 Solution

A cubic spline that vanishes for $x\le 0$ and is twice continuously differentiable can be written using truncated power functions as $s(x)={\sum }_{j=0}^k{a}_j(x-jh{)}_{+}^3$. Vanishing for $x\ge kh$ imposes the four moment equations ${\sum }_j{a}_j{j}^r=0$ for $r=0,1,2,3$. Their Vandermonde matrix has rank $\min (4,k+1)$, so a nonzero coefficient vector first exists at $\boxed{k=4}$.

A normalized compact cubic cardinal B-spline isIt has nodal values $0,1/6,2/3,1/6,0$ at $0,h,2h,3h,4h$, so their sum is one. The alternating binomial coefficients meet the four moment equations, and the factor $1/6$ fixes the requested normalization.

For interpolation on $[-1,1]$, centre the translates as ${\beta }_j(x)=B_3(x/h-j+2)$, with $j=-n-1,\dots ,n+1$. Their restrictions form a basis for the cubic $C^2$ spline space on these knots. Write $S={\sum }_j{c}_j{\beta }_j$. The $2n+1$ interpolation equations areThe recurrence determines all remaining coefficients after choosing $c_{-n-1},c_{-n}$, giving exactly two free parameters. The two natural cubic spline conditions becomebecause $S^{\prime \prime }(rh)=h^{-2}(c_{r-1}-2c_r+c_{r+1})$.

For uniqueness, take the difference $s$ of two candidates. It vanishes at every knot and has $s^{\prime \prime }(\pm 1)=0$. Integrating by parts twice on each cubic piece gives ${\int }_{-1}^1(s^{\prime \prime }{)}^2=0$: terms involving $s$ vanish at the knots, and those involving $s^{\prime \prime }{s}^{\prime }$ cancel at internal knots and vanish at the endpoints. Thus $s^{\prime \prime }=0$; the remaining affine function vanishes at both endpoints and is zero. The square linear system for the coefficients is therefore nonsingular, giving existence as well as uniqueness of the natural cubic spline interpolant.