Weierstrass Approximation Theorem

Weierstrass Approximation theorem in real analysis presents the notion of approximating continuous functions by polynomial functions. According to the theorem, every continuous function defined on a closed interval [a, b] can approximately be represented by a polynomial function. Polynomial functions are simple functions that even computers can easily process, hence the Weierstrass Approximation theorem has great practical as well as theoretical utility. Especially, when it comes to polynomial interpolations in numerical analysis.

The Weierstrass Approximation theorem is named after German mathematician Karl Theodor Wilhelm Weierstrass. He gave this result when he was 70 years old.

Statement of the Weierstrass Approximation Theorem

Let f: [a,b] → R be a real valued continuous function. Then we can find polynomials p_n(x) such that every p_n converges uniformly to x on [a,b].

In other words, if f is a continuous real-valued function on [a, b] and if any ε > 0 is given, then there exist a polynomial P on [a, b] such that |f(x) – P(x)| < ε, for every x in [a, b].

The essence of this theorem is that no matter how much complicated the function f is given, we can always find a polynomial that is as close to f as we desire.

Proof of Weierstrass Approximation Theorem

There are several ways of proving this theorem. Here we shall see the proof by using Bernstein Polynomial.

Definition of Bernstein Polynomial: If f is a real valued function defined on [0, 1], then for n ∈ N, the nth Bernstein Polynomial of f is defined as

B_n(x, f) :=

Proof: To prove the theorem on closed intervals [a,b], without loss of generality we can take the closed interval as [0, 1]. Now consider f is a continuous real-valued function on [0,1]. Since [0, 1] is compact, the continuity of f implies uniform continuity. Now for a given ε > 0 there exist 𝛿 > 0 by the definition of uniform continuity of functions

|x – y| ≤ 𝛿 ⇒ |f(x) – f(y)| ≤ ε/2 for every x, y ∈ [0, 1].

Let M = ||f||_∞ exists as f is a continuous function on a compact set [0, 1]. Now, fix ζ ∈ [0, 1]. Then by uniform continuity of f we can have

|x – ζ| ≤ 𝛿 ⇒ |f(x) – f(ζ)| ≤ ε/2

Now, |f(x) – f(ζ)| ≤ 2M ≤ 2M [(x – ζ)/ 𝛿]² + ε/2

that is, |f(x) – f(ζ)| ≤ 2M [(x – ζ)/ 𝛿]² + ε/2 ∀ x ∈ [0, 1].

The Bernstein Polynomial is used to approximate f on [0, 1].

B_n(x, f – f(ζ)) =

= B_n(x, f) – f(ζ)B_n(x, 1)

And, B_n(x, 1) =

where the Binomial Theorem is used.

Since, if 0 ≤ f ⇒ B_n(x, f) and if g ≤ f ⇒ B_n(x, f).

Now,

So,

A simple calculation shows that on [0, 1], the maximum of z – z² is ¼. Thus,

Let N ≥ M/(2𝛿²ε), then for n ≥ N, we have

|| (B_n(ζ, f) – f(ζ)||_∞ ≤ ε.

This proves the theorem for continuous functions on [0, 1]. By application of the theorem for function on [0, 1], the case for an arbitrary interval [a, b] follows.

Related Articles

• Continuity of Functions
• Uniform Convergence
• Fatou’s Lemma
• Limits

Solved Example on Weierstrass Approximation Theorem

Example:

If

Solution:

We may assume that

By Weierstrass Approximation Theorem, there exists a sequence of polynomials p_n on C[0, 1], that is, continuous functions on [0, 1], which converges uniformly to f.

||p_n – f||_∞ → 0

Since the given integral is convergent, we have

But we have

then

Since, if f ≥ 0 and

Hence, f = 0.