Let
\(F,G\in \overline{\mathit{HL}}_{p}(\alpha,\lambda )\), then, from (
1.8), we know that, in order to prove
\(F\ast G \in \overline{\mathit{HL}}_{p}(\alpha,\lambda )\), we need to show that
$$ \sum_{k=1}^{p}\sum _{j=1}^{\infty } \biggl((k-1)+\frac{j^{\lambda }(j- \alpha )}{2(1-\alpha )} \biggr) \bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b _{j,p-k+1} \vert \bigr)\leq 1. $$
(2.9)
Since
\(F,G\in \overline{\mathit{HL}}_{p}(\alpha,\lambda )\), using (
1.8), we have
$$ \sum_{k=1}^{p}\sum _{j=1}^{\infty } \biggl((k-1)+\frac{j^{\lambda }(j- \alpha )}{2(1-\alpha )} \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr) \leq 1 $$
(2.10)
and
$$ \sum_{k=1}^{p}\sum _{j=1}^{\infty } \biggl((k-1)+\frac{j^{\lambda }(j- \alpha )}{2(1-\alpha )} \biggr) \bigl( \vert A_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \bigr) \leq 1. $$
(2.11)
From (
2.10) and (
2.11), we obtain
$$ \sum_{j=1}^{\infty } \biggl((k-1)+ \frac{j^{\lambda }(j-\alpha )}{2(1- \alpha )} \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr)\leq 1 $$
(2.12)
and
$$ \sum_{j=1}^{\infty } \biggl((k-1)+ \frac{j^{\lambda }(j-\alpha )}{2(1- \alpha )} \biggr) \bigl( \vert A_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \bigr)\leq 1. $$
(2.13)
Using the Cauchy–Schwarz inequations, from (
2.12) and (
2.13), we get
$$ \sum_{j=1}^{\infty } \biggl((k-1)+ \frac{j^{\lambda }(j-\alpha )}{2(1- \alpha )} \biggr) \sqrt{\bigl( \vert A_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \bigr) \bigl( \vert a_{j,p-k+1} \vert + \vert b _{j,p-k+1} \vert \bigr)}\leq 1, $$
(2.14)
because
$$\begin{aligned} &\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr) \\ &\quad\leq \bigl( \vert A_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \bigr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr)\quad (1 \leq k\leq 1,j\in \mathbb{N}). \end{aligned}$$
(2.15)
So from (
2.14) and (
2.15), we have
$$ \sum_{j=1}^{\infty } \biggl((k-1)+ \frac{j^{\lambda }(j-\alpha )}{2(1- \alpha )} \biggr)\sqrt{\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b _{j,p-k+1} \vert \bigr)}\leq 1, $$
and hence
$$ \sum_{k=1}^{p}\sum _{j=1}^{\infty } \biggl((k-1)+\frac{j^{\lambda }(j- \alpha )}{2(1-\alpha )} \biggr) \sqrt{\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B _{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr)}\leq p, $$
(2.16)
which implies that
$$ \sqrt{\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr)} \leq \frac{p}{ ((k-1)+ \frac{j^{\lambda }(j-\alpha )}{2(1-\alpha )} )}. $$
(2.17)
In addition, if
$$ \bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr)\leq \frac{1}{p} \sqrt{\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr)}, $$
that is,
$$ \sqrt{\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr)} \leq \frac{1}{p}, $$
(2.18)
then we obtain the conditions of satisfaction (
2.9). Again, combining (
2.17) and (
2.18) with
\(k=1\) and
\(j=2\), we can get
$$ \frac{p}{ (\frac{2^{\lambda }(2-\alpha )}{2(1-\alpha )} )} \leq \frac{1}{p}, $$
which deduces condition (
2.8). The proof is completed. □