Translations of the half-order kernel are continuous in the improper L1 norm

If $x=x^{\mathrm{\prime }}$, then the integrand is zero away from the omitted singular point and $T(x,x^{\mathrm{\prime }})=0$. Suppose $x\mathrm{\ne }{x}^{\mathrm{\prime }}$. Split $[0,1]$ at $x$ and $x^{\mathrm{\prime }}$. On the interior of each resulting interval the integrand is continuous by [F4]. Near a singular endpoint $c\in \{x,x^{\mathrm{\prime }}\}$, take a cutoff $\rho <\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}3$. The other singular term is bounded there by $(2\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}3{)}^{-1\mathrm{/}2}$, while the term singular at $c$ has tail integral at most $2\sqrt{\rho }$ by [F2] after the lower cutoff tends to zero. Thus each omitted endpoint tail is at most $2\sqrt{\rho }+2\rho (2\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}3{)}^{-1\mathrm{/}2}$, which tends to zero. Taking cutoffs ${\rho }_m=\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}(4(m+1))$ for integers $m\ge 0$ gives a Cauchy sequence of truncated sums by additivity and comparison [F5]; [F3] gives a finite limit. The same tail bound compares any sufficiently small cutoff with this cofinal sequence, so the improper limit exists. The argument also applies when a singular point is 0 or 1, with only one adjacent interval.

For $t\in [-1,1]$, define $s_{\eta }(t):=\min \{{\eta }^{-1\mathrm{/}2},\mathrm{\mid }t{\mathrm{\mid }}^{-1\mathrm{/}2}\}$ when $t\mathrm{\ne }0$ and $s_{\eta }(0):={\eta }^{-1\mathrm{/}2}$. This is continuous. Direct integration using [F2] gives ${\int }_{-1}^1\mathrm{\mid }s(t)-s_{\eta }(t)\mathrm{\mid }dt=2{\int }_0^{\eta }(t^{-1\mathrm{/}2}-{\eta }^{-1\mathrm{/}2})dt=2\sqrt{\eta }\mathrm{.}$ The two translated tails on $y\in [0,1]$ are each bounded by this full tail.

For every $y$ outside the finitely many singular points, the triangle inequality gives $\mathrm{\mid }s(x-y)-s(x^{\mathrm{\prime }}-y)\mathrm{\mid }\le \mathrm{\mid }s(x-y)-s_{\eta }(x-y)\mathrm{\mid }+\mathrm{\mid }{s}_{\eta }(x-y)-s_{\eta }(x^{\mathrm{\prime }}-y)\mathrm{\mid }+\mathrm{\mid }{s}_{\eta }(x^{\mathrm{\prime }}-y)-s(x^{\mathrm{\prime }}-y)\mathrm{\mid }\mathrm{.}$ Integrating truncated intervals and taking the limits from step 1.1, the first and third terms each contribute at most $2\sqrt{\eta }$ by step 1.2. The middle term is at most ${\omega }_{\eta }(d):={\sup }_{\mathrm{\mid }t-t^{\mathrm{\prime }}\mathrm{\mid }\le d}\mathrm{\mid }{s}_{\eta }(t)-s_{\eta }(t^{\mathrm{\prime }})\mathrm{\mid }$, since the interval has length one. Thus $T(x,x^{\mathrm{\prime }})\le 4\sqrt{\eta }+{\omega }_{\eta }(d)\mathrm{.}$ By [F8], the domain $[-1,1]$ is compact, so [F6] gives ${\omega }_{\eta }(d)\to 0$ as $d\downarrow 0$ for fixed $\eta$; also ${\omega }_{\eta }(d)\le 2{\eta }^{-1\mathrm{/}2}$, so the preceding bound bounds every $T(x,x^{\mathrm{\prime }})$. The set in the definition of $\tau (d)$ is nonempty because it contains $T(x,x)=0$, so [F7] gives its finite supremum. Given $ϵ>0$, first choose $\eta$ with $4\sqrt{\eta }<ϵ\mathrm{/}2$, then choose $d_0>0$ with ${\omega }_{\eta }(d_0)<ϵ\mathrm{/}2$. For every $d\le d_0$, the same bound gives $\tau (d)\le ϵ$, proving $\tau (d)\to 0$.