 Solution

Consider a competing interpolant $f$ and set $h=f-g$. At each spline knot, $h(x_i)=0$. If its second derivative roughness penalty is infinite there is nothing to prove, so assume $f^{\prime \prime }\in L^2$. Its absolutely continuous first derivative makes $h^{\prime }$ absolutely continuous as well, with $h^{\prime \prime }$ defined almost everywhere.

On each knot interval the natural cubic spline has constant third derivative. Integrating by parts once givesThe last term vanishes because $h$ is zero at both endpoints. Summing over intervals cancels the interior boundary terms, since $g^{\prime \prime }$ and $h^{\prime }$ are continuous at the knots. The exterior terms vanish because $g^{\prime \prime }(x_1)=g^{\prime \prime }(x_n)=0$. Hence $\int g^{\prime \prime }{h}^{\prime \prime }=0$. Expanding the square now givesSince $\alpha >0$, equality requires $h^{\prime \prime }=0$ almost everywhere. Absolute continuity then makes $h$ affine. It has at least two distinct zeros because $n\ge 2$, so $h=0$. The natural cubic spline is the unique minimizer. This proves the minimum roughness property with absolutely continuous first derivatives, covering the weaker regularity in the question rather than assuming competitors are twice continuously differentiable. The case of two knots is included: the minimizing spline is their straight-line interpolant.