Riesz potential

Remarks

So $I_\alpha f$ is the convolution of $f$ with the kernel $\abs{x}^{\alpha - n} / c_\alpha$. If $f \in L^p(\R^n)$ with $1 \le p < n/\alpha$, the integral converges absolutely for almost every $x$. With the Fourier transform $\hat f(\xi) = \int f(x) e^{-2\pi i x \cdot \xi} \, dx$, the constant $c_\alpha$ is chosen so that $\widehat{I_\alpha f}(\xi) = (2\pi \abs{\xi})^{-\alpha} \hat f(\xi)$ for Schwartz functions $f$, that is, $I_\alpha = (-\Delta)^{-\alpha/2}$. The semigroup law $I_\alpha I_\beta = I_{\alpha + \beta}$ holds for $\alpha + \beta < n$.

Example: for $n \ge 3$ and $\alpha = 2$, one has $c_2 = 4 \pi^{n/2} / \Gamma(n/2 - 1) = (n - 2) \sigma_{n-1}$, where $\sigma_{n-1} = 2 \pi^{n/2} / \Gamma(n/2)$ is the area of the unit sphere. Indeed $\Gamma(n/2) = (n/2 - 1) \Gamma(n/2 - 1)$, so $(n - 2) \cdot 2 \pi^{n/2} / \Gamma(n/2) = 4 \pi^{n/2} / \Gamma(n/2 - 1)$. So $I_2 f(x) = \frac{1}{(n-2) \sigma_{n-1}} \int f(y) \abs{x - y}^{2-n} \, dy$, and $-\Delta I_2 f = f$ for smooth $f$ with compact support.

Non-example: the constant function $f = 1$ has no Riesz potential. For every $x$ the integral $\int \abs{x - y}^{\alpha - n} \, dy$ diverges, since $\abs{y}^{\alpha - n}$ is not integrable at infinity when $\alpha > 0$.