Question: Is it valid to do something like this? My idea is to "take out" the function $g$ to form the Fourier transform of $f$. If it is incorrect. Is there a legal way to do something similar? Thank you.
If $g(x)=|g(x)|\leq M$:
\begin{align} |\widehat{fg}(x)|&=\left|\int_{\mathbb{R}}\mathrm{e}^{ix\xi}f(\xi)g(\xi)\,d\xi\right|\\ &=|g(x)| \left|\int_{\mathbb{R}}\mathrm{e}^{ix\xi}f(\xi)\,d\xi\right|\\ &\leq M|\widehat{f}(x)| \end{align}
Here's a counterexample: pick $f(x) = \mathbf1_{x\in [-\pi,\pi]} \sin x$ and $g(x) = \mathbf1_{x\in [0,\pi]}$. Then $g(x) = |g(x)|\le 1$ and \begin{align} |\widehat{fg}(0)| &= \int_0^\pi \sin x dx > 0, \\ |g(0)| \left|\int_{\mathbb R} e^0 f(x)dx\right| &= 1\cdot \int_{-\pi}^\pi \sin x dx = 0 \end{align} so $$ |\widehat{fg}(0)| \not\le |g(0)| \left|\int_{\mathbb R} e^0 f(x)dx\right|.$$
I don't think there is any similar legal step.