I was trying to show that:
If $f \in C[a,b]$ is best approximated by $p^* \in \mathbb{P_n[a,b]}$ then,
$\exists \text{ } t_1, t_2 \in [a,b]$ such that
$$\epsilon(t_1) = ||\epsilon||_\infty \text{ }, \text{ } \epsilon(t_2)=-||\epsilon||_\infty$$
for $\epsilon=f-p^*$, the error function.
I know that, by defintion
$$p^* = min_{p\in \mathbb{P_n}} (||f-p||_\infty) = min_{p\in \mathbb{P_n}}(max_{x \in [a,b]}( |f(x)-p(x)|))$$
I also know that what I'm trying to show is equivalent to finding an alternating set for the function $\epsilon$ of at least two points, but I don't know how to combine this idea with the information given.
Thanks in advance!
Best Answer
Yes, this is true. Since $\epsilon= f-p^*$ is continuous, we know that either $-\|\epsilon\|$ or $\|\epsilon\|$ is assumed. Suppose, without loss of generality, that $\|\epsilon\|$ is assumed and $-\|\epsilon\|$ is not. Since $\epsilon$ attains its minimal value in $[a,b]$, there exists $\delta>0$ such that $$ -\|\epsilon\|+\delta \leq \epsilon(x) \leq \|\epsilon\|. $$ But then, consider the polynomial $q^*=p^*+ \delta/2 \in P_n[a,b]$. It would follow that $$ -\|\epsilon\|+\frac{\delta}{2} \leq f-q^* \leq \|\epsilon\|-\frac{\delta}{2}, $$ so $\|f-q^*\|<\|\epsilon\|=\|f-p^*\|$, contradicting the fact that $p^*$ is the best approximation of $f$ in $\mathbb P_n[a,b]$.