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?
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.