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`].
We have previously defined:

`d`_{n} = inf{||`f` − `p`||: `p` is a polynomial on [`a`, `b`] of degree ≤ `n`}

Our goal is to show that this minimum is actually achieved: there exists a polynomial
`p`_{n}
for which
||`f` − `p`_{n}|| = `d`_{n}.
Such a polynomial
`p`_{n}
is called a polynomial of best approximation of degree `n` to `f`. Our proof follows the argument in Isaacson and Keller (1994). Later we will show that for every `f` and `n`, the polynomial of best approximation is unique.

Let `f`, `g` be continuous real-valued functions on the interval
[`a`, `b`].
Then

- ||
`f`|| ≥ 0 and ||`f`|| = 0 iff`f`= 0. - ||
`f`+`g`|| ≤ ||`f`|| + ||`g`||. - For any constant
`c`, ||`c``f`|| = |`c`| · ||`f`||.

The proof is easy. The lemma says that the supremum norm satisfies the requirements of a vector space norm of the space of all continuous real-valued functions on
[`a`, `b`].
Thus, we can define a metric on this space: the distance between `f` and `g` is
`d`(`f`, g) = ||`f` − `g`||.

Let
`a` = (`a`_{0}, `a`_{1}, ..., `a`_{n})
be a vector in
`R`^{n + 1}.
We define the polynomial associated with `a`:

`p`_{a}(`x`) = `a`_{0} + `a`_{1} `x`^{1} + ··· + `a`_{n} `x`^{n}

and the ordinary Euclidean length of `a`:

The Euclidean length defines a metric on
`R`^{n + 1}
where the distance between `a` and `b` is
`d`(`a`, `b`) = |`b` − `a`|.

The function mapping each `a` in
`R`^{n + 1}
to
||`p`_{a}||
is continuous.

We will show this when the interval is
[0, 1].
Given
`ε` > 0,
let
`δ` = `ε` / (`n` + 1).
Let
`a` = (`a`_{0}, ..., `a`_{n})
and
`b` = (`b`_{0}, ..., `b`_{n}),
and suppose that
|`b` − `a`| ≤ `δ`.
Then for every
`i` = 0, ..., `n`,

|`b`_{i} − `a`_{i}| ≤ |`b` − `a`| < `ε` / (`n` + 1).

Hence for any
`x` [0, 1],

|`p`_{b}(`x`) − `p`_{a}(`x`)| ≤ |`b`_{0} − `a`_{0}| + |`b`_{1}-`a`_{1}| · |`x`| + ··· + |`b`_{n} − `a`_{n}| · |`x`^{n}| < `ε` / (`n` + 1) + ··· + `ε` / (`n` + 1) = `ε`.

We now can state and prove the main result of this section:

Let `f` be a continuous function on
[`a`, `b`].
Then for every `n`, there is a polynomial
`p`_{n}(`x`)
of degree
≤ `n`
such that:

||`f` − `p`_{n}|| = `d`_{n} = inf {||`f` − `q`|| : `q` is a polynomial of degree } ≤ `n`}

Let
`S` = {`a` `R`^{n + 1} : |`a`| = 1}
and let
`ε` = inf{|| `p`_{a}|| : `a` `S`}.
Note that `ε` is the minimum value of a continuous function on `S`, a closed bounded set in
`R`^{n + 1}.
Hence there is an
`a` `S`
such that
`ε` = ||`p`_{a}(`x`)||.
Further, we must have
`ε` > 0,
for otherwise
`p`_{a} = 0
and
`a` = `0`, and so `a`
cannot be in `S`.

Next, observe that
||`p`_{a}||
tends to infinity as
|`a`|
tends to infinity, for

||`p`_{a}|| = |`a`| · || `p`_{a / |a|} || ≥ |`a`| · `ε`

which grows arbitrarily large as
|`a`|
tends to infinity.

Now choose `M` so large that
||`p`_{a}|| ≥ `d`_{n} + 1 + ||`f`||
whenever
|`a`| ≥ `M`.
Since

||`f` − `p`|| ≥ ||`p`|| − ||`f`|| ≥ `d`_{n} + 1 + ||`f`|| − ||`f`|| = `d`_{n} + 1

it follows that a polynomial
`p`_{c}
of best approximation, if it exists, must satisfy
|`c`| ≤ `M`.

Therefore

`d`_{n} = inf{||`f` − `p`_{a}|| : `a` `R`^{n + 1} } = inf{||`f` − `p`_{a}|| : |`a`| ≤ `M`}

and so
`d`_{n}
is the minimum of a continuous function defined on the closed ball
{`a` `R`^{n + 1} : |`a`| ≤ `M`}.
By compactness, there is a vector `c` within the closed ball that achieves the minimum value. For that vector `c`, we have
`d`_{n} = ||`f` − `p`_{c}|| and thus
`p`_{c}
is a polynomial of best approximation.