The main results in this section read as follows.
Theorem 4.1
Let
\(0<\alpha<1\), \(\frac{1}{\alpha}< p\leq q<\infty\)
and
\(\beta\geq0\). Let
u
be a non-increasing function on
I. Then the operator
\(T_{\alpha,\beta}\)
is compact from
\(L_{p,w}\)
to
\(L_{q,v}\)
if and only if
\(A_{\alpha, \beta}<\infty\)
and
$$\lim_{z\rightarrow a^{+}}A_{\alpha, \beta}(z)=\lim_{z\rightarrow b^{-}}A_{\alpha, \beta}(z)=0, $$
where
$$A_{\alpha, \beta}(z)= \biggl( \int_{a} ^{z}u^{p'}(s) W^{p'\beta }(s)w(s)\,ds \biggr)^{\frac{1}{p'}} \biggl( \int_{z} ^{b}W^{q(\alpha -1)}(x)v(x)\,dx \biggr)^{\frac{1}{q}}. $$
Theorem 4.2
Let
\(0<\alpha<1\), \(p>{\frac{1}{ \alpha}}\)
and
\(\beta\geq0\). Let
u
be a non-increasing function on
I. If
\(b<\infty\)
and
\(0< q<p<\infty\)
or
\(b=\infty\)
and
\(1< q<p<\infty \), then the operator
\(T_{\alpha,\beta}\)
is compact from
\(L_{p,w}\)
to
\(L_{q,v}\)
if and only if
\(B_{\alpha,\beta}<\infty\).
Proof of Theorem 4.1
Necessity. Let the operator \(T_{\alpha,\beta}\) be compact from \(L_{p,w}\) to \(L_{q,v}\). Then it is bounded and consequently, by Theorem 3.1, we have \(A_{\alpha, \beta}<\infty\). First we need to show that \(\lim_{z\rightarrow a^{+}}A_{\alpha, \beta }(z)=0\). Consider the family of functions \(\{f_{t}\}_{t\in I}\), where
$$ f_{t}(x)=\chi_{(a,t)}(x)u^{p'-1}(x) W^{(p'-1)\beta}(x) \biggl( \int _{a} ^{t}u^{p'}(s) W^{p'\beta}(s)w(s)\,ds \biggr)^{-\frac{1}{p}},\quad x\in I. $$
(4.1)
We note that
$$\begin{aligned} \biggl( \int_{a} ^{b}\bigl\vert f_{t}(x)\bigr\vert ^{p}w(x)\,dx \biggr)^{\frac {1}{p}} =& \biggl( \int_{a} ^{t}\bigl\vert f_{t}(x)\bigr\vert ^{p}w(x)\,dx \biggr)^{\frac{1}{p}} \\ =& \biggl( \int_{a} ^{t}u^{p'}(s) W^{p'\beta}(s)w(s)\,ds \biggr)^{-\frac {1}{p}} \\ &{}\times\biggl( \int_{a} ^{t}u^{p'}(s) W^{p'\beta}(s)w(s)\,ds \biggr)^{\frac{1}{p}}=1. \end{aligned}$$
(4.2)
Next we show that the family of functions \(\{f_{t}\}_{t\in I}\) defined by (4.1) converges weakly to zero in \(L_{p,w}\). Let \(g\in L_{p',w^{1-p'}}= (L_{p,w} )^{*}\). Then, by Hölder’s inequality and (4.2), we find that
$$\begin{aligned} \begin{aligned}[b] \int_{a} ^{b}f_{t}(x)g(x)\,dx&\leq \biggl( \int_{a} ^{t}\bigl\vert f_{t}(x)\bigr\vert ^{p}w(x)\,dx \biggr)^{\frac{1}{p}} \biggl( \int_{a} ^{t}\bigl\vert g(s)\bigr\vert ^{p'}w^{1-p'}(s)\,ds \biggr)^{\frac{1}{p'}} \\ &= \biggl( \int_{a} ^{t}\bigl\vert g(s)\bigr\vert ^{p'}w^{1-p'}(s)\,ds \biggr)^{\frac{1}{p'}}. \end{aligned} \end{aligned}$$
(4.3)
Since \(g\in L_{p',w^{1-p'}}\), the last integral in (4.3) converges to zero as \(t\rightarrow a^{+}\), which means weak convergence of the family of functions \(\{f_{t}\}\) to zero as \(t\rightarrow a^{+}\). Therefore, from the compactness of the operator \(T_{\alpha,\beta}\) from \(L_{p,w}\) to \(L_{q,v}\) it follows that
$$ \lim_{t\rightarrow a^{+}}\|T_{\alpha, \beta}f_{t} \|_{q,v}=0. $$
(4.4)
Moreover,
$$\begin{aligned} \|T_{\alpha, \beta}f_{t}\|_{q,v}^{q} =& \int_{a} ^{b}v(x) \biggl( \int _{a} ^{x}\frac{u(s)W^{\beta}(s)f_{t}(s)w(s)\,ds}{ (W(x)-W(s) )^{1-\alpha}} \biggr)^{q}\,dx \\ \geq& \int_{t} ^{b}v(x) \biggl( \int_{a} ^{t}\frac{u(s)W^{\beta }(s)f_{t}(s)w(s)\,ds}{ (W(x)-W(s) )^{1-\alpha}} \biggr)^{q}\,dx \\ \geq& \int_{t} ^{b}\frac{v(x)\,dx}{W^{q(1-\alpha)}(x)} \biggl( \int _{a} ^{t}u^{p'}(s)W^{p'\beta}(s)w(s) \,ds \biggr)^{-\frac{q}{p}} \\ &{}\times\biggl( \int_{a} ^{t}u^{p'}(s)W^{p'\beta}(s)w(s) \,ds \biggr)^{q} \\ =&A_{\alpha, \beta}^{q}(t). \end{aligned}$$
(4.5)
From (4.4) and (4.5) it follows that \(\lim_{t\rightarrow a^{+}}A_{\alpha, \beta}(t)=0\).
Now, we show that \(\lim_{t\rightarrow b^{-}}A_{\alpha, \beta}(t)=0\).
From the compactness of the operator \(T_{\alpha,\beta}\) from \(L_{p,w}\) to \(L_{q,v}\) compactness of the conjugate operator follows:
$$T_{\alpha,\beta}^{*}g(s)=u(s)W^{p}(s)w(s) \int_{s} ^{b}\frac{g(x)\, dx}{ (W(x)-W(s) )^{1-\alpha}} $$
from \(L_{q',v^{1-q'}}\) to \(L_{p',w^{1-p'}}\).
For \(t\in I\) we introduce the family \(\{g_{t}\}_{t\in I}\) of functions:
$$ g_{t}(x)=\chi_{[t,b)}(x) \biggl( \int_{t} ^{b}W^{q(\alpha-1)}(x)v(x)\, dx \biggr)^{-\frac{1}{q'}}W^{(q-1)(\alpha-1)}(x)v(x). $$
(4.6)
The family \(\{g_{t}\}_{t\in I}\) of functions defined by (4.6) is correctly defined, since due to condition \(A_{\alpha, \beta }<\infty\) the involving integrals are finite. We show that for all \(t\in I\) the functions \(g_{t}\in L_{q',v^{1-q'}}\) converge weakly to zero as \(t\rightarrow b^{-}\).
Indeed,
$$\begin{aligned} \|g_{t}\|_{q',v^{1-q'}} =& \biggl( \int_{t} ^{b}\bigl\vert g_{t}(x)\bigr\vert ^{q'}v^{1-q'}(x)\,dx \biggr)^{\frac{1}{q'}} \\ =& \biggl( \int_{t} ^{b}W^{q(\alpha-1)}(x)v(x)\,dx \biggr)^{-\frac {1}{q'}} \biggl( \int_{t} ^{b}\bigl\vert W^{(q-1)(\alpha -1)}(x)v(x) \bigr\vert ^{q'}v^{1-q'}(x)\,dx \biggr)^{\frac{1}{q'}} \\ =& \biggl( \int_{t} ^{b}W^{q(\alpha-1)}(x)v(x)\,dx \biggr)^{-\frac {1}{q'}} \biggl( \int_{t} ^{b}W^{q(\alpha-1)}(x)v(x)\,dx \biggr)^{\frac {1}{q'}}=1. \end{aligned}$$
(4.7)
By using (4.7) with \(f\in L_{q,v}= (L_{q',v^{1-q'}} )^{*}\) we obtain
$$\begin{aligned} \int_{a} ^{b}g_{s}(x)f(x)\,dx \leq& \biggl( \int_{t} ^{b}\bigl\vert g_{t}(x)\bigr\vert ^{q'}v^{-\frac{q'}{q}}(x)\,dx \biggr)^{\frac{1}{q'}} \biggl( \int_{t} ^{b}\bigl\vert f(x)\bigr\vert ^{q}v(x)\,dx \biggr)^{\frac{1}{q}} \\ \leq&\|g_{t}\|_{q',v^{1-q'}} \biggl( \int_{t} ^{b}\bigl\vert f(x)\bigr\vert ^{q}v(x)\, dx \biggr)^{\frac{1}{q}}= \biggl( \int_{t} ^{b}\bigl\vert f(x)\bigr\vert ^{q}v(x)\,dx \biggr)^{\frac{1}{q}}. \end{aligned}$$
Since \(f\in L_{q,v}\), the last integral tends to zero as \(t\rightarrow b^{-}\), which gives the weak convergence to zero of \(\{g_{t}\}_{t \in I}\) in \(L_{q',v^{1-q'}}\) as \(t\rightarrow b^{-}\). By compactness of \(T_{\alpha,\beta}^{*}: L_{q',v^{1-q'}} \rightarrow L_{p',w^{1-p'}}\) it follows that
$$ \lim_{s\rightarrow b^{-}}\bigl\Vert T_{\alpha,\beta}^{*}g_{t} \bigr\Vert _{p',w^{1-p'}}=0. $$
(4.8)
Furthermore, we note that
$$\begin{aligned} \bigl\| T_{\alpha,\beta}^{*}g_{t}\bigr\| _{p',w^{1-p'}} =&\biggl( \int_{a} ^{b}\bigl|u(s)W^{\beta}(s)w(s)\bigr|^{p'} \\ &{}\times \biggl( \int_{s} ^{b}\frac{g_{t}(x)\,dx}{ (W(x)-W(s) )^{1-\alpha}} \biggr)^{p'}w^{1-p'}(s)\,ds\biggr)^{\frac{1}{p'}} \\ \geq& \biggl( \int_{a} ^{t}u^{p'}(s)W^{p'\beta}(s)w(s) \biggl( \int_{t} ^{b}\frac{g_{t}(x)\,dx}{ (W(x)-W(s) )^{1-\alpha}} \biggr)^{p'}\,ds \biggr)^{\frac{1}{p'}} \\ \geq& \biggl( \int_{a} ^{t}u^{p'}(s)W^{p'\beta}(s)w(s) \,ds \biggr)^{\frac{1}{p'}} \biggl( \int_{t} ^{b}W^{q(\alpha-1)}(x)v(x)\,dx \biggr)^{-\frac{1}{q'}} \\ &{}\times \biggl( \int_{t} ^{b}\frac{W^{(q-1)(\alpha -1)}(x)v(x)\,dx}{W^{1-\alpha}(x)} \biggr)^{q}= A_{\alpha,\beta}(t). \end{aligned}$$
Hence, according to (4.8) we have \(\lim_{s\rightarrow b^{-}}A_{\alpha,\beta}(s)=0\). The proof of the necessity is complete.
Sufficiency. For \(a< c< d< b\) we define
$$P_{c}f:= \chi_{(a,c]}f,\qquad P_{cd}f:= \chi_{(c,d]}f,\qquad Q_{d}f:= \chi_{(d,b)}f. $$
Then
$$f=P_{c}f+P_{cd}f+Q_{d}f $$
and since \(P_{c}T_{\alpha, \beta}P_{cd}\equiv0\), \(P_{c}T_{\alpha, \beta}Q_{d}\equiv0\), \(P_{cd}T_{\alpha, \beta}Q_{d}\equiv0\), we have
$$ T_{\alpha, \beta}f=P_{cd}T_{\alpha, \beta}P_{cd}f+P_{c}T_{\alpha, \beta}P_{c}f+P_{cd}T_{\alpha, \beta}P_{c}f+ Q_{d}T_{\alpha, \beta}f. $$
(4.9)
We show that the operator \(P_{cd}T_{\alpha, \beta}P_{cd}\) is compact from \(L_{p,w}\) to \(L_{q,v}\). Since \(P_{cd}T_{\alpha, \beta}P_{cd}f(x)=0\) for \(x\in I\setminus(c,d)\), it is enough to show that the operator \(P_{cd}T_{\alpha, \beta }P_{cd}\) is compact from \(L_{p,w}(c,d)\) to \(L_{q,v}(c,d)\). This, in turn, is equivalent to compactness of the operator
$$Tf(x)= \int_{c} ^{d}K(x,s)f(s)\,ds $$
from \(L_{p}(c,d)\) to \(L_{q}(c,d)\) with the kernel
$$K(x,s)=\frac{u(s)W^{\beta}(s)v^{\frac{1}{q}}(x)\chi _{(c,d)}(x-s)w^{\frac{1}{p'}}(s)}{ (W(x)-W(s) )^{1-\alpha}}. $$
Let \(\{x_{k}\}_{k\in\mathbb{Z}}\) be the sequence of points defined in the proof of Theorem 3.1. There are points \(x_{i}\), \(x_{n+1}\), \(x_{i}< x_{n+1}\) such that \(x_{i}\leq c< x_{i+1}\), \(x_{n}< d\leq x_{n+1}\). We assume that the numbers c, d are chosen so that \(x_{i+1}< x_{n}\). Similarly to obtaining estimates of \(J_{1}\) and \(J_{2}\) in Theorem 3.1, we have
$$\begin{aligned}& \int_{c}^{d} \biggl( \int_{c}^{d}\bigl\vert K(x,s)\bigr\vert ^{p'}\,ds \biggr)^{\frac{q}{p'}}\,dx \\& \quad = \int_{c}^{d}v(x) \biggl( \int_{c}^{x} \frac{u^{p'}(s)W^{p'\beta}(s)w(s)\,ds }{(W(x)-W(s))^{p'(1-\alpha)}} \biggr)^{\frac{q}{p'}}\,dx \\& \quad \leq \sum_{k=i}^{n} \int_{x_{k}}^{x_{k+1}}v(x) \biggl[ \biggl( \int _{a}^{x_{k-1}}+ \int_{x_{k-1}}^{x} \biggr)\frac{u^{p'}(s)W^{p'\beta }(s)w(s)\,ds }{(W(x)-W(s))^{p'(1-\alpha)}} \biggr]^{\frac{q}{p'}}\,dx \\& \quad \leq \mu(n-i+1) A^{q}_{\alpha,\beta}< \infty, \end{aligned}$$
where the constant μ does not depend on i, n.
Therefore, on the basis of the Kantarovich condition ([16], p.420), the operator T is compact from \(L_{p}(c,d)\) to \(L_{q}(c,d)\), which is equivalent to compactness of the operator \(P_{cd}T_{\alpha, \beta}P_{cd}\) from \(L_{p,w}\) to \(L_{q,v}\).
From (4.9) it follows that
$$ \|T_{\alpha,\beta}-P_{cd}T_{\alpha, \beta}P_{cd} \|\leq\| P_{c}T_{\alpha, \beta}P_{c}\| + \|P_{cd}T_{\alpha, \beta}P_{c}\|+\| Q_{d}T_{\alpha, \beta} \|. $$
(4.10)
We will show that the right-hand side of (4.10) tends to zero at \(c\rightarrow a\) and \(d\rightarrow b\). Then the operator \(T_{\alpha ,\beta}\) as the uniform limit of compact operators is compact from \(L_{p,w}\) to \(L_{q,v}\).
By using Theorem 3.1 we find that
$$\begin{aligned} \|P_{c}T_{\alpha, \beta}P_{c}f\|_{q,v} =& \biggl( \int_{a} ^{c}v(x) \biggl\vert \int_{a} ^{x}\frac{u(s)W^{\beta}(s)f(s)w(s)\,ds}{ (W(x)-W(s) )^{1-\alpha}}\biggr\vert ^{q}\,dx \biggr)^{\frac{1}{q}} \\ \ll&\sup_{a< z< c} \biggl( \int_{a} ^{z}u^{p'}(s)W^{p'\beta}(s)w(s) \, ds \biggr)^{\frac{1}{p'}} \\ &{}\times \biggl( \int_{z} ^{c}v(x)W^{q(\alpha -1)}(x)\,dx \biggr)^{\frac{1}{q}}\|f\|_{p,w} \\ \leq&\sup_{a< z< c}A_{\alpha, \beta}(z)\|f\|_{p,w}. \end{aligned}$$
Consequently, \(\|P_{c}T_{\alpha, \beta}P_{c}\|\ll\sup_{a< z< c}A_{\alpha, \beta}(z)\). Hence,
$$ \lim_{c\rightarrow a^{+}}\|P_{c}T_{\alpha, \beta}P_{c} \|\ll\lim_{c\rightarrow a^{+}}\sup_{a< z< c}A_{\alpha, \beta}(z)= \lim_{z\rightarrow a^{+}}A_{\alpha, \beta}(z)=0. $$
(4.11)
To estimate \(\|P_{cd}T_{\alpha, \beta}P_{c}\|\) we assume that \(v_{\varepsilon}(x)=v(x)\) for \(x\in(c,d]\) and \(v_{\varepsilon }(x)=\varepsilon^{q} v(x)\) for \(x\in(a,c]\), \(u_{\varepsilon }(s)=u(s)\) for \(s\in(a,c]\) and \(u_{\varepsilon}(s)=\varepsilon u(s)\) for \(s\in(c,d]\), where \(1>\varepsilon>0\). Obviously, the function \(u_{\varepsilon}\) is non-increasing on I. Then, according to Theorem 3.1, we obtain
$$\begin{aligned} \|P_{cd}T_{\alpha, \beta}P_{c}\|_{q,v} =& \biggl( \int_{c} ^{d}v(x)\biggl\vert \int_{a} ^{c}\frac{u(s)W^{\beta}(s)f(s)w(s)\, ds}{ (W(x)-W(s) )^{1-\alpha}}\biggr\vert ^{q}\,dx \biggr)^{\frac{1}{q}} \\ \leq& \biggl( \int_{a} ^{d}v_{\varepsilon}(x)\biggl\vert \int_{a} ^{x}\frac {u_{\varepsilon}(s)W^{\beta}(s)f(s)w(s)\,ds}{ (W(x)-W(s) )^{1-\alpha}}\biggr\vert ^{q}\,dx \biggr)^{\frac{1}{q}} \\ \ll& A^{\varepsilon}_{\alpha,\beta}\|f\|_{p,w}, \end{aligned}$$
(4.12)
where
$$A_{\alpha,\beta}^{\varepsilon}=\sup_{a< z< d} \biggl( \int_{z} ^{d}W^{q(\alpha-1)}(x)v_{\varepsilon}(x) \,dx \biggr)^{\frac{1}{q}} \biggl( \int_{a} ^{z}u^{p'}_{\varepsilon}(s)W^{p'\beta}(s)w(s) \, ds \biggr)^{\frac{1}{p'}}. $$
We estimate the expression \(A_{\alpha,\beta}^{\varepsilon}\) from the above as follows:
$$\begin{aligned} A_{\alpha,\beta}^{\varepsilon} \leq&\sup_{a< z< c} \biggl( \int_{c} ^{d}W^{q(\alpha-1)}(x)v(x)\,dx+ \varepsilon^{q} \int_{z} ^{c}W^{q(\alpha-1)}(x)v(x)\,dx \biggr)^{\frac{1}{q}} \\ &{}\times \biggl( \int_{a} ^{z}u^{p'}(s)W^{p'\beta}(s)w(s) \,ds \biggr)^{\frac{1}{p'}} \\ &{}+\sup_{c< z< d} \biggl( \int_{z} ^{d}W^{q(\alpha-1)}(x)v(x)\,dx \biggr)^{\frac{1}{q}} \\ &{}\times \biggl( \int_{a} ^{c}u^{p'}(s)W^{p'\beta}(s)w(s) \, ds+\varepsilon^{p'} \int_{c} ^{z}u^{p'}(s)W^{p'\beta}(s)w(s) \, ds \biggr)^{\frac{1}{p'}} \\ \leq&\sup_{a< z< c} \biggl( \int_{c} ^{d}W^{q(\alpha-1)}(x)v(x)\, dx \biggr) \biggl( \int_{a} ^{z}u^{p'}(s)W^{p'\beta}(s)w(s) \,ds \biggr)+\varepsilon A_{\alpha,\beta} \\ &{}+\sup_{c< z< d} \biggl( \int_{z} ^{d}W^{q(\alpha-1)}(x)v(x)\,dx \biggr)^{\frac{1}{q}} \biggl( \int_{a} ^{c}u^{p'}(s)W^{p'\beta}(s)w(s) \, ds \biggr)^{\frac{1}{p'}}+\varepsilon A_{\alpha,\beta} \\ \leq&2 \bigl(A_{\alpha,\beta}(c)+\varepsilon A_{\alpha,\beta} \bigr). \end{aligned}$$
(4.13)
Since the left side of (4.12) does not depend on \(\varepsilon >0\), substituting (4.13) in (4.12) and letting \(\varepsilon\rightarrow0\), we get
$$\|P_{cd}T_{\alpha,\beta}P_{c}f\|\ll A_{\alpha,\beta}(c) \|f\|_{p,w}. $$
Therefore \(\|P_{cd}T_{\alpha,\beta}P_{c}\|\ll A_{\alpha,\beta}(c)\) and we conclude that
$$ \lim_{c\rightarrow a^{+}}\|P_{cd}T_{\alpha,\beta}P_{c} \|\ll\lim_{c\rightarrow a^{+}}A_{\alpha,\beta}(c)=0. $$
(4.14)
Next, arguing as above we find that
$$\begin{aligned} \|Q_{d}T_{\alpha,\beta}f\|_{q,v} =& \biggl( \int_{d} ^{b}v(x) \biggl\vert \int_{a} ^{x}\frac{u(s)W^{\beta}(s)f(s)w(s)\,ds}{ (W(x)-W(s) )^{1-\alpha}} \biggr\vert ^{q}\,dx \biggr)^{\frac{1}{q}} \\ \ll&\sup_{d< z< b}A_{\alpha,\beta}(z)\|f\|_{p,w}. \end{aligned}$$
Consequently,
$$ \lim_{d\rightarrow b^{-}}\|Q_{d}T_{\alpha,\beta}\| \leq\lim_{d\rightarrow b^{-}}\sup_{d< z< b}A_{\alpha,\beta}(z)= \lim_{z\rightarrow b^{-}}A_{\alpha,\beta}(z)=0. $$
(4.15)
From (4.11), (4.14) and (4.15) it follows that the right-hand side of (4.10) tends to zero as \(c\rightarrow a^{+}\) and \(d\rightarrow b^{-}\). The proof is complete. □
Proof of Theorem 4.2
In the case \(b<\infty\) and \(0< q<p<\infty\) the statement of Theorem 4.2 follows from the Ando theorem and its generalizations [17]. Therefore, we only need to prove Theorem 4.2 in the case \(a=0\), \(b=\infty\) and \(1< q<p<\infty\).
Necessity. Let the operator \(T_{\alpha,\beta}\) be compact from \(L_{p,w}\) to \(L_{q,v}\). Then the operator is bounded. Hence, by Theorem 3.2, \(B_{\alpha,\beta}<\infty\).
Sufficiency. Let \(B_{\alpha,\beta}<\infty\). Here \(T_{\alpha,\beta }f=P_{d}T_{\alpha,\beta}P_{d}f+Q_{d}T_{\alpha,\beta}f\). Therefore
$$ \|T_{\alpha,\beta}-P_{d}T_{\alpha,\beta}P_{d} \|\leq\| Q_{d}T_{\alpha,\beta}\|. $$
(4.16)
Since \(d<\infty\), the operator \(P_{d}T_{\alpha,\beta}P_{d}\) is compact from \(L_{p,w}(0, d)\) to \(L_{q,v}(0, d)\), which is equivalent to its compactness from \(L_{p,w}\) to \(L_{q,v}\). We show that the right-hand side of (4.16) tends to zero as \(d\rightarrow\infty\). Then the operator \(T_{\alpha,\beta}\) is compact from \(L_{p,w}\) to \(L_{q,v}\) as the uniform limit of compact operators.
Let \(1>\varepsilon>0\). To estimate \(\|Q_{d}T_{\alpha,\beta}f\|\) we suppose that \(v_{\varepsilon}(x)=v(x)\) for \(x\in[d,\infty)\) and \(v_{\varepsilon}(x)=\varepsilon^{q}v(x)\) for \(x\in(0, d)\). Using the relations \(B_{\alpha,\beta}\approx \widetilde{B}_{\alpha,\beta}\) (see Remark 2.3), in view of Theorem 3.2, we have
$$\begin{aligned} \|Q_{d}T_{\alpha,\beta}f\| \leq &\biggl( \int_{a} ^{\infty }v_{\varepsilon}(x)\biggl\vert \int_{a} ^{x}\frac{u(s)W^{\beta }(s)f(s)w(s)\,ds}{ (W(x)-W(s) )^{1-\alpha}}\biggr\vert ^{q}\, dx \biggr)^{\frac{1}{q}} \\ \ll&\widetilde{B}^{\varepsilon}_{\alpha,\beta}\|f\|_{p,w} \end{aligned}$$
or
$$ \|Q_{d}T_{\alpha,\beta}\|\ll\widetilde{B}^{\varepsilon}_{\alpha ,\beta}, $$
(4.17)
where
$$\begin{aligned} \widetilde{B}^{\varepsilon}_{\alpha, \beta} =&\biggl( \int_{a}^{\infty } \biggl( \int_{z}^{\infty}W^{q(\alpha-1)}(x)v_{\varepsilon}(x) \,dx \biggr)^{\frac{q}{p-q}} \\ &{}\times \biggl( \int _{a}^{z}u^{p'}(s)W^{p'\beta}(s)w(s) \,ds \biggr)^{\frac {q(p-1)}{p-q}}W^{q(\alpha-1)}(z)v_{\varepsilon}(z)\,dz \biggr)^{\frac {p-q}{pq}}. \end{aligned}$$
Passing to the limit \(\varepsilon\rightarrow0^{+}\), from (4.17) it follows that
$$\begin{aligned} \|Q_{d}T_{\alpha,\beta}\| \ll&\biggl( \int_{d}^{\infty} \biggl( \int _{z}^{\infty}W^{q(\alpha-1)}(x)v(x)\,dx \biggr)^{\frac{q}{p-q}} \\ &{}\times \biggl( \int_{a}^{z}u^{p'}(s)W^{p'\beta}(s)w(s) \, ds \biggr)^{\frac{q(p-1)}{p-q}}W^{q(\alpha-1)}(z)v(z)\,dz\biggr)^{\frac{p-q}{pq}}. \end{aligned}$$
Hence,
$$ \lim_{d\rightarrow\infty}\|Q_{d}T_{\alpha,\beta} \|=0. $$
(4.18)
Obviously, (4.18) implies that the right-hand side of (4.16) tends to zero as \(d\rightarrow\infty\). The proof is complete. □