into itself, for every $1<p<\infty$, and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$

Proof

We define $K_t(x,y) =\,_\lambda \tau _x(\phi _t)(y)$, $t,x,y\in (0,\infty )$. We decompose $K_t(x,y)$ as follows

$$\begin{aligned} K_t(x,y)=K_{t,1}(x,y)+K_{t,2}(x,y),\quad t,x,y \in (0,\infty ), \end{aligned}$$

being

$$\begin{aligned} K_{t,1}(x,y)= & {} \frac{\Gamma (\lambda + \frac{1}{2})}{\sqrt{\pi }\Gamma (\lambda )t^{2\lambda +1}}\int _0^{\frac{\pi }{2}} \phi \left( \frac{\sqrt{x^2+y^2-2xy\cos \theta }}{t}\right) \sin ^{2\lambda -1}\theta d\theta ,\\&\quad t,x,y \in (0,\infty ). \end{aligned}$$

We consider, for $j=1,2$,

$$\begin{aligned} T_{t,j}(f)(x) =\int _0^\infty K_{t,j}(x,y) f(y)y^{2\lambda }dy,\;\; t,x \in (0,\infty ). \end{aligned}$$

It follows that

$$\begin{aligned} V^\lambda _\rho (\{\phi _t\}_{t>0})(f)\le V_{\rho }(\{T_{t,1}\}_{t>0})(f) + V_\rho (\{T_{t,2}\}_{t>0})(f), \end{aligned}$$

where $V_\rho (\{T_{t,j}\}_{t>0})$, $j=1,2$, are defined in the obvious way. We are going to study $V_\rho (\{T_{t,j}\}_{t>0})$, $j=1,2$.

Suppose that $t_1>t_2> \ldots> t_k>0$, with $k \in {\mathbb {N}}$. We have that

$$\begin{aligned}&\left( \sum _{j=1}^{k-1}\left| T_{t_j,2}(f)(x)- |T_{t_{j+1,2}}(f)(x)\right| ^\rho \right) ^{\frac{1}{\rho }}\\&\quad = \left( \sum _{j=1}^{k-1}\left| \int _{t_{j+1}}^{t_j} \partial _tT_{t,2}(f)(x)dt\right| ^{\rho }\right) ^{\frac{1}{\rho }} \\&\quad \le C\sum _{j=1}^{k-1}\int _{t_{j+1}}^{{t_j}}\left| \partial _tT_{t,2}(f)(x)\right| dt\le C\int _0^\infty \left| \partial _tT_{t,2}(f)(x)\right| dt \\&\quad \le C\int _0^\infty |f(y)|y^{2\lambda }\int _0^\infty \left| \partial _tK_{t,2}(x,y)\right| dtdy, \quad x\in (0,\infty ). \end{aligned}$$

We get

$$\begin{aligned} V_\rho (\{T_{t,2}\}_{t>0})(f)(x) \le C\int _0^\infty {\mathbb {K}}_2(x,y) |f(y)|y^{2\lambda }dy,\quad x\in (0,\infty ) \end{aligned}$$

where

$$\begin{aligned} {\mathbb {K}}_2(x,y)= \int _0^\infty \left| \partial _tK_{t,2}(x,y)\right| dt,\quad x,y \in (0,\infty ). \end{aligned}$$

Since (a) for $k=0,1$ hold we obtain

$$\begin{aligned} |{\mathbb {K}}_2(x,y)|\le & {} C\int _0^\infty \frac{1}{t^{2\lambda +2}} \int _{\frac{\pi }{2}}^\pi (|\phi (u)| +|\phi '(u)|u)_{|{u=\frac{\sqrt{x^2+y^2-2xy\cos \theta }}{t}}}\sin ^{2\lambda -1}\theta d\theta dt\\\le & {} C\int _{\frac{\pi }{2}}^\pi \frac{\sin ^{2\lambda -1}\theta }{(\sqrt{x^2+y^2-2xy\cos \theta })^{2\lambda +1}}\int _0^\infty (|\phi (u)| +|\phi '(u)|u)u^{2\lambda }du d\theta \\\le & {} \frac{C}{(x^2+y^2)^{\lambda +\frac{1}{2}}} \le C\left\{ \begin{array}{ll} y^{-2\lambda -1}, &{} 0<x\le y< \infty ,\\ x^{-2\lambda -1}, &{} 0<y\le x <\infty . \end{array}\right. \end{aligned}$$

We consider the Hardy type operators defined by

$$\begin{aligned} H_0^\lambda (g)(x)= \frac{1}{x^{2\lambda +1}}\int _0^x g(y) y^{2\lambda }dy,\quad x \in (0,\infty ), \end{aligned}$$

and

$$\begin{aligned} H_\infty (g)(x)=\int _x^\infty g(y)\frac{dy}{y}, \quad x \in (0,\infty ). \end{aligned}$$

According to [10, Lemmas 2.6 and 2.7], $H_0^\lambda$ and $H_\infty$ are bounded from $L^p((0,\infty ),m_\lambda )$ into itself for every $1<p<\infty$, and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$.

We deduce that

$$\begin{aligned} V_\rho (\{T_{t,2}\}_{t>0})(f)(x) \le C(H_0^\lambda (|f|)(x)+H_\infty (|f|)(x)),\quad x\in (0,\infty ). \end{aligned}$$

We now study $V_\rho (\{T_{t,1}\}_{t>0})$. We write

$$\begin{aligned}T_{t,1}(f)(x)= & {} \left( \int _0^{\frac{x}{2}} + \int _{\frac{x}{2}}^{2x}+\int _{2x}^\infty \right) K_{t,1}(x,y)f(y)y^{2\lambda }dy\\= & {} \sum _{j=1}^3 T_{t,1,j}(f)(x),\;\; t,x \in (0,\infty ).\end{aligned}$$

By proceeding as above we get

$$\begin{aligned} V_\rho (\{T_{t,1,1}\}_{t>0})(f)(x)\le C\int _0^{\frac{x}{2}}{\mathbb {K}}_1(x,y) |f(y)|y^{2\lambda }dy,\quad x\in (0,\infty ), \end{aligned}$$

where

$$\begin{aligned} {\mathbb {K}}_1(x,y)=\int _0^\infty |\partial _tK_{t,1}(x,y)|dt,\quad x,y \in (0,\infty ). \end{aligned}$$

By using again property (a) for $k=0,1$ we have that

$$\begin{aligned}{\mathbb {K}}_1(x,y)\le & {} C\int _0^\infty \frac{1}{t^{2\lambda +2}}\int _0^{\frac{\pi }{2}}(|\phi (u)|+|\phi '(u)|u)_{|u=\frac{\sqrt{x^2+y^2-2xy\cos \theta }}{t}}\sin ^{2\lambda -1}\theta d\theta dt\\\le & {} C \int ^{\frac{\pi }{2}}_0 \frac{\sin ^{2\lambda -1}\theta }{((x-y)^2+2xy(1-\cos \theta ))^{\lambda +\frac{1}{2}}}d\theta \\\le & {} C\frac{1}{|x-y|^{2\lambda +1}},\quad x,y \in (0,\infty ),\;x \not =y.\end{aligned}$$

Then,

$$\begin{aligned} V_\rho (\{T_{t,1,1}\}_{t>0})(f)(x) \le \frac{C}{x^{2\lambda +1}}\int _0^x|f(y)|y^{2\lambda }dy=CH_0^\lambda (|f|)(x),\quad x\in (0,\infty ).\nonumber \\ \end{aligned}$$

(2.5)

In a similar way we can see that

$$\begin{aligned} V_\rho (\{T_{t,1,3}\}_{t>0})(f)(x) \le C\int _x^\infty |f(y)| \frac{dy}{y}=CH_\infty (|f|)(x),\quad x \in (0,\infty ). \end{aligned}$$

(2.6)

On the other hand, we write

$$\begin{aligned} K_{t,1}(x,y)= & {} \frac{\Gamma (\lambda +\frac{1}{2})}{\sqrt{\pi }\Gamma (\lambda )} \frac{1}{t^{2\lambda +1}}\int _0^{\frac{\pi }{2}}\phi \left( \frac{\sqrt{x^2+y^2-2xy\cos \theta }}{t}\right) \sin ^{2\lambda -1}\theta d\theta \\= & {} \frac{\Gamma \left( \lambda +\frac{1}{2}\right) }{\sqrt{\pi }\Gamma (\lambda )}\frac{1}{t^{2\lambda +1}}\left[ \int _0^{\frac{\pi }{2}}\phi \left( \frac{\sqrt{x^2+y^2-2xy\cos \theta }}{t}\right) (\sin ^{2\lambda -1}\theta -\theta ^{2\lambda -1})d\theta \right. \\&+ \int _0^{\frac{\pi }{2}}\theta ^{2\lambda -1} \left( \phi \left( \frac{\sqrt{x^2+y^2-2xy\cos \theta }}{t}\right) - \phi \left( \frac{\sqrt{(x-y)^2+xy\theta ^2}}{t}\right) \right) d\theta \\&-\frac{1}{2}\left( \frac{t}{\sqrt{xy}}\right) ^{2\lambda } \int _{\frac{\pi ^2xy}{4t^2}}^\infty u^{\lambda -1}\phi \left( \sqrt{\frac{(x-y)^2}{t^2}+u}\right) du\\&+\left. \frac{1}{2}\left( \frac{t}{\sqrt{xy}}\right) ^{2\lambda }\int _0^\infty u^{\lambda -1}\phi \left( \sqrt{\frac{(x-y)^2}{t^2}+u}\right) du\right] \\= & {} \sum _{j=1}^4K_{t,1}^j(x,y),\quad t,x,y \in (0,\infty ).\end{aligned}$$

We define, for every $j=1,2,3,4$,

$$\begin{aligned} {\mathfrak {K}}_{t,j}(f)(x) = \int _{\frac{x}{2}}^{2x} K_{t,1}^j(x,y)f(y) y^{2\lambda } dy,\quad t,x\in (0,\infty ). \end{aligned}$$

We get

$$\begin{aligned} V_\rho (\{T_{t,1,2}\}_{t>0})(f) \le \sum _{j=1}^4 V_\rho (\{{\mathfrak {K}}_{t,j}\}_{t>0})(f). \end{aligned}$$

As above we can obtain that, for $j=1,2,3$,

$$\begin{aligned} V_\rho (\{{\mathfrak {K}}_{t,j}\}_{t>0})(f)(x) \le C\int _{\frac{x}{2}}^{2x}|f(y)|y^{2\lambda }\int _0^\infty |\partial _tK^j_{t,1}(x,y)|dtdy ,\quad x\in (0,\infty ). \end{aligned}$$

Since $|\sin ^{2\lambda -1}\theta - \theta ^{2\lambda -1}| \le C\theta ^{2\lambda +1}$, $\theta \in (0,\frac{\pi }{2})$, we get

$$\begin{aligned} \int _0^\infty |\partial _tK^1_{t,1}(x,y)|dt\le & {} C\int _0^\infty \frac{1}{t^{2\lambda +2}}\int _0^{\frac{\pi }{2}}\theta ^{2\lambda +1} (|\phi (u)|+|\phi '(u|)u)_{|u=\frac{\sqrt{x^2+y^2-2xy\cos \theta }}{t}}d\theta dt\\\le & {} C\int _0^{\frac{\pi }{2}}\frac{\theta ^{2\lambda +1}}{((x-y)^2+2xy (1-\cos \theta ))^{\lambda +\frac{1}{2}}}d\theta \\\le & {} C\int _0^{\frac{\pi }{2}} \frac{\theta ^{2\lambda +1}}{(xy(1-\cos \theta ))^{\lambda +\frac{1}{2}}} d\theta \\\le & {} \frac{C}{(xy)^{\lambda +\frac{1}{2}}} \le \frac{C}{x^{2\lambda +1}},\quad 0<\frac{x}{2}<y<2x.\end{aligned}$$

We have used that $\phi$ satisfies (a) for $k=0,1$. Then

$$\begin{aligned} V_\rho (\{{\mathfrak {K}}_{t,1}\}_{t>0})(f)(x) \le \frac{C}{x^{2\lambda +1}}\int _{\frac{x}{2}}^{2x}|f(y)|y^{2\lambda }dy\,\quad x \in (0,\infty ). \end{aligned}$$

(2.7)

We have that

$$\begin{aligned} \partial _tK^2_{t,1}(x,y)&= -\frac{\Gamma \left( \lambda +\frac{1}{2}\right) }{\sqrt{\pi }\Gamma (\lambda )} \frac{1}{t^{2\lambda +1}}\\&\qquad \times \left[ \frac{2\lambda +1}{t} \int _0^{\frac{\pi }{2}}\theta ^{2\lambda -1}\left( \phi \left( \frac{\sqrt{(x-y)^2+2xy(1-\cos \theta )}}{t}\right) \right. \right. \\&\qquad \left. -\phi \left( \frac{\sqrt{(x-y)^2+xy\theta ^2}}{t}\right) \right) d\theta \\&\qquad +\int _0^{\frac{\pi }{2}}\theta ^{2\lambda -1}\left( \phi '\left( \frac{\sqrt{(x-y)^2+2xy(1-\cos \theta )}}{t}\right) \right. \\&\left. \qquad \frac{\sqrt{(x-y)^2+2xy(1-\cos \theta )}}{t^2}\right. \\&\qquad \left. \left. -\phi '\left( \frac{\sqrt{(x-y)^2+xy\theta ^2}}{t}\right) \frac{\sqrt{(x-y)^2+xy\theta ^2}}{t^2}\right) d\theta \right] \\&\quad =K_{t,1}^{2,1}(x,y) + K_{t,1}^{2,2}(x,y),\quad t,x,y \in (0,\infty ). \end{aligned}$$

We can write

$$\begin{aligned} K^{2,1}_{t,1} (x,y)= & {} -\frac{\Gamma \left( \lambda + \frac{3}{2}\right) }{\sqrt{\pi }\Gamma (\lambda )t^{2\lambda +2}}\int _0^{\frac{\pi }{2}}\theta ^{2\lambda -1}\\&\int _{\theta ^2}^{2(1-\cos \theta )}\phi '\left( \frac{\sqrt{(x-y)^2+xyz}}{t}\right) \frac{xy}{t\sqrt{(x-y)^2+xyz}}dz d\theta . \end{aligned}$$

Then, since $\phi$ satisfies (a) for $k=1$, we get

$$\begin{aligned}&\int _0^\infty \left| K^{2,1}_{t,1}(x,y)\right| dt \\&\quad \le C \int _0^\infty \frac{1}{t^{2\lambda +3}}\int _0^{\frac{\pi }{2}}\theta ^{2\lambda -1}\int _{2(1-\cos \theta )}^{\theta ^2} \left| \phi '\left( \frac{\sqrt{(x-y)^2+xyz}}{t}\right) \right| \\&\qquad \times \frac{xy}{\sqrt{(x-y)^2+xyz}}dz d\theta dt\\&\quad \le Cxy\int _0^{\frac{\pi }{2}}\int _{2(1-\cos \theta )}^{\theta ^2}\frac{\theta ^{2\lambda -1}}{((x-y)^2+xyz)^{\lambda +\frac{3}{2}}}\int _0^\infty u^{2\lambda +1}|\phi '(u)|dudzd\theta \\&\quad \le Cxy\int _0^{\frac{\pi }{2}}\frac{\theta ^{2\lambda -1}(\theta ^2-2(1-\cos \theta ))}{((x-y)^2+xy\theta ^2)^{\lambda +\frac{3}{2}}}d\theta \le Cxy\int _0^{\frac{\pi }{2}}\frac{\theta ^{2\lambda +3}}{((x-y)^2+xy\theta ^2)^{\lambda +\frac{3}{2}}}d\theta \\&\quad \le \frac{C}{(xy)^{\lambda +\frac{1}{2}}} \le \frac{C}{x^{2\lambda +1}}, \quad 0<\frac{x}{2}<y<2x. \end{aligned}$$

We have that

$$\begin{aligned} K_{t,1}^{2,2}(x,y)= & {} -\frac{\Gamma \left( \lambda +\frac{1}{2}\right) }{2\sqrt{\pi }\Gamma (\lambda )t^{2\lambda -1}}\int _0^{\frac{\pi }{2}}\theta ^{2\lambda -1}\int _{\theta ^2}^{2(1-\cos \theta )}\left[ \phi ''\left( \frac{\sqrt{(x-y)^2+xyz}}{t}\right) \frac{xy}{t^3}\right. \\&\left. +\phi '\left( \frac{\sqrt{(x-y)^2+xyz}}{t}\right) \right] \frac{xy}{\sqrt{(x-y)^2+xyz}t^2}dzd\theta ,\quad t,x,y \in (0,\infty ).\end{aligned}$$

Since $\phi$ verifies (a) for $k=1,2$, we obtain

$$\begin{aligned}&\int _0^\infty |K_{t,1}^{2,2}(x,y)|dt \\&\quad \le Cxy\int _0^\infty \frac{1}{t^{2\lambda +1}}\int _0^{\frac{\pi }{2}}\theta ^{2\lambda -1} \int _{2(1-\cos \theta )}^{\theta ^2}\left[ \left| \phi ''\left( \frac{\sqrt{(x-y)^2+xyz}}{t}\right) \right| \frac{1}{t^3}\right. \\&\qquad \left. +\left| \phi '\left( \frac{\sqrt{(x-y)^2+xyz}}{t}\right) \right| \frac{1}{t^2\sqrt{(x-y)^2+xyz}}\right] dzd\theta dt\\&\quad \le Cxy\int _0^{\frac{\pi }{2}}\theta ^{2\lambda -1}\int _{2(1-\cos \theta )}^{\theta ^2}\left[ \int _0^\infty |\phi ''(u)|\left( \frac{u}{\sqrt{(x-y)^2+xyz}}\right) ^{2\lambda +4}\right. \\&\qquad \left. \times \frac{\sqrt{(x-y)^2+xyz}}{u^2}du+\int _0^\infty |\phi '(u)|\left( \frac{u}{\sqrt{(x-y)^2+xyz}}\right) ^{2\lambda +3}\frac{du}{u^2}\right] dzd\theta \\&\quad \le Cxy \int _0^{\frac{\pi }{2}}\theta ^{2\lambda -1}\int _{2(1-\cos \theta )}^{\theta ^2}\frac{dzd\theta }{((x-y)^2+xyz)^{\lambda +\frac{3}{2}}} \left( \int _0^\infty |\phi ''(u)|u^{2\lambda +2}du\right. \\&\qquad \left. +\int _0^\infty |\phi '(u)|u^{2\lambda +1}du \right) \\&\quad \le \frac{C}{x^{2\lambda +1}}, \quad 0< \frac{x}{2}<y<2x. \end{aligned}$$

We conclude that

$$\begin{aligned} V_\rho (\{{\mathfrak {K}}_{t,2}\}_{t>0})(f)(x) \le \frac{C}{x^{2\lambda +1}}\int _{\frac{x}{2}}^{2x}|f(y)|y^{2\lambda }dy,\quad x\in (0,\infty ). \end{aligned}$$

(2.8)

We have that

$$\begin{aligned} \partial _tK^3_{t,1}(x,y)= & {} \frac{\Gamma (\lambda +\frac{1}{2})}{2\sqrt{\pi }\Gamma (\lambda )(xy)^\lambda }\left[ \frac{1}{t^2}\int _{\frac{\pi ^2xy}{4t^2}}^\infty u^{\lambda -1}\phi \left( \sqrt{\frac{(x-y)^2}{t^2}+u}\right) du\right. \\&-\frac{\pi ^2xy}{2t^4}\left( \frac{\pi ^2xy}{4t^2}\right) ^{\lambda -1}\phi \left( \sqrt{\frac{(x-y)^2}{t^2}+\frac{\pi ^2xy}{4t^2}}\right) \\&\left. -\frac{1}{t}\int _{\frac{\pi ^2xy}{4t^2}}^\infty u^{\lambda -1}\phi '\left( \sqrt{\frac{(x-y)^2}{t^2}+u}\right) \frac{(x-y)^2}{t^3\sqrt{\frac{(x-y)^2}{t^2}+u}}du\right] \\= & {} \sum _{j=1}^3 K^{3,j}_{t,1}(x,y),\quad t,x,y\in (0,\infty ).\end{aligned}$$

It follows that

$$\begin{aligned} \int _0^\infty |K^{3,1}_{t,1}(x,y)|dt\le & {} \frac{C}{(xy)^{\lambda }}\int _0^\infty \frac{1}{t^2}\int _{\frac{\pi ^2xy}{4t^2}}^\infty u^{\lambda -1}\left| \phi \left( \sqrt{\frac{(x-y)^2}{t^2} +u}\right) \right| dudt\nonumber \\\le & {} \frac{C}{|x-y|(xy)^\lambda }\int _0^\infty \frac{1}{\sqrt{s}}\int _{\frac{\pi ^2xys}{4|x-y|^2}}^\infty u^{\lambda -1}|\phi (\sqrt{s+u})|duds\nonumber \\=: & {} H(x,y),\quad x,y\in (0,\infty ),\,x\not =y. \end{aligned}$$

(2.9)

Since $\phi$ satisfies (b) for $k=0$ we obtain, for every $x\in (0,\infty )$,

$$\begin{aligned} \int _{\frac{x}{2}}^{2x}H(x,y) y^{2\lambda }dy=C\int _{\frac{1}{2}}^2\frac{v^\lambda }{|1-v|}\int _0^\infty \frac{1}{\sqrt{s}}\int _{\frac{\pi ^2vs}{4(1-v)^2}}^\infty u^{\lambda -1}|\phi (\sqrt{s+u})|dudsdv < \infty .\nonumber \\ \end{aligned}$$

(2.10)

By using Jensen’s inequality, we get from (2.10) that the operator $T_\lambda$ defined by

$$\begin{aligned} T_\lambda (f)(x) = \int _{\frac{x}{2}}^{2x} H(x,y)f(y)y^{2\lambda }dy,\quad x\in (0,\infty ), \end{aligned}$$

is bounded from $L^p((0,\infty ),m_\lambda )$ into itself for every $1 \le p<\infty$.

We also have that

$$\begin{aligned} \int _0^\infty |K^{3,2}_{t,1}(x,y)|dt&\le C\int _0^\infty \frac{1}{t^{2\lambda +2}}\phi \left( \sqrt{\frac{(x-y)^2+\pi ^2xy}{t^2}}\right) dt\nonumber \\&\le \frac{C}{((x-y)^2+\pi ^2xy)^{\lambda +\frac{1}{2}}}\int _0^\infty u^{2\lambda }|\phi (u)|du \le \frac{C}{x^{2\lambda +1}},\nonumber \\&\quad 0< \frac{x}{2}<y < 2x. \end{aligned}$$

(2.11)

By proceeding as in the $K^{3,1}_{t,1}$ case we deduce that

$$\begin{aligned} \int _0^\infty |K^{3,3}_{t,1}(x,y)| dt&\le C\frac{|x-y|^2}{(xy)^\lambda }\int _0^\infty \frac{1}{t^4}\int _{\frac{\pi ^2xy}{4t^2}}^\infty u^{\lambda -1}\left| \phi '\left( \sqrt{\frac{(x-y)^2}{t^2}+u}\right) \right| \nonumber \\&\qquad \times \frac{1}{\sqrt{\frac{(x-y)^2}{t^2}+u}}dudt\nonumber \\&\le \frac{C}{|x-y|(xy)^{\lambda }}\int _0^\infty \sqrt{s}\int _{\frac{\pi ^2xys}{4|x-y|^2}}^\infty u^{\lambda -1}|\phi '(\sqrt{s+u})|\frac{du}{\sqrt{s+u}}ds\nonumber \\&=:F(x,y),\quad x,y\in (0,\infty ),\,x\not =y. \end{aligned}$$

(2.12)

Since $\phi$ satisfies (b) for $k=1$ we obtain, for every $x \in (0,\infty )$,

$$\begin{aligned} \int _{\frac{x}{2}}^{2x}F(x,y)y^{2\lambda }dy = C\int _{\frac{1}{2}}^2\frac{v^\lambda }{|1-v|}\int _0^\infty \sqrt{s}\int _{\frac{\pi ^2vs}{4(1-v)^2}}^\infty u^{\lambda -1}\frac{|\phi '(\sqrt{u+s})|}{\sqrt{u+s}}dudsdv < \infty .\nonumber \\ \end{aligned}$$

(2.13)

Again by Jensen’s inequality, (2.13) implies that the operator ${\mathcal {S}}_\lambda$ defined by

$$\begin{aligned} {\mathcal {S}}_\lambda (f)(x) = \int _{\frac{x}{2}}^{2x}F(x,y) f(y) y^{2\lambda }dy,\quad x \in (0,\infty ), \end{aligned}$$

is bounded from $L^p((0,\infty ),m_\lambda )$ into itself for every $1 \le p < \infty$. By using (2.9), (2.11) and (2.12) we conclude that

$$\begin{aligned} V_\rho (\{{\mathfrak {K}}_{t,3}\}_{t>0})(f)(x)\le & {} C\left( T_\lambda (|f|)(x) + {\mathcal {S}}_\lambda (|f|)(x) + \frac{1}{x^{2\lambda +1}}\int _{\frac{x}{2}}^{2x}|f(y)|y^{2\lambda }dy\right) , \nonumber \\&\quad x\in (0,\infty ), \end{aligned}$$

(2.14)

From (2.7), (2.8) and (2.14) it follows that

$$\begin{aligned} \sum ^3_{j=1}V_\rho (\{{\mathfrak {K}}_{t,j}\}_{t>0})(f)(x)\le & {} C\left( T_\lambda (|f|)(x) + {\mathcal {S}}_\lambda (|f|)(x) + \frac{1}{x^{2\lambda +1}}\int _{\frac{x}{2}}^{2x}|f(y)|y^{2\lambda }dy\right) , \nonumber \\&\quad x\in (0,\infty ). \end{aligned}$$

(2.15)

Since $\displaystyle \frac{1}{x^{2\lambda +1}}\int _{\frac{x}{2}}^{2x}y^{2\lambda }dy = \int _{\frac{1}{2}}^2v^{2\lambda }dv$, $x\in (0,\infty )$, Jensen’s inequality implies that the operator ${\mathfrak {D}}_\lambda$ defined by

$$\begin{aligned} {\mathfrak {D}}_\lambda (f)(x)=\frac{1}{x^{2\lambda +1}}\int _{\frac{x}{2}}^{2x}f(y) y^{2\lambda }dy,\quad x\in (0,\infty ), \end{aligned}$$

is bounded from $L^p((0,\infty ),m_\lambda )$ into itself for $1\le p<\infty$.

By using (2.5), (2.6) and (2.15) we can see that the operator

$$\begin{aligned} V_\rho (\{T_{t,1,1}\}_{t>0}) + V_\rho (\{T_{t,1,3}\}_{t>0})+\sum _{j=1}^3V_\rho (\{{\mathfrak {K}}_{t,j}\}_{t>0}) \end{aligned}$$

is bounded from $L^p((0,\infty ),m_\lambda )$ into itself, for every $1<p<\infty$, and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$.

We now study the operator $V_\rho (\{{\mathfrak {K}}_{t,4}\}_{t>0})$. We recall that

$$\begin{aligned} {\mathfrak {K}}_{t,4}(f)(x)=\int _{\frac{x}{2}}^{2x}K^4_{t,1}(x,y)f(y)y^{2\lambda }dy,\quad x\in (0,\infty ), \end{aligned}$$

where

$$\begin{aligned} K^4_{t,1}(x,y)= & {} \frac{\Gamma (\lambda +\frac{1}{2})}{2\sqrt{\pi }\Gamma (\lambda )}\frac{1}{t(xy)^\lambda }\int _0^\infty u^{\lambda -1}\phi \left( \sqrt{\left( \frac{x-y}{t}\right) ^2 +u}\right) du,\\&\quad t,x,y \in (0,\infty ). \end{aligned}$$

We consider the function

$$\begin{aligned} \Phi (z)=\int _0^\infty u^{\lambda -1}\phi \left( \sqrt{z^2+u}\right) du,\quad z \in {\mathbb {R}}. \end{aligned}$$

We define, for every $t>0$, the classical convolution operator ${\mathcal {S}}_t$ by

$$\begin{aligned} {\mathcal {S}}_t(f) = f*\Phi _{(t)}, \end{aligned}$$

where $\Phi _{(t)}(x) = \frac{1}{t}\Phi \left( \frac{x}{t}\right)$, $x \in {\mathbb {R}}$. We observe that $\Phi$ is an even function, and by property (c), $\lim _{|z|\rightarrow \infty }\Phi (z)=0$. Furthermore, if $\varphi (|z|)=\Phi (z)$, $z\in {\mathbb {R}}$, by making a change of variables and using (a) for $k=1$ we have that

$$\begin{aligned} \int _0^\infty |\varphi '(t)|tdt&=\int _0^\infty \Big |\int _0^\infty u^{\lambda -1}\phi '\left( \sqrt{t^2+u}\right) \frac{t^2}{\sqrt{t^2+u}}du\Big |dt\\&=2\int _0^\infty t^2\Big |\int _t^\infty (v^2-t^2)^{\lambda -1}\phi '(v)dv\Big |dt\\&\le 2\int _0^\infty |\phi '(v)|\int _0^t(v^2-t^2)^{\lambda -1}t^2dtdv\\&=B(\lambda ,\frac{3}{2})\int _0^\infty |\phi '(v)|v^{2\lambda +1}dv<\infty . \end{aligned}$$

Hence, according to [14, Lemma 2.4] the $\rho$-variation operator $V_\rho (\{{\mathcal {S}}_t\}_{t>0})$ is bounded from $L^p({\mathbb {R}},dx)$ into itself, for every $1<p<\infty$, and from $L^1({\mathbb {R}},dx)$ into $L^{1,\infty }( {\mathbb {R}},dx)$.

We can write

$$\begin{aligned} K^4_{t,1}(x,y)= \frac{\Gamma \left( \lambda +\frac{1}{2}\right) }{2\sqrt{\pi }\Gamma (\lambda )} \frac{1}{(xy)^\lambda }\Phi _{(t)}(x-y),\quad t,x,y \in (0,\infty ), \end{aligned}$$

and

$$\begin{aligned} {\mathfrak {K}}_{t,4}(f)(x)=\frac{\Gamma \left( \lambda +\frac{1}{2}\right) }{2\sqrt{\pi }\Gamma (\lambda )}x^{-\lambda }{\mathcal {S}}_t\left( y^{\lambda } f(y)\chi _{(\frac{x}{2},2x)}(y)\right) (x),\quad t,x\in (0,\infty ). \end{aligned}$$

Let $j \in {\mathbb {Z}}$. We have that

$$\begin{aligned} {\mathfrak {K}}_{t,4}(f)(x)= & {} \left( \int _{2^{j-1}}^{2^{j+2}}-\int _{2^{j-1}}^{\frac{x}{2}}- \int _{2x}^{2^{j+2}}\right) K^4_{t,1}(x,y)f(y)y^{2\lambda }dy,\\&\quad x \in [2^j,2^{j+1}),\;t\in (0,\infty ). \end{aligned}$$

Then,

$$\begin{aligned} {\mathfrak {K}}_{t,4}(f)(x)= & {} \sum _{j \in {\mathbb {Z}}} \chi _{[2^j,2^{j+1})}(x)\frac{\Gamma \left( \lambda +\frac{1}{2}\right) }{2\sqrt{\pi }\Gamma (\lambda )}x^{-\lambda }{\mathcal {S}}_t (y^\lambda f(y)\chi _{[2^{j-1},2^{j+2})}(y))(x)\\&-\sum _{j \in {\mathbb {Z}}}\chi _{[2^j,2^{j+1})}(x) \frac{\Gamma (\lambda +\frac{1}{2})}{2\sqrt{\pi }\Gamma (\lambda )}\int _{2^{j-1}}^{\frac{x}{2}}(xy)^{-\lambda } \frac{1}{t}\Phi \left( \frac{x-y}{t}\right) f(y) y^{2\lambda }dy\\&-\sum _{j \in {\mathbb {Z}}}\chi _{[2^j,2^{j+1})}(x) \frac{\Gamma \left( \lambda +\frac{1}{2}\right) }{2\sqrt{\pi }\Gamma (\lambda )}\int _{2x}^{2^{j+2}}(xy)^{-\lambda } \frac{1}{t}\Phi \left( \frac{x-y}{t}\right) f(y) y^{2\lambda }dy\\= & {} \sum _{i=1}^3 L_{t,4}^i(f)(x),\quad t,x \in (0,\infty ). \end{aligned}$$

Let $1<p<\infty$. By using the $L^p$-boundedness of $V_\rho (\{{\mathcal {S}}_t\}_{t>0})$ we have that

$$\begin{aligned}&\int _0^\infty |V_\rho (\{L^1_{t,4}\}_{t>0})(f)(x)|^p x^{2\lambda }dx \\&\quad \le C\sum _{j\in {\mathbb {Z}}}\int _{2^j}^{2^{j+1}}x^{(2-p)\lambda }|V_\rho (\{{\mathcal {S}}_t\}_{t>0})(y^\lambda f(y)\chi _{[2^{j-1},2^{j+2})}(y))(x)|^pdx\\&\quad \le C\sum _{j\in {\mathbb {Z}}}2^{j(2-p)\lambda }\int _0^\infty |V_\rho (\{{\mathcal {S}}_t\}_{t>0})(y^\lambda f(y)\chi _{[2^{j-1},2^{j+2})}(y))(x)|^pdx\\&\quad \le C\sum _{j\in {\mathbb {Z}}}2^{j(2-p)\lambda }\int _{2^{j-1}}^{2^{j+2}}|y^\lambda f(y)|^pdy\le C\sum _{j \in {\mathbb {Z}}} \int _{2^{j-1}}^{2^{j+2}} |f(y)|^py^{2\lambda }dy \\&\quad \le C\int _0^\infty |f(y)|^py^{2\lambda }dy,\quad f \in L^p((0,\infty ),m_\lambda ). \end{aligned}$$

Also, since $V_\rho (\{{\mathcal {S}}_t)_{t>0})$ is bounded from $L^1({\mathbb {R}},dx)$ into $L^{1,\infty }({\mathbb {R}},dx)$ we get, for every $\alpha >0$,

$$\begin{aligned}&m_\lambda \Big (\big \{x\in (0,\infty ): V_\rho (\{L^1_{t,4}\}_{t>0})(f)(x)>\alpha \big \}\Big )\\&\quad \le \sum _{j\in Z}m_\lambda \Big (\big \{x\in [2^j,2^{j+1}):\frac{\Gamma (\lambda +\frac{1}{2})}{2\sqrt{\pi }\Gamma (\lambda )}x^{-\lambda } V_\rho (\{{\mathcal {S}}_t\}_{t>0})(y^\lambda f(y)\chi _{[2^{j-1},2^{j+2})}(y))(x)>\alpha \big \}\Big )\\&\quad \le C \sum _{j\in {\mathbb {Z}}}2^{2j\lambda }\left| \big \{x\in (0,\infty ): V_\rho (\{{\mathcal {S}}_t\}_{t>0})(y^\lambda f(y) \chi _{[2^{j-1},2^{j+2})}(y))(x) >C 2^{\lambda j}\alpha \big \}\right| \\&\quad \le \frac{C}{\alpha }\sum _{j \in {\mathbb {Z}}}2^{\lambda j}\int _{2^{j-1}}^{2^{j+2}}y^\lambda |f(y)|dy\le \frac{C}{\alpha }\int _0^\infty |f(y)|y^{2\lambda }dy,\quad f \in L^1((0,\infty ),m_\lambda ). \end{aligned}$$

On the other hand, by using property (d) we can write

$$\begin{aligned}&V_\rho (\{L^2_{t,4}\}_{t>0})(f)(x)\\&\quad \le C \sum _{j \in {\mathbb {Z}}}\chi _{[2^j,2^{j+1})}(x)\\&\qquad \int _{2^{j-1}}^{\frac{x}{2}}\left( \frac{y}{x}\right) ^\lambda |f(y)|\int _0^\infty \left| \partial _t \left( \frac{1}{t}\Phi \left( \frac{x-y}{t}\right) \right) \right| dtdy\\&\quad \le C\sum _{j\in {\mathbb {Z}}}\chi _{[2^j,2^{j+1})}(x) \int _{2^{j-1}}^{\frac{x}{2}}\left( \frac{y}{x}\right) ^\lambda |f(y)|\\&\qquad \int _0^\infty \left( \frac{1}{t^2}\left| \Phi \left( \frac{x-y}{t}\right) \right| + \frac{|x-y|}{t^3}\left| \Phi '\left( \frac{x-y}{t}\right) \right| \right) dtdy\\&\quad \le C\sum _{j\in {\mathbb {Z}}}\chi _{[2^j,2^{j+1})}(x)\int _{2^{j-1}}^{\frac{x}{2}}\left( \frac{y}{x}\right) ^\lambda \frac{|f(y)|}{|x-y|}dy\int _0^\infty (|\Phi (u)|+u|\Phi '(u)|)du\\&\quad \le C\sum _{j\in {\mathbb {Z}}}\chi _{[2^j,2^{j+1})}(x) \frac{1}{x^{\lambda +1}}\int _{2^{j-1}}^{\frac{x}{2}}\frac{y^{2\lambda }}{2^{\lambda j}}|f(y)|dy\\&\quad \le C\sum _{j\in {\mathbb {Z}}}\chi _{[2^j,2^{j+1})} (x)\frac{1}{x^{2\lambda +1}}\int _{2^{j-1}}^{\frac{x}{2}}y^{2\lambda }|f(y)|dy\\&\quad \le \frac{C}{x^{2\lambda +1}}\int _0^{\frac{x}{2}}y^{2\lambda }|f(y)|dy\le CH_0^\lambda (|f|)(x),\quad x\in (0,\infty ). \end{aligned}$$

In a similar way we get

$$\begin{aligned} V_\rho (\{L^3_{t,4}\}_{t>0})(f)(x) \le C\int _{2x}^\infty \frac{|f(y)|}{y}dy\le CH_\infty (|f|)(x), \quad x\in (0,\infty ). \end{aligned}$$

By using again the $L^p$-boundedness properties of Hardy type operators [10, Lemmas 2.6 and 2,7] we conclude that the operators $V_\rho ( \{L^j_{t,4}\}_{t>0})$, $j=2,3$, are bounded from $L^p((0,\infty ),m_\lambda )$ into itself, for every $1<p<\infty$, and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$.

It follows that the operator $V_\rho (\{{\mathfrak {K}}_{t,4}\}_{t>0})$ is bounded from $L^p((0,\infty ),m_\lambda$) into itself, for every $1<p<\infty$ and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$. By putting together all the properties we have just proved we establish that $V_\rho (\{\phi _t\}_{t>0})$ is bounded from $L^p((0,\infty ),m_\lambda )$ into itself, for every $1<p<\infty$, and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$. $\square$

We now consider the operator

$$\begin{aligned}&V_\rho ^{\lambda ,\#}(\{\phi _t\}_{t>0})(f)(x)\\&\quad =\sup _{\begin{array}{c} Q\,\mathrm{cube}\\ Q\ni x \end{array}}\mathop {\mathrm {ess\, sup \;}}_{y_1,y_2\in Q}\big |V_\rho ^\lambda (\{\phi _t\}_{t>0})(f{\mathcal {X}}_{(3Q)^c})(y_1)-V_\rho ^\lambda (\{\phi _t\}_{t>0})(f{\mathcal {X}}_{(3Q)^c})(y_2)\big |,\\&\qquad x\in (0,\infty ). \end{aligned}$$

Proposition 2.5

Let $\lambda >-1/2$ and $\rho >2$. Assume that $\phi$ is a twice differentiable function on $(0,\infty )$ such that there exists $C>0$ for which $|\Phi '(x)|\le C(1+x^2)^{-(\lambda +3/2)}$, $x\in (0,\infty )$, where $\Phi (x)=(2\lambda +1)\phi (x)+x\phi '(x)$, $x\in (0,\infty )$.

Then, the operator $V_\rho ^{\lambda ,\#}$ is bounded from $L^p((0,\infty ),m_\lambda )$ into itself, for every $1<p<\infty$ and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$.

Proof

Let $x\in (0,\infty )$. Suppose that Q is an interval in $(0,\infty )$ such that $x\in Q$.

By using Minkowski inequality we obtain

$$\begin{aligned}&\big |V_\rho ^\lambda (\{\phi _t\}_{t>0})(f{\mathcal {X}}_{(3Q)^c})(y_1)-V_\rho ^\lambda (\{\phi _t\}_{t>0})(f{\mathcal {X}}_{(3Q)^c})(y_2)\big |\\&\quad \le \int _{(0,\infty )\setminus (3Q)}|f(z)|z^{2\lambda } \Vert \{\,_\lambda \tau _{y_1}(\phi _t)(z)-\,_\lambda \tau _{y_2}(\phi _t)(z)\}_{t>0}\Vert _{v_\rho }dz,\quad y_1,y_2\in Q. \end{aligned}$$

We get as above that

$$\begin{aligned} \Vert \{\,_\lambda \tau _{y_1}(\phi _t)(z)-\,_\lambda \tau _{y_2}(\phi _t)(z)\}_{t>0}\Vert _{v_\rho } \le \int _0^\infty \big |\partial _t(\,_\lambda \tau _{y_1}(\phi _t)(z)-\,_\lambda \tau _{y_2}(\phi _t)(z))\big |dt. \end{aligned}$$

We can write

$$\begin{aligned} \partial _t\,_\lambda \tau _{y}(\phi _t)(z)&=\frac{\Gamma (\lambda +1/2)}{\sqrt{\pi }\Gamma (\lambda )}\partial _t\left[ \frac{1}{t^{2\lambda +1}}\int _0^\pi \phi \left( \frac{\sqrt{z^2+y^2-2zy\cos \theta }}{t}\right) \sin ^{2\lambda -1}\theta d\theta \right] \\&=-\frac{\Gamma (\lambda +1/2)}{\sqrt{\pi }\Gamma (\lambda )}\left( \frac{2\lambda +1}{t^{2\lambda +2}}\int _0^\pi \phi \left( \frac{\sqrt{z^2+y^2-2zy\cos \theta }}{t}\right) \sin ^{2\lambda -1}\theta d\theta \right. \\&\quad \left. +\frac{1}{t^{2\lambda +1}}\int _0^\pi \frac{\sqrt{z^2+y^2-2zy\cos \theta }}{t^2}\phi '\left( \frac{\sqrt{z^2+y^2-2zy\cos \theta }}{t}\right) \sin ^{2\lambda -1}\theta d\theta \right) \\&=-\frac{\Gamma (\lambda +1/2)}{\sqrt{\pi }\Gamma (\lambda )}\frac{1}{t^{2\lambda +2}}\int _0^\pi \Phi \left( \frac{\sqrt{z^2+y^2-2zy\cos \theta }}{t}\right) \sin ^{2\lambda -1}\theta d\theta ,\\&\quad z,y,t\in (0,\infty ), \end{aligned}$$

where $\Phi (z)=(2\lambda +1)\phi (z)+z\phi '(z)$, $z\in (0,\infty )$. Then,

$$\begin{aligned} \partial _t\,_\lambda \tau _{y}(\phi _t)(z)=-\frac{1}{t}\,_\lambda \tau _{y}(\Phi _t)(z),\quad z,y,t\in (0,\infty ). \end{aligned}$$

By proceeding as in the proof of [73, (2.8)], since $|\Phi '(x)|\le C(1+x^2)^{-\lambda -3/2}$, $x\in (0,\infty )$, we can see that

$$\begin{aligned} \big |\partial _t(\,_\lambda \tau _{y_1}(\phi _t)(z)-\,_\lambda \tau _{y_2}(\phi _t)(z))\big | =&\frac{1}{t}\big |\,_\lambda \tau _{y_1}(\Phi _t)(z)-\,_\lambda \tau _{y_2}(\Phi _t)(z)\big |\\ \le&\frac{C}{m_\lambda (B(y_1, t))+m_\lambda (B(y_1,|y_1-z|))}\frac{|y_1-y_2|}{(t+|y_1-z|)^2},\\&\quad z\in (0,\infty )\setminus (3Q),\;y_1,y_2\in Q,\;t>0. \end{aligned}$$

We obtain

$$\begin{aligned} \int _0^\infty \big |\partial _t(\,_\lambda \tau _{y_1}(\phi _t)(z)-\,_\lambda \tau _{y_2}(\phi _t)(z))\big |dt\le & {} C\frac{|y_1-y_2|}{m_\lambda (B(y_1,|y_1-z|))|y_1-z|},\\&\quad z\in (0,\infty )\setminus (3Q),\;y_1,y_2\in Q. \end{aligned}$$

It follows that

$$\begin{aligned}&\int _{(0,\infty )\setminus (3Q)} \Vert \{\,_\lambda \tau _{y_1}(\phi _t)(z)-\,_\lambda \tau _{y_2}(\phi _t)(z)\}_{t>0}\Vert _{v_\rho }|f(z)|z^{2\lambda }dz\\&\quad \le C|y_1-y_2|\int _{(0,\infty )\setminus (3Q)}\frac{|f(z)|z^{2\lambda }}{m_\lambda (B(y_1,|y_1-z|)|y_1-z|}dz\\&\quad \le C|y_1-y_2|\sum _{k=1}^\infty \int _{2^{k+1}Q\setminus 2^kQ}\frac{|f(z)|z^{2\lambda }}{m_\lambda (2^kQ)2^k|Q|}dz\\&\quad \le C|y_1-y_2|{\mathcal {M}}_\lambda (f)(x)\sum _{k=1}^\infty \frac{1}{2^k|Q|}\le C{\mathcal {M}}_\lambda (f)(x),\quad y_1,y_2\in Q. \end{aligned}$$

Then,

$$\begin{aligned} V_\rho ^{\lambda ,\#}(\{\phi _t\}_{t>0})(f)(x)\le C{\mathcal {M}}_\lambda (f)(x). \end{aligned}$$

The $L^p$-boundedness properties of $V_\rho ^{\lambda ,\#}(\{\phi _t\}_{t>0})$ follow from the corresponding $L^p$-boundedness properties of the Hardy-Littlewood maximal function ${\mathcal {M}}_\lambda$. $\square$

The proof of the Theorem 1.1 for $V_\rho ^\lambda (\{\phi _t\}_{t>0})$ can be finished as the proof of Theorem 1.1 for $g_\phi ^\lambda$.

2.4 Applications of theorem 1.1.

Note that if $\phi \in C^2 (0,\infty )$ is such that $|\phi (x)|\le C(1+x)^{-2\lambda -2}$, $|\phi '(x)|\le C(1+x)^{-2\lambda -3}$ and $|\phi ''(x)|\le C(1+x)^{-2\lambda -4}$, $x\in (0,\infty )$, then $\phi$ satisfies all the conditions $(a)-(d)$ in Proposition 2.4.

We now present some special functions $\phi$ satisfying the properties in the propositions in this section.

(I) The Poisson semigroup $\{P_t^\lambda \}_{t>0}$ associated with the operator $\Delta _\lambda$ is given by

$$\begin{aligned} P_t^\lambda (f)=P _t^\lambda \# _\lambda f,\quad t>0, \end{aligned}$$

where $P^\lambda (x)=\frac{2\lambda \Gamma (\lambda )}{\Gamma (\lambda +1/2)\sqrt{\pi }}(1+x^2)^{-\lambda -1}$, $x\in (0,\infty )$.

It is clear that

$$\begin{aligned} \left| \frac{\partial ^k}{\partial x^k}P^\lambda (x)\right| \le \frac{C}{(1+x)^{2\lambda +2+k}},\quad x\in (0,\infty ),\;k\in {\mathbb {N}}. \end{aligned}$$

We have that

$$\begin{aligned} t\partial _tP_t^\lambda (x)=\frac{1}{t^{2\lambda +1}}\left( -(2\lambda +1)P^\lambda \Big (\frac{x}{t}\Big )-\frac{x}{t}(P ^\lambda )'\Big (\frac{x}{t}\Big )\right) =\psi _t^{\lambda ,1}(x),\quad t,x\in (0,\infty ), \end{aligned}$$

where $\psi ^{\lambda ,1} (x)=-(2\lambda +1)P^\lambda (x)-x(P^\lambda )'(x)$, $x\in (0,\infty )$. Then,

$$\begin{aligned} \left| \frac{\partial ^k}{\partial x^k}\psi ^{\lambda ,1} (x)\right| \le \frac{C}{(1+x)^{2\lambda +2+k}},\quad x\in (0,\infty ),\;k\in {\mathbb {N}}. \end{aligned}$$

By proceeding inductively we conclude that, for every $m\in {\mathbb {N}}$,

$$\begin{aligned} t^m\partial _t^mP_t^\lambda (x)=\psi _t^{\lambda ,m}(x),\quad t,x\in (0,\infty ), \end{aligned}$$

where, for certain $c_k^{\lambda ,m}\in {\mathbb {R}}$, $k=0,...,m$,

$$\begin{aligned} \psi ^{\lambda ,m}(x)=\sum _{k=0}^mc_k^{\lambda ,m}x^k(P^\lambda )^{(k)}(x),\quad x\in (0,\infty ). \end{aligned}$$

We have that

$$\begin{aligned} \left| \frac{\partial ^r}{\partial x^r}\psi ^{\lambda ,m} (x)\right| \le \frac{C}{(1+x)^{2\lambda +2+r}},\quad x\in (0,\infty ),\;r\in {\mathbb {N}}. \end{aligned}$$

Let $m\in {\mathbb {N}}$. According to [29, (19) p. 24]

$$\begin{aligned} h_\lambda (\psi ^{\lambda ,m})(x)&=h_\lambda (t^m\partial _t^m(P_t^\lambda )_{|t=1})(x)=t^m\partial _t^mh_\lambda (P _t^\lambda )(x)_{|t=1}\\ {}&=\frac{2^{1/2-\lambda }}{\Gamma (\lambda +1/2)}t^m\partial _t^m(e^{-tx})_{|t=1}=p(x)e^{-x},\quad x\in (0,\infty ), \end{aligned}$$

where p is a polynomial and $p(0)=0$ provided that $m\ge 1$. Hence

$$\begin{aligned} \int _0^\infty |h_\lambda (\psi ^{\lambda ,m})(x)|^2\frac{dx}{x}<\infty ,\quad \text{ when } m\ge 1. \end{aligned}$$

From Theorem 1.1 we deduce the following result.

Corollary 2.6

Let $\lambda >0$, $\rho >2$, $m\in {\mathbb {N}}$, $1<p<\infty$ and $w\in A_p^\lambda (0,\infty )$.

(a) The maximal operator $P_*^{\lambda ,m}$ defined by

$$\begin{aligned} P_*^{\lambda ,m}(f)=\sup _{t>0}|t^m\partial _t^mP_t^\lambda (f)| \end{aligned}$$

is bounded from $L^p((0,\infty ),w,m_\lambda )$ into itself and

$$\begin{aligned} \Vert P_*^{\lambda , m}(f)\Vert _{L^p((0,\infty ),w,m_\lambda )}\le C[w]_{A_p^\lambda }^{\max \{1/(p-1), 1\}}\Vert f\Vert _{L^p((0,\infty ),w,m_\lambda )}, \end{aligned}$$

for every $f\in L^p((0,\infty ) ,w,m_\lambda )$.

(b) The Littlewood-Paley function $g_{\{P_t^\lambda \}_{t>0}}^{\lambda ,m}$ defined by

$$\begin{aligned} g_{\{P_t^\lambda \}_{t>0}}^{\lambda ,m}(f)(x)=\left( \int _0^\infty \Big |t^m\partial _t^mP_t^\lambda (f)(x)\Big |^2\frac{dt}{t}\right) ^{1/2},\quad x\in (0,\infty ), \end{aligned}$$

with $m\ge 1$, is bounded from $L^p((0,\infty ) ,w,m_\lambda )$ into itself and

$$\begin{aligned} \Vert g_{\{P_t^\lambda \}_{t>0}}^{\lambda , m}(f)\Vert _{L^p((0,\infty ),w,m_\lambda )}\le C[w]_{A_p^\lambda }^{\max \{1/(p-1), 1\}}\Vert f\Vert _{L^p((0,\infty ),w,m_\lambda )}, \end{aligned}$$

for every $f\in L^p((0,\infty ) ,w,m_\lambda )$.

(c) The variation operator $V_\rho (\{t^m\partial _t^mP_t^\lambda \}_{t>0})$ is bounded from $L^p((0,\infty ) ,w,m_\lambda )$ into itself and

$$\begin{aligned} \Vert V_\rho (\{t^m\partial _t^mP_t^\lambda \}_{t>0})(f)\Vert _{L^p((0,\infty ),w,m_\lambda )}\le C[w]_{A_p^\lambda }^{\max \{1/(p-1), 1\}}\Vert f\Vert _{L^p((0,\infty ),w,m_\lambda )}, \end{aligned}$$

for every $f\in L^p((0,\infty ) ,w,m_\lambda )$.

Here $C>0$ does not depend on w.

(II) Suppose that $\phi \in S(0,\infty )$, the Schwartz class in $(0,\infty )$. It is clear that, for every $k\in {\mathbb {N}}$, $|\phi ^{(k)}(x)|\le C(1+x)^{-2\lambda -2-k}$, $x\in (0,\infty )$. According to [2, Satz 5 and p. 201], $h_\lambda (\phi )\in S(0,\infty )$. Then, by [2, Lemma 3, p. 203] there exists $C>0$ such that $|h_\lambda (\phi )(x)|\le Cx^2$, $x\in (0,1)$, provided that $h_\lambda (\phi )(0)=0$.

Corollary 2.7

Let $\lambda >0$, $\rho >2$, $1<p<\infty$ and $w\in A_p^\lambda (0,\infty )$. Then,

$$\begin{aligned} \Vert Tf\Vert _{L^p((0,\infty ),w,m_\lambda )}\le C[w]_{A_p^\lambda }^{\max \{1/(p-1),1\}}\Vert f\Vert _{L^p((0,\infty ),w,m_\lambda )}, \end{aligned}$$

for every $f\in L^p((0,\infty ),w,m_\lambda )$, where C does not depend on w, when $T=\phi _*^\lambda$, $T=V_\rho ^\lambda (\{\phi _t\}_{t>0})$, with $\phi \in S(0,\infty )$, and $T=g_\phi ^\lambda$, with $\phi \in S(0,\infty )$ and $h_\lambda (\phi )(0)=0$.

The heat semigroup generated by $-\Delta _\lambda$ is $\{W_t^\lambda \}_{t>0}$ where

$$\begin{aligned} W_t^\lambda (f)=W_{\sqrt{2t}}^\lambda \# _\lambda f,\quad t>0, \end{aligned}$$

being $W^\lambda (x)=\frac{2^{(1-2\lambda )/2}}{\Gamma (\lambda +1/2)}e^{-x^2/2}$, $x\in (0,\infty )$. We have that $W ^\lambda \in S(0,\infty )$ and according to [29, (10) p. 29], $h_\lambda (W^\lambda )(x)=\frac{2^{(1-2\lambda )/2}}{\Gamma (\lambda +1/2)}e^{-x^2/2}$, $x\in (0,\infty )$. By proceeding as in the Poisson case we deduce that $h_\lambda ((t^m\partial _t^mW_t^\lambda ) _{|t=1})(0)=0$, for every $m\in {\mathbb {N}}$, $m>0$.

Corollary 2.8

Let $\lambda >0$, $\rho >2$, $m\in {\mathbb {N}}$, $1<p<\infty$ and $w\in A_p^\lambda (0,\infty )$. Assertions (a), (b) and (c) in Corollary 2.6 hold when the Poisson semigroup $\{P_t^\lambda \}_{t>0}$ is replaced by the heat semigroup $\{W_t^\lambda \}_{t>0}$.

(III) We define, for every $t>0$ the Bochner-Riesz means $B_t^{\lambda ,\alpha }$ associated with the operator $\Delta _\lambda$ by

$$\begin{aligned} B_t^{\lambda ,\alpha }(f)=h_\lambda \left( \Big (1-\frac{y^2}{t^2}\Big )_+^\alpha h_\lambda (f)\right) , \end{aligned}$$

where $(z)_+=\max \{0,z\}$, $z\in {\mathbb {R}}$. We can write

$$\begin{aligned} B_t^{\lambda ,\alpha }(f)=\phi _{1/t}^{\lambda ,\alpha }\# _\lambda f,\quad t>0, \end{aligned}$$

being $\phi ^{\lambda ,\alpha }(z)=2^\alpha \Gamma (\alpha +1)z^{-\alpha -\lambda -1/2}J_{\alpha +\lambda +1/2}(z)$, $z\in (0,\infty )$. We can see that

$$\begin{aligned} \psi _*^\lambda (f)=\sup _{t>0}|B_t^{\lambda ,\alpha }(f)|=:B_*^{\lambda ,\alpha }(f), \end{aligned}$$

and, for every $\rho >2$,

$$\begin{aligned} V_\rho (\{\psi _t\}_{t>0})=V_\rho (\{B_t^{\lambda ,\alpha }\}_{t>0}), \end{aligned}$$

where $\psi =\phi ^{\lambda ,\alpha }$.

According to [52, (5.4.3) and (5.11.6)] we have that, for every $\nu >-1$,

$$\begin{aligned} |J_\nu (x)|\le C\left\{ \begin{array}{ll} x^{-1/2},&{}x\in (1,\infty ),\\ x^\nu ,&{}x\in (0,1). \end{array} \right. \end{aligned}$$

(2.16)

Thus, we have that

$$\begin{aligned} |\phi ^{\lambda , \alpha }(x)|\le \frac{C}{(1+x)^{\alpha +\lambda +1}},\quad x\in (0,\infty ). \end{aligned}$$

Hence,

$$\begin{aligned} |\phi ^{\lambda , \alpha }(x)|\le \frac{C}{(1+x)^{2\lambda +2}},\quad x\in (0,\infty ), \end{aligned}$$

provided that $\alpha \ge \lambda +1$.

Since $\frac{d}{dz}(z^{-\nu } J_\nu (z))=-z^{-\nu }J_{\nu +1}(z)$, $z\in (0,\infty )$ and $\nu >-1$ ([52, (5.3.5)]), we get

$$\begin{aligned} \frac{d}{dx}(\phi ^{\lambda ,\alpha }(x))=2^\alpha \Gamma (\alpha +1)x^{-\alpha -\lambda -1/2}J_{\alpha +\lambda +3/2}(x),\quad x\in (0,\infty ). \end{aligned}$$

By using (2.16) we obtain

$$\begin{aligned} \Big |\frac{d}{dx}\phi ^{\lambda ,\alpha } (x)\Big |&\le C\left\{ \begin{array}{ll} x,&{}x\in (0,1)\\ x^{-\alpha -\lambda -1},&{}x\in (1,\infty ) \end{array} \right. \le C\frac{x}{(1+x)^{\alpha +\lambda +2}}\\&\le C\frac{x}{(1+x)^{2\lambda +4}},\quad x\in (0,\infty ), \end{aligned}$$

when $\alpha \ge \lambda +2$. According to [52, (5.3.5)] we get

$$\begin{aligned} \frac{d^2}{dx^2}\phi ^{\lambda ,\alpha } (x)= & {} 2^\alpha \Gamma (\alpha +1)\left( x^{-\alpha -\lambda -3/2}J_{\alpha +\lambda +3/2}(x)-x^{\alpha -\lambda -1/2}J_{\alpha +\lambda +5/2}(x)\right) ,\\&\quad x\in (0,\infty ). \end{aligned}$$

Then,

$$\begin{aligned} \Big |\frac{d^2}{dx^2}\phi ^{\lambda ,\alpha } (x)\Big |\le \frac{C}{(1+x)^{2\lambda +4}},\quad x\in (0,\infty ), \end{aligned}$$

provided that $\alpha \ge \lambda +3$.

The Stein-square function associated with the operator $\Delta _\lambda$ is defined by

$$\begin{aligned} G_\lambda ^\alpha (f)(x)=\left( \int _0^\infty \Big |t\partial _tB_t^{\lambda ,\alpha }(f)(x)\Big |^2\frac{dt}{t}\right) ^{1/2},\quad x\in (0,\infty ). \end{aligned}$$

We have that

$$\begin{aligned} G_\lambda ^\alpha (f)(x)=\left( \Big |(\Phi _{1/t}\# _\lambda f)(x)\Big |^2\frac{dt}{t}\right) ^{1/2}=g_\Phi ^\lambda (f)(x),\quad x\in (0,\infty ), \end{aligned}$$

where $\Phi (z)=\phi ^{\lambda ,\alpha }(z)+z\frac{d}{dz}\phi ^{\lambda ,\alpha }(z)$, $z\in (0,\infty )$.

We get

$$\begin{aligned} \frac{d}{dz}\Phi (z)=2\frac{d}{dz}\phi ^{\lambda ,\alpha }(z)+z\frac{d^2}{dz^2}\phi ^{\lambda ,\alpha }(z),\quad z\in (0,\infty ). \end{aligned}$$

As above it follows that

$$\begin{aligned} \Big |\frac{d}{dz}\Phi (z)\Big |\le \frac{C}{(1+x)^{\lambda +\alpha }}\le \frac{C}{(1+x)^{2\lambda +3}},\quad x\in (0,\infty ), \end{aligned}$$

when $\alpha \ge \lambda +3$.

On the other hand, since $h_\lambda (\phi ^{\lambda ,\alpha })(x)=(1-x^2)_+^\alpha$, $x\in (0,\infty )$, ([29, (7) p. 48]) we obtain that $h_\lambda (\Phi )(x)=2\alpha x^2(1-x^2)_+^{\alpha -1}$, $x\in (0,\infty )$. Then, when $\alpha >1/2$,

$$\begin{aligned} \int _0^\infty |h_\lambda (\Phi )(x)|^2\frac{dx}{x}<\infty . \end{aligned}$$

From Theorem 1.1 we deduce the following result.

Corollary 2.9

Let $\lambda >0$, $\rho >2$, $1<p<\infty$ and $w\in A_p^\lambda (0,\infty )$.

(a) If $\alpha \ge \lambda +1$, the maximal operator

$$\begin{aligned} B_*^{\lambda ,\alpha }(f)=\sup _{t>0}|B_t^{\lambda ,\alpha } (f)| \end{aligned}$$

is bounded from $L^p(0,\infty ),w,m_\lambda )$ into itself and

$$\begin{aligned} \Vert B_*^{\lambda , \alpha }(f)\Vert _{L^p((0,\infty ),w,m_\lambda )}\le C[w]_{A_p^\lambda }^{\max \{1/(p-1), 1\}}\Vert f\Vert _{L^p((0,\infty ),w,m_\lambda )}, \end{aligned}$$

for every $f\in L^p((0,\infty ) ,w,m_\lambda )$.

(b) If $\alpha \ge \lambda +3$ and T denotes the square function $G_\lambda ^\alpha$ or the $\rho$-variation operator $V_\rho (\{B_t^{\lambda ,\alpha } \}_{t>0})$, defined in the obvious way by

$$\begin{aligned} V_\rho (\{B_t^{\lambda ,\alpha } \}_{t>0})(f)(x)=V_\rho (\{B_t^{\lambda ,\alpha }(f)(x)\}_{t>0}),\quad x\in (0,\infty ), \end{aligned}$$

then T is bounded from $L^p((0,\infty ) ,w,m_\lambda )$ into itself and

$$\begin{aligned} \Vert T(f)\Vert _{L^p((0,\infty ),w,m_\lambda )}\le C[w]_{A_p^\lambda }^{\max \{1/(p-1), 1\}}\Vert f\Vert _{L^p((0,\infty ),w,m_\lambda )}, \end{aligned}$$

for every $f\in L^p((0,\infty ) ,w,m_\lambda )$.

Here $C>0$ does not depend on w.

3 Proof of theorem 1.2

In order to prove Theorem 1.2 we can adapt the procedure used in the proof of [24, Theorems 1.3 and 3.7] to a vector-valued setting. Actually the arguments in the proof of those theorems can be seen as a version in spaces of homogeneous type of the ones developed in [55] and [56]. In [60, §9] Lorist commented that the arguments presented in [56] can be extended to the vector-valued framework and to spaces of homogeneous type.

Suppose that the triple $(\Omega ,d,\mu )$ is a space of homogeneous type where d is a quasimetric on $\Omega$ and $\mu$ is a positive doubling measure on $\Omega$. Assume that T is a linear and bounded operator from $L^1(\Omega ,\mu )$ into $L^{1,\infty }(\Omega , \mu , E)$ where E is a Banach space. Given a ball $B_0$ in $\Omega$ we define the maximal type operator ${\mathcal {M}}_{T,B_0}$ by

$$\begin{aligned} {\mathcal {M}}_{T,B_0}(f)(x)=\sup _{\begin{array}{c} B\ni x,B\subset B_0\\ B \text{ ball } \end{array}}\Vert T(f{\mathcal {X}}_{3B_0\setminus 3B})\Vert _{L^\infty (B,\mu ,E)},\quad x\in \Omega . \end{aligned}$$

By proceeding as in the proof of [52, Lemma 3.2] (see [24, Lemma 3.6]) we can see that, for almost all $x\in B_0$,

$$\begin{aligned} \Vert T(f{\mathcal {X}}_{3B_0})(x)\Vert _E\le C\Vert T\Vert _{L^1(\Omega ,\mu , E)\rightarrow L^{1,\infty }(\Omega ,\mu , E)}|f(x)|+{\mathcal {M}}_{T,B_0}(f)(x). \end{aligned}$$

This property is used in the proof of Theorem 1.2 for each one of the operators we consider.

3.1 Proof of theorem 1.2 for $\phi _{*,b}^{\lambda ,m}$

We see the properties we need in order for the arguments developed in the proof of [24, Theorems 1.3 and 3.7] to work for $\phi _{*,b}^{\lambda ,m}$.

By proceeding as in Sect.  2.1 we can see that

$$\begin{aligned} \phi _{*,b}^{\lambda ,1}(f)(x)\le {\mathcal {M}}_\lambda (|b(x)-b(\cdot )|f)(x),\quad x\in (0,\infty ). \end{aligned}$$

Then, Theorem 1.2 for $\phi _{*,b}^{\lambda ,1}$ can be deduced from [34, Theorem 1.2]. In [31] $L^p$-weighted inequalities were obtained for the maximal commutator operator ${\mathcal {M}}_b^m$ defined by

$$\begin{aligned} {\mathcal {M}}_b^m(f)(x)={\mathcal {M}}(|b(x)-b(\cdot )|^mf)(x),\quad x\in (0,\infty ), \end{aligned}$$

where ${\mathcal {M}}$ is the Euclidean Hardy-Littlewood operator. In [31] they did not study how the $L^p(w)$-norm of the operators depends on the weight w.

The results in [24] (specifically, [24, Theorems 1.3 and 3.7]) work for commutators involving Calderón-Zygmund operators. Here we are concerned with the operator $\phi _{*,b}^{\lambda ,m}$ that can be written as

$$\begin{aligned} \phi _{*,b}^{\lambda ,m}(f)(x)=\big \Vert (\phi _t\#_\lambda [(b(\cdot )-b(x))^mf])(x)\big \Vert _{L^\infty ((0,\infty ),dt)},\quad x\in (0,\infty ). \end{aligned}$$

In order to use the procedure developed in [24, Theorems 1.3 and 3.7] we consider the operator T defined by

$$\begin{aligned} T(f)(x,t)=(\phi _t\#_\lambda f)(x),\quad x,t\in (0,\infty ). \end{aligned}$$

It is clear that $\phi _*^\lambda (f)(x)=\Vert T(f)(x,\cdot )\Vert _{L^\infty ((0,\infty ),dt)}$, $x\in (0,\infty )$. It was established in Sect. 2.1 that $\phi _*^\lambda (f)\le {\mathcal {M}}_\lambda (f)$, and then, $\phi _*^\lambda$ is bounded from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$. In the next proposition we establish that the operator T can be seen as a $L^\infty ((\frac{1}{\ell },\ell ),dt)$-valued Calderón-Zygmund operator, for every $\ell \in {\mathbb {N}}$. This property is uniform in $\ell \in {\mathbb {N}}$. This means that the bounds appearing (see (3.1) and (3.2) below) do not depend on $\ell \in {\mathbb {N}}$.

Proposition 3.1

Let $\lambda >0$. Assume that $\phi$ is a differentiable function on $(0,\infty )$ such that $|\phi ^{(k)}(x)|\le C(1+x)^{-2\lambda -2-k}$, $x\in (0,\infty )$ and $k=0,1$. Let $\ell \in {\mathbb {N}}$. We consider the operator

$$\begin{aligned}{}[T_\ell (f)(x)](t)=(\phi _t\# _\lambda f)(x),\quad t\in \Big [\frac{1}{\ell },\ell \Big ] \text{ and } x\in (0,\infty ). \end{aligned}$$

We have that

(i) For every $f\in C_c^\infty (0,\infty )$, the space of smooth functions with compact support on $(0,\infty )$,

$$\begin{aligned} T_\ell (f)(x)=\int _0^\infty \,_\lambda \tau _x(\phi _\centerdot )(y)f(y)y^{2\lambda }dy,\quad x\not \in \mathrm{supp }f, \end{aligned}$$

where the integral is understood in the Banach space $(C([1/\ell ,\ell ]),\Vert \cdot \Vert _\ell )$ of the continuous functions on $[1/\ell ,\ell ]$ where

$$\begin{aligned} \Vert g\Vert _\ell =\max _{t\in [1/\ell ,\ell ]}|g(t)|,\quad g\in C\Big (\Big [\frac{1}{\ell },\ell \Big ]\Big ). \end{aligned}$$

(ii) We define

$$\begin{aligned} K_t(x,y)=\,_\lambda \tau _x(\phi _t)(y),\quad t,x,y\in (0,\infty ). \end{aligned}$$

Then, there exists $C>0$ such that

$$\begin{aligned} \Vert K_{\centerdot } (x,y)\Vert _\infty \le \frac{C}{m_\lambda (B(x,|x-y|))},\quad x,y\in (0,\infty ),\;x\not =y, \end{aligned}$$

(3.1)

and

$$\begin{aligned}&\Vert \partial _xK_\centerdot (x,y)\Vert _\infty +\Vert \partial _yK_\centerdot (x,y\Vert _\infty \le \frac{C}{|x-y|m_\lambda (B(x,|x-y|))},\nonumber \\&\quad x,y\in (0,\infty ),\;x\not =y. \end{aligned}$$

(3.2)

Proof

Firstly we prove (3.1). Since $|\phi (x)|\le C(1+x)^{-2\lambda -2}$, $x\in (0,\infty )$, as in [73, (2.4)] we get

$$\begin{aligned} |K_t(x,y)|\le \frac{C}{m_\lambda (B(x,t))+m_\lambda (B(x,|x-y|))}\frac{t}{t+|x-y|},\quad t,x,y\in (0,\infty ). \end{aligned}$$

Then,

$$\begin{aligned} \Vert K_\centerdot (x,y)\Vert _\infty \le \frac{C}{m_\lambda (B(x,|x-y|))},\quad x,y\in (0,\infty ),x\not =y. \end{aligned}$$

We have, for every $t,x,y\in (0,\infty )$,

$$\begin{aligned} \partial _xK_t(x,y)= & {} \frac{\Gamma (\lambda +1/2)}{\Gamma (\lambda ) \sqrt{\pi }t^{2\lambda +2}}\int _0^\pi \phi '\\&\left( \frac{\sqrt{x^2+y^2-2xy\cos \theta }}{t}\right) \frac{x-y\cos \theta }{\sqrt{x^2+y^2-2xy\cos \theta }}\sin ^{2\lambda -1}\theta d\theta . \end{aligned}$$

Since $|\phi '(x)|\!\le \! C(1+x)^{-2\lambda -3}$, $x\in (0,\infty )$, and $|x-y\cos \theta |\!\le \! \sqrt{x^2+y^2-2xy\cos \theta }$, $x,y\in (0,\infty )$ and $\theta \in (0,\pi )$, we get

$$\begin{aligned} |\partial _xK_t(x,y)|\le Ct\int _0^\pi \frac{\sin ^{2\lambda -1}\theta }{(t^2+x^2+y^2-2xy\cos \theta )^{\lambda +3/2}}d\theta ,\quad t,x,y\in (0,\infty ). \end{aligned}$$

By proceeding as in [73, (2.8)] we obtain

$$\begin{aligned} |\partial _xK_t(x,y)|&\le C\frac{1}{m_\lambda (B(x,t))+m_\lambda (B(x,|x-y|))}\frac{t}{(t+|x-y|)^2}\\&\le \frac{C}{|x-y|m_\lambda (B(x,|x-y|))},\quad t,x,y\in (0,\infty ),\;x\not =y. \end{aligned}$$

Then,

$$\begin{aligned} \Vert \partial _xK_\centerdot (x,y)\Vert _\infty \le \frac{C}{|x-y|m_\lambda (B(x,|x-y|))},\quad x,y\in (0,\infty ),\;x\not =y. \end{aligned}$$

By simmetry, (3.2) is established.

Let $f\in C_c^\infty (0,\infty )$ and $\ell \in {\mathbb {N}}$. According to (3.1) the integral defining $T_\ell (t)(x)$ is $\Vert \cdot \Vert _\ell$-Bochner convergent for every $x\not \in \mathrm{supp }f$. The dual space of $C([1/\ell ,\ell ])$ is the space ${\mathbb {M}}([1/\ell ,\ell ])$ of complex measures in $[1/\ell ,\ell ]$. Let $\mu \in {\mathbb {M}}([1/\ell ,\ell ])$. We have that

$$\begin{aligned} \int _0^\infty \int _{1/\ell }^\ell K_t(x,y)d\mu (t)f(y)y^{2\lambda }dy=\int _{1/\ell }^\ell (\phi _t\#_\lambda f)(x)d\mu (t),\quad x\not \in \mathrm{supp }f. \end{aligned}$$

This equality is justified because, from (3.1) we get

$$\begin{aligned}&\int _0^\infty \int _{1/\ell }^\ell |K_t(x,y)|d|\mu |(t)|f(y)|y^{2\lambda }dy\\&\quad \le C|\mu |\Big (\Big [\frac{1}{\ell },\ell \Big ]\Big )\int _{\mathrm{supp }f}\frac{dy}{m_\lambda (B(x,|x-y|))}<\infty ,\quad x\not \in \mathrm{supp }f. \end{aligned}$$

Then, for every $x\not \in \mathrm{supp }f$

$$\begin{aligned} \left( \int _0^\infty \,_\lambda \tau _x(\phi _\centerdot )(y)f(y)y^{2\lambda }dy\right) (t)=(\phi _t\# _\lambda f)(x),\quad \text{ a.e. } t\in \Big (\frac{1}{\ell },\ell \Big ). \end{aligned}$$

$\square$

The proof of [24, Theorem 3.7] is an adaptation to spaces of homogeneous type of the proof of the sparse domination property in [41, §5] stated in [41, Theorem 1.1]. The operators considered in [41, Theorem 1.1] satisfy a Hörmander type condition that is weaker than Calderón-Zygmund property when T is a singular integral (see [41, §4.3]). We prefer in this case to establish in Proposition 3.1 a Calderón-Zygmund property for our operator. Furthermore, the estimate in (3.2) allows us to establish the following proposition.

We define the operator

$$\begin{aligned} {\mathbb {M}}_\phi ^{\lambda ,\#}(f)(x)= & {} \sup _{\begin{array}{c} Q\ni x\\ Q \text{ interval } \end{array}}\mathop {\mathrm {ess\, sup \;}}_{y_1,y_2\in Q}\Big |\phi _*^\lambda (f{\mathcal {X}}_{(0,\infty )\setminus (3Q)})(y_1)-\phi _*^\lambda (f{\mathcal {X}}_{(0,\infty )\setminus (3Q)})(y_2)\Big |,\\&\quad x\in (0,\infty ). \end{aligned}$$

According to (3.2) by proceeding as in the proof of Proposition 2.3 we can see the following.

Proposition 3.2

Let $\lambda >0$. Suppose that $\phi$ is a differentiable function on $(0,\infty )$ such that $|\phi ^{(k)}(x)|\le C(1+x)^{-2\lambda -1-k}$, $x\in (0,\infty )$ and $k=0,1$. Then, the operator ${\mathbb {M}}_\phi ^{\lambda ,\#}$ is bounded from $L^p((0,\infty ),m_\lambda )$ into itself, for every $1<p<\infty$, and from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$.

By arguing as at the beginning of Sect. 2, from Proposition 3.2 we can deduce that the operator ${\mathcal {M}}_\phi ^\lambda$ defined by

$$\begin{aligned} {\mathcal {M}}_\phi ^\lambda (f)(x)=\sup _{\begin{array}{c} Q\ni x\\ Q \text{ interval } \end{array}}\Vert \phi _*^\lambda (f{\mathcal {X}}_{(0,\infty )\setminus (3Q)})\Vert _{L^\infty (Q)},\quad x\in (0,\infty ), \end{aligned}$$

is bounded from $L^1((0,\infty ),m_\lambda )$ into $L^{1,\infty }((0,\infty ),m_\lambda )$. This property is other of the central points in the proof of our result.

We have established all the properties that we need to prove Theorem 1.2 by using the arguments in the proofs of [24, Theorems 1.3 and 3.7] (see also [41,  §5]).

3.2 Proof of theorem 1.2 for $g_{\phi ,b}^{\lambda , m}$

After pointing out the key results to adapt the proof of [24, Theorem 3.7] for our operator in the previous Sect. 3.1, we can see that the properties in Propositions 2.2 and 2.3 allow us to prove Theorem 1.2 for $g_{\phi ,b}^{\lambda , m}$ by proceeding as in [24,  Theorem 3.7].

3.3 Proof of theorem 1.2 for $V_{\rho ,b}^{\lambda ,m}(\{\phi _t\}_{t>0})$

By taking into account the properties established in Sect. 2.3 and the comments in the Sect.  3.1, in order to prove Theorem 1.2 for $V_{\rho ,b} ^{\lambda ,m} (\{\phi _t\}_{t>0})$ it is sufficient to show that

$$\begin{aligned} \Vert \{\,_\lambda \tau _x(\phi _t)(y)\}_{t>0}\Vert _{v_\rho }\le \frac{C}{m_\lambda (B(x,|x-y|))},\quad x,y\in (0,\infty ),\;x\not =y. \end{aligned}$$

We have that

$$\begin{aligned} \Vert \{\,_\lambda \tau _x(\phi _t)(y)\}_{t>0}\Vert _{v_\rho }\le \int _0^\infty |\partial _t\,_\lambda \tau _x(\phi _t)(y)|dt,\quad x,y\in (0,\infty ). \end{aligned}$$

Since $\partial _t\,_\lambda \tau _x(\phi _t)(y)=-\frac{1}{t}\,_\lambda \tau _y(\Phi _t)(x)$, $t,x,y\in (0,\infty )$, where $\Phi (x)=(2\lambda +1)\phi (x)+x\phi '(x)$, $x\in (0,\infty )$, we deduce that

$$\begin{aligned} \Vert \{\,_\lambda \tau _x(\phi _t)(y)\}_{t>0}\Vert {v_\rho }&\le C\int _0^\infty \Big |\,_\lambda \tau _y(\Phi _t)(x)\Big |\frac{dt}{t}\\&\le C\int _0^\pi \sin ^{2\lambda -1}\theta \int _0^\infty \Big |\Phi \left( \frac{\sqrt{x^2+y^2-2xy\cos \theta }}{t}\right) \Big |\frac{dt}{t^{2\lambda +2}}d\theta \\&\le C\int _0^\pi \frac{\sin ^{2\lambda -1}\theta }{(x^2+y^2-2xy\cos \theta )^{\lambda +1/2}}\int _0^\infty u^{2\lambda }|\Phi (u)|dud\theta . \end{aligned}$$

Since $|\Phi (u)|\le C(1+u)^{-2\lambda -2}$, $u\in (0,\infty )$, according to [6, Lemma 3.1] it follows that

$$\begin{aligned} \Vert \{\,_\lambda \tau _x(\phi _t)(y)\}_{t>0}\Vert _{v_\rho }\le \frac{C}{m_\lambda (B(x,|x-y|))},\quad x,y\in (0,\infty ),\;x\not =y. \end{aligned}$$

3.4 Applications of theorem 1.2

The special cases of Theorem 1.1 presented in Sect. 2.4 also can be seen as special cases of Theorem 1.2.