The Journal of Online Mathematics and Its Applications, Volume 6 (2006)

The Chebyshev Equioscillation Theorem, Robert Mayans

Let `f` be a continuous function on the interval
[`a`, `b`],
and let `p` be a polynomial approximation. An *alternating set* on
`f`, `p`
is defined to be a sequence of points
`x`_{0}, ..., `x`_{n − 1}
such that:

`a`≤`x`_{0}<`x`_{1}< ··· <`x`_{n − 1}≤`b`,`f`(`x`_{i}) −`p`(`x`_{i}) = (−1)^{i}`E`for`i`= 0, 1, ...,`n`− 1.

where either
`E` = ||`f` − `p`||
or
`E` = −||`f` − `p`||.
The number `n` is the *length* of the alternating set.

Let
`f`(`x`) = `x`^{2}
on the interval
[−1, 1],
and let
`p`(`x`)
be the constant function
`p`(`x`) = 1 / 2.

Clearly
||`f` − `p`|| = 1 / 2,
and the points
(−1, 0, 1)
form an alternating set of length 3.

Given two alternating sets
`X` = (`x`_{0}, ..., `x`_{n})
and
`Z` = (`z`_{0}, ..., `z`_{m})
for
`f`, `p`, we say that `Z`
* extends * `X` if, as sets,
{`x`_{0}, ..., `x`_{n}}{`z`_{0}, ..., `z`_{m}}.

The Chebyshev theorem states that if `p` is a polynomial of degree
≤ `n`
of best approximation to `f`, then `f`, `p` has an alternating set of length
`n` + 2.
We start by proving a simple partial result.

Let `f` be a continuous real-valued function on
[`a`, `b`],
and suppose that `p` is a polynomial of best approximation of degree `n` to `f`. Then there is an alternating set on
`f`, `p`
of length 2.

Let
`x`_{0}
be a minimum and
`x`_{1}
be a maximum of the function
`f`(`x`) − `p`(`x`)
on
[`a`, `b`],
and let
`m`_{0} = `f`(`x`_{0}) − `p`(`x`_{0})
and
`m`_{1} = `f`(`x`_{1}) − `p`(`x`_{1}).
They must be of opposite sign and equal in magnitude, for otherwise we could add a constant to `p`(`x`) and get a better approximation. Specifically, if
`q`(`x`) = `p`(`x`) + (`m`_{1} + `m`_{0}) / 2,
then:

So
||`f` − `q` || ≤ (`m`_{1} − `m`_{0}) / 2 ≤ ||`f` − `p`||,
and equality holds if and only if
`m`_{0} = −`m`_{1}.
If
`x`_{0} < `x`_{1},
then
(`x`_{0}, `x`_{1})
is the required alternating set; otherwise,
(`x`_{1}, `x`_{0})
is the alternating set.