1
$\begingroup$

Does the equation $$\partial^{\alpha} (Ff)(y) = (-i)^{|\alpha|} F(x^{\alpha} f)$$

still hold for $|\alpha| \le m$ and $f \in H^m(\mathbb{R}^n) = W^{m,2}(\mathbb{R}^n)$$?

$F$ denotes the Fourier trafo. It definitely holds for Schwartz functions, but does it also hold in this more general case?- In case that it does, I would like to see a proof of course :-). The problem is that for Schwartz functions we know that the $x^{\alpha} f$ is still a Schwartz function, so i.e. $x^{\alpha}f \in L^1$ and the Fourier transform exists and we can differentiate under the integral sign. Despite, I don't see why this is true for $H^m$ functions.

Please ask me if anything is unclear.

$\endgroup$
  • $\begingroup$ It should hold. $\endgroup$ – science Mar 19 '15 at 5:40
  • $\begingroup$ @science can you give a proof or a reference? $\endgroup$ – RealAnalysis Mar 19 '15 at 15:57
2
$\begingroup$

This should be a straightforward computation if you use tempered distributions. The Fourier transform of a tempered distribution $T$ is defined by $$(FT)(\phi) = T(F\phi)$$ and its partial derivative $\partial^\alpha T$ is defined by $$ (\partial^\alpha T)(\phi) = (-1)^{|\alpha|} T(\partial^\alpha \phi)$$for each Schwarz function $\phi$. When $g$ is a locally integrable function, it is regarded as a tempered distribution via the formula $$g (\phi) = \int g \phi \, dx.$$

Begin with $T = x^\alpha f$. Then $$ (FT)(\phi) = T(F\phi) = \int x^\alpha f F(\phi) \, dx = \int f x^\alpha F(\phi) \, dx = f(x^\alpha F(\phi)).$$

If $\phi$ is a Schwarz function then you have the differentiation formula $$ F(\partial^\alpha \phi) = i^{|\alpha|} x^\alpha F(\phi)$$ so that $$i^{|\alpha|} f(x^\alpha F(\phi)) = f(F(\partial^\alpha \phi)) = (Ff)(\partial^\alpha \phi) = (-1)^{|\alpha|}\partial^{\alpha}(Ff)(\phi).$$

Put these together to obtain $i^{|\alpha|}FT = (-1)^{|\alpha|}\partial^{\alpha}(Ff)$, as required.

$\endgroup$

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.