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Boundary-compatible clamped cubic interpolation is fourth order in the maximum norm

Statement

Let $a = x_{0} < \dots < x_{m} = b$ , $m \geq 1$ , with $h_{i} = x_{i + 1} - x_{i}$ and $H = \max_{i} h_{i}$ . Suppose $f$ has continuous derivatives through order four on an open interval containing $[a, b]$ , and let $M \geq 0$ satisfy $∣ f^{(4)} (x) ∣ \leq M$ there. Let $s$ be the clamped cubic interpolant at $y_{i} = f (x_{i})$ with the boundary-compatible slopes $s^{'} (a) = f^{'} (a)$ and $s^{'} (b) = f^{'} (b)$ . Then

∥ s - f ∥_{L^{\infty} ([a, b])} \leq \frac{83}{324} M H^{4} \leq M H^{4} .

The conclusion uses the exact endpoint derivatives; it is not a rate claim for the natural boundary condition or for estimated endpoint slopes.

Facts & Assumptions

Given: A strictly increasing partition, $f$ with continuous derivatives through order four near $[a, b]$ , a bound $M$ on $∣ f^{(4)} ∣$ , and the clamped interpolant with endpoint slopes equal to those of $f$ .

[F1]

Clamped cubic spline interpolation exists uniquely gives the second-derivative system, the cubic piece formula, and the endpoint-slope formulae for the clamped interpolant.

[F2]

A uniform derivative bound gives a uniform Taylor remainder bound: if a function has derivatives through order $q + 1$ on the closed segment between $c$ and $x$ , with endpoint continuity, and its $(q + 1)$ st derivative is bounded by $M$ , its degree- $q$ Taylor remainder is at most $M ∣ x - c ∣^{q + 1} / (q + 1)!$ . Here the proof uses $q = 1$ for $f^{''}$ , $q = 2$ for $f^{'}$ , and $q = 3$ for $f$ ; the resulting factorials are $2$ , $6$ , and $24$ .

[A1]

$h_{i} > 0$ , $H = \max_{i} h_{i}$ , and $∣ f^{(4)} ∣ \leq M$ on a neighborhood of the partition.

Proof

1.1A1F1F2algebra

Put $g_{i} = f^{''} (x_{i})$ and $q_{i} = f^{'''} (x_{i})$ . By [F2], on each adjacent interval of length $h$ , Taylor expansion gives $f^{''} (x_{i} + h) = g_{i} + h q_{i} + R_{g}^{+}$ and $f^{''} (x_{i} - h) = g_{i} - h q_{i} + R_{g}^{-}$ with $∣ R_{g}^{\pm} ∣ \leq M h^{2} / 2$ ; the corresponding secant slopes satisfy $d_{i} = f^{'} (x_{i}) + h_{i} g_{i} / 2 + h_{i}^{2} q_{i} / 6 + R_{i}^{+}$ and $d_{i - 1} = f^{'} (x_{i}) - h_{i - 1} g_{i} / 2 + h_{i - 1}^{2} q_{i} / 6 + R_{i}^{-}$ , with $∣ R_{i}^{\pm} ∣ \leq M h_{i or i - 1}^{3} / 24$ . Substitution in an interior row of [F1] shows that the residual obtained by inserting $g_{i}$ in place of $M_{i}$ has magnitude at most $(3 / 4) M H^{2} (h_{i - 1} + h_{i})$ : the first-derivative terms cancel, and the remaining terms are bounded by $(1 / 2 + 1 / 4) M (h_{i - 1}^{3} + h_{i}^{3})$ . The same expansions in the first row give residual at most $(3 / 4) M H^{2} h_{0}$ ; expansion backward from $b$ gives at most $(3 / 4) M H^{2} h_{m - 1}$ in the last row.

1.2A1F2constructalgebra

Fix an interval $[x_{i}, x_{i + 1}]$ of length $h_{i}$ and let $H_{i}$ be the cubic Hermite interpolant matching $f$ and $f^{'}$ at both endpoints. Let $T_{3}$ be the Taylor polynomial of degree three for $f$ at $x_{i}$ , and put $g = f - T_{3}$ . By [F2], $g (x_{i}) = g^{'} (x_{i}) = 0$ , $∣ g (x) ∣ \leq M h_{i}^{4} / 24$ , $∣ g (x_{i + 1}) ∣ \leq M h_{i}^{4} / 24$ , and $∣ g^{'} (x_{i + 1}) ∣ \leq M h_{i}^{3} / 6$ . With $t = (x - x_{i}) / h_{i}$ , the Hermite basis functions for the right value and right slope are $3 t^{2} - 2 t^{3}$ and $t^{3} - t^{2}$ ; their absolute values on $[0, 1]$ are at most $1$ and $4 / 27$ . Therefore $∣ H_{i} g (x) ∣ \leq (43 / 648) M h_{i}^{4}$ , and $∣ f (x) - H_{i} (x) ∣ = ∣ g (x) - H_{i} g (x) ∣ \leq (35 / 324) M h_{i}^{4}$ .

2.1A1F1step 1.1algebrachoose

In the clamped system matrix, each diagonal exceeds the sum of the absolute off-diagonal entries by respectively $h_{0}$ , $h_{i - 1} + h_{i}$ , and $h_{m - 1}$ . Let $e_{i} = M_{i} - g_{i}$ and choose an index with maximal $∣ e_{i} ∣$ . Since the residual equation is $A e = b - A g$ , its row at that index has magnitude at least the row's diagonal-dominance margin times $\max_{j} ∣ e_{j} ∣$ . Step 1.1 bounds that same row by $(3 / 4) M H^{2}$ times the margin. Hence $\max_{i} ∣ M_{i} - g_{i} ∣ \leq (3 / 4) M H^{2}$ .

3.1A1F1step 1.1step 2.1algebra

At an interior node $x_{i}$ , [F1] gives $s^{'} (x_{i}) = d_{i - 1} + h_{i - 1} (M_{i - 1} + 2 M_{i}) / 6$ . Substituting the backward Taylor expansions in step 1.1 and $M_{j} = g_{j} + e_{j}$ cancels the terms in $g_{i}$ and $q_{i}$ . The remaining Taylor terms are bounded by $(1 / 24 + 1 / 12) M h_{i - 1}^{3} = M H^{3} / 8$ , and the $e$ terms by $h_{i - 1} (3 / 4) M H^{2} / 2 \leq 3 M H^{3} / 8$ . Thus $∣ s^{'} (x_{i}) - f^{'} (x_{i}) ∣ \leq M H^{3} / 2$ . At $a$ and $b$ this difference is zero by the boundary conditions.

4.1A1step 3.1step 1.2algebra

The difference $s - H_{i}$ is a cubic with zero endpoint values and with endpoint slope differences $ϵ_{i} = s^{'} (x_{i}) - f^{'} (x_{i})$ and $ϵ_{i + 1} = s^{'} (x_{i + 1}) - f^{'} (x_{i + 1})$ . Its Hermite form is $h_{i} (ϵ_{i} t (1 - t)^{2} + ϵ_{i + 1} t^{2} (t - 1))$ . Each displayed shape function has absolute value at most $4 / 27$ on $[0, 1]$ . Step 3.1 therefore gives $∣ s - H_{i} ∣ \leq (4 / 27) M H^{4}$ . Combining with step 1.2 yields $∣ s - f ∣ \leq (35 / 324 + 4 / 27) M H^{4} = (83 / 324) M H^{4}$ on this interval.

5.1step 4.1algebra∎

The intervals cover $[a, b]$ , and $83 / 324 < 1$ . Taking the supremum over all intervals proves the stated maximum-norm estimate.

Depends on

Used by

The natural interpolant of x squared has a second-order endpoint error Example

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Sources

Carl de Boor, A Practical Guide to Splines, revised edition, Chapter V, pp. 51–56 (standard reference, not scraped)
Brian LeMesurier, Introduction to Numerical Methods and Analysis with Python, §3.4.3, Theorem 3.5 (standard reference, not scraped)