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.

  • $\begingroup$ 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

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.


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