Suppose that \(F(x)\) is a monotone nondecreasing function on \([0,\infty)\) such that \(\lim_{x\to 0}F(x)=F(0)=0\) and \(F(x)>0\) for \(x>0\). Call \(F\) an increasing function superpolynomial at 0 if \(\liminf_{x\to 0}F(\lambda x)/F(x)>0\) for some \(\lambda\), \(0<\lambda<1\). If \(F\) is an increasing function superpolynomial at 0, then the function \(G(x)=F(1/x)\) is called a decreasing function superpolynomial at infinity. Suppose that the Laplace integral \(\widehat{F}(t)=\int_0^{\infty}\exp \{-tx\} d F_1(x)\) is finite for \(t>0\), where \(F_1(t)=\sup_{0\leq y<x}F(y)\) for \(x>0\) and \(F_1(0)=0\). The aim of the article is to prove that the following statements are equivalent:
(1) \(F(x)\) is an increasing function superpolynomial at 0;
(2) \(\widehat{F}(t)\) is a decreasing function superpolynomial at infinity;
(3) for some (all) \(c>0\), \(0<\liminf_{x\to 0} \frac{F(x)}{\widehat{F}(c/x)}\leq \limsup_{x\to 0} \frac{F(x)}{\widehat{F}(c/x)}<\infty\).