= Solution
Fix $i$, write $\alpha=x_{i-1}$, $\beta=x_i$, and let $E=T-S$ on this <interval>. The <Hermite interpolation> conditions give both
$$
E(\alpha)=E(\beta)=0,\qquad E'(\alpha)=E'(\beta)=0.
$$
Two applications of <integration by parts> therefore give
$$
\int_\alpha^\beta E T^{(4)}\,dx
=[ET'''-E'T'']_\alpha^\beta+\int_\alpha^\beta E''T''\,dx
=\int_\alpha^\beta E''T''\,dx.
$$
On this single cubic piece, the same calculation also gives
$$
\int_\alpha^\beta E''S''\,dx
=[E'S''-ES''']_\alpha^\beta+\int_\alpha^\beta ES^{(4)}\,dx=0.
$$
Here $S^{(4)}=0$ inside the <interval>, and all four endpoint values of $E,E'$ are zero. Since $T''=E''+S''$, we obtain the exact identity
$$
\boxed{\int_{x_{i-1}}^{x_i}(T-S)T^{(4)}\,dx
=\int_{x_{i-1}}^{x_i}(T''-S'')^2\,dx\geq0.}
$$
This is an <integral> inequality, with no assertion that the <integrand> on the left has one sign pointwise. The stronger nodal <derivative> conditions make the boundary terms vanish on each <interval> separately; part (i) only needed their cancellation after summing.
The stated <Hermite interpolating polynomial> exists uniquely in $\Pi_{2n+1}$. Indeed, a <polynomial> in the <kernel of a linear map> of the value-and-<derivative> data map has a double <root of a polynomial> at each of $n+1$ distinct <interpolation nodes>. A nonzero such <polynomial> would have <polynomial degree> at least $2n+2$, which is impossible. The map is therefore an <injective function> and a <linear map> between <vector spaces> of the same finite <dimension of a vector space>, $2n+2$, and hence is a <linear isomorphism>. No higher smoothness of the original <function> $f$ is required to construct it.
An explicit illustration uses <interpolation nodes> $0,1,2$, values $0,1,0$, and zero <derivatives> at every <interpolation node>. The <clamped cubic spline> is $S(x)=3x^2-2x^3$ on $[0,1]$ and its reflection on $[1,2]$. Its <Hermite interpolating polynomial> is the <quartic polynomial> $T(x)=x^2(2-x)^2$. On $[0,1]$, $T-S=x^2(1-x)^2$, and the corresponding reflected formula holds on $[1,2]$. Direct <integration> gives $J(S)=24$, $J(T)=128/5$, and $J(T-S)=8/5$, exactly as the energy identity predicts.
\Image[/past-exam-of-the-numerical-analysis-course-of-the-university-of-oxford/2017-spline-energy.png]
{title=A <clamped cubic spline> and a <Hermite interpolating polynomial> with the same nodal values and <derivatives>: the excess <second derivative roughness penalty> is $8/5$}
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