Question: Let $S$ be a spline function of degree 1 that interpolates $f$ at $t_0, t_1....t_n$. Let $t_0<t_1<...t_n$ and let $\delta=max (t_{i+1}-t_i)$,where $0 \leq i \leq{n-1}$. Then $|f(x)-S(x)|$$\leq C\delta/2$, where $C$ is an upper bound of $|f'(x)|$ on $(t_o,t_n)$. Assume $f'$ exists and is continuous.
Facts given/we know:
Theorem: Let p be a first-degree spline having knots $a=x_0<x_1<....x_n$=b. If $p$ interpolates a function $f$ at these knots, then with $h=max(x_i-x_{i-1})$, we have: |$f(x)-p(x)|$ $\leq \omega(f;h)$ ,$a\leq x \leq b$:
$\omega(f;h)$ is given to be: $Sup${$|f(u)-f(v)|:a \leq u \leq v \leq b, |(u-v)| \leq h$}, h is just the width of a small interval on $[a,b]$
My attempt at a proof we know that: By the Mean Value Theorem: $|f(u)-f(v)|$=$|f'(c)| \leq C|u-v|$ $\leq C \delta$. (taking our $h$ to be $\delta $ where $\delta=max (t_{i+1}-t_0)$. Thus $\omega(f;\delta)$ $\leq $ $C \delta$. Thus: by the Theorem, we have:$|f(x)-S(x)|$ $\leq \omega(f;\delta) \leq C \delta$.
Problem: I'm unable to get the tighter inequality which the problem asks for with the $\delta/2$. That is I'm unable to get this:$|f(x)-S(x)|$$\leq C\delta/2$. Maybe I'm supposed to use something else instead The book gives the inequality: $|f(x)-S(x)|$$\leq D\delta/8$, where $D$ is an upper bound of $|f''(x)|$ on $(t_o,t_n)$ for reference.
In Particular I'm not sure how they go from:
How do they deduce the inequality $|f(t)-p(t)|$ $\leq M_1/2(t_{i+1}-t_i)$? See also this picture:
Thanks.
I'm not sure how they are deducing the $|f(x)-S(x_i)|$=$|f''(c)|/2(x-x_i)(x-x_{i+1})$