Chapter 3

Stone Weierstrass Approximation The-orem3.1 The Bernstein Polynomials

These polynomials give an explicit description of a sequence of polynomials which con-verge uniformly to a continuous function. Recall that if you have a bounded functiondefined on some set S with values in Y.

∥ f∥∞≡ sup{∥ f (x)∥ : x ∈ S}

This is one way to measure distance between functions.

Lemma 3.1.1 The following estimate holds for x ∈ [0,1] and m≥ 2.

m

∑k=0

(mk

)(k−mx)2 xk (1− x)m−k ≤ 1

4m

Proof: First of all, from the binomial theorem

m

∑k=0

(mk

)(et(k−mx)

)xk (1− x)m−k = e−tmx

m

∑k=0

(mk

)(etk)

xk (1− x)m−k

= e−tmx (1− x+ xet)m ≡ e−tmxg(t)m , g(0) = 1,g′ (0) = g′′ (0) = x

Take a derivative with respect to t twice.

m

∑k=0

(mk

)(k−mx)2 et(k−mx)xk (1− x)m−k

= (mx)2 e−tmxg(t)m +2(−mx)e−tmxmg(t)m−1 g′ (t)

+e−tmx[m(m−1)g(t)m−2 g′ (t)2 +mg(t)m−1 g′′ (t)

]Now let t = 0 and note that the right side is m(x− x2)≤ m/4 for x ∈ [0,1] . Thus

m

∑k=0

(mk

)(k−mx)2 xk (1− x)m−k = mx−mx2 ≤ m/4 ■

With this preparation, here is the first version of the Weierstrass approximation theorem.I will allow f to have values in a complete normed linear space. Thus, f ∈ C ([0,1] ;X)where X is a Banach space, Definition 2.5.31. Thus this is a function which is continuouswith values in X as discussed earlier with metric spaces.

Theorem 3.1.2 Let f ∈C ([0,1] ;X) and let the norm be denoted by ∥·∥ .

pm (x)≡m

∑k=0

(mk

)xk (1− x)m−k f

(km

).

Then these polynomials converge uniformly to f on [0,1].

79

Chapter 3Stone Weierstrass Approximation The-orem3.1 The Bernstein PolynomialsThese polynomials give an explicit description of a sequence of polynomials which con-verge uniformly to a continuous function. Recall that if you have a bounded functiondefined on some set S with values in Y.I[fllc. = sup {|| f (x)|| sx € S}This is one way to measure distance between functions.Lemma 3.1.1 The following estimate holds for x € [0,1] and m > 2.& (72) & my? 1a" <k=0mAleProof: First of all, from the binomial theoremm m» (‘.) (em) (1 xy" _ amy @ (e) (1 —xyn*=e ™ (1—x+4xe')" =e" g(t)”, g(0) = 1,8’ (0) =9" (0) =xTake a derivative with respect to ¢ twice.yo (™ k— mx)? em) € (1 — xy¥ (iz) ) (=x)= (mx)? eg (t)" +2 (—mx) e™mg (t)" g! (t)+e"™ Im(m—1) 8 (1)"* 8! (0)? +mg (1)! 8" (|Now let t = 0 and note that the right side is m(x —x?) < m/4 for x € [0, 1]. Thusv(m 2 m—k 2y? k (k— mx)? x* (1 —x)"“* = mx — mx? <m/4k=0With this preparation, here is the first version of the Weierstrass approximation theorem.I will allow f to have values in a complete normed linear space. Thus, f € C((0, 1];X)where X is a Banach space, Definition 2.5.31. Thus this is a function which is continuouswith values in X as discussed earlier with metric spaces.Theorem 3.1.2 Let f € C((0,1];X) and let the norm be denoted by |\-|| .Pm (x) = y ( A )xa -ayte (Z).k=0Then these polynomials converge uniformly to f on {0, 1].79