# Close fourier transforms implies close time domain functions?

What conditions do we need so that $A(f)\approx B(f) \Rightarrow \mathcal{F}^{-1}\left\{A\right\}(t)\approx\mathcal{F}^{-1}\left\{B\right\}(t)$

The Fourier transforms of my two things look alike. In what sense do I need them to be similar so that their inverse Fourier transforms are close in the time (i.e. non-frequency) domain sense (say, that they are less than $\varepsilon$ apart)?

It can't be that the Fourier transforms are 'close almost everywhere:' what if $a(t)=0$ and $b(t)=\cos(x)$. Then their Fourier transforms are alike almost everywhere, but the functions are alike almost nowhere!

Your best bet is to consider an $L^2$ metric, i.e., mean-square error. in this case, by Parseval's/Plancherel's theorem, we have
$$\int_{-\infty}^{\infty} dx \left |f_1(x)-f_2(x) \right |^2 = \frac1{2 \pi}\int_{-\infty}^{\infty} dk \left |F_1(k)-F_2(k) \right |^2$$
for FT pairs $f_j$ and $F_j$. In this way, you know what the "distance" between two functions and their respective transforms must be.