The Journal of Online Mathematics and Its Applications, Volume 6 (2006)
The Chebyshev Equioscillation Theorem, Robert Mayans

## 8. Proof of the Theorem

#### Theorem 6 (The Chebyshev Equioscillation Theorem)

Let f be a continuous real-valued function on [a, b]. Then pn(x) is a polynomial of best approximation of degree n if and only if f, pn has an alternating set of length n + 2. Furthermore, the polynomial of best approximation is unique.

#### Proof

The theorem is trivially true if f is itself a polynomial of degree n. We assume not, and so dn > 0.

##### Step 1

Suppose that f, pn has an alternating set of length n + 2. By Theorem 4, we have || fpn || ≤ dn. As dn ≤ || fpn || by the definition of dn, it follows that pn is a polynomial of best approximation to f.

##### Step 2

Now suppose that pn is a polynomial of best approximation to f. By Lemma 4, f, pn has an alternating set of length 2, and by Theorem 5, it can be extended into a sectioned alternating set of length m. We must have mn + 2, for if mn + 1 then by Lemma 6, we could add a polynomial q of degree n to pn and get a better approximation than pn, which is impossible. Thus every polynomial of best approximation has an alternating set of length at least n + 2.

##### Step 3

To show uniqueness, suppose that pn and qn are both polynomials of best approximation, and we will show that they are equal.

Note that (pn + qn) / 2 is a polynomial of best approximation, as:

Therefore, there are n + 2 alternating points at which (fpn) / 2 + (fqn) / 2 = ± dn.

At each of these alternating points, fpn and fqn are both dn or both dn. So fpn and fqn agree on n + 2 points, and so (fpn) − (fqn) = qnpn = 0 at these n + 2 points. Since qnpn is a polynomial of degree n, qn and pn must be identical. Therefore the polynomial pn of best approximation is unique.

#### Notes

The polynomial pn of best approximation may have degree < n, and may produce an alternating set of length > n + 2. As an example of both, consider the polynomial of best approximation of degree m to the function y = f(x) = cos(x) on the interval [−n / π, n / π]. There is an obvious alternating set of length 2n + 1 when p(x) = 0. By the Chebyshev theorem, the polynomial pm of best approximation for m ≤ 2n − 1 is pm(x) = 0.