Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 23J / c

By the Cauchy-Schwarz inequality, the integral is bounded by $\Vert f{\Vert }_2\Vert k_x{\Vert }_2$, where $k_x(y)=|x-y{|}^{-\alpha }$. For $0\le x\le 1$,For $x$ outside the interval, the integral is bounded by the value at the nearest endpoint; the endpoint value is finite as well. Since $2\alpha <1$, the kernel is square-integrable for every real $x$, proving absolute convergence for every $x$.

Subtract the two absolutely convergent integrals and apply the Cauchy-Schwarz inequality to $f$ and the difference of the kernels. This gives precisely

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 $x$.