I am wondering if there is a theorem that states something like the following?
If $$\big|\;\tilde f(\omega)-\tilde g(\omega)\,\big| < \varepsilon\qquad \forall\omega$$ then there exists a $\delta(\varepsilon)$ such that $$\big|\; f(x)- g(x)\,\big| < \delta(\varepsilon)\qquad \forall x $$ where $\tilde f(\omega)$ and $\tilde g(\omega)$ are the Fourier transforms of the $L^2(\mathbb R)$ functions $f(x)$ and $g(x)$.
It is just an intuition, since a the fourier transform is invertible and unique. The extrem situation $$\big|\;\tilde f(\omega)-\tilde g(\omega)\,\big| =0 \quad \Rightarrow \quad \big|\; f(x)- g(x)\,\big| =0$$ holds trivially.
A naive way to find such a theorem is to say that we have a function $\eta(\omega)$ with $|\eta(\omega)|<\varepsilon$ and $\tilde g(\omega)=\tilde f(\omega)+\eta(\omega)$. Then one can easily derive $$\big|\;\tilde f(\omega)-\tilde g(\omega)\,\big| <\varepsilon \quad \Rightarrow \quad \big|\; f(x)- g(x)\,\big| < \frac{1}{\sqrt{2\pi}}\int|\eta(\omega)|\;\text d \omega$$
This is false. For example, take a smooth compactly supported function $f(x)$ and set $F_\epsilon(x)=\epsilon f(\epsilon x)$. Then $|F_\epsilon(x)|\le \epsilon\max|f(x)|$, but $$ \widetilde F_\epsilon(p)=\frac1{2\pi}\int e^{-ipx}\epsilon f(\epsilon x)dx= \frac1{2\pi}\int e^{-ipy/\epsilon}f(y)dy=\widetilde f(p/\epsilon), $$ and the maximum of $|\widetilde F_\epsilon(p)|$ is independent of $\epsilon$.