Riesz potential of order alpha

Assume the Axiom of Countable Choice for the Euclidean Lebesgue framework (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Fix an integer $n\ge 1$ and a real order $0<\alpha <n$, and work on ${\mathbb{R}}^n$ with Lebesgue measure ${\lambda }_n$ on the Lebesgue sigma-algebra (Lebesgue measurable sets, the family $\mathcal{L}({\mathbb{R}}^n)$, and the restricted set function ${\lambda }_n$).

The kernel. Put $K_{\alpha }(z):=\mathrm{\mid }z{\mathrm{\mid }}^{\alpha -n}(z\mathrm{\ne }0),K_{\alpha }(0):=0.$ Since $\alpha -n<0$, the kernel is strictly positive and continuous on ${\mathbb{R}}^n\setminus \{0\}$ and has a singularity at the origin. The assigned value $K_{\alpha }(0)=0$ is a normalization convention: it changes the integrand ${\mathbb{R}}^n\ni y\mapsto K_{\alpha }(x-y)f(y)$ only at the single point $y=x$, which is the diagonal point of the domain; every statement in this pair is unchanged if another finite value is assigned instead, and the value at the origin is never used as a bound on the kernel everywhere.

The potential. Let $f:{\mathbb{R}}^n\to \mathbb{C}$ be measurable, with real and imaginary parts measurable in the sense of Integrable real and complex functions, and their integrals. The Riesz potential of order $\alpha$ of $f$ is defined at a point $x\in {\mathbb{R}}^n$ precisely when the nonnegative integral is finite, ${\int }_{{\mathbb{R}}^n}{K}_{\alpha }(x-y)\mathrm{\mid }f(y)\mathrm{\mid }d{\lambda }_n(y)<\mathrm{\infty },$ and at every such point it is $I_{\alpha }f(x):={\int }_{{\mathbb{R}}^n}{K}_{\alpha }(x-y)f(y)d{\lambda }_n(y)\mathrm{.}$ The integral displayed in the definition is the Lebesgue integral of the complex function $y\mapsto K_{\alpha }(x-y)f(y)$, which is absolutely convergent exactly at the points where the first display holds; the value $I_{\alpha }f(x)$ is then a complex number.

Scope of the definition. The set of points at which $I_{\alpha }f$ is defined may be empty, all of ${\mathbb{R}}^n$, or anything in between, and this definition asserts nothing about which case occurs: in particular it makes no claim that $I_{\alpha }f$ exists at every point, at almost every point, or for every $f$ belonging to any Lebesgue space. The strict-range theorem of this pair proves almost-everywhere absolute existence for every $f\in L^p({\mathbb{R}}^n;\mathbb{C})$ when $1<p<n\mathrm{/}\alpha$.

Normalization. The unit normalization is fixed once and for all by $c_{n,\alpha }=1$ in $I_{\alpha }f(x)=c_{n,\alpha }\int K_{\alpha }(x-y)f(y)dy$, the convention of both cited sources. A different positive constant rescales every potential and every estimate of this pair by that constant; no constant carrying a Fourier multiplier identity is asserted, and the identification of $I_{\alpha }$ with a power of $-\mathrm{\Delta }$ is not used here.