What is the image of Laplace transform given domain $\mathcal{D}$ In this post on MO https://mathoverflow.net/questions/44713/when-i-can-safely-assume-that-a-function-is-a-laplace-transform-of-other-functio
The top answers says we can assume that the domain of laplace transform $\mathcal{L}$ is 
$\mathcal{D} = \{\phi(t): \mathbb{R_{\geq 0}}\to \mathbb{R}| \exp(-ct)\phi(t) \in \mathbb{L}^2(0, \infty), c \in \mathbb{R}\}$
Given this assumption, what would be the image of $\mathcal{D}$ under $\mathcal{L}$?
 A: The Laplace transform of $f \in L^2[0,\infty)$ is a function $F(s)$ that is holomorphic in the open right half plane and satisfies
$$
             \sup_{u > 0} \int_{-\infty}^{\infty}|F(u+iv)|^2dv < \infty.
$$
This is the $H^2$ Hardy class on the right half plane. Conversely, every $F\in H^2$ is the Laplace transform of a unique $f \in L^2[0,\infty)$. This is the content of the
Paley-Wiener Theorem.
Similarly, if you start with a function $f(t)$ where $e^{-ct}f(t)\in L^2[0,\infty)$, then the Laplace transform of $e^{-ct}f(t)$ is a Holomorphic function $F(s)$ in the half plane where $\Re s > c$, and which satisfies
$$
            \sup_{u > c} \int_{-\infty}^{\infty}|F(u+iv)|^2dv < \infty
$$
Example: For a given $c > 0$, the function $F(s)=\frac{1}{s}$ satisfies
$$
             \sup_{u > c}\int_{-\infty}^{\infty}|F(u+iv)|^2dv < \infty.
$$
So $F(s)$ is the Laplace transfor of a function $f(t)$ such that $e^{-ct}f(t)\in L^2$ for every $c > 0$. But $f(t)$ cannot be in $L^2$, which can be checked directly because $\mathscr{L}\{1\}=\frac{1}{s}$.
Added Note: For $1/2 \le r < 1$, there is no $c > 0$ such that $t^{-r}e^{-ct} \in L^2$. However, the Laplace transform of $f(t)=t^{-r}$ extends to the right half-plane from the real axis
\begin{align}
 \int_{0}^{\infty}e^{-st}t^{-r}dt 
   & = s^{-1+r}\int_{0}^{\infty}e^{-st}(st)^{-r}d(ts) \\
   & = s^{-1+r}\int_{0}^{\infty}e^{-u}u^{-r}du \\
   & = \Gamma(1-r)s^{-1+r}.
\end{align}
So, I'm not sure why someone could assume the domain is as described.
