6. The Weierstrass Theorem on Uniform Approximation and the Separability of C[0, 1]

Theorem 7 (Weierstrass). If a function f(x) is continuous on the interval [a, b], then there exists a sequence of polynomials {Pn(x)}, converging uniformly on the interval [a, b] to f(x), i.e., for any e > 0 there is a polynomial Pn(x) with index n, depending on e, such that |Pn(x) - f(x)| < e simultaneously for all x from the interval [a, b].

In other words, a function f(x) continuous on the segment [a, b] can be uniformly approximated on this interval by a polynomial with a prescribed accuracy e.

Proof. Without loss of generality, instead of the segment [a, b] we may consider the segment [0, 1], since by the linear transformation x = (b – a)t + a one interval is carried into the other. Moreover, it suffices to prove the theorem for a continuous function f(x), vanishing at the endpoints of the interval [0, 1], i.e., satisfying the conditions f(0) = 0 and f(1) = 0. Indeed, if f(x) did not satisfy these conditions, then, setting g(x) = f(x) – f(0) - x[f(l) -f(0)] we would obtain a function g(x), continuous on the interval [0, 1] and satisfying the conditions g(0) = 0 and g(1) = 0. Then from the possibility of representing g(x) as the limit of a uniformly convergent sequence of polynomials, it would follow that f(x) is also representable as the limit of a uniformly convergent sequence of polynomials (since the difference f(x) – g(x) is a polynomial of the first degree).

Thus, let a function f(x) be continuous on the interval [0, 1] and satisfy the conditions f(0) = 0, f(1) = 0. Such a function f(x) can be extended to the whole line by setting it equal to zero outside the interval [0, 1], and it can be asserted that the function so extended is, by Cantor's theorem, uniformly continuous on the whole line.

Let us consider the following sequence of polynomials of degree 2n:

Qn(x) = cn(1 – x2)n (n = 1, 2, …), (2)

for each of which the constant cn is chosen so that the following equality holds

(n = 1, 2, …). (3)

It is obvious that all the polynomials take non-negative values. Without computing the exact value of the constant cn, let us estimate it from above. For this, note that for any index n = 1, 2, ... and for all x from the interval [0, 1] the following inequality holds

(1 - x2)n ≥ 1 - nx2. (4)

(this inequality is easily proved if one shows that the function h(x) = (1 - x2)n - 1 + nx2 is equal to zero at zero and increases on the interval – the derivative is positive).

Applying inequality (4) and taking into account that for any n ≥1, we obtain

. (5)

From (2), (3) and (5) we conclude that for all indices n = 1, 2, ... the following upper estimate holds for the constant cn

cn (6)

From (6) and (2) it follows that for any d > 0, for all x from the segment d ≤ |x| ≤1, the following inequality holds

0 ≤ Qn(x) ≤ (1 - d2)n. (7)

From (7) it follows that for any fixed d > 0 the sequence of non-negative polynomials {Qn(x)} converges to zero uniformly on the segment d ≤ |x| ≤1.

Let us now set, for any x from the interval [0, 1]

Pn(x) = (8)

and let us verify that for any n = 1, 2, ..., the function Pn(x) is a polynomial of degree 2n, and moreover that {Pn(x)} is indeed the desired sequence of polynomials, converging uniformly on the interval [0, 1] to the function f(x).

Since the function f(x) under study is equal to zero outside the segment [0, 1], then for any x from the segment [0, 1] the integral (8) can be written in the form

Pn(x) =

Replacing in the last integral the variable t by t–x, we bring it to the form

Pn(x) = (9)

From (9) and (2) it is clear that the function Pn(x) represents a polynomial of degree 2n.

It remains to prove that the sequence {Pn(x)} converges to f(x) uniformly on the interval [0, 1]. Let us fix an arbitrary e > 0. For the fixed e, by virtue of the uniform continuity of f(x) on the whole real line, there is found d > 0 such that

|f(x) - f(y)| < e/2 for |x – y| < d. (10)

Let us also note that, since f(x) is continuous on the interval [0, 1], it is bounded on this interval, and hence also everywhere on the real line. This means that there exists a constant A such that for all x

|f(x)| ≤ A. (11)

Using (3), (6), (10) and (11), and taking into account the non-negativity of Qn(x), let us estimate the difference Pn(x) – f(x). For all x from the interval [0, 1] we shall have

| Pn(x) – f(x)| =

.

To complete the proof of the theorem it suffices to note that for all sufficiently large indices n the inequality holds.

Theorem 8. The space C[a, b] is separable.

Proof. Let us consider, in the space C[a, b], the set of all polynomials Z whose coefficients are rational numbers. This set can be regarded as a countable union of countable sets (by coordinates). Let us show that this set is everywhere dense in C[a, b].

Let f(x) ∈ C[a, b]. By virtue of the Weierstrass theorem, for an arbitrary e > 0 there is a polynomial Pn(x) = a0 + a1x + …+anxn such that d(f(x), Pn(x)) < e/2. Let us choose in the set Z a polynomial Zn = b0 + b1x + … + bnxn such that |ak – bk| < e/(2n×max(|a|, |b|, |a|n, |b|n)). Then, as is easily verified, d(Pn(x), Zn) < e/2. By the triangle inequality this proves the everywhere density of the countable set Z in C[a, b].