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The cubic spline is even, so the symmetry of the normalized B-splines gives $N_{j,4}(-x)=N_{6-j,4}(x)$. It therefore suffices to extract three coefficients. The two polynomial pieces join with matching values and first two derivatives at zero, so the function belongs to the stated cubic spline space. At zero its right-hand derivatives areUse the supplied De Boor–Fix spline coefficient functional with the unnormalized knot polynomial ${\psi }_j(x)={\prod }_{r=j+1}^{j+3}(t_r-x)$. In this sign convention the coefficient isFor $j=1$, ${\psi }_1=(-1-x{)}^3$. Taking the right-hand limit at $-1$ kills the first three terms and gives $c_1=f(-1)=1$. For the next two coefficients take the right-hand limit at zero. Their knot polynomials and derivatives there areConsequently $c_2=(-27+6)/6=-7/2$ and $c_3=(27+6)/6=11/2$. Although the third derivative jumps at zero, its multiplier ${\psi }_j(0)$ vanishes for these two functionals; taking either one-sided limit gives the same coefficients. Symmetry gives $c_4=c_2$ and $c_5=c_1$. ThusFor a normalized B-spline basis, define the synthesis linear map $Tc={\sum }_j{c}_j{N}_{j,4}$ from the maximum coefficient norm into the spline space with its supremum norm. Nonnegativity and partition of unity give $\Vert T\Vert =1$, including equality on the all-ones vector. The coefficient condition number of a normalized B-spline basis isOn $[0,1]$, $f^{\prime }(x)=(27/2)x(3x-2)$. The only interior extremum is at $2/3$, where $f=-1$; at zero and one, $f=1$. Reflection therefore gives $\Vert f{\Vert }_{\infty }=1$. Its largest absolute coefficient is $11/2$, proving

A cubic spline belongs to $C^2[a,b]$ and is a polynomial of degree at most three on each interval separated by the spline knots. A natural cubic spline is additionally an affine function on the exterior intervals $[a,x_1]$ and $[x_n,b]$. On $[x_1,x_n]$ this is equivalent to the endpoint conditions $s^{\prime \prime }(x_1)=s^{\prime \prime }(x_n)=0$; these conditions remain part of the definition when an outer knot coincides with $a$ or $b$.

Let $s$ be the natural cubic spline interpolant of the prescribed values, and let $v$ be any other interpolating $C^2$ function. Set $u=v-s$, so $u(x_i)=0$. On the interval $[x_i,x_{i+1}]$, integration by parts twice givesbecause the fourth derivative of each cubic piece is zero. The second boundary term is zero because $u$ vanishes at the spline knots. After summing the first terms, the interior contributions cancel by continuity of $s^{\prime \prime }$ and $u^{\prime }$, and the two remaining terms vanish by the natural endpoint conditions. On the exterior intervals $s^{\prime \prime }=0$. Thus ${\int }_a^b{s}^{\prime \prime }{u}^{\prime \prime }=0$, and the second derivative roughness penalty satisfiesIf equality holds, the continuous nonnegative function $(u^{\prime \prime }{)}^2$ has zero integral, so $u^{\prime \prime }=0$ everywhere. Hence $u$ is an affine function, and its zeros at two distinct spline knots force $u=0$. This proves the minimum roughness property of the natural cubic spline interpolant and its equality case: the unique minimum-roughness interpolant is $s$.

For the fixed-design nonparametric regression fit, use the penalized least squares criterionLet $s_z$ be the natural cubic spline interpolant with value vector $z=(z_1,\dots ,z_n{)}^T$. Replacing $v$ by $s_z$ for $z_i=v(x_i)$ leaves the residual sum unchanged and reduces the second derivative roughness penalty, strictly unless $v=s_z$. It therefore remains to minimize the finite-dimensional quadratic formThe supplied spline roughness penalty matrix $K$ is positive semidefinite. For every nonzero vector $w$,Consequently $I+\lambda K$ is a positive-definite matrix, and $Q_{\lambda }$ is strictly convex. Its normal equations for linear least squares have the unique solutionThe preceding strict reduction also proves uniqueness over the whole class $C^2[a,b]$, not just over the natural cubic splines. This is the cubic smoothing spline. If the residual criterion is divided by $n$ instead, the corresponding formula is $(I+n\lambda K{)}^{-1}Y$, reflecting a change in the numerical convention for the smoothing parameter.

The independence and specified error mean and variance are not needed for existence or uniqueness of this minimizer. They do imply, with smoothing matrix $A_{\lambda }=(I+\lambda K{)}^{-1}$ and $g_X=(g(x_i){)}_i$, that $\mathbb{E}\hat{z}=A_{\lambda }{g}_X$ and $\mathrm{Cov}(\hat{z})={\sigma }^2{A}_{\lambda }^2$. The null space of $K$ consists of sampled affine functions: these have zero roughness, and conversely zero roughness forces their natural cubic spline interpolant to be an affine function. Thus these functions are reproduced without bias of an estimator, while other components are smoothed.

Use the natural cubic spline convention appropriate to the interior design knots: the spline is linear on $[a,x_1]$ and $[x_n,b]$. We also need $n\ge 2$. These points will be important for uniqueness.

A cubic spline is a function in $C^2[a,b]$ whose restriction to each of $[a,x_1]$, $[x_1,x_2]$, through $[x_n,b]$ is a polynomial of degree at most three. A spline knot is a junction between consecutive pieces. For a natural cubic spline, the two exterior pieces are affine functions; equivalently, one uses the natural endpoint conditions $g^{\prime \prime }(x_1)=g^{\prime \prime }(x_n)=0$ on the spline over $[x_1,x_n]$ and then extends linearly with matching derivatives. Merely prescribing $g^{\prime \prime }(a)=g^{\prime \prime }(b)=0$ on otherwise arbitrary exterior cubic pieces is a different condition and does not provide the unique interpolant invoked in this question.

The natural cubic spline interpolant $N\mathbf{v}$ to values $\mathbf{v}=(v_1,\dots ,v_n{)}^T$ is the natural cubic spline satisfying $(N\mathbf{v})(x_i)=v_i$. Its existence and uniqueness for $n\ge 2$ are the interpolant fact allowed in the PDF. The interpolation map $N$ is a linear map: a linear combination of interpolants satisfies the same spline, boundary and value conditions, so uniqueness identifies it with the interpolant to the corresponding linear combination of value vectors.

Let $g=N\mathbf{v}$, let $\tilde{g}\in C^2[a,b]$ have the same values, and put $r=\tilde{g}-g$. Then $r(x_i)=0$. We prove the orthogonalityApply integration by parts twice on each polynomial interval. On an interior interval, $g^{\prime \prime \prime \prime }=0$, andThe $g^{\prime \prime \prime }r$ terms vanish because $r$ vanishes at the knots. The $g^{\prime \prime }{r}^{\prime }$ terms cancel at internal knots because $g^{\prime \prime }$ and $r^{\prime }$ are continuous, and vanish at $x_1,x_n$ because $g^{\prime \prime }$ is zero there. On the exterior intervals, $g^{\prime \prime }=0$ identically. Summing therefore gives the claimed orthogonality without assuming that $g^{\prime \prime \prime }$ is continuous across knots.

Expanding the square now gives the minimum roughness property of the natural cubic spline interpolantEquality forces $r^{\prime \prime }=0$ everywhere, since $r^{\prime \prime }$ is continuous. Thus $r$ is an affine function. It vanishes at two distinct design points, so $r=0$, proving equality if and only if $\tilde{g}=g$.

For the penalized fit, take the allowed spline roughness penalty matrix $\Gamma$ to be a symmetric matrix with ${\mathbf{z}}^T\Gamma \mathbf{z}\ge 0$; thus it is a positive semidefinite matrix. If its supplied representative were not symmetric, its symmetric part defines the same quadratic form and is all that is needed. For any $\tilde{g}$, let $\mathbf{v}=(\tilde{g}(x_i){)}_{i=1}^n$. Replacing it by $N\mathbf{v}$ preserves every fitted value and can only decrease the second derivative roughness penalty. Therefore it is enough to minimizeover ${\mathbb{R}}^n$. The matrix $A_{\lambda }=I_n+\lambda \Gamma$ is a positive-definite matrix, sinceCompleting the quadratic or differentiating gives the unique solutionThis is the cubic smoothing spline. More explicitly, the normal equations are $(I_n+\lambda \Gamma ){\hat{\mathbf{v}}}_{\lambda }=\mathbf{Y}$, andFor an arbitrary function with fitted vector $\mathbf{v}$, the earlier orthogonality also givesBoth terms are nonnegative. Equality forces $\mathbf{v}={\hat{\mathbf{v}}}_{\lambda }$ and, by the proved interpolation equality case, $\tilde{g}=N\mathbf{v}={\hat{g}}_{\lambda }$. This proves both existence in $C^2[a,b]$ and uniqueness over the entire stated function class, not just over splines.

The argument is deterministic for each observed $\mathbf{Y}$. The fixed-design nonparametric regression and homoscedasticity assumptions motivate the squared-error loss, but Gaussian errors are not required.

At least two distinct design points are required for uniqueness. Question 4 does not explicitly restate $n\ge 2$, and this hypothesis is necessary. With one design point, every function $\tilde{g}(x)=Y_1+c(x-x_1)$ has zero residual and zero second derivative roughness penalty, for any real $c$. Thus the printed unrestricted uniqueness conclusion would be false for $n=1$. The corrected theorem assumes at least two distinct knots and the stated linear-tail convention. At $\lambda =0$, uniqueness over all of $C^2[a,b]$ also fails; the stipulated $\lambda >0$ eliminates that degeneracy.

Fit the Cox proportional-hazards model with the other explanatory variables, initially omitting the variable $z$ whose form is being assessed, and plot the resulting martingale residuals against $z$. Add a smooth trend, for example using a cubic spline or a local smoother. A systematic pattern indicates that $z$ affects the hazard function; the shape helps suggest a linear term, a transformation, or a nonlinear spline term in the log hazard.

Refit with the proposed form and inspect martingale residuals against $z$ again. Under an adequate functional form, their conditional mean should have no systematic trend, allowing for model fitting and the other covariates. Compare plausible alternative forms using the partial likelihood and suitable model checks.

Use the smoothed pattern to assess the covariate's functional form, rather than treating these residuals as normal regression errors. Martingale residuals are asymmetric, bounded above by 1, and can be arbitrarily negative; the raw residual plot can therefore be misleading without smoothing. This diagnostic concerns the covariate form, and does not by itself test the proportional hazards assumption.

A cubic spline with knots $x_1<\cdots <x_n$ is twice continuously differentiable and restricts to a polynomial of degree at most three between consecutive knots. It is a natural cubic spline when it is linear outside $[x_1,x_n]$, equivalently when its second derivative vanishes at the two outer knots and on the exterior intervals.

Let $s$ be the natural cubic spline interpolating prescribed values and let $\tilde{g}=s+h$ be any other interpolant in $S_2[a,b]$. Then $h(x_i)=0$. Piecewise integration by parts, using that $s^{\prime \prime \prime \prime }=0$ between knots, $s^{\prime \prime }$ is continuous, $s^{\prime \prime }=0$ on the exterior intervals, and every jump of $s^{\prime \prime \prime }$ is multiplied by $h(x_i)=0$, givesConsequentlyEquality forces $h^{\prime \prime }=0$ almost everywhere. Then $h$ is affine and its zeros at at least two distinct knots force $h=0$, proving uniqueness.

For the penalized problem, fix a vector $u=(u_1,\dots ,u_n{)}^T$ of fitted values. The preceding variational result says that its natural spline interpolant has the least roughness among all functions taking those values, and by assumption that roughness is $u^{T}Ku$. The infinite-dimensional problem therefore reduces toBecause $K$ is positive semidefinite, $I+\lambda K$ is positive definite. The unique fitted-value vector isand ${\hat{g}}_{\lambda }$ is its unique natural cubic spline interpolant.

For Leave-one-out cross-validation, let ${\hat{g}}_{-i,\lambda }$ minimizeand defineIf $u_j^{(-i)}={\hat{g}}_{-i,\lambda }(x_j)$, its normal equations sayThus $u^{(-i)}=A_{\lambda }{\tilde{Y}}^{(i)}$, whereTaking the $i$th coordinate and writing $\hat{u}=A_{\lambda }Y$ givesTherefore one spline fit and the diagonal of its smoothing matrix give

A junction between adjacent polynomial pieces of a cubic spline or other spline. Continuity conditions at these junctions determine which piecewise polynomial functions are splines.