\lambda^\rho.\end{aligned}$$

We say that a function $f\colon (0, \infty) \mapsto (0, \infty)$ is regularly varying at the infinity with parameter $\rho$, if for all $\lambda>0$, $$\begin{aligned}
\lim_{x \to \infty} \frac{f(\lambda x)}{f(x)} = \lambda^\rho.\end{aligned}$$ When $\rho = 0$ we say that a function $f$ is slowly varying at $0$ (or at $\infty$). Clearly, if $f$ is regularly varying at $0$ and at $\infty$ then it satisfies global weak lower scaling condition as well (with the same scaling parameter).

Next, we recall Potter’s lemma, which is a very useful property of regularly varying functions ([@RegularVariation89 Theorem 1.5.6]). If a function $f$ is regularly varying at infinity with a parameter $\rho$, then for any $C>1, \delta > 0$ exists $X = X(C, \delta),$ such that $$\begin{aligned}
\label{potter_lem}
\frac{f(y)}{f(x)} {\leqslant}C \max \left[\left(\frac{y}{x} \right)^{\rho + \delta}, \left(\frac{y}{x} \right) ^{\rho - \delta} \hspace{0.1cm}  \right ], \quad x,y>X.\end{aligned}$$ Let $k\colon (0, \infty) \mapsto {\mathbb{R}}$. We introduce the Mellin transform $\mathcal{M}$ of the function $k$ by $$\begin{aligned}
\mathcal{M}\{k\}(z) = \int_0^\infty t^{-z} k(t) \frac{dt}{t} ,$$ for $z \in \mathbb{C}$ such that the integral converges.

Next, for $f,g\colon (0, \infty) \mapsto {\mathbb{R}}$ we define its *Mellin convolution* by $$\begin{aligned}
g\stackrel{m}{*} f(x)= \int_0^\infty g(x/t)f(t)\frac{dt}{t}, \quad x>0. \end{aligned}$$ Finally, by $$\begin{aligned}
f\stackrel{s}{*} 