Closed forms for the half-order piecewise-linear product weights

Fix $r>0$ and put $q:=\sqrt{r}$, which is positive by [F3] because $q^2=r>0$. For $s>0$, rationalizing gives $\mathrm{\mid }\sqrt{s}-q\mathrm{\mid }=\mathrm{\mid }s-r\mathrm{\mid }\mathrm{/}(\sqrt{s}+q)\le \mathrm{\mid }s-r\mathrm{\mid }\mathrm{/}q$, so $\sqrt{s}\to q$ as $s\to r$. For $s\mathrm{\ne }r$ the difference quotient is $(\sqrt{s}-q)\mathrm{/}(s-r)=1\mathrm{/}(\sqrt{s}+q)$, and $\mid 1\mathrm{/}(\sqrt{s}+q)-1\mathrm{/}(2q)\mid =\mathrm{\mid }q-\sqrt{s}\mathrm{\mid }\mathrm{/}(2q(\sqrt{s}+q))\le \mathrm{\mid }s-r\mathrm{\mid }\mathrm{/}(2q^3)$, so $(\sqrt{r}{)}^{\mathrm{\prime }}=1\mathrm{/}(2\sqrt{r})$. By the product rule [F4], $P_{x,u,v}^{\mathrm{\prime }}(r)=C{r}^{-1\mathrm{/}2}+v\sqrt{r}=r^{-1\mathrm{/}2}(C+vr)$ and $Q_{x,u,v}^{\mathrm{\prime }}(r)=C{r}^{-1\mathrm{/}2}-v\sqrt{r}=r^{-1\mathrm{/}2}(C-vr)$. Since $u+vy=C+v(y-x)$, [F8] makes $P_{x,u,v}(y-x)$ an antiderivative of $(y-x{)}^{-1\mathrm{/}2}(u+vy)$ for $y>x$. Since $u+vy=C-v(x-y)$, [F8] makes $-Q_{x,u,v}(x-y)$ an antiderivative of $(x-y{)}^{-1\mathrm{/}2}(u+vy)$ for $y<x$.

By [F3], for every $ϵ>0$, $0\le \delta <{ϵ}^2$ implies $0\le \sqrt{\delta }<ϵ$; hence $\sqrt{\delta }\to 0$ and both $P_{x,u,v}(\delta )$ and $Q_{x,u,v}(\delta )$ tend to zero as $\delta \downarrow 0$. If $x>\alpha$, evaluate the left integral on $[\alpha ,\min (\beta ,x-\delta )]$ for sufficiently small $\delta >0$ using [F6], [F5], and step 1.1. As $\delta \downarrow 0$, its value tends to $Q_{x,u,v}((x-\alpha {)}_{+})-Q_{x,u,v}((x-\beta {)}_{+})$. If $x\le \alpha$, the left contribution and this expression are zero. Likewise, if $x<\beta$, evaluate the right integral on $[\max (\alpha ,x+\delta ),\beta ]$ for sufficiently small $\delta >0$; its limit is $P_{x,u,v}((\beta -x{)}_{+})-P_{x,u,v}((\alpha -x{)}_{+})$. If $x\ge \beta$, the right contribution and this expression are zero. Thus both one-sided integrals exist and have the displayed values. Splitting truncated intervals by [F7] and passing to their finite limits gives exactly ${\mathcal{J}}_x[\alpha ,\beta ;u,v]$.

On $[0,h]$, ${\varphi }_{n,0}(y)=1-y\mathrm{/}h$; for an interior index $j$, its left and right affine pieces are $(y-y_{j-1})\mathrm{/}h$ and $(y_{j+1}-y)\mathrm{/}h$; on $[y_{n-1},1]$, ${\varphi }_{n,n}(y)=(y-y_{n-1})\mathrm{/}h$. Substituting these coefficients into the integral formula of step 2.1 gives the three displayed formulas for $w_{n,j}$. The endpoint cases use positive parts in $\mathcal{J}$, so a singularity at a split point contributes a zero primitive value, and when $n=1$ the interior-index range is empty while both endpoint formulas remain valid.