I'm trying to understand this theorem:
Theorem 11.5
For each $n \geq 1,$ the function $S_{(n)}$ defined by $$S_{(n)}(x)=\sum_{k=0}^{n+1}(-1)^{k}\left(\begin{array}{c} n+1 \\k\end{array}\right)(x-k h)_{+}^{n}$$
is a spline of degree $n$ with equally spaced knots $k h, k=0,1, \ldots, n+1$. It has a continuous derivative of order $n-1$ and is identically 0 outside the interval $(0,(n+1) h)$
In the proof they only explain why $S_{(n)}(x)$ is identically $0$ outside the interval $(0,(n+1)h)$. But it is not clear to me that this is a spline of degree $n$. And that it has a continuous derivative of order $n-1$. How can I prove this to myself ?
Let $f_k(x) = (x - kh)^n_+$. If $k$ is even, then $f_k(x) = (x - kh)^n$, so it's just a polynomial. If $k$ is odd, then $f_k(x)$ is a piecewise polynomial with a "break" or "knot" at $x=kh$. In fact $$ f_k(x) = 0 \quad \text{for } x \le kh \\ f_k(x) = (x - kh)^n \quad \text{for } x > kh \\ $$ Since $S_{(n)}(x)$ is just a sum of these $f_k(x)$ functions, it is again a piecewise polynomial, therefore it's a spline.
It's easy to check that $f_k(x)$ has $n$ continuous derivatives at $x=kh$ -- just compare derivatives from the left and right. And it's infinitely differentiable everywhere else, of course.
So, $S_{(n)}(x)$ has $n$ continuous derivatives, also. The only places where $S_{(n)}(x)$ might not be infinitely differentiable are at the points $x=kh$ for $k=0,\ldots,n+1$. At these places, $S_{(n)}(x)$ might inherit a discontinuity from one of the $f_k(x)$ functions.