I was looking for a straightforward proof for Bedrosian's theorem which says the Hilbert transform of the baseband signal times the passband signal is the original baseband signal times the Hilbert transform of the passband signal in which by assuming $x(t)$ is a passband signal and and $m(t)$ is a baseband signal, then we have : $$c(t)=\widehat{x(t)m(t)}=\widehat {x(t)}m(t)$$ Now for the sake of proof since we know Fourier transform of Hilbert equals to : $$\hat X(f)=-j\operatorname{sgn}(f)X(f)$$ Then we say : $$\hat c(f)=-j\operatorname{sgn}(f) [x(f) \star m(f)]$$ Now can we say( because $x(t)$ is a passband signal) $X(f)$ is only in positive f part and $\operatorname{sgn}(f)$ value is constant and bring it in? and then we have: $$[-j\operatorname{sgn}(f)x(f)] \star m(f)=\widehat {x(f)}\star m(f)$$ And so we can say
$$\hat c(f)=\widehat {x(f)}\star m(f)$$ $$\hat c(t)=\widehat{x(t)m(t)}=\widehat {x(t)}m(t)$$