= Solution
A cubic <spline> that vanishes for $x\leq0$ 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 $x\geq kh$ imposes the four moment equations $\sum_j a_jj^r=0$ 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> is
$$
\boxed{B_3(x/h)=\frac16\sum_{j=0}^4(-1)^j\binom4j(x/h-j)_+^3.}
$$
It 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 $j=-n-1,\ldots,n+1$. Their restrictions form a <basis> for the cubic $C^2$ <spline> space on these knots. Write $S=\sum_jc_j\beta_j$. The $2n+1$ interpolation equations are
$$
\boxed{\frac{c_{r-1}+4c_r+c_{r+1}}6=u(rh),\qquad r=-n,\ldots,n.}
$$
The recurrence determines all remaining coefficients after choosing $c_{-n-1},c_{-n}$, giving exactly two free parameters. The two <natural cubic spline> conditions become
$$
\boxed{c_{-n-1}-2c_{-n}+c_{-n+1}=0,\qquad c_{n-1}-2c_n+c_{n+1}=0,}
$$
because $S''(rh)=h^{-2}(c_{r-1}-2c_r+c_{r+1})$.
For uniqueness, take the difference $s$ of two candidates. It vanishes at every knot and has $s''(\pm1)=0$. Integrating by parts twice on each cubic piece gives $\int_{-1}^1(s'')^2=0$: terms involving $s$ vanish at the knots, and those involving $s''s'$ cancel at internal knots and vanish at the endpoints. Thus $s''=0$; 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>.
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