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

## 4. Alternating Sets

#### Definition

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 x0, ..., xn − 1 such that:

• ax0 < x1 < ··· < xn − 1b,
• f(xi) − p(xi) = (−1)i E for i = 0, 1, ..., n − 1.

where either E = ||fp|| or E = −||fp||. The number n is the length of the alternating set.

#### Example

Let f(x) = x2 on the interval [−1, 1], and let p(x) be the constant function p(x) = 1 / 2.

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

Given two alternating sets X = (x0, ..., xn) and Z = (z0, ..., zm) for f, p, we say that Z extends X if, as sets, {x0, ..., xn}{z0, ..., zm}.

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.

#### Lemma 3

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.

#### Proof

Let x0 be a minimum and x1 be a maximum of the function f(x) − p(x) on [a, b], and let m0 = f(x0) − p(x0) and m1 = f(x1) − p(x1). 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) + (m1 + m0) / 2, then:

So ||fq || ≤ (m1m0) / 2 ≤ ||fp||, and equality holds if and only if m0 = −m1. If x0 < x1, then (x0, x1) is the required alternating set; otherwise, (x1, x0) is the alternating set.