Chapter Sequences and Series of Functions Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.1, Slide Section 9.1 Pointwise and Uniform Convergence Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.1, Slide Definition 9.1.1 Let ( f ) be a sequence of functions defined on a subset S of n Then ( f ) converges pointwise on S if for each x ∈ S the sequence of n numbers ( f (x)) converges If ( f ) converges pointwise on S, then we define f : S → n n by f ( x) = lim f n ( x) n→∞ for each x ∈ S, and we say that ( f ) converges to f pointwise on S n Question: What properties are preserved in taking the limit? If each fn is continuous on a set S, is f continuous on S? If each fn is integrable, is f integrable? And is ∫ f = lim ∫ fn ? No, not in general If each fn is differentiable, is f differentiable? And is f ′ = lim ? Copyright © 2013, 2005, 2001 Pearson Education, Inc f n′ Section 9.1, Slide Suppose S is an interval with x ∈ S at x means that To say that the functions fn are continuous f n ( x) = lim f n (t ) t→x for all n If ( f ) converges to f pointwise on S, then n f ( x) = lim f n ( x) = lim lim f n (t ) n → ∞ t → x n→∞ If the limit function f is continuous at x, then equal?? f ( x) = lim f (t ) = lim lim f n (t ) t→x t→x n →∞ Thus the limit of a sequence of continuous functions will be continuous at x when these limits are equal That is, we wish to know when the order of the limit processes may be reversed Since derivatives and integrals are also defined in terms of limits, a similar question applies: Can the order of the limits be reversed? Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.1, Slide Example 9.1.2 For x ∈ [0, 1] and n ∈ n , define f (x) = x n When n = we have f1(x) = x When n = we have f2(x) = x When n = we have f3(x) = x y etc f 1 f 2 f 3 Copyright © 2013, 2005, 2001 Pearson Education, Inc x Section 9.1, Slide Example 9.1.2 For x ∈ [0, 1] and n ∈ n Then for each x ∈ [0,1) we have , define f (x) = x n lim f (x) = n → ∞ n and lim f (1) = n→∞ n Thus ( f ) is pointwise convergent on S = [0, 1], and the limit function f is given by n 0, f ( x) = 1, if ≤ x < 1, if x = y (1, f (1)) Each f is continuous n (and differentiable) on [0, 1], but the limit function f is neither continuous nor f 1 f 2 f 3 differentiable at x = ) f Copyright © 2013, 2005, 2001 Pearson Education, Inc x Section 9.1, Slide In terms of the limit operations, we have lim lim f n (t ) = lim [0] = t → 1− t →1− n → ∞ Go down at t < 1, then across toward x = lim lim f n (t ) = lim [1] = n→∞ n → ∞ t → 1− First, go across each fn to where x = y (1, f (1)) f 1 f 2 f 3 ) f Copyright © 2013, 2005, 2001 Pearson Education, Inc x Section 9.1, Slide Example 9.1.3 For n ≥ and x ∈ [0, 1], construct a sequence of functions fn as follows: f2 graphs as a peak of height and width f3 graphs as a peak of height and width 2/3 f4 graphs as a peak of height and width 2/4, etc The limit function f is the constant function f (x) = for all x For each n we have f ∫ f n ( x) dx = 1, But f ∫ f ( x) dx = f So the integral of the limit is not the limit of the integrals Copyright © 2013, 2005, 2001 Pearson Education, Inc f Section 9.1, Slide To reverse the limit operations, we need a stronger kind of convergence Definition 9.1.6 Let ( fn ) be a sequence of functions defined on a subset S of Then ( fn ) converges uniformly on S to a function f if for each ε > there exists a natural number N such that for all x ∈ S, n ≥ N implies that | fn(x) – f (x)| < ε The difference between pointwise and uniform convergence: If ( fn ) converges pointwise to f on S, then for every ε > and each x ∈ S, there exists N ∈ (depending on both ε and x) such that | fn(x) – f (x)| < ε whenever n > N If ( fn ) converges uniformly to f on S, then it is possible for each ε > to find one number N that will work for all x ∈ S This is the same type of quantifier reversal that produced uniform continuity from continuity Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.1, Slide Geometrically, uniform convergence means that the entire graph of fn must lie within the strip bounded by the graphs of f – ε and f + ε y y = fn (x) y = f (x) + ε y = f (x) y = f (x) – ε S Copyright © 2013, 2005, 2001 Pearson Education, Inc x Section 9.1, Slide 10 Back to the first example: And the second example: y f (1, f (1)) f f 1 f 2 f ) f+ε ) f ) f+ε x f–ε f–ε In neither case can we get fn within ε of f uniformly Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.1, Slide 11 As we might expect, there is a Cauchy criterion for sequences of functions Theorem 9.1.10 Let ( fn ) be a sequence of functions defined on a subset S of There exists a function f such that ( fn ) converges to f uniformly on S iff for each ε > there exists N ∈ such that | fn(x) – fm(x)| < ε for all x ∈ S and all m, n ≥ N Series of functions are treated similar to series of constants and power series ∞ If ( fn ) is a sequence of functions defined on a set S, the series is said ∑ n=0 fn to converge pointwise (resp uniformly) on S iff the sequence of partial sums sn ( x) = n ∑ k =0 f k ( x) Copyright © 2013, 2005, 2001 Pearson Education, Inc converges pointwise (resp uniformly) on S Section 9.1, Slide 12 There is a very useful test for establishing the uniform convergence of a series of functions: Theorem 9.1.11 (Weierstrass M-test) Suppose that ( fn ) is a sequence of functions defined on S and (Mn) is a sequence of nonnegative numbers such that | fn(x)| ≤ Mn for all x ∈ S and all n ∈ If Σ Mn converges, then Σ fn converges uniformly on S In the next section we will see that uniform convergence does what we want it to do—namely, we can interchange the order of the limit operations Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.1, Slide 13 Example 9.1.12 Consider ∑ the series of functions ∞ f n=0 n where fn(x) = x In Example 8.3.7(a) we saw that this series is (pointwise) convergent on We now show that the convergence is not uniform on the convergence is uniform on the closed interval [− To show that the convergence of the series Σ fn For example, let ε = Given any N ∈ , but given any t ∈ for all x ∈ n /n! , t, t] is not uniform on , we show that the sequence (sn) of partial sums does not satisfy the Cauchy criterion of Theorem 9.1.10 , we must find m, n ≥ N and x ∈ such that | sn(x) − sm(x)| ≥ = ε It follows that To this end, let m = N, n = N + 1, and x = n nn | sn (n) − sm (n) | = | f n (n) | = ≥ n! Hence the series is not uniformly convergent on Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.1, Slide 14 Example 9.1.12 Consider ∑ the series of functions ∞ f n=0 n where fn(x) = x In Example 8.3.7(a) we saw that this series is (pointwise) convergent on We now show that the convergence is not uniform on the convergence is uniform on the closed interval [− On the other hand, if t ∈ , then let Mn = t n /n! , but given any t ∈ for all x ∈ n /n! , t, t] For any x ∈ [−t, t] we have xn tn f n ( x) = ≤ = M n n! n! Since Σ Mn is convergent (by the ratio test), it follows from the Weierstrass M-test that Σ x n/n! is uniformly convergent on [− t, t] Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 9.1, Slide 15 ... Section 9.1, Slide Suppose S is an interval with x ∈ S at x means that To say that the functions fn are continuous f n ( x) = lim f n (t ) t→x for all n If ( f ) converges to f pointwise on S, then... converges to f uniformly on S iff for each ε > there exists N ∈ such that | fn(x) – fm(x)| < ε for all x ∈ S and all m, n ≥ N Series of functions are treated similar to series of constants and power... difference between pointwise and uniform convergence: If ( fn ) converges pointwise to f on S, then for every ε > and each x ∈ S, there exists N ∈ (depending on both ε and x) such that | fn(x) –