1
$\begingroup$

(This may sound too simple a question to be asked in MO. But it is a research problem but not an exercise, so I post it here. If it is no appropriate, I will move it to Stackexchange.)

Let $0 < a < b$ be real numbers and $p$ a polynomial. The Fourier transforms of $\exp(-ax^2/2)$ and $\exp(-bx^2/2)$ are (up to constant factors) $\exp(-\xi^2/(2a))$ and $\exp(-\xi^2/(2b))$ respectively. The Fourier transform of $p(x) \exp(-ax^2/2)$ is $q(\xi) \exp(-\xi^2/(2a))$ where $q$ is a polynomial of the same degree as $p$'s.

Suppose we know that a function $f$ has its Fourier transform $$ \hat{f}(\xi) = \left( q(\xi) e^{-\frac{\xi^2}{2a}} \right) e^{-\frac{\xi^2}{2b}}, $$ then $f$ has a simple expression as the convolution (up to a constant) $$ f(x) = \int^{\infty}_{-\infty} \left( p(y) e^{-\frac{ay^2}{2}} \right) e^{-\frac{b(x - y)^2}{2}} dy. $$

My question is: if a function $g$ has its Fourier transform $$ \hat{g}(\xi) =\left( q(\xi) e^{-\frac{\xi^2}{2a}} \right) e^{\frac{\xi^2}{2b}}, $$ is there a simple formula of $g$ in $p$, analogous to the convolution formula above? A naive extrapolation $$ \int^{\infty}_{-\infty} \left( p(y) e^{-\frac{ay^2}{2}} \right) e^{\frac{b(x - y)^2}{2}} dy. $$ does not converge at all.

$\endgroup$
6
  • $\begingroup$ I don't understand the notation. You have the variable $x$ for the function and $\xi$ for the Fourier-transform, but what does $\zeta$ mean? $\endgroup$
    – user1688
    Nov 7, 2014 at 8:25
  • $\begingroup$ I did not use the notation $\zeta$. Did I? $\endgroup$
    – Dong Wang
    Nov 7, 2014 at 8:29
  • $\begingroup$ Oh I see, there's a TeX=problem here: when you use $\xi$ in the exponent, the upper loop gets killed and it looks like a $\zeta$. $\endgroup$
    – user1688
    Nov 7, 2014 at 9:29
  • $\begingroup$ When your function is in the Schwartz space, you only need to apply the Fourier transform and multiply the variable by (-1). That's what the inversion formula tells you. $\endgroup$
    – user1688
    Nov 7, 2014 at 9:48
  • $\begingroup$ The well-definedness of $g$ is not a problem, but I need a handy formula. The last integral in my post is a formal solution expected by me, while the divergence is incurable. $\endgroup$
    – Dong Wang
    Nov 7, 2014 at 11:05

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.