I'm trying to show the fourier transform of $$t\cdot f(t)$$ is $$i \cdot \frac {d\widetilde{f}(\omega)}{d\omega} $$
I have tried to follow this example but I get lost when $$\frac{dG(f)}{df} = \frac{dF\{ g(t)\} }{df}$$
I know that $ g(t)\cdot t$ is the function being transformed and this first step is working to prove that the transform of $g(t)$ is $G(t)$ and it exists but what allows me to define $G(f) = F\{ g(t)\}$ ?
For a good function $f$, \begin{align*} \widehat{f}(\omega)=\int_{-\infty}^{\infty}f(t)e^{-it\omega}dt, \end{align*} so \begin{align*} \dfrac{d}{d\omega}\widehat{f}(\omega)=\int_{-\infty}^{\infty}\dfrac{d}{d\omega}(f(t)e^{-it\omega})dt=\int_{-\infty}^{\infty}f(t)(-it)e^{-it\omega}dt=-i\int_{-\infty}^{\infty}tf(t)e^{-it\omega}dt, \end{align*} so \begin{align*} i\dfrac{d}{d\omega}\widehat{f}(\omega)=\int_{-\infty}^{\infty}tf(t)e^{-it\omega}dt=\hat{(tf(t))}(\omega). \end{align*}