What function has $x \hat{f} (x)$ as its Fourier transform?

Question

I have a function $f$ such that $$ f(y) = \int_{- \infty}^{\infty} \hat{f}(x)\,e^{2 \pi i x y} \,{\rm d}x. $$ I know the function $f$, but what I have is $$ G(y) = \int_{- \infty}^{\infty}x \hat{f}(x)\,e^{2 \pi i x y} \,{\rm d}x. $$ I am trying to understand this function $G$. Searching around on Wikipedia, it looks like $G$ is the convolution of $f$ and the derivative of Delta function, but since this is not a function I am struggling to understand. Any explanation would be appreciated! thank you

Answer 1

The inverse Fourier transform of the product of two functions is equal to the convolution of the inverse Fourier transforms of the two functions.

Assume the inverse Fourier transform of $\hat{f}(x)$ is defined as:

$$f(y)=\mathcal{F}_x^{-1}[\hat{f}(x)](y)=\int\limits_{-\infty}^\infty \hat{f}(x)\, e^{2 \pi i x y}\,dx\tag{1}$$

The inverse Fourier transform of $x$ is a distribution:

$$\mathcal{F}_x^{-1}[x](y)=\int\limits_{-\infty }^\infty x\, e^{2 \pi i x y}\,dx=-\frac{i\, \delta'(y)}{2 \pi}\tag{2}$$

So the inverse Fourier transform of $x\, \hat{f}(x)$ is:

$$G(y)=\mathcal{F}_x^{-1}[x\, \hat{f}(x)](y)=\int\limits_{-\infty }^\infty x\, \hat{f}(x)\, e^{2 \pi i x y}\,dx=$$ $$=-\frac{i\, \delta'(y)}{2 \pi} * f(y) =\int\limits_{-\infty }^\infty -\frac{i\, \delta'(y')}{2 \pi}\, f(y-y') \, dy'=-\frac{i\, f'(y)}{2 \pi }\tag{3}$$

Answer 2

So let me write the more basic way to prove the result. Indeed, it just follows noticing that $x\,e^{2i\pi x y} = \frac{1}{2i\pi}\partial_y(e^{2i\pi x y})$ and using the Fourier inversion theorem, which gives $$ \int_{\Bbb R} x\,e^{2i\pi x y}\,\widehat{f}(x)\,{\rm d}x = \frac{1}{2i\pi}\partial_y\int_{\Bbb R} e^{2i\pi x y}\,\widehat{f}(x)\,{\rm d}x = \frac{1}{2i\pi}\, f' $$