Conditions to have an inverse laplace transform? If I had a function $\widehat f(s)$ how would I know if there exists a function $f(t)$ so that the laplace transform of $f$ is $\widehat f$?
From looking at the formula for finding the laplace transform
$$\widehat f(s) = \int_o^\infty e^{-st}f(t)dt$$
it seems that $\widehat f$ must be a non-increasing function since as $s$ increases the integrand approaches $0$ would this be correct?
However I have found that
$$\mathcal{L}^{-1}\{s\}=\delta'(t)$$
even though $s$ is increasing. Although I am not quite sure what the derivative of the dirac delta function means since the dirac delta function not exactly a function and is called a 'generalised function'?
Are there any conditions on $\widehat f$ would prevent it from having an inverse laplace transform?
 A: Maybe it helps to consider the complex Laplace transform. Thus
$$
\hat{f}(z)=\int_{0}^{\infty }dt\exp [izt]f(t),\;{Im}z>0
$$
This object will exist for $f(t)$ absolutely integrable or square
integrable. Now put $z=\omega +i\delta $. Then ($\theta (t)$ is the
Heaviside step function)
$$
\hat{f}(\omega +i\delta )=\int_{0}^{\infty }dt\exp [i(\omega +i\delta
)t]f(t)=\int_{-\infty }^{+\infty }dt\theta (t)\exp [-\delta t]f(t)\exp
[i\omega t]
$$
so it is the Fourier transform of $\theta (t)\exp [-\delta t]f(t)$ and%
\begin{eqnarray*}
\theta (t)\exp [-\delta t]f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty
}d\omega \exp [-i\omega t]\hat{f}(\omega +i\delta ) \\
f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty }d\omega \exp [-i(\omega
+i\delta )t]\hat{f}(\omega +i\delta ),\;t>0
\end{eqnarray*}
In the literature the integral is usually given in terms of a straight
integration path parallel to the real axis. You see that in your case $%
s=\delta $ but $\omega $ vanishes. What you need is a requirement about the
analytic continuation in a strip including the real axis. 
Your special case  requires an extension of the theory of Fourier
transforms to distributions. But that is a different story.
A: I don't think it has to be a non increasing function of s for consider f(t) ->delta(t=0) - delta(t=1) then as s ->0  f(s) -> 0 while as s -> inf  f(s) -> 1. And for example f=exp(t^2) i don't think there exists a Laplace transform. And in general for any function of t that asymptotically rises of order > exponentially as t -> inf Laplace transform does not exist. While otherwise  it generally does exist. I think f(t) has to be of 'exponential' order is the usual terminology.
I know that is not a full answer because it is kind of the 'inverse' of your question but may help.
