7
$\begingroup$

The Schwartz space (of rapidly decreasing functions) is the set of all $C^{\infty}$ functions $f\colon\mathbb{R}\to\mathbb{C}$ such that $$ x^jf^{(k)}(x) \to 0 $$ for all integers $j,k\geq0$ as $x \to {\pm\infty}$.

Let $f$ be a rapidly decreasing function in Schwartz space and let $\mathcal{F}(f)$ its Fourier transform. If $$ \| f \|_2 = \| \mathcal{F}(f) \|_2 \quad, $$ can we deduce that $\mathcal{F}(f)$ is a Schwartz function?

$\endgroup$
11
$\begingroup$

You don't need their norms to be equal - this is a consequence of the normalization factor chosen for the Fourier transform, and is called the Plancherel Theorem.

To show that the Fourier transform of a Schwartz function is Schwartz, it suffices to show the following:

  • $k^\alpha\hat{f}(k)=i^{-\alpha}\mathcal{F}\left[f^{(\alpha)}\right]$
  • $\frac{d^\alpha}{dk^\alpha}\hat{f}(k)=(-ik)^\alpha\hat{f}(k)$

Combine these two facts with the Riemann-Lebesgue lemma and you've got it.

Note: I'm using the following convention for the Fourier transform:

$$ \hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-ikx}dx $$

$\endgroup$
  • $\begingroup$ Why do those those two facts and the Riemann-Lebesgue lemma imply the result? $\endgroup$ – Robert Cardona Nov 3 '14 at 3:55
  • 1
    $\begingroup$ Since $f$ is Schwartz, all its derivatives will be $L^1$, so Riemann-Lebesgue gives that $\mathcal{F}[f^{(\alpha)}]$ goes to zero for any $\alpha$. The two facts above relate this to decay of $\hat{f}$ and its derivatives, which is needed to prove that $\hat{f}$ is Schwartz. $\endgroup$ – icurays1 Nov 3 '14 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.