# Invertibility of Fractional Laplacian in Lebesgue spaces

I'm trying to digest the following statements, which are taken from §2 of Carlo Morpurgo's paper Sharp inequalities for functional integrals and traces of conformally invariant operators.

For $$0< \alpha < n$$, up to a positive constant multiple, one can define the fractional Laplacian of order $$n$$ on the Schwartz class $$\mathcal S(\mathbb R^n)$$ as $$\begin{equation*} \mathcal F((-\Delta)^{\alpha/2}\varphi)(\xi) = |\xi|^\alpha \mathcal F\varphi(\xi), \qquad \text{ for }\varphi\in \mathcal S(\mathbb R^n), \end{equation*}$$ where $$\mathcal F$$ is the Fourier transform. This operator can be extended by duality to the space of tempered distributions $$\mathcal S'(\mathbb R^n)$$.

If $$p\geq 1$$ and $$f\in L^p(\mathbb R^n)$$ is given, the equation $$(-\Delta)^{\alpha/2} u = f$$ has a unique solution in $$L^q(\mathbb R^n)$$, where $$\frac 1 q = \frac 1 p - \frac \alpha n$$. Up to constant multiple, this solution is given explicitly by convolution with the Riesz potential: $$\begin{equation*} u(x) = I_\alpha (f):=\int_{\mathbb R^n}\frac{f(y)}{|x - y|^{n - \alpha}}\; \mathrm d y. \end{equation*}$$

My first question is in regards to the extension of $$(-\Delta)^{\alpha/2}$$ to $$\mathcal S'(\mathbb R^n)$$ via duality. Heuristically, for $$v\in \mathcal S'(\mathbb R^n)$$ one would like to define $$\begin{equation*} \langle (-\Delta)^{\alpha/2}v, \varphi\rangle := \langle v, (-\Delta)^{\alpha/2}\varphi\rangle = \langle \mathcal F v , |\xi|^\alpha \mathcal F\varphi\rangle \end{equation*}$$ whenever $$\varphi \in \mathcal S(\mathbb R^n)$$. The problem with this heuristic is that $$\mathcal F v$$, being a tempered distribution, acts on functions in $$\mathcal S(\mathbb R^n)$$. However, $$|\xi|^\alpha \mathcal F\varphi\notin \mathcal S(\mathbb R^n)$$ due to the lack of differentiability at $$\xi = 0$$. What is the correct interpretation of the action of $$\mathcal F v$$ on $$|\xi|^\alpha \mathcal F\varphi$$?

My second question is in regard to the statement that $$I_\alpha f$$ satisfies $$(-\Delta )^{\alpha/2} u = f$$ whenever $$f\in L^p$$ (of particular interest is $$p = 2n/(n + \alpha)$$ and $$q = 2n/(n - \alpha)$$, but that doesn't matter for the question at hand). The Hardy-Littlewood-Sobolev inequality guarantees that $$I_\alpha f\in L^q$$, so $$((-\Delta)^{\alpha/2}\circ I_\alpha )(f)$$ should be interpreted in the distributional sense. In particular, for $$\varphi\in \mathcal S(\mathbb R^n)$$ I would like to perform the following computation (up to constant multiple): $$\begin{eqnarray*} \langle (-\Delta)^{\alpha/2}\circ I_\alpha f, \varphi\rangle & = & \langle I_\alpha f, (-\Delta)^{\alpha/2}\varphi\rangle\\ & = & \langle \mathcal F(I_\alpha f), \mathcal F((-\Delta)^{\alpha/2} \varphi)\rangle\\ & = & \langle |\xi|^{-\alpha}\mathcal F f, |\xi|^\alpha \mathcal F\varphi\rangle\\ & = & \langle\mathcal F f, \mathcal F\varphi\rangle\\ & = & \langle f, \varphi\rangle. \end{eqnarray*}$$ The main problem is the interpretation of the expression $$\langle \mathcal F(I_\alpha f), \mathcal F((-\Delta)^{\alpha/2} \varphi)\rangle$$. Since $$I_\alpha f\in L^q$$, we can interpret $$\mathcal F(I_\alpha f)$$ as a tempered distribution. However, with this interpretation, $$\mathcal F (I_\alpha f)$$ can not act on $$\mathcal F((-\Delta)^{\alpha/2}f)$$ since the latter of these two quantities is not in the Schwartz class. What is the correct interpretation of the action of $$\mathcal F(I_\alpha f)$$ on $$\mathcal F((-\Delta)^{\alpha/2}\varphi)$$?

To the first question, I don't think there is a way to extend the fractional Laplacian to all tempered distributions. The difficulty is exactly as you stated, that multiplication by $$|\xi|^{\alpha}$$ doesn't preserve the Schwartz class $$\mathcal{S}$$. That said, if the distribution is such that its Fourier transform agrees with a measure or function in a neighborhood of the origin, we can make things work by keeping the multiplication by $$|\xi|^{\alpha}$$ on the distribution rather than the Schwartz function, at least in that neighborhood.
To the second question, there's a slightly different way of looking at things that avoids the difficulty in the calculation. To begin, even though the fractional Laplacian $$(-\Delta)^{\alpha/2}$$ doesn't preserve the Schwartz space, it does map this space into $$L^r$$ for $$1 \leq r \leq \infty$$. (By looking at the inverse Fourier transform of $$|\xi|^{\alpha} \mathcal{F} \varphi(\xi)$$ and using integration by parts, we can obtain generic decay at infinity on the order of $$|x|^{-n - \alpha}$$.) Then for $$f \in L^p$$, and letting $$u = I_{\alpha} f \in L^q$$ (with $$1/p = 1/q - \alpha/n$$), one (weak type) sense in which $$(-\Delta)^{\alpha/2} u = f$$ is that, for $$\varphi \in \mathcal{S}$$, $$\int u(x) \overline{(-\Delta)^{\alpha/2} \varphi(x)} \, dx = \int f(x) \overline{\varphi(x)} \, dx.$$ Heuristically the same thing is going on as in the not-quite-formally-correct distributional calculation, of course. Perhaps the key thing to keep in mind is the basic fact that distributions given by $$L^p$$ functions act via integration, and so the integration point of view can help with some of the formalities.