Interpolation

One of the simplest (and most common) ways of approximating functions is to use interpolation. For example, consider linear interpolation in one dimension: given a function s(x) on an interval [a,b], we can find the linear interpolant p(x)=(s(a)(b-x)+s(a)(x-a))/(b-a). The error in the interpolant s(x)-p(x) is given by  

$$s(x)-p(x) = \frac12(x-a)(b-x)\cdot s^{\prime\prime}(\xi)$$ (1)

for some $\xi$ between a and b.

This gives a bound on the interpolation error:

$$\vert s(x)-p(x)\vert \le \frac12(x-a)(b-x)\cdot \max_{a\le\xi\le b}\vert s^{\prime\prime}(\xi)\vert.$$

In terms of the infinity norm,

$$\vert\vert s-p\vert\vert _{L^\infty} \le \frac18 \vert\vert s^{\prime\prime}\vert\vert _{L^\infty}.$$ (2)

Sometimes we use $\pi_h$ (for example) to mean the interpolation operation: $p(x) = \pi_h s(x)$.Then

$$\vert\vert s-\pi_hs\vert\vert _{L^\infty} \le \frac18(b-a)^2 \vert\vert s^{\prime\prime}\vert\vert _{L^\infty}.$$ (3)

This kind of error bound works well if have a pointwise bound on $s^{\prime\prime}(x)$.In finite element methods, however, we don't have this kind of a bound. More usually, we have a bound on $\int s^{\prime\prime}(x)^2 dx$.What do we do then?

Error bounds using integral bounds

To use integral bounds on $s^{\prime\prime}$ we need a different strategy. To start with, we know that e(x) = s(x)-p(x) is zero at the endpoints (i.e., e(a)=e(b)=0). It turns out for any twice-differentiable function e(x) with e(a)=e(b)=0, that

$$e(x) = \int_a^b G(x,\xi) e^{\prime\prime}(\xi) d\xi$$ (4)

where

$$G(x,\xi) = \left\{\begin{array} {l@{\qquad}l} -(b-\xi)(x-a)... ...le\xi\\ -(\xi-a)(b-x)/(b-a)&\xi\le x\le b \end{array}\right.$$

[Exercise: Prove this. You will need to break up the integral on $\xi$ to the intervals [a,x] and [x,b].]

Then we can apply the Cauchy inequality to get

$$\vert e(x)\vert \le \left[\int_a^b G(x,\xi)^2 d\xi\right]^{1/2} \left[\int_a^b e^{\prime\prime}(\xi)^2 d\xi\right]^{1/2}.$$

After some calculation we can show that

$$\int_a^b G(x,\xi)^2 d\xi = \frac{(x-a)^2(b-x)^2}{3(b-a)}.$$

So, $\vert e(x)\vert\le (x-a)(b-x)/\sqrt{3(b-a)}\vert\vert e^{\prime\prime}\vert\vert _{L^2}$.Note that e(x)=s(x)-p(x), so taking the second derivative gives $e^{\prime\prime}(x) = s^{\prime\prime}(x)$ as $p^{\prime\prime}(x)=0$for all x. So

$$\vert e(x)\vert\le \frac{(x-a)(b-x)}{\sqrt{3(b-a)}}\vert\vert s^{\prime\prime}\vert\vert _{L^2}.$$ (5)

Integrating different powers of |e(x)| over [a,b] will give us bounds on the interpolation error in different norms (or taking the maximum over x will give the $L^\infty$ norm of the error).

In particular, taking maximums gives  

$$\vert\vert e\vert\vert _{L^\infty} \le \frac{(b-a)^{3/2}}{4\sqrt{3}} \vert\vert s^{\prime\prime}\vert\vert _{L^2}.$$ (6)

Squaring and integrating gives  

$$\vert\vert e\vert\vert _{L^2} \le \frac{(b-a)^2}{\sqrt{90}}\vert\vert s^{\prime\prime}\vert\vert _{L^2}.$$ (7)

Derivative bounds

We can also use our integral representation of e(x), to get bounds on the derivatives in the L² norm. First, note that

$$e'(x) = \int_a^b \frac{\partial G}{\partial x}(x,\xi) e^{\prime\prime}(\xi) d\xi.$$ (8)

Using the Cauchy inequality again, we get

$$\vert e'(x)\vert \le \left[\int_a^b \frac{\partial G}{\parti... .../2} \left[\int_a^b e^{\prime\prime}(\xi)^2 d\xi\right]^{1/2}.$$

Similar calculations reveal that

$$\int_a^b \frac{\partial G}{\partial x}(x,\xi)^2 d\xi = \frac{b-a}3.$$

Now we can integrate |e'(x)|² to get the L² bound on e':

$$\int_a^b\vert e'(x)\vert^2 dx\le\frac{(b-a)^2}3\vert\vert e^{\prime\prime}\vert\vert _{L^2}^2.$$

Taking the square root gives:  

$$\vert\vert e'\vert\vert _{L^2}\le \frac{b-a}{\sqrt{3}} \vert... ...ac{b-a}{\sqrt{3}} \vert\vert s^{\prime\prime}\vert\vert _{L^2}.$$ (9)

This can also be written as  

$$\vert\vert s'-(\pi_h s)'\vert\vert _{L^2}\le \frac{b-a}{\sqrt{3}}\vert\vert s^{\prime\prime}\vert\vert _{L^2}.$$ (10)