On $L^2(\mathbb{R}^d)$, we have $T_m$ defined $\widehat{T_m f} = m \widehat{f}$ is a bounded operator on $L^2$ if and only if $m \in L^\infty$.
What can be said about the same problem for more general measures? For example, I am interested in weighted $L^2$ spaces where $$ \mathrm{d}\mu(x) = (1 + |x|^{2})^{s/2} \mathrm{d}x, $$ when is $T_m$ a bounded operator on $L^2(\mu)$?
Let's assume $s>0$. Then $\mathcal{F} (L^2(\mu))$ is the Sobolev space $H^{2,s}$ (functions with $s$ derivatives in unweighted $L_2$), and your weighted space norm is giving the Sobolev norm of $\widehat{f}$.
To be bounded, $T$ can't destroy the smoothness of $\widehat{f}$, so you would want $m$ to be in some space like the $L_\infty$ Sobolev space of the same smoothness.
If you want a more precise statement, you might check Multiplication in Sobolev spaces for details