# For which tempered distributions is the fractional derivative well-defined?

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.

• 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? – Goulifet May 5 at 23:30
• @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). – 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...

• 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? – Goulifet May 5 at 23:38
• 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. – paul garrett May 6 at 13:05