Riesz potential operator

Remarks

If $f \in L^p(\R^n)$ with $1 \le p < n/\alpha$, then $I_\alpha f(x)$ converges absolutely for almost every $x$. For $1 < p < n/\alpha$ and $1/q = 1/p - \alpha/n$, the Hardy–Littlewood–Sobolev inequality gives $\norm{I_\alpha f}_q \le A \norm{f}_p$ with $A$ depending only on $n$, $\alpha$ and $p$, see Lp space.

With the Fourier transform $\hat f(\xi) = \int_{\R^n} f(x) e^{-2\pi i x \cdot \xi} \, dx$, the constant $c_\alpha$ is chosen so that $\widehat{I_\alpha f}(\xi) = (2\pi \abs{\xi})^{-\alpha} \hat f(\xi)$ for nice $f$. So $I_\alpha = (-\Delta)^{-\alpha/2}$, the inverse of the fractional Laplacian, and $I_\alpha I_\beta = I_{\alpha + \beta}$ when $\alpha + \beta < n$.

Example: $n = 3$ and $\alpha = 2$. Then $c_2 = \pi^{3/2} \cdot 4 \cdot \Gamma(1) / \Gamma(1/2) = 4\pi$, since $\Gamma(1/2) = \sqrt{\pi}$. So $I_2 f(x) = \frac{1}{4\pi} \int_{\R^3} \frac{f(y)}{\abs{x - y}} \, dy$, the Newtonian potential, which solves $-\Delta u = f$ for smooth $f$ with compact support.

Non-example: for $f = 1$ the integral $\int_{\R^n} \abs{x - y}^{\alpha - n} \, dy$ diverges at every $x$, because $\alpha - n > -n$, so $I_\alpha 1$ is not defined.