Properties of $L_{2}$ Fourier transform I have a question regarding the $L_{2}$ Fourier transform.
I know the fourier operator can be extended to functions in $L_{2}$, and I know Plancherel's formula works as well for functions in $L_{2}$.
My question is this,
Do all other formulas extend to $L_{2}$?
For example, is it correct that
$$F\left \{ x^{n}f(x) \right \}=i^{n}\frac{d^{n}}{dw^{n}}F\left \{ f(x) \right \}$$
if $x^{n}f(x)\in L_{2}$ ?
Thank you in advance
 A: Whether "all" formulas extend is a little too vague to have an answer.
Suppose that $f$ and $xf(x)$ are in $L^2$. Does it follow that $$\widehat{xf(x)}=iD\hat f?$$
Yes, if $D$ refers to the derivative "in $L^2$".
Say $\tau_hg(t)=g(t-h)$. We say that $g$ is differentiable "in $L^2$", with derivative $Dg$, if $$\lim_{h\to0}\left|\left|\frac{\tau_hg-g}{h}-Dg\right|\right|_2=0.$$ The formula above is correct with that interpretation of $D$ (you can prove this from Plancherel).
This is the same as saying that the Fourier transform of $xf(x)$ lies in a certain "Sobolev space"; you can read up on Sobolev spaces for more about all this.
A: Define $\partial f=-i\tilde{f}'$ on the domain $\mathcal{D}(\partial)$ consisting of $f\in L^2$ that is equal a.e. to an absolutely continuous $\tilde{f} \in L^2$ for which $\tilde{f}' \in L^2$.
Define $Mf = sf(s)$ on the domain $\mathcal{D}(M)$ consisting of all $f \in L^2$ for which $sf(s) \in L^2$.
Then the Fourier transform establishes a unitary equivalence between $\partial$ and $M$. That is, in the strictest operator sense,
$$
                      \partial  = \mathcal{F}^{-1}M\mathcal{F}.
$$
More precisely, $\mathcal{F}\mathcal{D}(\partial)=\mathcal{D}(M)$ (equivalently, $\mathcal{D}(\partial)=\mathcal{F}^{-1}\mathcal{D}(M)$,) and the above holds on $\mathcal{D}(\partial)$. The operator $\partial$ is selfadjoint in the strictest sense, which implies two facts:


*

*The operator $\partial$ is symmetric:
$$
                      (\partial f,g) = (f,\partial g),\;\;\;f,g\in\mathcal{D}(\partial).
$$

*The adjoint is equal to itself. That is, $g\in\mathcal{D}(\partial)$ iff there exists $h\in L^2$ such that
$$
            (\partial f,g) = (f,h),\;\;\;f\in\mathcal{D}(\partial).
$$
In that case, $g \in\mathcal{D}(\partial)$ and $\partial g = h$ a.e..


Therefore, $s^{n}\hat{f}(s) \in L^2$ iff $f \in \mathcal{D}(\partial^n)$ and, in that case,
$$
                   \mathcal{F}(\partial^{n}f) = s^{n}\mathcal{F}(s).
$$
In effect, the integration by parts evaluation terms at $\pm\infty$ vanish.
These are consequences of the Spectral Theorem applied to the selfadjoint operator $\partial$. Though a little surprising, $s^{n}\hat{f}(s)\in L^2$ iff $f$ has $n-1$ continuous derivatives in $L^2$ and $f^{(n-1)}$ is equal a.e. to an absolutely continuous function in $L^2$ for which $f^{(n)} \in L^2$. There is nothing nearly so nice in any other $L^p$.
