Is the Laplace transform a vector space isomorphism? And what space is it isomorphic to? The laplace transform is a linear transformation, $\mathcal{L}: \mathcal{M} \rightarrow?$, where $\mathcal{M}$ is the set of exponentially bounded functions on $\mathbb{R},$since $\mathcal{L}(af(x)+bg(x))=a \mathcal{L}(f(x))+b\mathcal{g(x)}$   for $a,b\in \mathbb{R}$ and $f,g \in \mathcal{M}$. 
It seems to be injective since $\operatorname{Ker}(\mathcal{L})=0$ unless I've missed something. Therefore by the rank-nullity theorem $\mathcal{L}$ must surjective and so it is an isomorphism. So my questions are 1) Is this proof outline correct? and 2) what set is the laplace transform mapping into?
 A: The Laplace transform does not have kernel zero and is not injective on the domain you have specified. In particular, if $f$ and $g$ differ at countably many points they will have the same Laplace transform. Using the same approach as is typically used to define $L^2$, we can define $\mathcal{M}'$ to be the quotient space produced by quotenting out by "having the same Laplace Transform." However I do not know a nice way to describe $\mathcal{M}'$. In either case, I am also unsure what the range is. You may be interested in this wikipedia article which provides the range when the domain is $L^2(0,\infty)$
A: *

*If $F(s)$ is analytic on $\Re(s) > c$ and  $L^2$ on vertical lines  and as $\Im(s) \to \pm \infty$, $F(s) \to 0$ uniformly on strips $\Re(s) \in [u,v]$
then let
$$f_{\sigma,T}(x)= \frac1{2\pi} \int_{-T}^T F(\sigma+it) e^{itx}dt, \qquad x\in \Bbb{R} $$ 


*

*(From Parseval theorem for the Fourier  transform) $\lim_{T \to \infty}f_{\sigma,T}$ converges in $L^2(\Bbb{R})$ 

*(From Cauchy integral formula) the limit is $e^{-\sigma x}f$ for some function $f$ independent of $\sigma>c$.


*If also $$\lim_{\sigma \to \infty}\int_{-\infty}^\infty |F(\sigma+it)|^2dt = 0$$
then $f$ vanishes on $(-\infty,0)$.

*Conversely if $f$ is supported on $[0,\infty)$ and $f e^{-c x} \in L^2$ for some $c$ then $f e^{-\sigma x}$ is $L^1$ for $ \sigma >c$, its Laplace transform $$F(s) = \int_0^\infty f(x)e^{-sx}dx$$ is well-defined for $\Re(s) > c$ and it satisfies the above properties.

*If $F$ is $L^1$ on some vertical line then $fe^{-\sigma x}$ is not only bounded but also uniformly continuous. I don't think there is any easy way to ensure from $F$ that $f e^{-\sigma x}$ is only bounded for some $\sigma$.
