Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I would like to know if the the following exist or are defined

  1. 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.)

  2. 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.

share|improve this question

3 Answers 3

up vote 6 down vote accepted

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.

share|improve this answer
    
As I don't know enough to judge which answer is best, I'm awarding the points to the answer with the highest votes. –  Olumide Oct 25 '10 at 15:05

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.

share|improve this answer
    
for function $u \in L^p$,where$p>2$,it's fourier transform is some finite order distribution,which the orderof it has a lower bound of $n(\frac{1}{2}-\frac{1}{p})$(if i have remembered it corectly ) –  user23078 May 2 '12 at 9:21

I'd like to add a little bit to the answers by Dick Palais and Denis Serre.

  1. 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 $$\widehat{D_{-\infty}^{\alpha}f}(\xi)=(2\pi)^{1/2}\widehat{h}(\xi)\widehat{f^{(n)}}(\xi)=(i\xi)^{\alpha-n}\widehat{f^{(n)}}(\xi) =(i\xi)^{\alpha}\widehat{f}(\xi).$$

  2. 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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