= Solution
Use $s_j=t_{i+j}$ again and take $k\ge2$. Set $g(s)=(s-t)_+^{k-2}$, so $(s-t)g(s)=(s-t)_+^{k-1}$. Let
$$
H=[s_0,\ldots,s_{k-1}]g,\qquad K=[s_1,\ldots,s_k]g,\qquad G=[s_0,\ldots,s_k]g=\frac{K-H}{s_k-s_0}.
$$
The <Leibniz rule for divided differences>, applied to the linear factor $s-t$, reads
$$
[s_0,\ldots,s_k]\bigl((s-t)g(s)\bigr)=(s_0-t)G+K.
$$
Multiply by $s_k-s_0$ and substitute $G$ to get $(t-s_0)H+(s_k-t)K$. Converting $H$ and $K$ to the lower-order normalized <B-splines> proves the <Cox-de Boor recurrence>:
$$
\boxed{N_{i,k}(t)=\frac{t-t_i}{t_{i+k-1}-t_i}N_{i,k-1}(t)+\frac{t_{i+k}-t}{t_{i+k}-t_{i+1}}N_{i+1,k-1}(t).}
$$
It starts with $N_{i,1}=1_{[t_i,t_{i+1})}$. The coefficients are nonnegative wherever the corresponding lower-order <B-spline> is nonzero, so induction proves nonnegativity.
For later use, extend the finite knot sequence to a strictly increasing sequence in both directions. The full order-one sum is one. When the <Cox-de Boor recurrence> is summed, the two coefficients of each fixed lower-order <B-spline> add to one. Induction therefore gives a full partition of unity. Removing terms proves the <subpartition of unity for B-splines>, $\sum_{i=1}^nN_{i,k}(t)\le1$. For repeated knots, the same statement follows by a limit of strictly increasing knot sequences, with the usual convention that a recurrence term with zero denominator is zero; the bound holds on each open knot interval and for the standard one-sided endpoint values. The terminology “$L_\infty$-normalized” refers to this standard basis normalization; it does not claim that each individual <B-spline> has <supremum norm> exactly one.
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