Initial value theorem states that for a bounded function $f(t) = O(e^{ct})$ and an existing initial value, one-sided Laplace transform $F(s) = \int\limits^\infty_{0^-}f(\tau)e^{-s\tau}\mathrm{d\tau}$ can give the initial value of this function: $$ \lim_{t\rightarrow 0}{f(t)} = \lim_{s\rightarrow\infty}sF(s) \tag 1 $$
Is there a similar theorem for the Fourier transform $\mathcal{F}(\omega) = \int\limits^\infty_{-\infty}f(\tau)e^{-j\omega\tau}\mathrm{d\tau}$, something along the lines: $$ \lim_{t\rightarrow 0}{f(t)} = \lim_{j\omega\rightarrow\infty}j\omega \mathcal{F}(j\omega) \tag 2 $$
I find myself lacking the knowledge of complex analysis. I have a problem with formulation of the above hypothesis. I am not sure whether it's legal to write $j\omega\rightarrow\infty$. Also, one of the potential issues I see that I am trying to formulate the same theorem for the two-sided transform. Could you help me prove or disprove the above statement for the Fourier transform? Any pointers are appreciated.
The stated is identity is false. If $f(t)=e^{-t^{2}/2}$ then the left side is $1$ and the right side is $0$ since $\mathcal F(t) =e^{-t^{2}/2}$.