A single new knot refines the B-spline basis

Since the knot vector is clamped, a nonempty span index satisfies $p\le k\le n-1$. The insertion map is $t_j^{\mathrm{\prime }}=t_j$ for $j\le k$, $t_{k+1}^{\mathrm{\prime }}=u$, and $t_j^{\mathrm{\prime }}=t_{j-1}$ for $j\ge k+2$. For $q=0$, ${\lambda }_i^{(0)}=1$ when $i\le k$ and $0$ when $i\ge k+1$. Every old half-open degree-zero cell other than $[t_k,t_{k+1})$ is unchanged or shifts by one index; the affected cell splits as $[t_k,u)\cup [u,t_{k+1})$. Thus the indicator identity holds for all degree-zero indices, including zero-length repeated-knot cells; at $u$ the right cell supplies the value. The value at $b$ follows by the stipulated left limit.

Fix $1\le q\le p$ and assume the identity holds in degree $q-1$. For $x\in [a,b)$, write $A_i=(x-t_i)\mathrm{/}(t_{i+q}-t_i)$ and $D_i=(t_{i+q+1}-x)\mathrm{/}(t_{i+q+1}-t_{i+1})$, with any zero-denominator term defined as zero as in [F1]; define $A_i^{\mathrm{\prime }},D_i^{\mathrm{\prime }}$ using $t^{\mathrm{\prime }}$. Expanding the old recurrence and applying the degree-$(q-1)$ identity gives the coefficient triple $(A_i{\lambda }_i^{(q-1)},A_i(1-{\lambda }_{i+1}^{(q-1)})+D_i{\lambda }_{i+1}^{(q-1)},D_i(1-{\lambda }_{i+2}^{(q-1)}))$ on $B_{i,q-1}^{\mathrm{\prime }},B_{i+1,q-1}^{\mathrm{\prime }},B_{i+2,q-1}^{\mathrm{\prime }}$. Expanding the proposed refined expression gives $({\lambda }_i^{(q)}{A}_i^{\mathrm{\prime }},{\lambda }_i^{(q)}{D}_i^{\mathrm{\prime }}+(1-{\lambda }_{i+1}^{(q)})A_{i+1}^{\mathrm{\prime }},(1-{\lambda }_{i+1}^{(q)})D_{i+1}^{\mathrm{\prime }})$. It remains to prove these three coefficient equalities for $k-q\le i\le k$.

For the first coefficient, at $i=k-q$ both ${\lambda }_i^{(q-1)}$ and ${\lambda }_i^{(q)}$ are $1$, and the insertion map gives $t_i^{\mathrm{\prime }}=t_i$ and $t_{i+q}^{\mathrm{\prime }}=t_{i+q}=t_k$. Hence $A_i^{\mathrm{\prime }}=A_i$; if this common denominator is zero, both terms are zero by [F1]. For $k-q+1\le i\le k$, put $a=t_i$, $b=t_{i+q}$ and $v=t_{i+q}^{\mathrm{\prime }}$. Then $a\le t_k<u<b$. At the lower index $i=k-q+1$, $v=u$ and ${\lambda }_i^{(q-1)}=1=(u-a)\mathrm{/}(v-a)$. At higher indices in this range, the definition gives the same ratio because $v=t_{i+q-1}$. All denominators here are positive, and $A_i{\lambda }_i^{(q-1)}=((x-a)\mathrm{/}(b-a))((u-a)\mathrm{/}(v-a))=((u-a)\mathrm{/}(b-a))((x-a)\mathrm{/}(v-a))={\lambda }_i^{(q)}{A}_i^{\mathrm{\prime }}$. This also covers $q=1$, where the lower-degree weight is the degree-zero value $1$.

For the last coefficient, first take $k-q\le i\le k-1$ and put $c=t_{i+1}$, $d=t_{i+q+1}$, and $w=t_{i+2}^{\mathrm{\prime }}$. Then $c\le t_k<u<d$ and $w\le t_k$ for $i\le k-2$, while $w=u$ for $i=k-1$. In both cases $d-c>0$, $d-w>0$, and ${\lambda }_{i+2}^{(q-1)}=(u-w)\mathrm{/}(d-w)$ (zero in the case $w=u$). Consequently $1-{\lambda }_{i+2}^{(q-1)}=(d-u)\mathrm{/}(d-w)$, $1-{\lambda }_{i+1}^{(q)}=(d-u)\mathrm{/}(d-c)$, and $D_{i+1}^{\mathrm{\prime }}=(d-x)\mathrm{/}(d-w)$, while $D_i=(d-x)\mathrm{/}(d-c)$. Multiplying proves $D_i(1-{\lambda }_{i+2}^{(q-1)})=(1-{\lambda }_{i+1}^{(q)})D_{i+1}^{\mathrm{\prime }}$. At $i=k$, both ${\lambda }_{k+2}^{(q-1)}$ and ${\lambda }_{k+1}^{(q)}$ are zero, and the insertion map gives $D_{k+1}^{\mathrm{\prime }}=D_k$. If their common denominator is zero, both terms are zero by [F1]; otherwise equality follows from the identical numerator and denominator.

For the middle coefficient at the lower boundary $i=k-q$, put $c=t_{i+1}$ and $d=t_{k+1}$. Here ${\lambda }_i^{(q)}=1$ and ${\lambda }_{i+1}^{(q-1)}=1$, so the old middle coefficient is $D_i=(d-x)\mathrm{/}(d-c)$. Also $1-{\lambda }_{i+1}^{(q)}=(d-u)\mathrm{/}(d-c)$, $D_i^{\mathrm{\prime }}=(u-x)\mathrm{/}(u-c)$, and $A_{i+1}^{\mathrm{\prime }}=(x-c)\mathrm{/}(u-c)$. All displayed denominators are positive, and the new middle coefficient is ${\lambda }_i^{(q)}{D}_i^{\mathrm{\prime }}+(1-{\lambda }_{i+1}^{(q)})A_{i+1}^{\mathrm{\prime }}=(u-x)\mathrm{/}(u-c)+(d-u)(x-c)\mathrm{/}((d-c)(u-c))=(d-x)\mathrm{/}(d-c)=D_i$. The omitted old term $A_i(1-{\lambda }_{i+1}^{(q-1)})$ is zero even if $A_i$ has zero denominator, because [F1] defines that term as zero and its multiplier is zero. At the upper boundary $i=k$, ${\lambda }_{i+1}^{(q-1)}={\lambda }_{i+1}^{(q)}=0$, so the old middle coefficient is $A_i$. Put $a=t_k$ and $b=t_{k+q}$; then $a<u<b$. The insertion map gives ${\lambda }_i^{(q)}=(u-a)\mathrm{/}(b-a)$, $D_i^{\mathrm{\prime }}=(b-x)\mathrm{/}(b-u)$, and $A_{i+1}^{\mathrm{\prime }}=(x-u)\mathrm{/}(b-u)$. Therefore the new middle coefficient is ${\lambda }_i^{(q)}{D}_i^{\mathrm{\prime }}+(1-{\lambda }_{i+1}^{(q)})A_{i+1}^{\mathrm{\prime }}=((u-a)(b-x)+(b-a)(x-u))\mathrm{/}((b-a)(b-u))=(x-a)\mathrm{/}(b-a)=A_i$. The old term $D_i{\lambda }_{i+1}^{(q-1)}$ is zero without division by its denominator. For the strict interior $k-q+1\le i\le k-1$, which is empty if $q=1$, put $a=t_i$, $c=t_{i+1}$, $b=t_{i+q}$, and $d=t_{i+q+1}$. The insertion span gives $a\le c\le t_k<u<t_{k+1}\le b\le d$, so every denominator below is positive. The old and refined middle coefficients are respectively $((x-a)(b-u)\mathrm{/}(b-a)+(d-x)(u-c)\mathrm{/}(d-c))\mathrm{/}(b-c)$ and $((u-a)(b-x)\mathrm{/}(b-a)+(d-u)(x-c)\mathrm{/}(d-c))\mathrm{/}(b-c)$. Their first numerator terms differ by $x-u$, since $(x-a)(b-u)-(u-a)(b-x)=(b-a)(x-u)$. Their second numerator terms differ by $u-x$, since $(d-x)(u-c)-(d-u)(x-c)=(d-c)(u-x)$. Thus their sum is equal. These lower, strict-interior, and upper cases prove the middle coefficient identity.

For $i<k-q$, both weights in the refinement formula are $1$, and the complete local knot list used by $B_{i,q,t}$ is unchanged in $t^{\mathrm{\prime }}$, so the recursively defined basis function is unchanged. For $i\ge k+1$, both weights are $0$, and the local knot list shifts by one, giving $B_{i,q,t}=B_{i+1,q,t^{\mathrm{\prime }}}$. For the transition range, steps 3.1, 3.2 and 3.3 establish all three coefficients in step 2.1, so the Cox–de Boor expansions agree. Induction from the degree-zero base in step 1.1 proves the formula for every $q\le p$. The argument is pointwise on $[a,b)$ under the right-cell convention, including repeated knots; taking left limits gives the equality at $b$. No continuity at knots is assumed.