Partial answer. If $g$ is known to be a non-negative function, Karameta Tauberian theorem relates the behavior of $G = \int_0^\cdot g$ at infinity (resp. at 0) to the behavior of $\mathcal{L}g$ at $0$ (resp. at infinity). Reference: William Feller's book Probability.

In particular, $G(t) \sim 2ct^{1/2}$ as $t \to \infty$ if and only if $\mathcal{L}g(s) \sim \Gamma(1/2)c s^{-1/2}$ as $s \to 0$.

From $G(t) \sim 2ct^{1/2}$ as $t \to \infty$, one needs regularity assumptions on $g$ to derive $g(t) \sim ct^{-1/2}$ as $t \to \infty$, assuming that $g$ is non-increasing may be sufficient.

Partial answer. If $g$ is known to be a non-negative function, Karameta Tauberian theorem relates the behavior of $G = \int_0^\cdot g$ at infinity (resp. at 0) to the behavior of $\mathcal{L}g$ at $0$ (resp. at infinity). Reference: William Feller's book Probability.

In particular, $G(t) \sim 2ct^{1/2}$ as $t \to \infty$ if and only if $\mathcal{L}g(s) \sim \Gamma(1/2)c s^{-1/2}$ as $s \to 0$.

From $G(t) \sim 2ct^{1/2}$ as $t \to \infty$, one needs regularity assumptions on $g$ to derive $g(t) \sim ct^{-1/2}$ as $t \to \infty$, Assuming that $g$ is non-increasing is sufficient.

Indeed, if $g$ is non-increasing, then for all $r \in~]0,1[$ and $t>0$, $$\frac{G(t)-G(t-rt)}{rt} \ge g(t) \ge \frac{G(t+rt)-G(t)}{rt}.$$ $$\frac{2ct^{1/2}-2c(t-rt)^{1/2}+O(t^{1/2})}{rt^{1/2}} \ge t^{1/2}g(t) \ge \frac{2c(t+rt)^{1/2}-2ct^{1/2}+O(t^{1/2})}{rt^{1/2}}.$$ $$\frac{2c-2c(1-r)^{1/2}}{r} \ge \limsup_{t \to \infty} t^{1/2}g(t) \ge \liminf_{t \to \infty} t^{1/2}g(t) \ge \frac{2c(1+r)^{1/2}-2c}{r}.$$ Letting $r$ go to $0$ yields $$\lim_{t \to \infty} t^{1/2}g(t) = c.$$

Partial answer. If $g$ is known to be a non-negative function, Karameta Tauberian theorem relates the behavior of $G = \int0^\cdot$ at infinity (resp. at 0) to the behavior of $\mathcal{L}g$ at $0$ (resp. at infinity). Reference: William Feller's book {\it Probability}.

In particular, $G(t) \sim 2ct^{1/2} as t \to \infty$ if and only if $\lc{g}(t) \sim \Gamma(1/2)c s^{-1/2}$ as $s \to 0$.

From $G(t) \sim 2ct^{1/2} as t \to \infty$, one needs regularity assumptions on $g$ to derive $g(t) \sim ct^{-1/2} as t \to \infty$. Assuming th

Partial answer. If $g$ is known to be a non-negative function, Karameta Tauberian theorem relates the behavior of $G = \int_0^\cdot g$ at infinity (resp. at 0) to the behavior of $\mathcal{L}g$ at $0$ (resp. at infinity). Reference: William Feller's book Probability.

In particular, $G(t) \sim 2ct^{1/2}$ as $t \to \infty$ if and only if $\mathcal{L}g(s) \sim \Gamma(1/2)c s^{-1/2}$ as $s \to 0$.

From $G(t) \sim 2ct^{1/2}$ as $t \to \infty$, one needs regularity assumptions on $g$ to derive $g(t) \sim ct^{-1/2}$ as $t \to \infty$, assuming that $g$ is non-increasing may be sufficient.