spline derivation Assume the following representation for cubic splines with $T$ interior knots is given. Let
$g(Y)=\sum_{j=0}^3 \alpha_j Y_j+\sum_{t=1}^T \gamma_t (Y-\zeta_t)_{+}^{3}$
where $(Y-\zeta_t)_{+}:= max\{0,Y-\zeta_t\}$
1) How can we show that the natural boundary conditions for natural cubic splines results in the following linear constraints on the
coefficients:
$\alpha_2 = 0, \sum_{t=1}^T \gamma_t=0$
$\alpha_3 = 0, \sum_{t=1}^T \zeta_t\gamma_t=0$
2) How can we use result of part (1) to derive the basis
$V_1(Y) = 1, V_2(Y) = Y, V_{t+2}(Y) = g_t(Y) − g_{t-1}(Y), t = \{1, ..., T − 2\}$
where
$g_t(Y)=\frac{(Y-\zeta_t)_{+}^{3}-(Y-\zeta_T)_{+}^{3}}{\zeta_T-\zeta_t}$
Thanks!
 A: I'm going to prove 1) using a slightly different notation. As for 2) it's my first time seeing people use such a basis, I need some time. Could you give the source where this is from?
A natural spline of degree $2m+1$ is a function $s\in C^{2m}(\mathbb{R})$
that reduces to a polynomial of degree $\leq2m+1$ in each inner interval
and to a polynomial of degree at most $m$ in $(-\infty,t_{1})$ and
$(t_{n},\infty)$.
Theorem Every natural spline can be represented using truncated power functions
as follows
$$
s(x)=\sum_{k=0}^{m}\alpha_{k}x^{k}+\sum_{i=1}^{n}\beta_{i}(x-x_{i})_{+}^{2m+1}
$$
with the coefficient conditions,
$$
\sum_{i=1}^{n}\beta_{i}x_{i}^{k}=0,\qquad k=0,1,\ldots m
$$
The coefficient conditions follow from the natural spline condition,
$$
0=s^{(m+1)}(x)\bigg|_{x\geq x_{n}}=\sum_{i=1}^{n}\beta_{i}(2m+1)(2m)\dots(m+1)(x-x_{i})^{m},
$$
the Binomial Theorem, $(x-x_{i})^{m}=\sum_{k=0}^{m}\binom{m}{k}x_{i}^{m-k}(-x_{i})^{k}$,
and demanding that coefficients for each $k$ of this zero-polynomial vanish.
For cubic spline, $m=1$, note that the front low-degree poly is linear
and the conditions are $\sum\beta_{i}=0$ and $\sum\beta_{i}x_{i}=0$.
A: Since part (1) has been solved, I just answer the part (2). The notation may be a little different. 
Based on the definition, 
$$f(x)=\beta_0+\beta_1x+\beta_2x^2+\beta_3x^3+\sum\limits_{i=1}^K\theta_i(x-\xi_i)^3_{+}$$
Using the conclusion in (1).
$$f(x)=\beta_0+\beta_1x+\sum\limits_{i=1}^K\theta_i(x-\xi_i)^3_{+}$$
Extract the last two elements from the summation
$$f(x)=\beta_0+\beta_1x+\sum\limits_{i=1}^{K-2}\theta_i(x-\xi_i)^3_{+}+\theta_{K-1}(x-\xi_{K-1})_{+}^3+\theta_{K}(x-\xi_{K})_{+}^3$$
Using the results in (1) to solve $\theta_K$ and $\theta_{K-1}$,
$$\theta_K+\theta_{K-1}=-\sum\limits_{i=1}^{K-2}\theta_i=a$$
$$\theta_K\xi_K+\theta_{K-1}\xi_{K-1}=-\sum\limits_{i=1}^{K-2}\theta_i\xi_i=b$$
then
$$\theta_K=\frac{b-a\xi_{K-1}}{\xi_K-\xi_{K-1}}$$
$$\theta_{K-1}=\frac{-b+a\xi_{K}}{\xi_K-\xi_{K-1}}$$
Apply this results to $f(x)$
$$f(x)=\beta_0+\beta_1x+\sum\limits_{i=1}^{K-2}\theta_i(x-\xi_i)^3_{+}+\frac{-b+a\xi_{K}}{\xi_K-\xi_{K-1}}(x-\xi_{K-1})_{+}^3+\frac{b-a\xi_{K-1}}{\xi_K-\xi_{K-1}}(x-\xi_{K})_{+}^3$$
Since $d_{K-1}(X)=\frac{(x-\xi_{K-1})_{+}^3-(x-\xi_{K})_{+}^3}{\xi_K-\xi_{K-1}}$
Using this condition,
$$f(x)=\beta_0+\beta_1x+\sum\limits_{i=1}^{K-2}\theta_i(x-\xi_i)^3_{+}-b\cdot d_{K-1}(X)+\frac{a\xi_{K}}{\xi_K-\xi_{K-1}}(x-\xi_{K-1})_{+}^3+\frac{-a\xi_{K-1}}{\xi_K-\xi_{K-1}}(x-\xi_{K})_{+}^3$$
then
$$f(x)=\beta_0+\beta_1x+\sum\limits_{i=1}^{K-2}\theta_i(x-\xi_i)^3_{+}-b\cdot d_{K-1}(x)+\frac{a\xi_{K}}{\xi_K-\xi_{K-1}}(x-\xi_{K-1})_{+}^3+\frac{-a\xi_{K-1}+a\xi_{K}-a\xi_{K}}{\xi_K-\xi_{K-1}}(x-\xi_{K})_{+}^3$$
$$f(x)=\beta_0+\beta_1x+\sum\limits_{i=1}^{K-2}\theta_i(x-\xi_i)^3_{+}-b\cdot d_{K-1}(x)+a\xi_Kd_{K-1}(x)+a(x-\xi_K)^3_{+}$$
Extract $\theta_i$
$$f(x)=\beta_0+\beta_1x+\sum\limits_{i=1}^{K-2}\theta_i[(x-\xi_i)^3_{+}-(x-\xi_K)^3_{+}+(\xi_i-\xi_K)d_{K-1}(x)]$$
Extract $(\xi_K-\xi_i)$
$$f(x)=\beta_0+\beta_1x+\sum\limits_{i=1}^{K-2}\theta_i(\xi_K-\xi_i)[\frac{(x-\xi_i)^3_{+}-(x-\xi_K)^3_{+}}{\xi_K-\xi_i}-d_{K-1}(x)]$$
Then get the result, 
$$f(x)=\beta_0+\beta_1x+\sum\limits_{i=1}^{K-2}\theta_i(\xi_K-\xi_i)[d_i(x)-d_{K-1}(x)]$$
Done. 
