How to obtain the convolution property in terms of f in hertz (from the formula in terms of $ \omega $)
Denoting
$ \mathcal{U}(\omega)=\mathcal{X}(\omega)\mathcal{Y}(\omega) \ $
$ U(f)=X(f)Y(f) \ $
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ \begin{align} U(f) &= \mathcal{U}(2\pi f) \\ &=\mathcal{X}(2\pi f)\mathcal{Y}(2\pi f) \\ &= X(f)Y(f) \end{align} $
$ Since\ X(\alpha)=\mathcal{X}(2\pi \alpha),Y(\alpha)=\mathcal{Y}(2\pi \alpha) $
