 Solution

Put $s_j=t_{i+j}$ for $0\le j\le k$, with the knots in increasing order. The explicit formula for a divided difference at distinct nodes givesEach truncated power function is a piecewise polynomial function of degree $k-1$ and is $C^{k-2}$ for $k\ge 2$. Therefore their sum has the same global smoothness. The jump in the $(k-1)$st derivative at $s_j$ equalswhich is nonzero. Thus each $s_j$ really is a knot, and generally there is no additional smoothness.

If $t<s_0$, the values being differenced are the values of the degree-$k-1$ polynomial $(s-t{)}^{k-1}$, so their $k$th divided difference vanishes. If $t>s_k$, every sampled truncated power function is zero. To check that both support endpoints occur, for $s_0<t<s_1$ subtract the vanished full polynomial divided difference to obtainFor $s_{k-1}<t<s_k$, only the last term survives and is positive. The nonnegativity from the recurrence in (b) actually gives positivity throughout the interior: for each $s_0<t<s_k$, at least one lower-order B-spline covering $t$ has a positive recurrence coefficient, and induction starts with positive order-one indicators. Hence $\mathrm{supp}{M}_i=[t_i,t_{i+k}]$ and $M_i\in C^{k-2}$ for $k\ge 2$, with degree $k-1$ on its knot intervals. For $k=1$, the B-spline is the step function $1_{[s_0,s_1)}(t)/(s_1-s_0)$, so no continuity is asserted.

The unit-integral normalization of a B-spline also follows directly: integrating $k(s_j-t{)}_{+}^{k-1}$ over $s_0\le t\le s_k$ gives $(s_j-s_0{)}^k$. The $k$th divided difference of this monic degree-$k$ polynomial is one. Thus $\int M_i(t)dt=1$, explaining its normalization.