Fourier transform - How can O prove it?

Question

If TF [$f(\vec{r}$)] = f($\vec{k}$), show TF [$(\vec{\nabla} f)]$ = $\vec{k}$i f($\vec{k}$)

Answer 1

Here is an approach to consider:

$$ FT[f(\vec{r})]=f(\vec{k})=\int_{-\infty}^{\infty}f(\vec{r})e^{-i\vec{k}\cdot \vec{r}}\mathrm{d\vec{r}} $$

Now:

$$ \nabla FT[ f(\vec{r})]=\nabla f(\vec{k})=0=\int_{-\infty}^{\infty}\nabla f(\vec{r})e^{-i\vec{k}\cdot \vec{r}}\mathrm{d\vec{r}} $$

$$ 0=\int_{-\infty}^{\infty}\left[f(\vec{r})\nabla e^{-i\vec{k}\cdot \vec{r}}+ e^{-i\vec{k}\cdot \vec{r}}\nabla f(\vec{r})\right]\mathrm{d\vec{r}} $$

$$ 0=\int_{-\infty}^{\infty}f(\vec{r})\nabla e^{-i\vec{k}\cdot \vec{r}}\mathrm{d\vec{r}}+ \int_{-\infty}^{\infty}e^{-i\vec{k}\cdot \vec{r}}\nabla f(\vec{r})\mathrm{d\vec{r}} $$

$$ 0=\int_{-\infty}^{\infty}f(\vec{r})\nabla e^{-i\vec{k}\cdot \vec{r}}\mathrm{d\vec{r}}+ FT[\nabla{f(\vec{r})}] $$

$$ FT[\nabla{f(\vec{r})}]=-\int_{-\infty}^{\infty}f(\vec{r})\nabla e^{-i\vec{k}\cdot \vec{r}}\mathrm{d\vec{r}} $$

$$ FT[\nabla{f(\vec{r})}]=-\int_{-\infty}^{\infty}f(\vec{r})\left(-ik_x\vec{x}-ik_y\vec{y}-ik_z\vec{z}\right)e^{-i\vec{k}\cdot \vec{r}}\mathrm{d\vec{r}} $$

$$ FT[\nabla{f(\vec{r})}]=-\int_{-\infty}^{\infty}f(\vec{r})\left(-i\vec{k}\right)e^{-i\vec{k}\cdot \vec{r}}\mathrm{d\vec{r}} $$

$$ FT[\nabla{f(\vec{r})}]=\left(i\vec{k}\right)\int_{-\infty}^{\infty}f(\vec{r})e^{-i\vec{k}\cdot \vec{r}}\mathrm{d\vec{r}} $$

$$ FT[\nabla{f(\vec{r})}]=\left(i\vec{k}\right)FT[f(\vec{r})]=\left(i\vec{k}\right)f(\vec{k}) $$

I hope this helps.