Fourier transform of fractional differential operator and Plancherel formula equivalent for fractional norms I would like to know if the the following exist or are defined


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*The Fourier transform $\mathcal{F}\left(\frac{d^{\frac{1}{2}}y}{dx^\frac{1}{2}}\right)$ of a fractional differential operator such as $\frac{d^{\frac{1}{2}}y}{dx^\frac{1}{2}}$. (I'm aware that the fractional Fourier transform exists, but this isn't quite the same thing.)

*An equivalent of the Plancherel theorem for fractional Lebesgue norms. I'm aware that the Plancherel theorem is defined for $\mathcal{L}_n$ where $n = 2$. What I'd like to know is if a similar theorem exists for positive non-integer values of $n \ne 2$.
Plancherel theorem where $n = 2$
$\int_{\mathbb{R}^m} \mid f({x}) \mid^n \; d{x} \; = \; \int_{\mathbb{R}^m} \mid \tilde{f}({\omega}) \mid^n d{\omega}$
Where $m$ is the number of dimensions. The answer to these questions would rule out or extend certain possibilities.
As always I'm sorry if these questions are total nonsense. I'm just a computer scientist teaching myself mathematics while writing my thesis.
 A: Regarding Plancherel formula, there is no hope to extend it in $L^p({\mathbb R})$, because 1- we don't know what is the image of $L^p$ under the Fourier transform if $n\ne 2$, 2- we know that this image is not $L^p$. It is a classical theorem that ${\mathcal F}:L^p\mapsto L^q$ is a continuous operator if and only if $1\le p\le2$ and $q=p'$ (the conjugate exponent, $p'=\frac{p}{p-1}$). That ${\mathcal F}:L^p\mapsto L^q$ is a continuous for $(p,q)=(1,\infty)$ or $(2,2)$ is classical. That it is true for pairs $(p,p')$ when $1\le p\le2$ follows from the Riesz--Thorin interpolation Theorem. That it is false for over pairs folows by an explicit calculation with Gaussian functions $f(x)=\exp(-z|x|^2)$ when $\Re z> 0$. Actually, the uniform boundedness principle (Banach--Steinhaus Theorem) even says that the image of $L^p$ under ${\mathcal F}$ is not contained in $L^q$, unless the cases of continuity described above.
A consequence is that, unless $p=q=2$, you cannot go forward and backward between an $L^p$ and an $L^q$ via the Fourier transform.
A: Regarding your first question, the fractional derivative operators are in fact DEFINED by how they act on the Fourier transform side. If the Fourier Transform of $f(x)$ is $\hat f(\xi)$ then the Fourier transform of its derivative $f'(x) = Df(x)$ is $2\pi i \xi \hat f(\xi)$ and hence for a positive integer $k$, the Fourier Transform of its $k$-th derivative $D^kf(x)$ is $(2\pi i \xi)^k \hat f(\xi)$. To define the $k$-th derivative operator when $k$ is a fraction or even an arbitrary real number, that same formula is used---i.e., Fourier Transform, multiply the resulting function of $\xi$ by $(2\pi i \xi)^k$, and then inverse Fourier Transform.
As for your second question, I do not know of anything that is in the nature of a generalization of the Plancherel Theorem from $L^2$ to other $L^p$ spaces, and certainly the obvious generalization is false.
A: I'd like to add a little bit to the answers by Dick Palais and Denis Serre.


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*The standard way to define a fractional derivative of order $\alpha>0$ on the real axis is via the Riemann-Liouville integral operator
$$D_{-\infty}^{\alpha}f(x)\equiv\frac{1}{\Gamma(n-\alpha)}\int^{x}_{-\infty}(x-y)^{\alpha+1-n}f^{(n)}(y)dy,$$
where $\alpha\in(n-1,n)$. 
This can be written as a convolution $D_{-\infty}^{\alpha}f = h*f^{(n)}$ with
$$h(t) = \begin{cases} \frac{t^{n-1-\alpha}}{\Gamma(n-\alpha)}, & \mbox{if } t > 0, \\\ \  \\\ 0, & \mbox{if } t\leq 0, \end{cases}$$
so a direct calculation yields 
$$\widehat{D_{-\infty}^{\alpha}f}(\xi)=(2\pi)^{1/2}\widehat{h}(\xi)\widehat{f^{(n)}}(\xi)=(i\xi)^{\alpha-n}\widehat{f^{(n)}}(\xi) =(i\xi)^{\alpha}\widehat{f}(\xi).$$

*W.H. Young obtained an earlier version of the Hausdorff-Young inequality  when he was trying to generalize Parseval's theorem for Fourier series to  other $L^p$-spaces. There is a sharper inequality due to W. Beckner ("Inequalities in Fourier analysis")
$$\|\widehat{f}\|_q \leq C(p,q)\|f\|_p,\qquad f\in L^p(\mathbb R^d),$$
where  $1< p <2$, $p^{-1}+q^{-1}=1$,  and
$$C(p,q)=\left[\left(\frac{p}{2\pi}\right)^{1/p}\left(\frac{q}{2\pi}\right)^{-1/q}\right]^{d/2}.$$
The inequality turns into equality on Gaussian functions $f(x)=\exp(-a|x|^2)$, $a>0$.
This is just a quantitative version of the continuity of the Fourier transform $\mathcal F: L^p\to L^q$. As Denis Serre explained in his answer, one basically cannot do better than this.
