Now I would like to show a kind of extension of the Laplace transform.
For $\lambda \geq 0$, let $f_\lambda :[0,\infty) \to [0,1] $ be strictly decreasing and continuous. We assume \begin{align} f(0) = 1, \ \lim_{x \to \infty}f_\lambda(x) = 0 \end{align} and $f_\lambda(x)$ is strictly decreasing as a function of $\lambda$ for every $x \geq 0$. We denote a linear space generated by $( f_\lambda )_{\lambda \geq 0}$ by $X$.
Question: Can we show that $X$ is dense in $C_{0}[0,\infty)$? Here $C_{0}[0,\infty)$ denotes a space of continuous functions on $[0,\infty)$ vanishing at $\infty$ with supremum norm.
If $f_{\lambda}(x) = \mathrm{e}^{-\lambda x}$, this is the usual Laplace transform and by applying Stone-Weierstrass theorem, the answer is yes. But I cannot show it for general $(f_\lambda)$ since $X$ may not comprise subalgebra of $C_{0}[0,\infty)$. I think it is necessary to assume more for $(f_\lambda)$, but I cannot find appropriate assumptions. I'd appreciate it if you could answer to my question or tell me some relevant known results.
This cannot be true: First, $f_\lambda$ can be independent of $\lambda$, so the linear space generated by $(f_\lambda)_{\lambda\ge0}$ is one-dimensional.
Here is another function that fulfills the assumptions: $$ f_\lambda(x) = \max(0, \ 1 - e^\lambda \cdot x). $$ The support of these functions is contained in $[0,1]$ for all $\lambda\ge0$. So, the span of $(f_\lambda)_{\lambda\ge0}$ cannot be dense in $C_0([0,+\infty))$.