18.100A: Complete Lecture Notes Lecture 24: Uniform Convergence, the Weierstrass M-Test, and Interchanging Limits Theorem 1  0 x ∈ [0, 1) Let fn (x) = xn , and let f (x) = . 1 x = 1 1. ∀0 < b < 1, fn → f uniformly on [0, b]. 2. fn does not converge to f uniformly on [0, 1]. Proof : 1. Let > 0. Since b ∈ (0, 1), bn → 0. Therefore, ∃M0 ∈ N such that for all n ≥ M0 , bn < . Choose M = M0 . Then, ∀n ≥ M, ∀x ∈ [0, b], |fn (x) − f (x)| = |fn (x)| = xn ≤ bn < . Thus, fn → f uniformly on [0, b]. Before proving the other part, we first note the following negation: Negation 2 (Not Uniformly Convergent) fn : S → R does not converge to f : S → R uniformly if ∃0 > 0 such that ∀M ∈ N, ∃n ≥ M and ∃x ∈ S with |fn (x) − f (x)| ≥ 0 . 2. Hence, for our example, choose 0 = 14 . Let M ∈ N and choose x = |fM (x) − f (x)| = fM (x) = let fj : S → R and suppose ∃Mj > 0 such that a) ∀x ∈ S, |fj (x)| ≤ Mj . P∞ b) j=1 Mj converges. Then, P∞ fj (x) converges absolutely. P∞ 2. Let f (x) = j=1 fj (x) for x ∈ S. Then, j=1 n X fj → f uniformly on S. j=1 1 M1 1 > 0 . 2 Theorem 3 (Weierstrass M-test) 1. ∀x ∈ S, 1 2 ∈ (0, 1). Thus, Proof : 1. The first part follows from a), b), and the Comparison Test. P 2. Let > 0. Since Mj converges, ∃N0 ∈ N such that ∀n ≥ N0 , ∞ X ∞ X Mj = Mj − j=1 j=n+1 n X Mj < . j=1 Choose N = N0 . Then, for all n ≥ N and ∀x ∈ S, f (x) − n X ∞ X fj (x) = fj (x) j=n+1 j=1 ∞ X ≤ ≤ j=n+1 ∞ X |fj (x)| Mj < . j=n+1 Thus, Pn j=1 fj → f uniformly on S. Interchange of Limits Remark 4. In general, limits cannot be interchanged. Example 5 For instance, consider the following example: n/k 0 = lim =0 n/k + 1 n→∞ 0 + 1 n/k lim lim = lim 1 = 1. k→∞ n→∞ n/k + 1 k→∞ lim lim n→∞ k→∞ Question 6. Hence, we ask three questions about interchanging limits: 1. If fn : S → R, fn continuous and fn → f pointwise or uniform, then is f continuous? 2. If fn : [a, b] → R, fn differentiable, and fn → f with fn0 → g, then is f differentiable and does g = f 0 ? 3. If fn : [a, b] → R, with fn and f continuous such that fn → f , then does Z b Z fn = a b f? a The answer to the above questions are all yes, if the convergence is uniform. Question 7. What if the convergence is only pointwise? If convergence is only pointwise, the answer to the above questions are all no, which we will show next time. 2