It is known that
\(I_{\alpha,\gamma} f\) is an operator of weak type
\(( 1, \frac{1}{1-\frac{\alpha}{n+|\gamma |}} )\) and is an operator of strong type
\(( r, \frac {1}{\frac{1}{r}-\frac{\alpha}{n+|\gamma|}} )\), where
\(1< r<\infty \). Refer to
\(I_{\alpha,\gamma} f\), Lemma
3, taking
$$p_{1}=1,\qquad q_{1}=\frac{1}{1-\frac{\alpha}{n+|\gamma|}}, \qquad p_{2}=r,\qquad q_{2}=\frac{1}{\frac{1}{r}-\frac{\alpha}{n+|\gamma|}}. $$
Then
$$\sigma_{1}=\frac{1}{q_{1}}-\frac{1}{q_{2}}=1- \frac{1}{r}, \qquad \sigma _{2}=\frac{1}{p_{1}}- \frac{1}{p_{2}}=1-\frac{1}{r}, \qquad \frac{\sigma _{1}}{\sigma_{2}}=1, $$
and
$$\begin{aligned}& \biggl[ \int_{0}^{\infty} \bigl( (I_{\alpha,\gamma} f )^{\ast,\gamma}(t) \bigr)^{q}u^{\ast,\gamma} (t)\,dt \biggr]^{\frac{1}{q}} \\& \quad \leq C \biggl[ \int_{0}^{\infty} \biggl(t^{\frac{\alpha}{n+|\gamma |}-1} \int_{0}^{t}f^{\ast,\gamma} (s)\,ds + t^{\frac{\alpha }{n+|\gamma|}-\frac{1}{r}} \int_{t}^{\infty}s^{\frac{1}{r}-1}f^{\ast ,\gamma} (s) \,ds \biggr)^{q}u^{\ast,\gamma} (t)\,dt \biggr]^{\frac{1}{q}}. \end{aligned}$$
Applying the Minkowski inequality we obtain
$$\begin{aligned}& \biggl[ \int_{0}^{\infty} \bigl( (I_{\alpha,\gamma} f )^{\ast,\gamma}(t) \bigr)^{q}u^{\ast,\gamma} (t)\,dt \biggr]^{\frac{1}{q}} \\& \quad \leq C \biggl[ \int_{0}^{\infty}u^{\ast,\gamma} (t) t^{(\frac{\alpha }{n+|\gamma|}-1)q} \biggl( \int_{0}^{t}f^{\ast,\gamma} (s)\,ds \biggr)^{q} \,dt \biggr]^{\frac{1}{q}} \\& \qquad {}+C \biggl[ \int_{0}^{\infty}u^{\ast,\gamma} (t)t^{ (\frac {\alpha}{n+|\gamma|}-\frac{1}{r} )q} \biggl( \int_{t}^{\infty}s^{\frac{1}{r}-1}f^{\ast,\gamma} (s) \,ds \biggr)^{q} \,dt \biggr]^{\frac{1}{q}}. \end{aligned}$$
(10)
If we take the notation
$$\xi(t)= u^{\ast,\gamma} (t) t^{(\frac{\alpha}{n+|\gamma|}-1)q},\qquad \psi(t)=f^{\ast,\gamma}(t), \qquad \theta(t)=\frac{1}{ (\frac{1}{v} )^{\ast,\gamma} (t)}, $$
then we have (
2) from (
8) and applying (
1)
$$\begin{aligned}& \biggl[ \int_{0}^{\infty}u^{\ast,\gamma} (t) t^{(\frac{\alpha }{n+|\gamma|}-1)q} \biggl( \int_{0}^{t}f^{\ast,\gamma} (s)\,ds \biggr)^{q} \,dt \biggr]^{\frac{1}{q}} \\& \quad \leq C \biggl( \int_{0}^{\infty}\frac{1}{ (\frac{1}{v} )^{\ast,\gamma} (t)} \bigl(f^{\ast,\gamma}(t) \bigr)^{p}\,dt \biggr)^{\frac{1}{p}}. \end{aligned}$$
(11)
Now if we take
$$\xi(t)= u^{\ast,\gamma} (t) t^{(\frac{\alpha}{n+|\gamma|}-\frac{1}{r})q},\qquad \psi(t)=t^{\frac{1}{r}-1}f^{\ast,\gamma}(t), \qquad \theta (t)=\frac{1}{ (\frac{1}{v} )^{\ast,\gamma} (t)}t^{p (\frac{1}{r}-1 )}, $$
then we have (
4) from (
9) and applying (
3) we can assert that
$$\begin{aligned}& \biggl[ \int_{0}^{\infty}u^{\ast,\gamma} (t)t^{ (\frac{\alpha }{n+|\gamma|}-\frac{1}{r} )q} \biggl( \int_{t}^{\infty}s^{\frac{1}{r}-1}f^{\ast,\gamma} (s) \,ds \biggr)^{q} \,dt \biggr]^{\frac{1}{q}} \\& \quad \leq \biggl( \int_{0}^{\infty} \bigl(t^{\frac{1}{r}-1}f^{\ast,\gamma }(t) \bigr)^{p}\frac{1}{ (\frac{1}{v} )^{\ast,\gamma} (t)}t^{p(\frac{1}{r}-1)}\,dt \biggr)^{\frac{1}{p}} \\& \quad = \biggl( \int_{0}^{\infty}\frac{1}{ (\frac{1}{v} )^{\ast ,\gamma} (t)} \bigl(f^{\ast,\gamma}(t) \bigr)^{p}\,dt \biggr)^{\frac{1}{p}} . \end{aligned}$$
(12)
Combining (
10), (
11), (
12) yields
$$ \biggl[ \int_{0}^{\infty} \bigl( (I_{\alpha,\gamma} f )^{\ast,\gamma}(t) \bigr)^{q}u^{\ast,\gamma} (t)\,dt \biggr]^{\frac{1}{q}} \leq C \biggl( \int_{0}^{\infty}\frac{1}{ (\frac{1}{v} )^{\ast,\gamma} (t)} \bigl(f^{\ast,\gamma}(t) \bigr)^{p}\,dt \biggr)^{\frac{1}{p}}. $$
(13)
Applying (
5), (
13), and (
6) we have
$$\begin{aligned}& \biggl[ \int_{ \mathbb{R}^{n}_{k,+}} \bigl( \bigl(I_{\alpha,\gamma} f(x) \bigr) \bigr)^{q}u (x) \bigl(x'\bigr)^{\gamma}\,dx \biggr]^{\frac{1}{q}} \\& \quad \leq \biggl[ \int_{0}^{\infty} \bigl( (I_{\alpha,\gamma} f )^{\ast,\gamma}(t) \bigr)^{q}u^{\ast,\gamma} (t)\,dt \biggr]^{\frac{1}{q}} \\& \quad \leq C \biggl( \int_{0}^{\infty}\frac{1}{ (\frac{1}{v} )^{\ast,\gamma} (t)} \bigl(f^{\ast,\gamma}(t) \bigr)^{p}\,dt \biggr)^{\frac{1}{p}} \\& \quad \leq C \biggl[ \int_{ \mathbb{R}^{n}_{k,+}} \bigl( f(x) \bigr)^{p}v (x) \bigl(x'\bigr)^{\gamma}\,dx \biggr]^{\frac{1}{p}}. \end{aligned}$$
Thus the proof the theorem is completed. □