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Let $a,b\in\mathbb{R}$, $a

The following statement is well known as Chebyshev Equioscillation Theorem:

$P\in\Pi_n[a,b]$ is a best approximation from $\Pi_n[a,b]$ to $f\in C[a,b]$, that is, \begin{equation} \|P-f\|_{C[a,b]} =\inf_{Q\in\Pi_n[a,b]}\|Q-f\|_{C[a,b]} \end{equation} if and only if there exists an increasing sequence of points $x_i\in[a,b]$, $0\le i\le n+1$, and $\sigma\in\{-1,1\}$ such that \begin{equation} P(x_i)-f(x_i) =\sigma\ (-1)^i\|P-f\|_{C[a,b]}\text{,$\quad0\le i\le n+1$.}\tag{*} \end{equation}

Does a similar statement hold if $C[a,b]$ is replaced by $L_{\infty}[a,b]$?

I am particularly interested in the case $n=1$ and the implication \begin{equation} \text{$P$ is a best approximation to $f$}\implies\text{A relation similar to (*) holds.} \end{equation}

Clearly, (*) cannot remain unchanged in the generalized setting for if $P\in\Pi_n[a,b]$ is a best approximation from $\Pi_n[a,b]$ to $f\in L_{\infty}[a,b]$ then $P$ is a best approximation from $\Pi_n[a,b]$ to each $g\in L_{\infty}[a,b]$ such that $\|g-f\|_{L_{\infty}[a,b]}=0$.

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    Something wrong with $(*)$. Let $\sigma =1$, then for $i=1$ lhsis greater than zero while rhs is less than zero.2012-08-25
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    @Norbert Thank you for the hint. Absolute value on the left hand side of (*) removed.2012-08-25

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