= Solution
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 $n\geq2$. 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 $g''(x_1)=g''(x_n)=0$ on the spline over $[x_1,x_n]$ and then extends linearly with matching <derivatives>. Merely prescribing $g''(a)=g''(b)=0$ 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 $\mathbf v=(v_1,\ldots,v_n)^T$ is the <natural cubic spline> satisfying $(N\mathbf v)(x_i)=v_i$. Its existence and uniqueness for $n\geq2$ 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 $\widetilde g\in C^2[a,b]$ have the same values, and put $r=\widetilde g-g$. Then $r(x_i)=0$. We prove the orthogonality
$$
\int_a^b g''r''\,dx=0.
$$
Apply <integration by parts> twice on each <polynomial> interval. On an interior interval, $g''''=0$, and
$$
\int_{x_i}^{x_{i+1}}g''r''\,dx=[g''r'-g'''r]_{x_i}^{x_{i+1}}.
$$
The $g'''r$ terms vanish because $r$ vanishes at the knots. The $g''r'$ terms cancel at internal knots because $g''$ and $r'$ are <continuous>, and vanish at $x_1,x_n$ because $g''$ is zero there. On the exterior intervals, $g''=0$ identically. Summing therefore gives the claimed orthogonality without assuming that $g'''$ is <continuous> across knots.
Expanding the square now gives the <minimum roughness property of the natural cubic spline interpolant>
$$
\boxed{\int_a^b(\widetilde g'')^2\,dx
=\int_a^b(g'')^2\,dx+\int_a^b(r'')^2\,dx
\geq\int_a^b(g'')^2\,dx.}
$$
Equality forces $r''=0$ everywhere, since $r''$ is <continuous>. Thus $r$ is an <affine function>. It vanishes at two distinct design points, so $r=0$, proving equality if and only if $\widetilde g=g$.
For the penalized fit, take the allowed <spline roughness penalty matrix> $\Gamma$ to be a <symmetric matrix> with $\mathbf z^T\Gamma\mathbf z\geq0$; 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 $\widetilde g$, let $\mathbf v=(\widetilde g(x_i))_{i=1}^n$. Replacing it by $N\mathbf v$ preserves every fitted value and can only decrease the <second derivative roughness penalty>. Therefore it is enough to minimize
$$
Q_\lambda(\mathbf v)=\|\mathbf Y-\mathbf v\|_2^2+\lambda\mathbf v^T\Gamma\mathbf v
$$
over $\mathbb R^n$. The <matrix> $A_\lambda=I_n+\lambda\Gamma$ is a <positive-definite matrix>, since
$$
\mathbf z^TA_\lambda\mathbf z
=\|\mathbf z\|_2^2+\lambda\mathbf z^T\Gamma\mathbf z>0
\qquad(\mathbf z\ne0).
$$
Completing the quadratic or differentiating gives the unique solution
$$
\boxed{\widehat{\mathbf v}_\lambda=(I_n+\lambda\Gamma)^{-1}\mathbf Y,\qquad
\widehat g_\lambda=N\widehat{\mathbf v}_\lambda.}
$$
This is the <cubic smoothing spline>. More explicitly, the normal equations are $(I_n+\lambda\Gamma)\widehat{\mathbf v}_\lambda=\mathbf Y$, and
$$
Q_\lambda(\mathbf v)-Q_\lambda(\widehat{\mathbf v}_\lambda)
=(\mathbf v-\widehat{\mathbf v}_\lambda)^TA_\lambda
(\mathbf v-\widehat{\mathbf v}_\lambda).
$$
For an arbitrary function with fitted vector $\mathbf v$, the earlier orthogonality also gives
$$
S_\lambda(\widetilde g)-S_\lambda(\widehat g_\lambda)
=(\mathbf v-\widehat{\mathbf v}_\lambda)^TA_\lambda
(\mathbf v-\widehat{\mathbf v}_\lambda)
+\lambda\int_a^b\bigl(\widetilde g''-(N\mathbf v)''\bigr)^2\,dx.
$$
Both terms are nonnegative. Equality forces $\mathbf v=\widehat{\mathbf v}_\lambda$ and, by the proved interpolation equality case, $\widetilde g=N\mathbf v=\widehat g_\lambda$. 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.
\b[At least two distinct design points are required for uniqueness.] Question 4 does not explicitly restate $n\geq2$, and this hypothesis is necessary. With one design point, every function $\widetilde g(x)=Y_1+c(x-x_1)$ 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.
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