How to obtain the multiplication property in terms of f in hertz (from the formula in terms of $ \omega $)
Denoting
$ \mathcal{Z}(\omega)=\mathcal{X}(\omega)*\mathcal{Y}(\omega) \ $
$ Z(f)=X(f)*Y(f) \ $
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ Z(f)=\mathcal{Z}(2\pi f)=\frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{X}(\theta)\mathcal{Y}(2\pi f-\theta)d\theta \ $
$ Let\ \varphi =\frac{\theta}{2\pi},\ then\ \theta=2\pi \varphi \ $
$ \begin{align} Z(f) &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{X}(2\pi \varphi)\mathcal{Y}(2\pi f-2\pi \varphi)d2\pi \varphi \\ &= \int_{-\infty}^{\infty} \mathcal{X}(2\pi \varphi)\mathcal{Y}(2\pi (f-\varphi))d\varphi \\ &= \int_{-\infty}^{\infty} X(\varphi)Y(f-\varphi)d\varphi \end{align} $
$ Since\ X(\alpha)=\mathcal{X}(2\pi \alpha),Y(\alpha)=\mathcal{Y}(2\pi \alpha) $
