A Note on the Inverse Laplace Transformation of $f(t)$

Let $\mathcal{L}\{f(t)\}={\int }_0^{\mathrm{\infty }}{e}^{-st}f(t)dt$ denote the Laplace transform of $f$. It is well-known that if $f(t)$ is a piecewise continuous function on the interval $t:[0,\mathrm{\infty })$ and of exponential order for $t>N$; then ${\lim }_{s\to \mathrm{\infty }}F(s)=0$, where $F(s)=\mathcal{L}\{f(t)\}$. In this paper we prove that the lesser known converse does not hold true; namely, if $F(s)$ is a continuous function in terms of $s$ for which ${\lim }_{s\to \mathrm{\infty }}F(s)=0$, then it does not follow that $F(s)$ is the Laplace transform of a piecewise continuous function of exponential order.