What is the Fourier transform of $f(x)e^{2\pi ix^2}$ with $f\in L^2({\mathbb{R}})$?

What is the Fourier transform of $f(x)e^{2\pi ix^2}$ with $f\in L^2({\mathbb{R}})$?

Attempt.

Suppose $f\in L^2$ and let $g(x):=e^{2\pi ix^2}$. Clearly, $fg$ is in $L^2(\mathbb R)$, so its Fourier transform is well defined and in $L^2(\mathbb R)$, but can it be written in terms of the Fourier transform of $f$?

• This post is being discussed on the meta. – Stella Biderman Dec 8 '17 at 17:37
• @user243301 What is your point exactly? What don't you like about the question? – amsmath Dec 11 '17 at 19:53
• @user243301 You wrote "This is a ephemeral message, that I am going to delete in next minute. With my last message I have concluded all that I want to say, good luck." Seems a little childish to me, sorry. – amsmath Dec 11 '17 at 20:04

The Gaussian is its own Fourier transform $$\int_{-\infty}^\infty e^{-\pi x^2} e^{-2i \pi \xi x}dx = e^{-\pi \xi^2}$$

(complete the square assuming $2i \pi \xi \in \mathbb{R}$ and extend the obtained result to $2i \pi \xi \in \mathbb{C}$ by analytic continuation)

Thus for $a > 0$

$$H_a(\xi) = \int_{-\infty}^\infty e^{-\pi a x^2} e^{-2i \pi \xi x}dx =\frac{1}{a^{1/2}} e^{-\pi \xi^2/a}$$

Again we can use analytic continuation to extend the result to $a \in \mathbb{C}, \Re(a) > 0$.

Therefore $$\mathcal{F}[f e^{- \pi a x^2}](\xi)=\widehat{f} \ast \frac{e^{-\pi \xi^2/a}}{a^{1/2}}$$ And by continuity in the $L^2$ sense $$\mathcal{F}[f e^{- \pi (-2i) x^2}]=\lim_{b \to 0^+}\mathcal{F}[f e^{- \pi (b-2i) x^2}] =\lim_{b \to 0^+} \widehat{f} \ast \frac{e^{-\pi \xi^2/(b-2i)}}{(b-2i)^{1/2}}$$ If $\widehat{f} \in L^1$ then the limits goes out and this is $$\mathcal{F}[f e^{- \pi (-2i) x^2}](\xi)=\widehat{f} \ast \frac{e^{-\pi \xi^2/(-2i)}}{(-2i)^{1/2}}$$

• Dear reuns, thanks, very interesting. This confuses me a little. From the last term I can only see that it is in $L^\infty$, right? But actually, it should be in $L^2$. How can I see it from there? And what if $\hat f\notin L^1$? – amsmath Dec 9 '17 at 22:01
• @amsmath If $\hat{f} \in L^2$ but $\not \in L^1$ then $\lim_{b \to 0^+} \widehat{f} \ast \frac{e^{-\pi \xi^2/(b-2i)}}{(b-2i)^{1/2}}$ converges in $L^2$ but not pointwise. To convince yourself it has good reasons to be $L^2$ beside Parseval's theorem, you can try with ... $\hat{f}(x) = e^{-c (x-d)^2}$, of course (and use that the linear combinations of Gaussians are dense in $L^2$). – reuns Dec 9 '17 at 22:10
• So, if $f\notin L^1$, I can only write the Fourier transform as an $L^2$-limit and not explicitly. Thank you. – amsmath Dec 11 '17 at 1:24