 Solution

For $h\ne 0$, enlarge the nonnegative integral $I(h)$ to all of $\mathbb{R}$, and substitute $y=x+|h|t$. Reflection if necessary givesNear $0$ and $1$, its integrand is bounded by a constant times $1+|t{|}^{-2\alpha }$ or $1+|t-1{|}^{-2\alpha }$, which is integrable. For large $|t|$, the mean value theorem bounds the difference by a constant times $|t{|}^{-\alpha -1}$, whose square is integrable. Thus $C_{\alpha }<\infty$. The preceding estimate yieldsIn particular $g$ is continuous on all of $\mathbb{R}$; the proof establishes Hölder continuity of a one-dimensional fractional kernel integral of exponent $1/2-\alpha$ uniformly in $x$.