Lecture 10. δ-sequences and approximation theorems

Transcription

1 Lecture 10. δ-sequences and approximation theorems 1 Dirac δ-function In 1930 s a great physicist Paul Dirac introduced the δ-function which has the following properties: δ(x) = 0 for x 0 = for x = 0 δ(x) = 1. Surely, there is no such a function. Nevertheless, the following theorem may be proved Theorem 1 For any continuous function f C(), f(x)δ(x)dx = f(0). Proof By the integral mean value theorem, ε f(x)δ(x)dx = f(c) ε δ(x)dx = f(c). For ε ε ε 0, c 0. On one hand, the integral stays unchanged, on the other hand it tends to f(0). δ-sequences The idea of δ-function may be formalized by the notion of δ-sequences. Definition 1 A sequence of continuous or piecewise continuous functions ( n ) on the line is called a δ-sequence provided that the following holds: a) δ n 0 b) δ n 1 c) ε > 0 \[ ε,ε] δ n 0. By default, the integration domain is. Theorem For any finite continuous function f, and any δ-sequence, f(x) n (x)dx := (f, n ) f(0) as n. 1

2 Proof (f, n ) = I n + J n, I n = f(x) \[ ε,ε] n(x)dx, J n = f(x) [ ε,ε] n(x)dx I n supp f max f n (x)dx 0, n. \[ ε,ε] Here supp f is the length of the support of f. By the mean value theorem which is applicable because n 0, we have: J n = f(c) ε ε n (x)dx. The value f(c) is close to f(0) for small ε, and the value of ε ε n(x)dx is close to 1 for a large n. Therefore, I n is small, and J n is close to f(0). 3 Uniform convergence Corollary 1 Let f C 0 (), n be a δ-sequence. Then f n (x) = f(x) n (x y)dx f(y). Theorem 3 In assumptions of Corollary 1, the convergence f n fis uniform. Proof We want tot prove that α > 0 N: f n (y) f(y) < α n > N. (1) Take ε such that osc [y ε,y+ε] f < α y. Such an ε exists because the function f is finite. Take N such that \[ ε,ε] n(x)dx < α n > N. Such an N exists because of Assumption c). Hence (1) holds. 4 Important δ-sequences Let ϕ C,0 () be a unimodular function, ϕ(0) = 1. Unimodality means that sign ϕ = sign x. The graph of ϕ resembles a bell; the function ϕ has a maximum value at zero. Theorem 4 Let n = ϕ n ϕn (x)dx. () Then n is a δ sequence.

3 Proof The unimodality of the function ϕ implies that it is non-negative: ϕ 0. Moreover, ϕ(x) < 1 x 0. Assumption b of Definition 1: n(x)dx 1 holds by definition of n. Let us check the only remaining Assumption c. For any ε max ϕ(x) = q(ε), 0 < q(ε) < 1. x ε Let I n = ϕ n (x)dx. On the set x ε, we have: n (x) qn (ε) I n := λ n (ε). (3) Then x ε n (x)dx λ n (ε) supp ϕ. Proposition 1 In assumptions of the theorem, λ n (ε) 0 for any ε > 0. Theorem 4 is an immediate consequence of Proposition 1. Proof [of Proposition 1.] Fix ε > 0. The numerator of the fraction above tends to zero exponentially. Let us prove that the denominator is greater than some negative power of n: there exists c such that ϕ n (x)dx c. (4) n for any n sufficiently large. The n the ratio (3) tends to zero because the exponential decreases faster than any power. The unimodality of the function ϕ implies that for small x : x x 0, there exists CC such that ϕ(x) 1 Cx. For the large n, we have 1 n < x 0. On the interval x 1 n we get: ϕ(x) 1 C n. Hence, for x 1 n, We have: ϕ n (x) ( 1 C n ) n = i n. i n e C при n 3

4 Hence, for sufficiently large n, where This proves the proposition. I n i n n > c = 3 C. c n 5 Examples. The following two δ-sequences are used in the proof of Weierstrass approximation theorems. Take ϕ(x) = (1 x )χ [ 1,1]. (5) The sequence () with this ϕ is a δ-sequence. This sequence is remarkable because on the interval x y < 1, the function n (x y) is a polynomial in y. Take ϕ(x) = (cos x 1)χ [ π,π]. (6) The sequence () with this ϕ is also a δ-sequence. Any function in this sequence is a trigonometric polynomial in y for x y < π. 6 Polynomial approximations. Theorem 5 (Weierstrass) Any continuous function on a segment may be uniformely approximated by polynomials. Proof Let us bring the given segment to [0, 1] by an affine transformation. Consider the δ-sequence n (x) = P n (x)χ [ 1,1] constructed above; P n (x) = c n (1 x ) n. This sequence is given by formulas (), (5). Then n (x y) = P n (x y)χ [ 1+y,1+y] For fixed x, P n (x y) = Σ n 0 a k (x)y k. Moreover, n (x y) = Σa k (x)y k, provided that x y 1. Let us take a finite continuous function f C 0, supp f [0, 1]. For x [0, 1], y [0, 1] we have: x y 1. Hence, for these x and y, n (y x) = Σ n 0 a k (x)y k. Moreover, for any y, 1 1 f n (y) = f(x) n (x y)dx = f(x)σ n 0 a k (x)y k dx = Σ n 0 b k Y k, b k = f(x)a k (x)dx. 0 By Theorem 3, f n (y) f on. But on the segment [0, 1] the functions f n are polynomials. 0 4

5 Theorem 6 The same result, but polynomials trigonometric polynomials. (1+cos x)n Proof The proof is the same, but we take n (x) = I n, where I n = π (1 + cos π x)n dx. In the previous proof the segment [ 1, 1] should be replaced by [ π, π], and [0, 1] by [0, π]. 5