Splines

**Sogan** · May 12th 2011, 03:57 PM

Hello,

i try to solve this problem:

Let $a=x_0 < x_1 <...<x_n =b$ be a partition of [a,b]. and let $V={v \in C^0[a,b] : v_{|[a,b]} \in C^4[x_i-1,x_i]$ with $v(x_i)=y_i$ for some $y_i \in \mathbb{R}$ . Show that every solution of $argmin_{v \in V} \int |v'(x)|^2 dx$ must be a spline of degree 1.

Do you have a idea, how i can solve this?
It is obvious, that there is a 1 Spline in V (the only one).
Now we have to show that any other element in V is greater than this 1-spline. But how can i do this?

Regards

**jamesdt** · May 13th 2011, 03:32 PM

The only spline of degree one that interpolates the points $(x_i,y_i)$ is in fact the piecewise linear interpolant. Let's assume that the data points are taken from some function $f$ such that $y_i = f(x_i)$ and denote the piecewise linear interpolant by $g = If$ . Now, let $h \in V$ be some other function that interpolates the data points, and define $e = h - g$ . Then,

$\int_a^b (h')^2 dx = \int_a^b (e'+g')^2 dx = \int_a^b (e')^2 dx + 2\int_a^b e'g' dx + \int_a^b (g')^2 dx$

If you can show that the second term on the right, i.e. $\int_a^b e'h' dx$ is zero, then you may conclude that
$\int_a^b (g')^2 dx \leq \int_a^b (h')^2 dx$ .

In other words, now you must show that $e'$ and the degree zero spline $g'$ are orthogonal. This is quite easy to show:
$\int_a^b e'g' dx = \sum_{i=1}^{n-1} \int_{x_i}^{x_{i+1}} e'g' dx$
First, we have divided the interval $[a,b]$ into parts $[x_i,x_{i+1}]$ , then we use integration by parts
$\sum_{i=1}^{n-1} \int_{x_i}^{x_{i+1}} e'g' dx = \sum_{i=1}^{n-1} \left( eg' \mid_{x_i}^{x_{i+1}} - \int_{x_i}^{x_{i+1}} eg'' dx \right)$
Since $g'$ is a degree zero spline, its derivative is obviously zero, so the integral on the right is zero. Only the sum remains, and now we use the fact that $g=Ih=If$ , so we may write $e=h-Ih$ . We find that
$\sum_{i=1}^{n-1} e(x)g'(x) \mid_{x_i}^{x_{i+1}} = \sum_{i=1}^{n-1} (h(x_{i+1}) - g(x_{i+1}))g'(x_{i+1}) - (h(x_{i}) - g(x_{i}))g'(x_{i}) = 0$
This last part is obviously zero because both $h$ and $g$ satisfy the same interpolation conditions at the points $x_i$ .

I based this on a similar proof in a compendium that was written for a course in splines which I'm taking. Here the proof was about the cubic spline interpolant which minimizes the integral of the second derivative squared. If you're curious, send me a PM and I'll see if I can find a link.

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