Math 402/502 Final Project: Approximation of compactly supported continuous functions by polynomials Charles Clauss and Joel Upston May 6, 2015 1 Abstract In this report we will be following Section 3.8 from Terrence Tao's Analysis II as he presents the Weier- strass approximation theorem. The goal will be to provide a fully fledged explanation and proof of every piece that Tao leaves as exercises for the reader. The theorem requires three main points, those being an approximation to the identity, convolution, and the relation between any chosen function and a carefully constructed polynomial. In essence, a continuous function on a closed interval that is convolved with a polynomial approximation to the identity will yield a new polynomial function that is uniformly close to the original function. After showing all of the machinery necessary to achieve this result we will also make a few remarks regarding extensions of this theorem, however the reader should note the original hypotheses that require a compactly supported, continuous function. 2 Compactly supported functions As we build toward the theorem we will need a few different pieces. A logical place to start will be to state some important properties of compactly supported functions that we will need. We would like to approximate continuous functions with polynomials, but are polynomials continuous? As a matter of fact, they are. If we consider the constant functions f(x) = c for all c 2 R, and the identity function f(x) = x, we can see that: limx!x0 f(x) = c = f(x0) and limx!x0 f(x) = x0 = f(x0). Therefore both of these functions are continuous everywhere, and additionally, arithmetic of functions pre- serves continuity, so we can build any polynomial out of these two functions, and it will also be continuous. A function f : R ! R is supported on [a; b] iff f(x) = 0 8 x2 = [a; b]. It is compactly supported iff 9 [a; b] such that f(x) is supported. Finally, if f is continuous and compactly supported on [a; b] then it should be R 1 R b clear that −∞ f(x) = a f(x). Given that we are considering functions that are continuous and of the form f :[a; b] ! R, we are on a compact set, and it follows immediately that they are uniformly continuous and bounded. 3 Polynomial approximation to the identity We would like an identity function that we can approximate with polynomials that will serve as a tool for approximation of more complicated functions. Given > 0 and 0 < δ < 1, a function f : R ! R is an (, δ) approximation if: (a) f is supported on [−1; 1], and f(x) ≥ 0 for all −1 ≤ x ≤ 1. R 1 (b) f is continuous, and −∞ f = 1. (c) f(x) ≤ for all δ ≤ jxj ≤ 1. 1 Given our definition of the identity, we will see that we will want it to be a polynomial. We can do this by following a few careful steps. First we will compare two functions by differentiation to show that (1 − x2)n ≥ 1 − nx for x 2 R; n 2 Z; 0 ≤ x ≤ 1; n ≥ 1 (1 − x2)n ≥ 1 − nx −2nx(1 − x2)n−1 ≥ −n x(1 − x2)n−1 ≤ 1. This holds for x positive and bounded above by 1. 0 ≤ x ≤ 1 0 ≤ x2 ≤ 1 0 ≥ −x2 ≥ −1 1 ≥ 1 − x2 ≥ 0 1 ≥ (1 − x2)n ≥ 0 x ≥ x(1 − x2)n ≥ 0. A symmetric argument holds for −1 ≤ x ≤ 0 using 1 + nx instead. Now we want to show that the function c(1 − x2)n can approximate the identity. Therefore we must show that each of the listed properties of an approximation holds. First, define the function f to be our indicated function on the interval [−1; 1] and 0 everywhere else. Then f is supported on [−1; 1] by definition of support, and we just showed previously that it is also positive. Continuity has already been established for all polynomials, so it remains to show that the integral is unit and that the tails are sufficiently small. R 1 Consider −∞ f since the function is zero outside the interval [−1; 1], we can say immediately that R 1 R 1 −∞ f = −1 f R 1 2 N Examining the integral −1 c(1 − x ) for some finite N, we can use binomial theorem to expand our polynomial, and exchange sum and integral due to integration on a compact set. N R 1 2 N R 1 P N n 2n −1 c(1 − x ) dx = c −1 n (−1) x dx n=0 N R 1 2 N P N n R 1 2n −1 c(1 − x ) = c n (−1) −1 x dx n=0 N R 1 2 N P N n 2n+1 1 −1 c(1 − x ) = c n (−1) x =(2n + 1) j−1 n=0 N R 1 2 N P N n −1 c(1 − x ) = 2c n (−1) =(2n + 1). n=0 N P N n −1 Thus the sum is finite and we can define c(N) := (2 ∗ n (−1) =(2n + 1)) such that the integral is 1. n=0 Notice that the shape of these functions becomes very narrow as N becomes large, so the constant c depends on N, and must also become very large to compensate. 1 1 N 2 Lastly, for any < c we can choose δ = (1 − ( c ) ) , and if we look again at the derivative of this functional form, we can see that from [−1; 0] it is increasing and from [0; 1] it is decreasing. Therefore for δ < jxj < 1 we can conclusively say that f(x) < . 4 Convolutions We will also need a few important properties of convolution in order to construct needed polynomials. Definition of convolution is given by, Z 1 f ∗ g(x) = (f(y)g(x − y))dy. −∞ Then we can show properties of convolutions given by the Proposition 3.8.11. 2 Proposition 1. Let f : R ! R; g : R ! R; h : R ! R be continuous and compactly supported functions then the following properties hold, 1. The convolution of f ∗ g(x) is also a continuous and compactly supported function. 2. The convolution is commutative f ∗ g(x) = g ∗ f(x). 3. The convolution is a linear operator such that f ∗ (g + h)(x) = f ∗ g(x) + f ∗ h(x) and let c 2 R then f ∗ (cg)(x) = c(f ∗ g(x)). Proof of 1. First we will show that the convolution of f ∗g(x) is continuous. First let us use that g : R ! R is continuous. Therefore 80 > 0 9δg such that for every x; y 2 R that satisfy jx − yj < δg, then jg(x) − g(y)j < .Therefore 8 > 0, we let = then 9δ such that 8x; y 2 , jx − yj < δ , so that 0 0 M(b − a) g R g jg(x) − g(y)j < 0 . Then we can use the fact that jf(x)j < M for all x 2 [a; b], since it is continuous on a compact set so it is uniformly bounded. Therefore we use these facts and triangle inequality to see that 8 > 0 9δ = δg such that for every x; y 2 R that satisfy jx − yj < δ , Z 1 Z 1 jf ∗ g(x) − f ∗ g(x0)j = j f(y)g(x − y)dy − f(y)g(x0 − y)dyj −∞ −∞ Z 1 Z 1 = j f(y)(g(x − y) − g(x0 − y))dyj ≤ jf(y)jj(g(x − y) − g(x0 − y))jdy −∞ −∞ Z 1 Z b ≤ jf(y)jdy = jf(y)jdy ≤ M(b − a) = . M(b − a) −∞ M(b − a) a M(b − a) Now to show that f ∗ g(x) is compactly supported we just need to look at the definition. So WLOG let a < b and c < d and let f be compactly supported on [a; b] and let g be compactly supported on [c; d]. Then since f is compactly supported on [a; b], Z 1 Z b f ∗ g(x) = f(y)g(x − y)dy = f(y)g(x − y)dy: −∞ a Therefore since g is also compactly supported on [c; d] if x − y 62 [c; d] then g(x − y) = 0. So then since y 2 [a; b] so if x 62 [a+c; b+d] then g(x−y) = 0 therefore f ∗g(x) is compactly supported on [a+c; b+d]. Z 1 Proof of 2. By definition, f ∗ g(x) = f(y)g(x − y)dy. So then, −∞ Z 1 f ∗ g(x) = f(y)g(x − y)dy −∞ du We let u = x − y so y = u − x and then = −1. Therefore dy Z 1 Z −∞ Z 1 Z 1 f ∗ g(x) = f(y)g(x − y)dy = − f(u − x)g(u)du = f(u − x)g(u)du = g(y)f(y − x)dy: −∞ 1 −∞ −∞ Proof of 3. Z 1 Z 1 f ∗ c(g + h)(x) = f(y)(cg(x − y) + ch(x − y)dy = c f(y)(g(x − y) − h(x − y))dy = −∞ −∞ Z 1 Z 1 c f(y)g(x − y)dy + c f(y)h(x − y)dy = c(f ∗ g(x)) + c(f ∗ h(x)) −∞ −∞ 3 5 Building the Weierstrass approximation theorem We will show that we can create polynomials that are very close to our original function on the interval [0; 1]. After which we will show that we can do this for any interval [a; b]. Lemma 1. Let f : R ! R be a continuous function supported on [0; 1] and let g : R ! R be a continuous function supported on [-1,1] which is a polynomial on [-1,1].