Bessel potential

Remarks

The kernel $G_\alpha$ exists, is positive, and has integral $\widehat{G_\alpha}(0) = 1$. So $J_\alpha$ is a bounded operator on each $L^p$ with norm at most $1$, by Young's inequality. On Schwartz functions $J_\alpha = (I - \Delta)^{-\alpha/2}$, where $\Delta$ is the Laplacian, since $-\Delta$ has symbol $4 \pi^2 \abs{\xi}^2$. The semigroup law $J_\alpha J_\beta = J_{\alpha + \beta}$ holds for all $\alpha, \beta > 0$. Near $0$, $G_\alpha(x)$ is comparable to $\abs{x}^{\alpha - n}$ when $0 < \alpha < n$, and $G_\alpha(x)$ decays like $e^{-c\abs{x}}$ for some $c > 0$ as $\abs{x} \to \infty$.

Example: for $n = 1$ and $\alpha = 2$, the kernel is $G_2(x) = \frac12 e^{-\abs{x}}$. Check: $\int_{\R} \frac12 e^{-\abs{x}} e^{-2\pi i x \xi} \, dx = \frac12 \cdot \frac{2}{1 + 4\pi^2 \xi^2} = (1 + 4\pi^2 \xi^2)^{-1}$. So $u = J_2 f$ solves $u - u'' = f$ for nice $f$.

Non-example: the kernel $\abs{x}^{\alpha - n}$ of the Riesz potential is not the Bessel kernel, since it is not integrable on $\R^n$, while $G_\alpha$ must be integrable with integral $1$.