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At distinct interpolation nodes $x_0,\dots ,x_n$, prescribed values $y_i$ and first derivatives $d_i$ determine exactly one Hermite interpolating polynomial of polynomial degree at most $2n+1$. The homogeneous data force a double root of a polynomial at every interpolation node. Hence the kernel of a linear map given by these $2n+2$ conditions is zero: a nonzero polynomial with those double roots of a polynomial would have polynomial degree at least $2n+2$. Equal finite dimensions of a vector space then give existence as well as uniqueness. This interpolation polynomial is distinct from the Hermite polynomials forming a classical orthogonal polynomial family.

Let $x_0<\cdots <x_n$ be distinct interpolation nodes, let $f\in C^{2n+2}[x_0,x_n]$, and let $H$ be the Hermite interpolating polynomial matching the values and first derivatives of $f$ at all nodes. For each $x$ in the interval there is an interior point $\xi$ for which the displayed formula holds. At a node the formula is immediate. Otherwise put $e=f-H$, $\pi (t)={\prod }_i(t-x_i)$ and $\phi (t)=e(t)-e(x)\pi (t{)}^2/\pi (x{)}^2$. This function vanishes at the $n+1$ nodes and at $x$, while its derivative vanishes at every node. Rolle's theorem gives $n+1$ additional zeros of its derivative between successive zeros of $\phi$. Repeating the theorem therefore gives ${\phi }^{(2n+2)}(\xi )=0$. Since $H^{(2n+2)}=0$ and $({\pi }^2{)}^{(2n+2)}=(2n+2)!$, rearrangement yields the formula. The extra point $x$ is a zero of $\phi$, not necessarily of its derivative.

For two nodes $a,b$, this gives the uniform bound $\Vert f-H{\Vert }_{\infty }\le \Vert f^{(4)}{\Vert }_{\infty }(b-a{)}^4/384$, since $(x-a{)}^2(b-x{)}^2\le (b-a{)}^4/16$. The squared nodal factors encode simultaneous matching of values and derivatives.

Fix $i$, write $\alpha =x_{i-1}$, $\beta =x_i$, and let $E=T-S$ on this interval. The Hermite interpolation conditions give bothTwo applications of integration by parts therefore giveOn this single cubic piece, the same calculation also givesHere $S^{(4)}=0$ inside the interval, and all four endpoint values of $E,E^{\prime }$ are zero. Since $T^{\prime \prime }=E^{\prime \prime }+S^{\prime \prime }$, we obtain the exact identityThis 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.

The printed product omits the node $x_0$: its lower index must be zero for the stated formula to hold. For example, with $n=1$, nodes $0,1$, and data $f_0=0,f_1=1,g_0=g_1=0$, the printed $L_{1,1}=1$ gives $h=1$, which fails the first value condition. Use the corrected Lagrange basis polynomialsThey satisfy $L_k(x_j)={\delta }_{kj}$. Define $H_k=L_k^2[1-2(x-x_k)L_k^{\prime }(x_k)]$ and $K_k=L_k^2(x-x_k)$. At a node with $j\ne k$, the double zero of $L_k^2$ gives $H_k(x_j)=H_k^{\prime }(x_j)=K_k(x_j)=K_k^{\prime }(x_j)=0$. At $x_k$, direct differentiation givesBoth polynomials have degree at most $2n+1$. Consequently $h={\sum }_k(f_k{H}_k+g_k{K}_k)$ has the specified values and derivatives, and is the required Hermite interpolating polynomial. The source product is retained literally in the private transcription; this solution identifies its necessary correction.

Use the corrected Hermite interpolating polynomial from part (a). The auxiliary function $\phi$ vanishes at all $n+1$ nodes and at the additional point $x$. Its derivative vanishes at all nodes, because both $e^{\prime }(x_i)$ and the derivative of ${\pi }^2$ vanish there. Rolle's theorem between the $n+2$ ordered zeros of $\phi$ gives $n+1$ further, distinct zeros of ${\phi }^{\prime }$. Thus ${\phi }^{\prime }$ has $2n+2$ distinct zeros on the closed interval. The printed phrase that these include $x$ is not generally true; the extra zeros lie in intervals between zeros of $\phi$, and ${\phi }^{\prime }(x)$ need not vanish.

Repeated application of Rolle's theorem gives ${\phi }^{(2n+2)}(\xi )=0$ for some interior $\xi$. Since $h$ has degree at most $2n+1$ and ${\pi }^2$ is monic of degree $2n+2$,Rearranging givesAt an interpolation node, both sides are zero, so the error formula extends to the whole closed interval. The false location claim about one zero is unnecessary for this result.

For nodes $0,1$ and $f(x)=e^x$, the cubic Hermite interpolating polynomial isIndeed its values at zero and one are $1,e$, and its derivatives there are also $1,e$. The Hermite interpolation error formula givessince $0\le \xi \le 1$ and $x(1-x)\le 1/4$ on $[0,1]$.