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With the convention that the residual term is a sum rather than an average, the penalised least squares criterion isMinimize over functions whose first derivative is absolutely continuous and whose second derivative is square-integrable, that is, the usual Sobolev space $H^2[x_1,x_n]$. The cubic smoothing spline ${\hat{g}}_{\lambda }$ is its minimizer, with natural boundary conditions and linear continuation outside the data range. It is a natural cubic spline with possible knots at the observed $x_i$, but usually does not interpolate the noisy $y_i$.

To see why the minimizer is a natural spline, for any $g$ take the natural cubic spline interpolant $s$ through its values $g(x_i)$ and write $h=g-s$. Then $h(x_i)=0$. Integrating $\int s^{\prime \prime }{h}^{\prime \prime }$ by parts on every knot interval twice gives zero: the terms involving $h$ vanish at knots, the terms involving $h^{\prime }$ cancel because $s^{\prime \prime }$ is continuous there, and the outer terms vanish by the natural boundary conditions. Hencewhile their residual sums of squares agree. Replacing $g$ by $s$ never increases the criterion. This proves the minimum roughness property of the natural cubic spline interpolant needed here.

For fitted values $u_i=s(x_i)$, write $\int (s^{\prime \prime }{)}^2=u^{T}Ku$, where $K$ is positive semidefinite and has the sampled affine functions as its two-dimensional nullspace. The finite-dimensional minimization becomes $\Vert y-u{\Vert }^2+\lambda u^{T}Ku$. Its strictly positive definite Hessian is $2(I+\lambda K)$, so it has the unique solutionThis is the roughness-matrix formula for a natural cubic smoothing spline. Larger $\lambda$ penalizes curvature more strongly. As $\lambda \downarrow 0$ the fit approaches interpolation; as $\lambda \to \infty$ it approaches the ordinary least-squares affine line.