4
$\begingroup$

Background: Given an increasing set of points $(x_i)_{i=0}^n \subset \mathbb [a,b]$, a cubic spline $S(x)\in C^2([a,b])$ is a piecewise cubic polynomial on each subinterval $(x_i, x_{i+1})$.

Given a set of real number $y_0, \ldots, y_n$, then if $S(x)$ is the natural cubic spline interpolant, it is also the minimizer of $\int\limits_a^b (u''(x))^2 \, dx$ over all $C^2$ functions with $u(x_i)=y_i$ for all $i$.

Question: I'm looking for a variational Euler-Lagrange kind of proof for theorems of this kind, i.e., how to build a $C^m$ interpolant that minimizes $\|Ku\|_2$ for some linear operator $K$.

Improve this question
$\endgroup$
2
  • $\begingroup$ I suspect that the property only holds for the natural cubic spline. $\endgroup$
    – Federico Poloni
    Commented May 15, 2017 at 21:46
  • 2
    $\begingroup$ @FedericoPoloni: in the clamped case you also have stationarity. Formally integrating by parts twice in the clamped case, with prescribed $u'$ at $x_0$ and $x_n$, gives you that the variation vanishes for piecewise cubic, $C^2$ functions that matches the data. $\endgroup$
    – Willie Wong
    Commented May 16, 2017 at 13:29

1 Answer 1

Reset to default
4
$\begingroup$

See pp.87~107 of Prenter, Paddy M. Splines and variational methods. Courier Corporation, 2008. especially p.100 where "uniqueness theorem" is proved and spline is defined as minimizer to $\displaystyle{\int}_{K}[f^{(m)}(u)]^2 du$ among $C^m$ functions over a set $K=[a,b]\subset\mathbb{R}$. The basic technique there is to realize $\int_K (Lf)^2=\int_K (L(f-g))^2 +\int_K (Lg)^2$ where $L=D^m$ is the differentiation operator by integration by parts.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Thanks! I read the proof, and I'm glad I did. However, I was told that for a more general linear $L$ operator one can solve it using the Euler-Lagrange variational method. $\endgroup$
    – Amir Sagiv
    Commented May 17, 2017 at 18:13
  • $\begingroup$ You can for this one...it depends on whether you use EL method for the minimization problem or not. And what did you mean by "more general $L$" $\endgroup$
    – Henry.L
    Commented May 17, 2017 at 20:38
  • $\begingroup$ I altered the original OP to clarify that, hopefully it worked. Do you know a reference which uses EL for this proof? $\endgroup$
    – Amir Sagiv
    Commented May 17, 2017 at 20:59
  • 1
    $\begingroup$ It is only possible if you know it is isomorphic to a reproducing kernel Hilbert space. $\endgroup$
    – Henry.L
    Commented May 17, 2017 at 21:07
  • 1
    $\begingroup$ Try to read Aronszajn's seminal paper. Aronszajn, Nachman. "Theory of reproducing kernels." Transactions of the American mathematical society 68.3 (1950): 337-404. $\endgroup$
    – Henry.L
    Commented May 17, 2017 at 21:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .