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

## 3. The Polynomial of Best Approximation

Let f be a continuous function on the interval [a, b]. We have previously defined:

dn = inf{||fp||: 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 pn for which ||fpn|| = dn. Such a polynomial pn 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.

#### Lemma 1

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, ||cf|| = |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) = ||fg||.

Let a = (a0, a1, ..., an) be a vector in Rn + 1. We define the polynomial associated with a:

pa(x) = a0 + a1 x1 + ··· + an xn

and the ordinary Euclidean length of a: The Euclidean length defines a metric on Rn + 1 where the distance between a and b is d(a, b) = |ba|.

#### Lemma 2

The function mapping each a in Rn + 1 to ||pa|| is continuous.

#### Proof

We will show this when the interval is [0, 1]. Given ε > 0, let δ = ε / (n + 1). Let a = (a0, ..., an) and b = (b0, ..., bn), and suppose that |ba| ≤ δ. Then for every i = 0, ..., n,

|biai| ≤ |ba| < ε / (n + 1).

Hence for any x [0, 1],

|pb(x) − pa(x)| ≤ |b0a0| + |b1-a1| · |x| + ··· + |bnan| · |xn| < ε / (n + 1) + ··· + ε / (n + 1) = ε.

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

#### Theorem 3

Let f be a continuous function on [a, b]. Then for every n, there is a polynomial pn(x) of degree n such that:

||fpn|| = dn = inf {||fq|| : q is a polynomial of degree } ≤ n}

#### Proof

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

Next, observe that ||pa|| tends to infinity as |a| tends to infinity, for

||pa|| = |a| · || pa / |a| || ≥ |a| · ε

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

Now choose M so large that ||pa|| ≥ dn + 1 + ||f|| whenever |a| ≥ M. Since

||fp|| ≥ ||p|| − ||f|| ≥ dn + 1 + ||f|| − ||f|| = dn + 1

it follows that a polynomial pc of best approximation, if it exists, must satisfy |c| ≤ M.

Therefore

dn = inf{||fpa|| : a Rn + 1 } = inf{||fpa|| : |a| ≤ M}

and so dn is the minimum of a continuous function defined on the closed ball {a Rn + 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 dn = ||fpc|| and thus pc is a polynomial of best approximation.