Let $\gamma \geq 0$ and consider the fractional derivative operator defined in Fourier domain by $$\mathcal{F} \{\mathrm{D}^{\gamma} \varphi \} (\omega) = (\mathrm{i} \omega)^{\gamma} \mathcal{F}\{\varphi\} (\omega),$$ where $\varphi \in \mathcal{S}(\mathbb{R})$ is a smooth and rapidly decaying function.

Of course, the definition can be extended to much more functions than $\varphi \in \mathcal{S}(\mathbb{R})$, including some, but not all, tempered distributions. It is for instance possible to extend $\mathrm{D}^{\gamma}$ to any compactly supported distribution (as for any convolution operator from $\mathcal{S}(\mathbb{R})$ to $\mathcal{S}'(\mathbb{R})$).

My question is the following: Is there a good notion of the "domain of definition" of the operator $\mathrm{D}^{\gamma}$, understood as the largest topological vector space $\mathcal{S}(\mathbb{R}) \subseteq \mathcal{X} \subseteq \mathcal{S}'(\mathbb{R})$ such that $\mathrm{D}^{\gamma} : \mathcal{X} \rightarrow \mathcal{S}'(\mathbb{R})$ is well-defined and continuous? Or at least, if the question is somehow meaningless, any natural construction that will include many tempered distributions in a satisfactory* manner?

*To give a bit of context, I am especially interested by the fractional case where $\gamma \notin \mathbb{N}$. The question is obvious for $\gamma = n \in \mathbb{N}$, since one can select $\mathcal{X} = \mathcal{S}'(\mathbb{R})$. However. when $\gamma$ is purely fractional, there is no hope to define the product $(\mathrm{i} \omega)^{\gamma} \mathcal{F}\{u\} (\omega)$ when $u \in \mathcal{S}'(\mathbb{R})$ is too irregular around the origin, which means morally that $u$ growth too fast at infinity. "In a satisfactory manner" would be a way of specifying properly a good "growth property" of $u \in \mathcal{X}$.


A preliminary remark. The operator $(d/dx)^\gamma$ is never continuous on the Schwartz space (and thus on tempered distributions) except if $\gamma$ is a non-negative integer, since you introduce a singularity at 0 by multiplication by $(i\omega)^\gamma=\exp(\gamma \log(i\omega))$ (you have to choose a determination of the logarithm on the imaginary axis and you get a singularity).

Now you can also define what is called an homogeneous distribution with degree $\lambda$, where $\lambda$ is a given complex number: a distribution $T$ on $\mathbb R$ is said to be homogeneous with degree $\lambda$ whenever $$ x\frac{d T}{dx}=\lambda T. $$ It is an exercise to prove that an homogeneous distribution is actually tempered. Examples are $$ \chi_{+,\lambda}=(x_+)^{\lambda}/\Gamma(\lambda +1),\quad \chi_{-,\lambda}=(x_-)^{\lambda}/\Gamma(\lambda +1), $$ and it is possible to prove that homogenous distributions of degree $\lambda\notin \mathbb Z_-$ are $$ c_{+}\chi_{+,\lambda}+c_{-}\chi_{-,\lambda} \quad\text{where $c_\pm$ are constants.} \tag{$\ast$} $$ Note that for $\lambda=-1$, homogeneous distributions of degree $-1$ on the real line are linear combinations of $ \text{pv}\frac{1}{x},\ \delta_0. $ With $\mathscr S'_\lambda$ standing for homogeneous distributions with degree $\lambda$, we get that for $\lambda, \lambda-\gamma\notin \mathbb Z_-$, $$ D^\gamma:\mathscr S'_\lambda\longrightarrow \mathscr S'_{\lambda-\gamma}. $$ To prove this you check that $ (d/dx)^\gamma\chi_{+,\lambda}=\chi_{+,\lambda-\gamma}. $

N.B. Multi-dimensional versions are available, more information in Lars Hörmander's ALPDO 256, Section 3.2.

  • $\begingroup$ Interesting, it suggest that one can extend to many tempered distributions with different growth. However, I am a bit curious about checking the relation for any $\gamma$ and $\lambda$. From the definition of the functions, there is for instance an issue with $\lambda = - 2n -1$ with $n \in \mathbb{N}$, for which the $\Gamma$ function is not defined. One can then remove the constant $\Gamma(\lambda + 1)$ but it should reappear when computing the fractional derivative from Fourier transform definition I guess. Do you have a restriction in mind, such as $\mathrm{Re} \lambda$ strictly positive? $\endgroup$ – Goulifet May 5 at 23:30
  • $\begingroup$ @Goulifet The function $1/\Gamma$ is entire, but you are right that to get $(\ast)$, you need a modification (I gave it in a new edit for $-1$ and a modification of the last statement). $\endgroup$ – Bazin May 6 at 15:23

Perhaps another way to approach a characterization of such tempered distributions is indeed to look at their Fourier transforms. First, the easy case is that distributions with support not containing $0$ admit fractional differentiation. Second, distributions $u$ with $\widehat{u}=\varphi\cdot v$ for a tempered distribution $v$ and $\varphi$ a smooth function vanishing to infinite order at $0$ (and perhaps identically $1$ outside an $\varepsilon$-ball around $0$). Then Fourier inversion gives something...

  • $\begingroup$ Agree when the support of $\widehat{u}$ does not contain 0. For your second example, to define $\mathrm{D}^\gamma u$ in this case, you use that $(\mathrm{i} \omega)^{\gamma} \varphi(\omega)$ is in $\mathcal{S}$, the smoothness coming from the conditions on $\varphi$, am I right? $\endgroup$ – Goulifet May 5 at 23:38
  • $\begingroup$ Well, I was specifically commenting about having such an action on tempered distributions, but, yes, the same idea would apply to Schwartz functions, sure: $u=\widehat{\varphi}*\widehat{v}$, etc. $\endgroup$ – paul garrett May 6 at 13:05

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