Approximation of Continuous Functions

Transcription

1 Approximation of Continuous Functions Francis J. Narcowich October Modulus of Continuit Recall that ever function continuous on a closed interval < a x b < is uniforml continuous: For ever ɛ > 0, there is a δ > 0 such that f(x f( < ɛ (1.1 as long as x, [a, b] satisf x < δ. This differs from the definition of continuit at a single point in that δ is independent of, and depends onl on ε. Let s turn around the roles of ε and δ. In uniform continuit, we start with ε and look for δ. What we want to do now, is start with δ and, essentiall, find ε. With this in mind we make the following definition 1 Definition 1.1. The modulus of continuit for f C[0, 1] and δ > 0 is defined to be ω(f, δ = sup{ f(x f( : x δ, x, [0, 1]}. (1.2 Example 1.2. Let f(x = x, 0 x 1. Show that ω(f, δ C δ. Proof. Let 0 < x < 1. We note that 0 < x = x x + = x ( x + x = x x δ 1 x 1 + x 1 + x = x 1 x 1 + x 1 From now on, without loss of generalit, we will work with the closed interval [0, 1]. 1

2 Hence, ω(f, δ δ. To get equalit, take x = 0 and = δ. Example 1.3. Suppose that f C (1 [0, 1]. Show that ω(f, δ f δ. Proof. Because f C (1, we can estimate f(t f(s this wa: f(t f(s t which immediatel gives ω(f, δ δ f. s f (x dx (t s f δ f, (1.3 2 Approximation with Linear Splines One ver effective wa to approximate a continuous function f C[0, 1], given a finite set of points in {x } [0, 1] and values {f(x } at these points, is to use a connect-the-dots approach. The idea is to form a piecewiselinear, continuous function b oining neighboring points (x, f(x and (x +1, f(x +1 with a straight line. This procedure results in a linear spline. Linear splines are used for generating plots in man standard programs, such as Matlab or Mathematica. Defining a space of linear splines starts with sequence of points (or partition of [0, 1] = {x 0 = 0 < x 1 < x 2 < < x n = 1}, which is called a knot sequence. Linear splines on [0, 1] with knot sequence are the set of all piecewise linear functions that are continuous on [0, 1] and that (possibl have corners at the knots, but nowhere else. As described above, we can interpolate continuous functions using linear splines: Let f C[0, 1] and let = f(x. The linear spline s f (x is constructed b oining pairs of points (x, and (x +1, +1 with straight lines. The resulting spline is the unique and satisfies s f (x =. The result below gives an estimate of the error made b replacing f b s f. Proposition 2.1. Let f C[0, 1] and let = {x 0 = 0 < x 1 < < x n = 1} be a knot sequence with norm = max x x +1, = 0,..., n 1. If s f is the linear spline that interpolates f at the x s, then, f s f ω(f,. (2.1 Proof. Consider the interval I = [x, x +1 ]. We have on I that s f (x is a line oining (x, f(x and (x +1, f(x +1 ; it has the form s f (x = x +1 x f(x + x x f(x +1 2

3 Also, note that we have x +1 x + x x = 1. Using these equations, we see that f(x s f (x for an x [x, x +1 ] can be written as f(x s f (x = f(x ( x +1 x + x x sf (x = (f(x f(x x +1 x x x + (f(x f(x +1. B the definition of the modulus of continuit, f(x f( ω(f, δ for an x, such that x δ. If we set δ = x +1 x, then we see that on the interval I we have f(x s f (x (f(x f(x x +1 x + f(x f(x +1 x t ( x+1 x + x x ω(f, δ = ω(f, δ. Because the modulus of continuit is non decreasing (exercise 5.2(c and δ, we have ω(f, δ ω(f,. Consequentl, f(x s f (x ω(f,, uniforml in x. Taking the supremum on the right side of this inequalit then ields ( The Weierstrass Approximation Theorem The completeness of various sets of orthogonal polnomials relies on being able to uniforml approximate continuous functions b polnomials. The Weierstrass Approximation Theorem does exactl that. The proof that we will give here follows the one Sergei Bernstein gave in The proof is not the slickest, but it does introduce a number of important things. Here is the statement. Theorem 3.1 (Weierstrass Approximation Theorem. Let f C[0, 1]. Then, for ever ε > 0 we can find a polnomial p such that f p C[0,1] < ε. 3

4 3.1 Bernstein polnomials Let n be a positive integer. The binomial theorem states that (x + n = x n. We define the Bernstein polnomial using the terms in the n ( n expansion with = 1 x: β,n (x := ( n x (1 x n. (3.1 Proposition 3.2. The Bernstein polnomials {β,n } n form a basis for P n, the space of polnomials of degree n or less. Proof. The dimension of P n is n + 1. Since there are n + 1 Bernstein polnomials, we need onl show that the span P n. We will show that 1, x are in the span of the Bernstein polnomials, and leave x 2,..., x n as an exercise. To get 1, set = 1 x in the binomial expansion we get (x + 1 x n = n ( n x (1 x n, so that 1 = n β,n(x. For x, we take the partial derivative of (x + n with respect to x to get n(x + n 1 = n =1 ( n x 1 n. Multipling this b x, setting = 1 x and dividing b n, we obtain x = n =1 n β,n(x. The others are obtained similarl. We will need several identities involving the Bernstein polnomials, which we now list. The last two identities start the sum at = 0, rather than = 1. 1 = β,n (x x = 1 n x + (1 1 n x2 = n β,n(x 2 n 2 β,n(x. 3.2 Proof of the Weierstrass Approximation Theorem (3.2 All of the Bernstein polnomials are positive, except at 0 and 1, where the are 0. When n is ver large, the Bernstein polnomial β,n is highl peaked near its maximum at x = /n. That is, a small distance awa from /n, the polnomial β,n is itself quite small. With this in mind, if f C[0, 1], define the polnomial f n (x := f(/nβ,n (x P n. 4

5 The idea here is that near each point /n the main contribution to the sum of Bernstein polnomials making up f n should come from the term f(/nβ,n (x, so f n should be a good approximation to f. Proof. (Weierstrass Approximation Theorem Choose n large; let δ > 0. Using the first identit in (3.2, we have f(x = f(x 1 = n f(xβ,n(x. Thus the difference between f and f n is E n (x := f(x f n (x = (f(x f(/nβ,n (x. We want to show that, for sufficientl large n, E n C[0,1] < ε. Fix x. We are now going to break the sum into two parts. The first will be all those for which x /n δ. This is F n below. The second, G n consists of all remaining s namel, all such that x /n > δ. Carring this out breaks E n into the sum E n = F n + G n, where F n (x = (f(x f(/nβ,n (x (3.3 G n (x = x /n δ x /n >δ (f(x f(/nβ,n (x. (3.4 Using the triangle inequalit on the sum in F n ields F n (x f(x f(/n β,n (x. x /n δ Because x /n δ, we also have f(x f(/n ω(f, δ. Consequentl, ( F n (x x /n δ ( β,n (x ω(f, δ β,n (x ω(f, δ. Using the first identit in (3.2 in the inequalit above ields this: F n (x ω(f, δ. (3.5 Estimating G n requires more care. We will treat the case in which x > /n; the other case is similar. The idea is to think of δ as a unit of measure like inches or centimeters. Then there will be a smallest k such that x will be between kδ and (k + 1δ units from /n. More precisel, let k be the 5

6 smallest integer such that kδ < x /n (k + 1δ. Write f(x f(/n this wa: f(x f( n = [ f(x f( n + kδ] + [ f( n + kδ] f( n + (k 1δ] + = [ f(x f( n + kδ] + + [ f( n + δ f(/n] k 1 m=0 [ f( n + (k mδ f( n + (k m 1δ] Since n + (m + 1δ n mδ = δ, for m = 1,..., k 1 each term satisfies f( n + mδ f( n + (m + 1δ ω(f, δ. Also, because and x n kδ δ, the first term satisfies f(x f( n + kδ ω(f, δ. There are k + 1 terms in the sum, so f(x f( n (k + 1ω(f, δ. Since kδ < x /n, we have (k + 1δ < 1 + (x /n/δ. As we mentioned earlier, a similar argument will give (k + 1δ < 1 + x /n /δ; consequentl, in either case we have f(x f(/n (k + 1ω(f, δ ( 1 + x n /δ ω(f, δ. What we do next is use a trick that will help us to get an explicit bound on G n (x. The trick is to note that, since kδ < x /n, we have x /n 1, and we thus also have x /n δ inequalit results in f(x f(/n < (1 + δ > < x /n 2 δ 2. Using this in the previous x /n 2 δ 2 ω(f, δ. (3.6 While we derived this for /n > x, essentiall the same argument holds for /n < x. Thus (3.6 holds for both cases. Using the triangle inequalit on the sum in (3.4 and bounding each term b (3.6, we obtain ( G n (x < (1 + ( < x /n 2 δ 2 β,n (x ω(f, δ ( 1 + x 2 δ 2 2x δ 2 n + 2 β,n δ 2 n 2 (x ω(f, δ (3.7 Using the three identities in (3.2 allows us to do all of the sums in (3.7. The result, after a little algebra, is G n (x < ( 1 + x x2 δ 2 n ω(f, δ. Since the maximum of x x 2 over [0, 1] is 1/4; consequentl, G n (x < ( nδ 2 ω(f, δ. (3.8 6

7 Combining (3.5, (3.8 and E n (x = F n (x + G n (x F n (x + G n (x ields E n (x < ( nδ 2 ω(f, δ. (3.9 The parameter δ > 0 is free for us to choose. Take it to be δ = n 1/2. With this choice we arrive at f(x f n (x = E n (x < 9 4 ω(f, n 1/2. (3.10 Taking the maximum of f(x f n (x over x [0, 1] gives us f f n < 9 4 ω(f, n 1/2. Choosing n so large that 9 4 ω(f, n 1/2 < ε then completes the proof. Previous: orthonormal sets and expansions Next: pointwise convergence of Fourier series 7